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Previously Asked Questions (FAQ) for Ask A Particle Physicist – by the High Energy Physics Group at Syracuse University

Questions previously asked and answered are given here. See if your own question has been asked before.

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Does light (photons) interact with protons? If so, at which frequency does light interact the most with protons. At which frequency does light gives the most energy to protons ?

Question by: Zack Hardwick

This is an interesting topic. Since a proton is charged it is effected by the electromagnetic force. The carrier of that force is a photon, so indeed light interacts with protons. Using low energy photons where the wavelength is much larger than the size of the proton (10-13 m) the interaction is the same as with an electron and a photon. When the photon energy gets large (i.e., it's wavelength is small compared to the proton's size), other interesting things happen. For example, the photon can directly interact with the constituents of the proton, the quarks. Furthermore, the photon may also act like a strongly interacting meson with the same quantum numbers as the photon, namely spin 1 and negative parity. Those mesons include the rho (r), omega(w), and phi (f). When these energies is reached the cross-sections grow and many particles are produced in the collision. This figure (see bottom plot) shows the measured total cross-section of photons on protons and deuterons. High energy photon beams have been made and used to study the structure of the proton and to produce other particles for study. Some of the best measurements of the properties of charm particles have recently been made with a photon beam at Fermilab by the FOCUS experiment (see http://www-focus.fnal.gov/<

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A proton has a mass of 1.7x10-27 [kg] which is equivalent to a rest energy of 1 GeV, that's a lot of energy. Since light can interact with protons, is there a possibility that a photon beam could cause the proton to release it's mass energy ? Could a photon beam cause protons to loose gluon cohesion ? Therefore could protons be a source of energy in the future ?

Question by: Zack Hardwick

No, total energy and total charge is always conserved, as is net baryon number (at least in today's world). This means that if a photon strikes a proton, whatever emerges must contain at least 1 baryon (such as a proton). It's possible at high energy for it to be other baryons, but the proton is the lightest, so you can't convert any of its mass into energy.

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What is the meaning of the Lie Groups SU and E?

Question by: Lazaro Blanco

The subject of Lie groups is quite broad, and there are a number of books devoted to it. There is really no way to do the subject justice via a short explanation. However, hopefully this short explanation might help. SU(N) is a Lie Group of N by N unitary unimodular matrices. The "U" stands for Unitary, and a Unitary Matrix U satisfies U times its hermitian conjugate is Unity (i.e., UU+=1). Hence, its determinant is equal +1 or -1. "S" stands for special in that the determinant is restricted to be +1. Such a Lie group is characterized by N continuous parameters. Its generators Ta , a = 1,..,N satisfy the commutation relations [Ta, Tb ] = fabc Tc.  The f's are known as structure constants. They are antisymmetric in their indices. 

If you'd like to learn more and get more details, a couple of good references are:
   "Lie Groups fro Pedestrians" by Lipkin
   "Lie Algebras in Particle Physics : from Isospin to Unified Theories" by  Howard Georgi

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Has any evidence been uncovered which might lend support to loop quantum gravity? I know that string theory is all the rage but the mathematical complexity and prediction of additional exotic particles and forces give me doubts. Do you personally have a preference between the two theories? Or as an experimental physicist do you not spend much time worrying over the theory?

Question by: John Chambers, Texas, Faculty Respondent: Steve

Thanks for your question. To date, there is no evidence for either quantum loop gravity or string theory. Both are ambitious and encompass one of the most intriguing areas of physics, coming up with a theory which unifies quantum mechanics and relativity. While loop quantum gravity attempts to unify General Relativity and Gravity, String Theory goes beyond and is
ambitiously trying to come up with a theory which not only solves the gravity issue, but unites all four forces under one umbrella. As an experimentalist, we don't have preference of one theory over another, we want the correct theory of everything. Moreover, we really need a theory which can provide testable predictions. Both quantum loop gravity and string theory don't quite pass that test yet. To remind you, a "successful" theory must:

(1) Make predictions which are consistent with currently existing data.
(2) Make new predictions for observables which have not as of yet been measured, but can be measured in the (near) future.
(3) The predictions from (2) should agree with the eventual measurements, or the theory should be flexible enough to accommodate the difference. Presumably this reduces the allowed parameter values in the theory, and then one makes more measurements to test the theory with the modified parameter set.

I think we are a long way from reaching the ambitious goals of these theories. For now, we (particle physicists) are working hard to find cracks in the Standard Model. The Standard Model is the model which describes the fundamental particles and their Strong, Weak and Electromagnetic interactions. Gravity is much weaker than these, and hence is not really addressed in the
Standard Model. So far, the Standard Model is holding up, but we are building more powerful machines and experiments to see if we can find inconsistencies, it might point the way to a more complete theory which clears up the inconsistencies. A candidate theory is Supersymmetry, but it also has it's "issues", such as a huge number of free parameters which are not predicted by the theory.

So, the road to a more complete theory of everything is a long and tough journey, but hopefully by attacking it from many angles (ie., particle physics experiments, space-based experiments, neutrino experiments, etc) we might come up with a more complete and unified theory.

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Recently I became interested in solar power production. In looking into this type of energy production I became curious. If we can produce energy from a very small section of the electromagnetic spectrum aka visible light then why is it not possible to harness all or at least some of the parts of the electromagnetic spectrum? For example, UV panels, gamma panels, infared panels (which is probably the most abundant source) these panels could convert what is all around us to electricity. Have materials been identified that could absorb different parts of the spectrum and convert such to electricity? Perhaps focusing part of the spectrum is the way? If either of these were developed there could be free clean power day or night.

Question by: Robert Kneberg, UTAH

This is very insightful. In fact there are many scientists & companies devoting a great deal of time to improving our ability to convert solar power to electrical power. As you probably know, solar panels can be purchased to put on your roof. It turns out that visible light has the proper amount of energy to liberate electrons from the material in the solar cell (mostly silicon), which can be used to generate electrical current.
UV, Infra-red and gamma do not have the proper energies to make this process work; the photons are either too energetic or not energetic enough. Moreover, the peak of the sun's spectrum is in the visible, so this is the best place to capture and convert the sunlight. Another problem is that solar energy conversion is not very efficient. Even today's state of the art gives a conversion efficiency of about 10%. That is, only 10% of the incident energy is converted to electrical energy. If one does a fairly straightforward calculation, you will find out that solar power is not very economical at this point, as compared to coal or nuclear power. I do feel though that there is hope in the future for this source of energy.

Here are some nice "friendly" references which may be of interest to you.

 http://www.re-energy.ca/
 http://www.astropower.com/faq_solar_basics.htm

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I understand that silicon is used for the visible light spectrum and that other parts of the spectra will not provide the desired results in silicon. I was just wondering if there were other materials that would produce an electric current from other parts of the electromagnetic spectrum or if a search for the desired materials was even in process?

Question by: Robert Kneberg, UTAH

The gist of it is that there is not very much UV energy from the sun to mine as compared to the visible part. So, I don't think it is being heavily pursued. Here is a JPEG file showing the radiant intensity from the sun (and earth), and you can see it peaks in the visible, and not much in the UV.

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I am a student and my instructor has posed this question. Is there any situation in which only 1 of the 4 fundamental forces (strong nuclear, weak nuclear, electromagnetic, gravity) is at work? If so, what is it?

Question by: Brian Jackson

Since your instructor posed this question, I assume he/she wants you to think about it. So this answer is in the spirit of helping you along this path. Can you isolate a situation where only one of the four forces apply? For example neutrally charged objects may not have any electromagetic forces. Can you find a situation where only two apply? Finally if one force is much much larger than all the others, does it matter if the others are present?

Follow up Question from Jackson: Yes he wants us to think about it and I have. I can't think of a situation when only one force is at work. It can be totally hypothetical. Do you know of any?

Answer: Actually, I don't. We thought of neutrino - neutrino scattering. This is almost pure weak interactions. However, since neutrino's have a tiny mass they also have an extremely small Gravitational interaction. One might also think of something very exotic, like the gravitational interaction between two neutron stars. In principle though, there could be quanta exchanged, although in practice its negligibly small. Since charged particles can be accelerated within the neutron star, they can emit radiation, which can be absorbed by the second neutron star. So, it's pretty tough to think of a case where strictly only 1 force is participating.

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What would I see inside a sphere with mirrored walls? (multi-part)

Question by: Malcolm Cooper

I have a strange question. I'm a artist and I came up with an idea the  other day, but I wanted some idea of it's outcome before attempting it. I've asked a number of people about it, but no one has an idea of what it would look like. Anyway here it is... I'm thinking of building a sphere that you can walk into. It would have a door (obviously) so that when you went inside
and closed the door you would be standing inside a perfect (or as good as I can get it) sphere. The inside of the sphere would be mirrored. I'm thinking that you would carry in some kind of light source. So here are some questions about this...

    1. What would you see when you were inside and had your light on?

       a) Would it just be a big blur that is the color of the sum of all of the  colors of yourself (i.e. if your skin was yellow and your clothes were blue you would see green all around    you.)
       b) Would you just see a bright light all around you because of the light you were carrying?
    2.
      a) This is probably a dumb question, but... If you were inside there for some time with your light on and then you turned it off would it still be light inside for some time? I mean, even though your light source was turned off would the light that had come from it continue to bounce around inside for a noticable amount of time?

      b) If you had something like a laser pointer (no other light) and you blinked it on and off a few times would you start seeing little red dots  appear and multiply all around you?

    I'm normally a oil painter so trying to figure this out without  building it is a bit daunting. As a painter I'm also financially strapped and don't want to spend the money to build such and object if the results are not going to be somewhat interesting.


This is a cute idea.
1    a) As far as color is concerned there is no color change. White light would stay white, yellow would stay yellow etc..
    b) You would see your reflection in portions of the mirror, possibly only multiple reflections. Sections of a sphere form what we call "spherical mirrors" which are focusing. What you see would depend if you were standing to one side or the other of the center of the sphere (actually what direction you look if you are standing off center) and how big the sphere was. A ray tracing program would be able to predict this precisely.

2   a) Light doesn't hang around it would be absorbed quickly. You would absorb it and the mirror reflectivity is only about 90%, so if you turned off the light you would be literally in the dark.
    b) I don't think you would see anything different then if you did the same things in a dark room

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How do lasers produce phase coherence?I would like to ask you a question related with laser theory. How could it be explained, physically speaking, that when radiation interacts with an excited atom, the emitted photon is in-phase with, has the polarization of, and propagates in the same direction as the stimulating radiation? Why not in any direction, with different phase?

Question by: Nilton Haramoni, Brazil, Faculty Respondent: Steve, Sheldon

Lasers operate as their names imply "Light Amplification by Stimulated Emission of Radiation". Atoms can be put in what is known as a meta-stable state. Atoms Atoms can reside for a longer period of time in these meta-stable states. However, In this meta-stable state, atoms can be "tickled" or stimulated into dropping down  to the ground state, and in doing so,
emitting a photon. A single photon causes many more atoms to emit photons, and thus an "amplification" effect. Also, since they have a common stimulus, they have a phase coherence with the initial photon. For more details, a nice elementary description can be found at:  http://hyperphysics.phy-astr.gsu.edu/hbase/optmod/qualig.html

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Photon propagation through space & time? Since a photon diverts all its motion through space at light speed and has no motion through time (i.e. photons do not age), what is the "something" that diverts all its motion at light speed through time and has no motion through space? Does this "something" exist even theoretically because isn't it impossible on a quantum level to have zero motion? If it doesn't exist theoretically, does that imply that motion itself creates time and without it there is no time? What is the opposite of "no passage of time" (photon) -- "all time", "no time"? Is this possibly an example of a broken symmetry?

Question by: Debbie Peters, Faculty Respondent: Steve, Sheldon

I forgot to inform you that I am not a scientist, but an attorney. I've referred to your excellent website for years. (Debbie)

Answer: I disagree with the premise that photons have "no motion through time." In fact, they do move "through time." All objects move in three dimensional space as well as time, which is often considered as a fourth dimension. As an aside, photons can be "bent" as they travel through space by the gravitational wells which are produced by massive objects. This is in the realm of General Relativity.  I hope this helps.

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CP Symmetry and the Matter-Antimatter Asymmetry

Question by: Tim Kelley, Faculty Respondent: Artuso, Steve

I understand that there is evidence that CP-Symmetry is broken (such as in the decay of a kaon) and understand some about the concept of CP-symmetry. ( CP-Symmetry is when there is an inversion of charge and parity the system remains the same...) However, I do not fully understand how CP-Symmetry breaking could allow a particle to decay resulting in more matter than antimatter... Could you explain or point out an information source that might help me in my understanding?
We believe that at the beginning of the universe there was an equal amount of matter and anti-matter, and only later on the matter dominated universe was produced. One of the key ingredient of this is CP violation. It basically produced a slight excess of matter at the time when most of matter and antimatter finally annihilated into light and this is the matter that is nowadays universe.
I hope that this helps, (Marina)

Let me expand a bit on this Tim, at the risk of being too detailed. (Steve)
   It "was" thought (before CP violation was discovered) that all interactions (at least the ones we currently know about) produce matter and antimatter equally. That is, it was thought that the rate at which a particle decays is equal to the rate at which its corresponding antiparticle decays. Becasue of CP violation, this is not true, and they do not decay away at the same rate. A very slight asymmetry is observed which favors matter over antimatter. This asymmetery can yield a sizeable matter-antimatter asymmetry over time.
However, we should tell you that the so called Standard Model does not fully account for the observed matter-antimatter asymmetry of the Universe, and therefore, many (including us) feel that there must be some new physics out there (ie.., new forces, interactions) which we have not yet observed. These interactions may only "kick in" at very high energy. The search for these new phenomenon at very high energy is one of the motivations for building more energetic particle colliders. Alternately, once can search for signatures of this new physics by other means by measuring precisely the decays of so-called B mesons and their corresponding antiparticles "Anti-B mesons". More information about the Large Hadron Collider can be found at: http://lhc-new-homepage.web.cern.ch.
More information about our experiment to study B mesons can be found at: http://lhcb.web.cern.ch/lhcb/.

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In SUSY, are neutralinos considered as force carriers?

Question by: Cyrus Bryant, Faculty Respondent: Artuso

I have read that a leading candidate for "cold dark matter" in supersymmetry theory is the neutralino, a superpartner of the photon, Zo, and Higgs bosons. That would make it a fermion, I suppose. My question is: should this particle exist (to be discovered by the LHC team?) would it also be a force transmitter n the realm of superpartner physics?
You are actually correct, the neutralino is a hypothetical electrically neutral particle, a superpartner of the photon, Z-boson, and Higgs Boson. There are 4 neutralinos and the lightest one is one of the leading candidates for cold dark matter in several theories. They are fermions, although in some theories they are considered "Majorana fermions", namely they can self-annihilate. Their production mechanism and coupling to other elements of the SUSY ensemble depend on the specific variant of supersymmetry that you may want to consider, but I believe that picturing them as additional building blocks of matter rather than mediators of new forces is more accurate.

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How are particles accelerated at Stanford Linear Accelerator (SLAC) and what happens?

Question by: Boris Milvich, Faculty Respondent: Steve

My textbook of physics states that the accelerator at SLAC accelerated electrons to a final speed of 0.999 999 999 948 of the speed of light, but does not provide any other information about it. Here is what I would like to know:
1. What were the voltages used to accelerate electrons to this speed?
2. Did this accelerated electron participate in a collision with a particle at rest, and what was that particle?
3. What was the end result of this reaction?


I am from the former Yugoslavia, living presently in Snowmass, Colorado. I am working on a writing project that touches upon particle collisions. I found you though a web-search. It is very nice to have the opportunity to submit questions directly to a particle physicist, as the answers are often difficult to find.

Hi Boris,
    #1)  At SLAC, the particles are accelerated to 9 giga-elecronvolts (GeV, for short), and positrons are accelerated to 3.1 GeV. This means that the total voltage required is 9,000,000,000 Volts for the electron beam and 3,100,000,000 for the positron beam.
    #2) In fact, at SLAC they collide the electrons into the positrons, both of which are moving toward each other.
    #3)  The end result of the collision can be many possibilities. In general, you could get:

     a) electron + positron   ---->   electron + positron     (not that interesting, but useful for various calibrations)
     b) electron + positron   ---->   muon+ antimuon        (muon is like a heavy electron ~200 times as heavy)
     c) electron + positron   ---->   tau + anti-tau             (tau is like a heavy electron ~3600 times as heavy)
     d) electron + positron   ---->   quark + anti-quark    
               - Here quark can be any of 5 quarks  (up, down, strange, charm or bottom)

Many other things can happen as well, but typically with smaller probability...

The last one (d) , the bottom, or simply "b" quark is the most interesting for physicists at the BaBar experiment for a number of reasons.
The primary reason is that B quarks exhibit a phenomeonon known as CP Violation, which can give clues into the preponderence of matter over antimatter in our Universe (see the previous question for more thoughts on CP Violation). SLAC does maintain an educational set of pages if you have more questions, see http://www.slac.stanford.edu/

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Is mass conserved in particle collisions?

Question by: Boris Milvich, Faculty Respondent: Steve

An electron that is moving at 0.999 999 999 948 c is supposed to have a mass of 8.93x10^–26 kg just before a collision, according to Einstein's equation for a moving particle. This is the mass of about 98,000 electrons at rest. However, the result of the collisions that you gave in your answers of such an electron with an accelerated positron (98,000 positrons at rest) does not come even close to what it should be produced according to Einstein's equation. The reaction that you listed, electron+positron yielding a tau and anti-tau particles, accounts only for the mass of about 7,200 electron masses at rest.
(1) How do you explain this discrepancy?
(2) What happened to the unaccounted mass of approximately 188,000 electron masses at rest?



Hi Boris,
 (1)  After the collision, the tau particles are not at rest. Their rest mass is 3600 times the electron mass. But, they also have energy associated with their motion (or soc-called kinetic energy). The total energy, E,  is given by E = sqrt(p^2c^2 + m^2 c^4). So you have to include the momentum of the tau to get the total energy. So, the lefft over energy goes into energy of motion!
  (2) The "unaccounted energy" goes into kinetic energy... Total energy after the collision is equal to total energy before.

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Does an electronvolt reflect the change in speed of an electron as it crosses 1 volt?

Question by: Boris Milovich, Faculty Respondent: Steve

An electronvolt is described as the energy that an electron gains as it passes through an electric field generated by the potential difference of 1 V. It is also described as "the kinetic energy acquired" by an electron. Since the speed of motion is the component of the kinetic energy, does this mean that the extra speed that the electron gains as it passes through the field is built into the electronvolt? In other words, does the electronvolt reflect the change in the speed of the electron? Yes, 1 electronVolt (or [eV], for short) is the energy gained as an electron crosses a potential difference of 1 volt. This gain in energy goes into increasing the electron's kinetic energy. Of course as one gets close to the speed of light, relativistic effects come into play and must be accounted for.. 
Cheers, Steve

Follow-up clarification: 
The total energy of an object, E, according to Einstien is E=mc2. The energy E can be expressed in *units* of electronVolts (eV), by using the conversion that 1 [eV] = 1.6x10^-19 [Joule]. The "correct" relativistic formula for the Energy of an electron is E = sqrt (p2c2 + m02c4). Here, "p" is the relativistic momentum, , m0 is the electron rest mass = 9.11x10-31 kg, and c=3x108 m/s is the speed of light. The relativistic momentum can be expressed as p=g*m0*v, where g = sqrt(1/(1-b2)), where b = v/c. Using this, you can easily show that:

E = m0c2*sqrt( (b*g)2 + 1 ) = 9.11x10-31* sqrt( (b*g)2 + 1 )

The 9.11x10-31 [J] is the rest mass energy of the electron. From this equation, you can see that if you increase the energy, E, it is the product

b*g = (v/c) / sqrt(1-(v/c)2) which changes.

So, if you have the (change in) energy, you can readily compute the (change in) velocity.
Note that the relativistic "mass", m, is given by

m = g*m0,

so, as the velocity increases, so does g, and therefore so does the relativistic mass. The "rest mass", m0=9.11x10-31 kg does not change though as this is the mass
of the electron when it is measured *at rest* (and therefore relativity is not necessary)

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Do gravitons exist? Isnt it true that we still don't know for sure if gravitons exist? It's just the best explanation we have but in reality we don't know what gravity is?

Question by: Brad Mills, Faculty Respondent: Steve

Yes, that is true. There is no experimental evidence for gravitons (yet). Gravitons are the "force carrier" of gravity, which is analogous to the "photon" being the force carrier of the electromagnetic force. Clearly the photon is on solid experimental ground. As of yet though, there is no successful quantum theory of gravity, which would be described
by interactions involving spin-2 gravitons. Our best theory of gravity is general relativity, which is at odds with quantum theory, so a deeper theory is needed. People are working hard on trying to come up with such a theory. String theory is one possibility. For more information on String Theory, take a look at Brian Greene's "An Elegant Universe"..

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What is an electron's mass and energy while being accelerated? What about radiation in circular accelerators?

Question by: Boris Milovich, Faculty Respondent: Steve

I read in the textbook by Ohanian (Principles of Physics) that at SLAC electrons are accelerated to a speed of 99.999 999 994 8% of the speed of light, or 0.999 999 999 948c. I would like to know the following:

1. The actual combined applied voltage used to attain this speed?
2. The mass of the electrons at this speed expressed in kg, as the electrons exist the accelerator, and just before a collision? 
3. The energy of the accelerated electrons expressed in MeV? (Or, is the energy of the accelerated electron deduced from applied voltage?)

I read somewhere that in linacs, as opposed to circular accelerators, electrons do not lose gained energy due to acceleration. Was this the case in this example? If there were a loss, what would have been its magnitude in MeV or in percentage?


1) The applied voltage is not done in one step. Simply put, we can boost the kinetic energy (KE) of an electron as it passes across a voltage DV, by qDV. If you do his many times, you can eventually reach a very large KE. The KE is realted to the velocity, v, by  KE=gmc2-mc2, where g=(1-v2/c2)-1/2 ., and c=speed of light=3x108 m/s. If you add up all the applied "voltage kicks", it will add up to a very high voltage, of order several billion volts !

2)  Their mass is increased dramatically, m=gm0, where m0 =9.11x10-31 kg is the rest mass of the electron and g was defined above.

3) Just use E=mc2, using m from (2) above.

4)  When electrons are accelerated, they radiate Em radiation, and in this case, x-rays. In going around a circle, they radiate energy, and the power (Energy/time) ~ g4.
Take a look at the web page: http://hyperphysics.phy-astr.gsu.edu/hbase/particles/synchrotron.html. You will get a nice explanation there.

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Several questions on the unit of electronVolt [eV], forces and interactions

Question by: Boris Milovich, Faculty Respondent: Steve

(Steve) Hopefully I can clarify your confusion. All my responses are in CAPITALS below.
You are getting into the realm of particle physics. I strongly reccommend
"The particle Adventure": See http://particleadventure.org/particleadventure/


The reason why I recently submitted a few related questions, including the last ones, is that I am intrigued by the electron-volt as a unit of energy. I would like to understand what is in it and what makes it what it is. I also want to find out whether the electron-volt is a natural unit or whether it arose from some pragmatic reasons. I am looking for a more intuitive and more visual approach to it.

THE ELECTRON-VOLE (OR EV) IS JUST A DEFINED UNIT OF ENERGY. IT IS, BY DEFINTION, THE ENERGY A SINGLE ELECTRON GAINS WHEN IT CROSSES 1 VOLT POTENTIAL DIFFERENCE. IT CAN EASILY BE CONVERTED TO THE MORE CONVENTIONAL UNIT OF [JOULES] BY USING THE CONVERSION THAT 1 [EV] = 1.6X10^-19 [JOULES]. IT IS CONVENIENT TO USE [EV] WHEN TALKING ABOUT A SINGLE CHARGED PARTICLE, WHOSE ENERGIES ARE VERY VERY SMALL BECAUSE THEIR MASSES ARE SO TINY.

I am particularly intrigued by the proportionality between the applied voltage that generates an electric field and the acquired energy by a charged particle as it passes through it. I believe that this is an extraordinary proportionality; yet it is never recognized as such.

IT IS SIMPLY BY DEFINITION. SEE EXPLANATION ABOVE.

An electron-volt is an amazing unit. For each volt of applied EMF, an electron gains 1 eV! That is amazing. It is ever more amazing that the electron would gain the same number of eVs whether it passes through a field generated by 5 V potential difference or if it passes through 5 successive fields (like in a linac) of 1 V each, or a combination of 1V and 4V, or 2V and 3V. What makes this proportionality possible? Where does it come from?

THIS IS BECAUSE FOR EACH VOLT IT CROSSES, THE ELECTRON GAINS 1 [EV] OF ENERGY. IT DOESN'T MATTER IF YOU DO IT ALL AT ONCE, OR IN SUCCESSIVE STEPS. ENERGY IS A SCALAR QUANTITY. WHETHER I DO 1 [EV]+1[EV]+2[EV] IN 3 STEPS, OR 4 [EV] IN ALL 1 SHOT, IT'S THE SAME. IN BOTH CASES, THE PARTICLE GAINS 4 [EV] OF ENERGY.

In order for an electron-volt to be a natural unit, the above-mentioned proportionality must be related to or caused by another proportionality. So far, I only found one: The strength of the electric field depends on the number of electrons on the negative plate. The electric charges only come in whole units of e, which is the charge of an electron. In other words, it cannot be one half of an e, or any part of it. Therefore, the strength of an electric field can increase of decrease by a fixed value associated with the electric charges, or the number of electrons on the negative plate. In this respect, an electric field is a quantum field.

NOTHING MAGICAL ABOUT THE [EV], IT'S JUST A CONVENIENT RESCALING. AN EXAMPLE WHICH YOU CAN PROBABLY RELATE TO IS TEMPERATURE. YOU CAN MEASURE IT IN CELCIUS OR FARENHEIT. 0 C = 32 F, BUT THE TEMPERATURE YOU FEEL IS THE SAME IN BOTH CASES. IT'S JUST THAT SOME THINK CELCIUS IS MORE CONVENEINT, OTHERS LIKE FARENHEIT. THE OBSERVABLE QUANTITY (TEMPERATURE) IS THE SAME!

As an accelerated particle passes through this quantum field it gains extra mass, speed and energy from it. There must be a relationship between the proportionality in the strength of the field and the acquired energy by the electron. How can we explain this relationship?

CHARGES DO CREATE ELECTRIC FIELDS, AND OTHER CHARGES CAN, AND DO ACCELERATE IN THAT FIELD AS A RESULT. IT'S JUST THE ELECTRIC FORCE AT WORK! AS THE CHARGE ACCELERATES (SINCE F=MA) IT SPEEDS UP, GAINING ENERGY. AS THE SPEED APPROACES THE SPEED OF LIGHT, REALTIVISTIC EFFECTS BECOM IMPORTANT, AND THE MASS INCREASES.

Perhaps an accelerated electron exchanges virtual photons in an equal manner with each electric charge (or electron) that generates the field. In other words, an accelerated electron might be exchanging and absorbing virtual photons in a equal manner with each electron on the negative plate or positive charge on the other plate.

YES, ONE CAN THINK OF IT THAT WAY. BUT GENERALLY WHEN TALKING ABOUT LARGE NUMBERS OF CHARGES, A CLASSICAL TREATMENT IS MUCH, MUCH, MUCH EASIER, AND WORKS JUST FINE. SO I WOULD NOT EVEN THINK ABOUT EXCHANGING VIRTUAL PHOTONS UNLESS YOU ARE TALKING ABOUT INTERACTIONS BETWEEN TWO AND ONLY TWO PARTICLES.

(The interactions between an electron and the field is probably much more complex. I am refering to the net results.) The energy that the accelerated electron gains is in the form of absorbed virtual photons, which are the carriers of the energy of an electric field.
If this is the case, then the energy acquired by an accelerated electron will be proportional to the applied voltage, and the applied voltage will be related to the number of electrons on the negative plate, or the potential difference between the two plates.
An accelerated electron also gains real mass by passing through a field, as more EMF is needed for further acceleration. This mass must come from the field, that is, from the absorbed virtual photons.
But if the energy gained by an accelerated electron is always the same for every applied volt, then the mass acquired must also be the same per every applied volt. This is probably the reason why the mass of an accelerated electron can be expressed in eV, or eV/c^2, or any other particle.


NO, I THINK YOU'RE INCORRECT HERE, THE ENERGY GAIN IS RELATE TO VELOCITY (V) BY:  
E2 = P2C2+M2C4= g2Mo2V2C2+g2Mo2C4       WHERE g2 = (1-V2/C2)-1   , P = MOMENTUM, AND C=SPEED OF LIGHT.

According to this reasoning, an electron-volt expresses the energy that the acquired mass of an electron can perform as it moves at the speed of light. In other words, an eV is equal to mc^2, where m is the mass picked up the electron in the interactions with the field amounting to 1.783x10^–36 kg. That is, the energy 1.602x10^-19 J of an eV represents the internal energy of that mass.
Hence, the mass of an electron accelerated to a speed of 0.999 999 999 948c, passing through a field generated by 5 billion volts will be 5GeV/c^2, or 8.933x10^–27 kg. In other words, an accelerated electron would gain mass equivalent to about 10,000 electron masses at rest.


I'M NOT SURE I UNDERSTAND YOUR LOGIC. IN SHORT, IF YOU ACCELERATE AN ELECTRON USING 5 BILLION VOLTS (TOTAL), IT WILL HAVE 5 GEV OF ENERGY. YOU CAN CONVERT THAT TO JOULES USING 1 [EV] = 1.6X10^-19 [J], AND THEN EQUATE THAT RESULTS TO MC^2. YOU CAN THEN COMPUTE THE MASS, AND COMPARE TO THE REST MASS OF 9.11X10^-31 [KG].

In the last set of questions, I ask you how to calculate the mass of an electron accelerated to a speed of 0.999 999 999 948c. Your answer was to use Einstein equation where the mass m is equal to the mass of an electron at rest times the gamma factor.
This equation yields the mass of the accelerated electron of 8.933x10^–26 kg. This mass is 1,000% greater than the one calculated using the equation eV/c^2. According to Einstein's equation, the mass of an electron accelerated to the above speed would be almost 100,000 times the mass of an electron at rest. Einstein's equation does not yield the same result. How do we explain this discrepancy?


YOU MUST HAVE MADE A MISTAKE IN YOUR CALCULATION.

In spite of a great effort from my part, I was unable to get precise data on the specifics about the results of the collisions of the electrons accelerated to the above-mentioned speed. The textbooks either elaborate upon the achieved speeds, applied EMF, or the end reaction, but never all at the same time.
Since I like to verify everything that I read, I was unable to verify if the experiments in particle accelerators agree with theories


YES, THEY GENERALLY DO. IF THEY DON'T THE THEORY MUST BE RE-EVALUATED AND FIXED SO TO AGREE WITH WHAT IS OBSERVED. THIS IS THE WAY MODERN SCIENCE WORKS! THE "STANDARD MODEL OF PARTICLE PHYSICS" INCORPORATES ALL THE KNOWN FUNDAMENTAL PARTICLES AND FORCES. IT IS VERY SUCCESSFUL, BUT YET WE LOOK FOR A DEEPER THEORY. THERE ARE MANY REASONS FOR THAT WHICH I WILL NOT GO INTO HERE.

It appears that in collisions of an electron that is accelerated to a speed of 0.999 999 999 948c, the final reactions yielded only a fraction of the calculated mass of either of the two calculating methods. The lost mass during acceleration is too small to account for the difference. It seems that at very very very high speeds, something altogether different happens. It seems that all theories fail at this enormous speed. To get an idea what this speed means, consider this:
If an electron traveling at the above speed is only 1 mm away from a virtual photon, the photon would need to travel about 19,000 km, or about 1.5 diameter of the earth, before it could catch up with the electron and give it its momentum. This distance was found from the time relationships. The time for an electron traveling at the speed v=0.999 999 999 948c to travel a distance D until it is caught up by a virtual photon is equal to the time the photon would travel the same distance D+1mm at speed c. That is, D/v=(D+1mm)/c. Thus, distance D=0.001v/(c-v).
If we are to add to the speed of the electron one more 9 among the collection of 9s after the decimal point, but retain the distance of 1 mm between the electron and the virtual photon, the virtual photon would need to travel a distance of 192,310 km before it could catch up with the electron. That is half the distance between the earth and the moon.
This tells us that the virtual photons would have a little chance of interacting with the electron that is moving at the above speed before it exists an electric field and collide with a target.
In other words, an electron traveling close to the speed of light might not interact with the field in the same manner as is the case at slower speeds. Therefore, there may not be an exact or perfect proportionality between the applied voltage and the gained energy (or mass) of a charged particle. At slower speeds, these impefections may not be distinguishable.
What facts could confirm or discredit these possibilities? As my research is along these lines, I would greately appreciate a reply.


YOUR ARGUMENTS ARE NOT VERY CLEAR TO ME, BUT I ASSURE YOU THAT IT ALL WORKS OUT. ELECTRONS AND POSITRONS CAN ANNIHILATE INTO PURE ENERGY. ELECTRONS AND PHOTONS DO INTERACT, ETC.

My attempts to visualize the events in the particle accelerators are influenced by Hans Ohanian and his textbook "Principle of Physics." He presented one of the most eloquent description of an electric field, which I like to share with you. He wrote:
“At the quantum level, we can picture the field of force generated by a particle as a swarm of quanta buzzing around the particle. For example, we can picture the electric field surrounding an electron, or any other charged particle, as a swarm of photons. The swarm is in a state of everlasting activity - the charge particle continuously emits and reabsorbs the photons of the swarm. Emission is creation of a photon; absorption is annihilation of the photon. Here we can say that the electric field arises from the continual interplay of three fundamental processes: creation, propagation, and annihilation of photons. The action of one charged particle on another involves a sequence of these three fundamental processes: a photon is emitted by one particle, propagates through the intervening distance, and is absorbed by the other particle. This exchange process can be represented graphically by a Feynman diagram invented by R. Feynman. The photon exchanged by between the two electrons is called a virtual photon, because it lasts only a very short time and, being reabsorbed by an electron, is undetectable by any direct experiment. The steady attractive or repulsive force between two charged particles is generated by continual repetition of this photon exchange process. This is action-by-contact with vengeance - at a fundamental level, all forces reduce to local acts of creation and destruction involving particles in direct contact.”


YES, IN SHORT HE'S SAYING THAT AT THE QUANTUM LEVEL, FORCES CAN BE REALIZED AS THE EXCHANGE OF QUANTA, IE., PHOTONS. IN FACT, THE MOST SUCCESSFUL THEORY KNOWN, CALLED QUANTUM ELECTRODYNAMICS, DESCRIBES THE INTERACTIONS OF CHARGED PARTICLES BY THE EXCHANGE OF PHOTONS. THIS IS OUR BEST DESCRIPTION OF FUNDAMENTAL FORCES! (GRAVITY STILL NEEDS A CORRESPONDING QUANTUM THEORY THOUGH)

“According to the modern view, there is indeed an entity that acts as mediator of force, conveying the force over the distance from one body to another. This entity is the field. A gravitating or electrically charged body generates gravitational or electric field which permeates (apparently) empty space around the body, and forms an invisible disturbance in the space around the body. Thus, fields convey forces from one body to another through action-by-contact ...
Fields are a form of matter — they are endowed with energy and momentum, and they therefore exist in a material sense. If we think of solids, liquids, gas and plasma as the first four states of matter, then fields are the fifth state of matter."

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How do physicists know that particles observed during collisions are actual particles and not random pieces of the nucleons flying apart? (Shawn)

Question by: Shawn, Faculty Respondent: Sheldon and Steve

How do physicists know that particles observed during collisions are actual particles and not random pieces of the nucleons flying apart? What if smashing protons and neutrons together was analagous to smashing two glass jars together at high speed? The pieces would give almost no information about how the nucleus works! Alternatively, what if new particles were created during collisions that do not exist in normal nuclei? Wouldn't particle physics then be way off course in it's understanding of the atom?

(Sheldon) Leon Lederman used to joke that smashing two protons together was similar to colliding two garbage cans and watching all the garbage spill out! (Leon won the Noble prize for his discoveries using proton-proton collisions.) Protons are composed of three quarks, as are neutrons. When these particles collide the quarks couple to strong force field through a kind of particle called a "colored gluon." These gluons can interact with each other and produce other gluons and even more quarks. Thus we are producing many such particles depending on the energy of the collision. Now if the collisions are at low energy then we can examine the nature of the nucleus, i. e. as you say how it works. At higher energy the aim is foten to produce new states of matter that usually don't exist in a nucleus but did exist in the early Universe. We need to understand all the states of matter in order to understand the underlying theory of how matter and forces are put together. This is a large topic. Perhaps you might want to look at
http://particleadventure.org/particleadventure/ for more information.

Followup (Shawn): What causes charge? I've looked around but haven't really found any answer. It's probably not known, but what is the actual cause of opposite charges attracting and like charges repelling? It would seem that some process is occurring that results in force being applied over a distance.

(Sheldon) The electromagnetic force acts through electrical charge. The strong force acts through "color" charge. Why there are such forces and charges is a question that science cannot address at this point. We are getting pretty good and figuring out HOW these forces work, however.

(Steve)  Let me also chime in on this:
Perhaps a "classical" picture might help clarify why opposite's attract and like charges repel. (although it might beg for more "why" questions). Anyway:

In physics, a quantity call "Potential Energy" is often useful. Between 2 charges, it is: U = k*q_1 q_2 / d,  where, k is a constant, q_1 and q_2 are the charges of the two particles, and d their separation. Nature, if not perturbed, "likes" to be in a state of minimal, or lowest potential energy, namely the smallest value of U allowed.

If q_1*q_2 is negative ==> (i.e., opposite charges) ==> U is negative.
==> to minimize U, it should become "more negative", meaning the distance
should decrease. So as separation, d-->0, U --> negative infinity (attraction)

If q_1*q_2 > 0 (like charges) ==> U is positive.
==> to minimize U, they should move far apart, i.e., U--> 0 as d--> infinity.   (they repel)

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What is the status of tachyons?

Question by: Marvin, PhD Candidate in Philosophy, Faculty Respondent: Sheldon

I am a PhD candidate in philosophy with an interest in philosophy of science. I found you through a google search. My question is what is the status of the tachion hypothesis? Has it been disproved? If not, might their existencec explain dark matter and dark energy?

Usually we call them "tachyons" Up to now there is no experimental evidence whatsoever for the existance of these particles. They can be searched for by looking for "Cherenkov" radiation which would be abundent as they are moving faster than the speed of light. However, we long ago learned, that you can never disprove the existence of anything, all you can do is say you didn't observe it and give some kind of ratio to the particles you did observe.

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What's the difference between "time" and "proper time" in relativity?

Question by: philosophy student, Faculty Respondent: Marina Artuso

Question1: I've been reading article on the Internet eg the Internet Encyclopaedia of Philosophy at http://www.iep.utm.edu/ancillaries/Proper-Time.htm trying to get a handle on time.

Answer: They open by saying "The essence of the Special Theory of Relativity (STR) is that it connects three distinct quantities to each other: space, time, and proper time. 'Time' is also called 'coordinate time' or 'real time', to distinguish it from 'proper time'. Proper time is also called clock time, or process time." but that's not really very helpful.

Question2: In relativity, (in simple layman's terms) is my "proper time" effectively just the time recorded by a clock that I always carry around with me? What's the difference between "time" and "proper time"?

Answer: In special relativity time is sort of a "4th dimension": what this means is that while in classical mechanics if my friend and I synchronize our watches at the airport, they will keep displaying the same time for her (staying) and for me (boarding the airplane), this will not be the case in special relativity, because clocks in relative motion do not tick at the same pace. Therefore time measured in different "reference frames" for the same event are not the same. For any object or particle or person we can define a unique reference frame where it or he or she is at rest. Proper time is the time measured in this particular reference frame. In case you are wondering, in special relativity a "reference frame" may be defined as a set of 3 rulers mutually perpendicular and a clock attached to them. If you want to learn more I would strongly suggest "Relativity: The Special and the General Theory" by A. Einstein.

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Is the proposed neutralino of Supersymmetry a force carrier?

Question by: Cyrus A. Bryant, Faculty Respondent: Marina Artuso

Question: I have read that a leading candidate for "cold dark matter" in supersymmetry theory is the neutralino, a superpartner of the photon, Zo, and Higgs bosons. That would make it a fermion, I suppose. My question is: should this particle exist (to be discovered by the LHC team?) would it also be a force transmitter in the realm of superpartner physics?

Answer: You are actually correct, the neutralino is a hypothetical electrically neutral particle, superpartner of the photon, Z-boson, and Higgs Boson. There are 4 neutralinos and the lightest one is one of the leading candidates for cold dark matter in several theories. They are fermions, although in some theories they are considered "Majorana fermions", namely they can self-annihilate. Their production mechanism and coupling to other elements of the SUSY ensamble depend on the specific variant of supersymmetry that you may want to consider, but I believe that picturing them as additional building blocks of matter rather than mediators of new forces is more accurate.

Followup question: Thank you for your response to my question concerning supersymmetry. I would like to ask you another, realizing that you must have plenty to do without fielding my questions. I have just bought a slim book titled "A World Witnout Time - The Forgotten Legacy of Goedel and Einstein" by Palle Yourgrau (Basic Books, 2005), but haven't yet begun it. Isn't it true that some cosmologists think the laws of physics can be framed without reference to a time variable? Any measure of a time interval depends on comparison with some periodic physical action, yes? And the "time" component of a four-vector in relativity is ict, also a distance measure so that all components of a metric have the same dimensionality. At the supposed pre-inflation beginning of our Universe we speak of the first picoseconds or Planck times or whatever, but what actions were present that could provide a measure of such time intervals? At the enormous density in that event, what would have been the gravitational effect on any clock? It seems to me that the frequently glib descriptions of the Beginning for the general publicare a bit too confident. I would be keenly interested in your comment.

Answer: I will try to help out with what I know about the intricate set of questions that you lay out, hoping that it helps. First of all, I have not read the book that you mention, but I think that its core theme is derived by a discovery performed in 1949 by Gödel: studying the the equations of general relativity, he found nexpected solutions to these equations corresponding to universes in which no universal temporal ordering is possible. A hypothetical inhabitant of such a universe could, in principle, travel to his own past. Most scientists do not thin that this is a viable option. Now you bring into this discussion the planck scale and quantum gravity, a theory not yet fully implemented satisfactorily. At a planck scale, classical general relativity does not hold, and i do not think that we can think of effects on ordinary objects such as clocks in this very peculiar state of the universe. Lastly, yes in general the measurement of time use as a reference a periodic phenomenon. In the past it used to be the swinging back and forth of a pendulum, now we use some well measured atomic transitions. I hope that I addressed most of your points, otherwise feel free to follow up. Marina Artuso ps. if you find Goedel interesting, may I suggest "Incompleteness: The Proof and Paradox of Kurt Gödel" by Rebecca Goldstein, W. W. Norton: 2005.

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Can you suggest a creative gift idea for my son who wants to be a particle physicist?

Question by: Hanna Hadley, N. Carolina, Faculty Respondent: Marina Artuso

My friend and I homeschool our children. Her son, Michael, is in 8th grade and knows he would like to be a Particle Physicist when he graduates from high school. What can he do now to prepare himself for this field of study? Also his birthday is next week and I would love to give him a gift relating to this. Can you help me with any creative gift ideas? Any information you can give me on the subject would be helpful, especially where schooling is concerned (special subjects he should study, experiments that can be done now or soon in the future, labs, etc.).

I am assuming that you have access to the web. If so, I would suggest that Michael experiments with the web site http://particleadventure.org/particleadventure/ this web site has also book suggestions, and other resources. It is also available as a CD, let me know if you need information on how to get it. For the birthday, for a kid of that age group I would need to know a bit more about the kid, if there is a discovery store nearby you, you may find something interesting. It may be way late for the birthday, but something that may be cool for another occasion is to get a t-shirt or something else from http://physics2005.org/merchandise.html [the american association of physics web site, selling small gadgets to celebrate the year of physics]. Lastly, if there is a research university close to where you live, I suggest that you try to get in touch with some local faculty, some of them invite high school students to participate in research internships. For example, I had high school student working in my laboratory a few hours per week in the past.

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Is there an excited state of the proton at 13 GeV ?

Question by: John Hagel, London, England, Faculty Respondent: Steve

I am actually a chemist by training, but, whilst doing research at Caltech in 1976-78, I became interested in particle physics. Whilst at Caltech, I seem to remember an infamous professor of physics by the name of Ricardo Gomez once telling me that the proton had an excited state at somewhere around 13-14 GeV. Is that true, or have I dreamt it? If it does, has a satisfactory explanation for this fact ever been put forward? I can find no reference to it in modern textbooks of particle physics.

I do not know of any excited state of the proton at that mass. There *is* an excited state, called the "Delta" resonances. The Delta resonances have a mass of 1.23 GeV (10 times less than you mention) and include 4 states: Delta- , Delta0, Delta+, and Delta++ (each differing by 1 unit of charge). The Delta+ has exactly the same quark content as a proton (ie., 2 up quarks and 1 down quark), but the spins of the quarks (each spin 1/2) are all aligned to give Spin = 3/2. For the proton, one has spin=1/2. So perhaps this is the excited state you are referring to?

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How can one explain attraction between charged particles using the concept of photon exchange?

Question by: Santosh Kumar, New Delhi, India, Faculty Respondent: Sheldon

I have just completed my masters in physics. For quite some time a question is bothering me, and that 's ''how the exchange of photons lead to attraction between two oppositely charged particles''. The repulsion is still easier to visualize, but in case of attraction it becomes slightly non trivial.

You could possibly think about it like this: the force results from not only the photon exchange but also the coupling "factors" that depend on the charge of the two particles. The photon exchange results in a - (repulsive if you wish) interaction and the coupling factors are the sign of the charge. So for two electrons we have - times - times - or an overall - sign or replusion. For proton plus electron we have - - + which results in + or overall attractive force.

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What is the connection between temperature and speed of particles?

Question by: Abbas Merchant, Faculty Respondent: Steve

Q:Is there an upper limit to the temperature of ordinary matter? Since temperature is measured as the average energy (or speed) of the particles, I would think that temperature would be limited by the speed of light. Is this so? And if additional energy is applied to these super energetic particles, where does it go? Do they become more massive? At what temperatures do these effects become significant?

A: Yes, temperature is a measure of the average kinetic energy of particles. At high enough energy, the particles begin to experience relativistic effects, which does include their masses increasing, as you suggest. Their total energy is given by
E=gmc2,
where m the rest mass of the particle, and g=sqrt(1-v2/c2) is the relativistic boost. The kinetic energy (KE) is the difference between the total energy (E) and the rest mass energy (E0=mc2)
KE = E - E0 = gmc2 - mc2,
Classically, the average KE of an ensemble of particles is KE = (3/2)kT, where k=1.38x10-23 [J/K] is Boltzmann's constant and T is the temperature in degrees Kelvin. However, this is a macroscopic quantity, and n the above we are talking about the energy of a single particle.

As you increase the temperature, the average KE of the particles increases, which means their speed's increase and also their mass increases. Of course you can continue to add energy, and the speeds will increase and they will get closer and closer to the speed of light (but never quite get there). If one roughly equates the thermal KE to the energy of an ensemble of electrons, the effective temperature of this "ideal gas of electrons" would be ~a billion degrees..

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How are wave energy and photon energy related?

Question by: Boris Milovich, Faculty Respondent: Steve

Q: The following question might be a silly one; however, I want to be 100% sure about it. All textbooks of physics state that the frequency of electromagnetic radiation indicates the number of waves emitted per second. However, none of the textbooks I consulted state unambiguously that frequency indicates the number of photons emitted per second. Therefore, my question is: Does the frequency of electromagnetic radiation indicate the number of individual photons released per second, so that each photon has energy hf (Planck’s constant times frequency), and that each wave represents a photon? If this is correct, then the total energy emitted by an electron in an atom per second would be hf times frequency f.

Answer:
* In classical electromagnetism, one deals with a WAVE picture. Here, frequency is the number of waves per sec which pass a fixed point. * In the particle (quantum) picture of light, a "wave" is composed of a very, very large number of photons. EACH PHOTON has energy, E=hf. If you have a wave of some known frequency whose energy is known, say E_wave, then the number of photons, N, is simply N = E_wave/hf. Perhaps an example would help: Suppose you had a 100 Watt bulb, which emitted blue light at 400 nanometers (400x10^-9 m). frequency = speed of light/wavelength = 3x108/400x10-9 = 7.5x1014 [hz]. Then, in 1 second, the energy emitted in the form of blue light is: E_wave = 100 W * 1 sec = 100 [Joules] Now, a SINGLE PHOTON has E_photon = h*f = 6.6x10-34 (J*s)*7.5x1014 [hz] = 4.95x10-19 [Joules/photon] So, the number of photons emitted from the sources in 1 second is: N_photon = E_wave / E_photon = 100 [Joules] / 4.95x10-19 [Joules/photon] = 2x1020 photons ! or 200,000,000,000,000,000,000 photons!!!! That's alot of photons!!

Followup Q:
Thank you very much for making an attempt to answer my questions. I am sorry but I cannot picture the answers. Your example of a 100W light bulb does not give a clear understanding of frequency of emission of photons because of numerous single emitting sources (atoms).
> Let us consider a singe oscillator, a single hydrogen atom and its single electron, for example. It is said that when an electron moves to a lower orbit it emits a photon of energy hf.
> Question #1. From the particle point of view, what is the frequency f in this case? Is the frequency associated with the number of emitted photons per second by a single electron, or does it simply indicate the energy content of photons, or both?


Answer: You can imagine a photon as a "packet of light". It has a wave function (quantum mechanics). The frequency does corresponds to a location in the EM spectrum, just as classcially. For the hydrogen atom example, the frequency of the photon, or light emitted, is given by (E_f - E_i) / h, where E_f and E_i are the final and intial energy levels of the electron in the hydrogen atom. Frequency is NOT the number of phtons emitted/sec.
> The ray of light is often represented as a series of sinusoidal waves. They are supposed to represent the number of emitted waves per second. Each individual complete wave is supposed to represent a single wavelength. >

> Question #2. In this case of a single oscillator (a single atom and a single electron), and these sinusoidal waves, what are the boundaries of a photon? That is, how does a photon possessing certain energy relate to these waves which also represent energy? > The sine waves represent something that occurs in time and space. Photons that are emitted by a single hydrogen atom are also emitted in the same time and in the same space. Therefore, the sine waves must correspond to the emitted photons in some fashion. > I understand that sine waves explain the nature of light from the wave view of light, and that the photons explain the particle nature of light. The sine waves do explain what the frequency is in the wave theory of light (number of waves per second is the frequency). But what is frequency in the particle aspect of the same radiation when emitted by a single oscillator, a hydrogen atom with its single electron?
>Answer: Frequency does refer to a location in the EM spectrum. But the photon is "localized" in space/time, just like a ball. The big deal here is that the photon has a wave function, which only describes the probability of finding the photon at some location X, at time t.
> To simplify my questions, draw a few wavelengths on a paper, then draw below it a few pockets or bundles representing photons emitted by a single oscillator. Both represent the same thing that occurs at the same time and in the same space. > In the first drawing, we can determine exactly where and when a wave starts and ends, and how many waves are emitted per second. > We should be able to do the same with photons from the particle view of light.

> Question #3. Does one wavelength correspond to a single emission of energy by an electron as it moves to a lower orbit?

>Answer: Yes, as an electron jumps from a higher energy level to a lower one, it emits a single photon. A wave would consist of many such photons.

> Question #4. In the wave theory of light, does each wavelength represent energy hf? If it does, then each wave would have to correspond to a photon, since the energy of a photon is also hf. If it does not, how many emitted waves amount to energy hf?
> Answer: No, as soon as you use "h" (Planck's constant), you are in the quantum world.Classically, the energy of a wave is proportional to E^2+B^2, where E and B are the electric and magnetic field which mek up the wave,

> Question #5. Since frequency indicates certain periodicity in time and space in the wave theory of light, what kind of periodicity do we have in the case of photons as particles?
>Answer: Again, you're mixing the wave and particle picture. I would not try and relate the periodicity of the wave to that of a photon. A photon is a quantum-mechanical object, and only has a probability amplitude associated with it.

Question #6: My impression is that in modern physics frequency is explained as a wave property, and left at that. But the energy of photons in the particle view of light is expressed by hf, where f is frequency, indicating certain periodicity. We need to define what is frequency in the particle theory of photons and how it differs from or corresponds to frequency in the wave theory of light, both occurring in time and space. For the purpose of clarity and simplicity, we have to start win a single oscillator, like an electron of a hydrogen atom continuously emitting photons of light of the same wavelength. Obviously, this electron can emit only one photon at a time. These photons fall on a frequency detector, which gives us a certain number indicating certain periodicity. What kind of periodicity does this number indicate in the particle view of light?
>Answer: They wil hit the detector randomly in time, but perhaps with some avergae rate. The averge rate is related to the INTENSITY, not the frequency of light. The frequency gives the place in the EM spectrum (ie., the color, if in the visible range)

> Questions #7: Here is the last question that might shed light on everything above: Suppose that individual photons of yellow light (577 nm) from a single source fall on a frequency detector one second apart (one second between each hit), then 10 seconds apart. Will the detector identify photons as photons of yellow light of frequency (5.20x10^14 Hz) in either case? >
>Answer: Yes. a photon has a specific frequency associated with it, which designates it's "color" or location in the EM spectrum.

>Question #8 If the answer is yes, then the frequency detector identifies photons according to their energy content. If the answer is that it detects the arrival of photons at the detector, one per second, or one per 10 seconds, then the detector indicates certain periodicity in time and space, that is, the number of photons per second. This would give us the answer what frequency is in particle view of light, if there are no other possible answers.
>Answer: In the wave picture, frequency describes both the number of waves/sec passing a single point AND the place in the EM spectrum. In the quantum picture, it only designates the place in the EM spectrum.

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Wave function of a deuteron: I am a high school physics teacher and I wonder will the wavelength of a travelling deuteron at a certain speed be calculated with the sum of the masses for the neutron and proton or is this treated as two waves coupled together in other words in quantum calculations does mass act like it does in gravitation calculations ?

Question by: James Tittle, MSHS, Faculty Respondent: Sheldon

I would treat the deuteron as an object with its own wavefunction; thus the deuteron mass would enter not the separate neutron and proton masses. I can do this because the deuteron is its own coherent quantum state.

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Degenerate Pressure of a Fermi gas: I have been teaching astronmy for 37 years at Calif. State Univ., Chico, CA and try to always appeal to the underlying physics - forces which explain the things that we observe in the universe. There are four fundamental forces in the universe: gravity, electric, nuclear and weak. White dwarf stars are held up by what is called degenerate gas pressure which comes from the Pauli Exclusion Principle, which says no two Fermions can be in the same place at the same time. What force keeps the two Fermions out of the same energy level? What is the underlying force which causes degenerate gas pressure?

Question by: Jim Regas, Calif. State Univ., Chico, CA, Faculty Respondent: Sheldon, Steve

Well you probably have a better understanding of this than we do, but let me give it a shot. Quantum Mechanics gives us rules which must be obeyed. That two fermions can't be in the same state is a rule. You are asking how is the rule enforced in terms of forces? I would say that the wave function is not allowed to have a finite probability to be in a region of space-time. There is no force per se that enforces the rule. (Sheldon)


(Steve)Just one additional point. There is no fundamental force involved here. One can think of the "pressure" as being caused by the kinetic energy of the fermions in the gas as they are forced to occupy higher energy levels (due to the Pauli Exclusion Principle).

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What do Particle Physicists do?: I am an aspiring physicst who is currently attending high school. My English teacher has given us a term paper to do. Would you tell me what your job consists of

Question by: Robert Dean, Faculty Respondent: Steve

Well, I could go on for a long time on what we do. Let me be brief. Our goals as particle physicists are to understand nature at its most fundamental level. That it, we seek to identify and understand the most fundamental constituents of matter. Nowadays, these are the quarks and leptons, although it is certainly plausible that they are not fundamental. All w can say is that their size is less than 10-18 meters. Our second goal is to understand how the fundamental particles interact with one another. Today, we know of four forces (from weakest to strongest): Gravity, Weak force, Electromagnetic force, and the Strong Force. There is some hope that with a more complete theory, these four forces may be in fact different manifestations of the same fundamental force (so called "Unification of the Forces"). All of this is going to get veryexciting next year as a powerful accelerator, called the Large Hadron Collider (LHC) turns on and smashes protons into protons at 7 TeV.

There's a lot of work to do, so we hope you continue to pursue physics. There are many fruits for your labors!

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Do protons, neutrons, or quarks have color? When I say color, I mean color that we could see. I know we see color because light is given off by electrons as they rise and fall in energy levels. If you had a proton under a microscope that would allow you to see it clearly, what color would it be? Would it have color? There are no electrons to "give" it color to see. Can quarks and protons do the same thing as electrons (without the movement into different energy levels)? I know that question sounds dumb because if you could see a proton with a microscope, you would see the protons in whatever other matter is there. Actually this makes me think of another question too. What shape are protons? Books always show a sphere. If they are made of three quarks wouldn't they be triangular? Unless the quarks move around inside a proton so fast that it would appear to be a sphere.

Question by: anonymous, Faculty Respondent: Sheldon

Lets start with color. When you see a certain color on an object that means it reflects light corresponding to the wavelength of the particular color and absorbs other wavelengths in the visible range. Now what cause this is the structure of the atomic electrons on the surface of the object. .
Now, protons and neutrons are made up of quarks but quarks cannot be separated out of these objects. In other words, they exist only as objects contained inside other particles. Now you could conceivably get quarks excited to other energy levels absorbing some wavelengths but not others, but the energies involved are many orders of magnitude higher than optical wavelengths visible to the human eye.
So it doesn't make sense to talk about a color for them. Now for quarks. They are considered to be fundamental objects with no internal structure, although that has been questioned. Experimentally we have not been able to measure the effect of any size of a quark or an electron, for that matter. You may have seen that quarks are labeled with a "color" quantum number. Here "color" does not mean color in the ordinary sense, but merely that quarks come in three different varieties, typically labeled as red, yellow, blue, or some other equivalent scheme.

You also ask if protons are round? I guess I would ask, what would make them not round? It's true they are "lumpy" due to the quarks, but they are moving around very quickly inside the proton, so overall effect is that they appear round to us experimentally.

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Black holes, neutron stars and their collapse: It seems like instead of a Black Hole, the collapse of a star should produce an emmitence of energy as the particles of matter destroy each other. Could a black hole be a temporary state of matter between a regular star and a neutron star? Why would a regular star collapse and not a neutron star or a magnetar? Do magnetars collapse? What do their collapses make?

Question by: Erik Traulsen, Faculty Respondent: Steve

Great question. Let me give you a few interesting points first:

1) Theoretically, all matter is composed of point-like (zero extent) particles.
(The experimental limit is that the fundamental particles are smaller than 0.0000000000000000001 meters in size)

2) All of matter takes on its shape, size, properties as a result of the forces. When you touch a table, you're NOT feeling the electrons on the outermost atoms, you're feeling the effects of the electromagnetic force!

3) Stars are nuclear reactors, and rely on the radiated energy to balance the mutual gravitational attraction of all the matter in it. That balance between the gravitational force trying to collapse the star, and the nuclear reactions pushing outward, accounts for a star's size.

But, what happens when all the nuclear fuel (hydrogen, helium, etc) in the star is used up?

Well, the star collapses under the force of gravity!

If the star is ~2-3 solar masses, it typically forms a neutron star.

If it is larger, the star often collapses into a black hole. In principle, most of the mass of the original star collapses to a single point in space. This point can have zero size, in principle, because of point (1). In reality, we don't have a successful theory that describes what this "singularity" looks like, since it requires combining General Relativity & Quantum Mechanics, which is a very difficult problem in physics. So, a black hole is not an exotic form of matter. It is the ultimate fate of stars with mass more than 3 or so times the mass of our sun. Neutron stars do not become black holes simply because the gravitational force associated with the mass of the neutron star is not strong enough to "crush" the neutrons. In the field, they call this "pressure" the neutron degenerate pressure. In a black hole , there *is* enough mass to overcome the
neutron degenerate pressure and the neutrons collapse under gravity's immense force.

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Questions about light :1) What causes light to slow down in a bose-einstein condensate? 2) Does light gain mass? 3) Does light generate gravity?

Question by: Tommy Lee Bartles, Faculty Respondent: Marina

1) Light slows down in a Bose-Einstein condensate if it is tuned to the right frequency: the atoms in the condensate respond all together to that light, absorbing it and re-emitting it, slowing it down. This happens to a lesser extent when light passes through glass or any other transparent medium: the light is shifted and slowed by the atoms in the material.

2) Light is always massless - it has energy, but all experiments show that to a very high precision, it has no mass. Any particle that travels as fast as space-time allows will have no mass, like light.

3) Yes, light does generate gravity, even though it has no mass. Einstein proposed a relation between the "stress-energy tensor" and the gravitational field. Energy is part of that equation, so since light has energy, it distorts space-time. For energies of light that we can see and produce, that effect is very small. However, any light energy trapped by a black hole, for example, makes it stronger.

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Are the weak and strong forces mediated by force carriers?

Question by: Brad Mills, Faculty Respondent: Mariana

Being that the strong and weak nuclear forces are not attributed to the exchange of particles why is gravity attributed to supposed gravitons? It seems to me that gravitons are a myth.
Brad Mills

All the fundamental forces are presently seen as "exchange forces": the weak forces are mediated by the W and Z bosons, the strong forces are interpreted as gluon exchange and the electromagnetic forces are interpreted as photon exchanges. We do not have empirical evidence for gravitons, but they are motivated from a scientific theory, based on the analogy with the other forces that I have listed, and not by a myth

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The Cosmic Bathtub Drain: I realize this is outside your expertise, but I'm sure it's general enough to be within the scope of your training. I just hope it's not SO general as to be a waste of your time. If so, perhaps you can forward it to someone with copious free time to fritter away.... :-) Essentially, the question is this: How can it be determined that, instead of the universe expanding, it is actually being sucked into a very powerful singularity? It seems to me that if this were true, the galaxies "closer" to the event horizon of such a singularity would be accelerated so that they would be red-shifted relative to the Milky Way. Likewise, we would be accelerated beyond those that are "farther" from such an event than we, and thus they would be red-shifted too. I suppose that this hypothetical singularity would not have a specific position in current 3D space, just as the Big Bang's hypothetical singularity did not. In fact, the more I think about this Cosmic Bathtub Drain, the more it seems to be just the other side of the Big Bang coin. In other words, it seems that since the two ideas are reciprocal, maybe both (or neither) are true....

Question by: Steve Barger: neurobiologist in Little Rock, Arkansas, Faculty Respondent: Sheldon

I suppose it could well be a singularity as you propose. However, the question then is what causes the singularity?

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Light reflection in a sphere: I am in the concrete industry, and my question in no way relates to my work. It is a discussion that some friends and I have been tossing around for a couple of weeks now. We are all college graduates, stemming in business, biology, and a few elementary education majors. Our basic experiment would involve the properties of a two-way mirror, or one-way glass, as it is called. The idea originated thinking that if you were to make a sphere out of this material, that light would be able to travel into the sphere, but not out, causing some sort of energy buildup, maybe ending up in an explosion, or bright light flash, or cracked glass, or nothing at all. (all ideas that we have come up with) After researching the topic as much as I could, with the limited amount of time I have here at work, I have found that there in fact is no such thing as a true two-way mirror. There is just a thin reflective coating on the glass called a 'half-silvered' surface. What would happen to this sphere when shining a laser into it. What if it were set outside to sit in the sun for minutes, or hours. Would the slight percentage of reflection of light build up to anything? When looking at this sphere, there would obviously be a void of light, due to the lack of light reflecting, but what would that look like? After discovering this, I in turn tried to come up with another idea that would be along the same lines. If you were to make a sphere out of a true mirrored surface, allowing no light transference, but had a laser of some sort inside, what would happen? It could be as small as a baseball, or as large as one could imagine. Would the effects of the casing be different if it were a cube shape? Or any other shape for that matter? Around the office, it has become known as the 'Light Grenade Theory'. We have been told by people that have no idea what would really happen that the effect would be anywhere from a flash bang to the next big bomb device, all the way down to nothing happening. I thought that I would get an opinion of someone who can place a scientific background

Question by: Joe Degitz , Faculty Respondent: Sheldon


Let me start with the second part of your questions. If you had a reflecting "ball," shape immaterial, and put a light source in it, what would happen.

The reflecting material on the inside of the ball is not 100% reflective. Even if its 99.999% reflective, and you pulse the light source on and off, within a very short amount of time all the light inside will be absorbed into the walls of the ball. Therefore the ball will absorb energy and will heat up. Given enough light the heat could be large enough that the ball will explode, but generally it would just crack.

On the first part, you are right there is no such thing as a one way mirror. Generally if you left such a ball out in the sun, it would trap some light but would quickly reach an equilibrium. You could see it because it is partially reflective. It would heat up from the outside due to the Sun....

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Is there any chance that mini black holes created by things such as the LHC could destroy the Earth?: I have been traveling so was not able to get to this before now. A good answer is "The black holes produced at CERN will be millions of times smaller than the nucleus of an atom; too small to swallow much of anything. And they'll only live for a tiny fraction of a second, too short a time to swallow anything around them even if they wanted to.

Question by: anonymous, Faculty Respondent: Sheldon

If it makes you feel any more comfortable, we're pretty sure that if the LHC can produce black holes, then so can cosmic rays, high-energy particles that smash into our atmosphere every day. There are probably a few tiny black holes forming and dying far above you right now. So I think we should all relax, fire up the Large Hadron Collider, and get ready for a view of the universe that we've never seen before."

For more information see http://www.csmonitor.com/2003/0523/p25s02-stss.html

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Why does lunar cycle / speed of light approximately equal the strength of gravity, g?

Question by: Robert Elliot, Faculty Respondent: Sheldon, Steve

why is it that the exact speed of light divided by the exact lunar year is exactly the acceleration of gravity???? 983571087.90472 ft/sec (speed of light), divided by 354.357 days (30616444.8 seconds)(lunar year) dividing speed of light by time of a lunar year -- 983571087.90472 / 30616444.8 = 32.1256 ft/sec/sec gravity varies from 32.116 to 32.228 ft/sec/sec if this is a coincidence, it is a precise coincidence.

(Sheldon) I would argue its a coincidence. Furthermore, its not exact. The value of does vary as you point out, so why pick 32.156?


(Steve) Also, note that the value of "g", gravity is not a fundamental quantity, but rather depends on the gravitational constant, G, the earth's mass (Me), and the distance from the center, r2, as: g = GMe/r2. Since many celestial bodies, such as the sun, other planets, etc have different masses, "g" will be different for those objects as well. So, I agree, that this is a mere coincidence that they are close.

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Education for a particle physicist.

Question by: Joe Abrams

What schools to attend before becoming a physicist (What types of schools, ie: 4 year college, or specialty schools):

You can obtain some physics related jobs with a 4 year undergraduate degree, but to do research you will need a Ph.d. degree which takes an additional 5-6 years and would be done at a University, such as Syracuse.
What is needed to progress to top of field: Continuing education required?

Ph.d. degree with research activities.
What is required or recommended?

This depends on the subfield in physics you choose to work in. Some subfields are more company oriented, while others are more academically oriented. For example, theoretical particle physics would be mostly at Universities.
High school: What can an interested person start doing now (ie: volunteering), or what courses should he/she take in high school?

Take as much math as possible, specifically calculus. If there is a good physics course you should take that also.

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Several questions on matter and antimatter differences

Question by: Theresa Mann, H.S. teacher, Faculty Respondent: sheldon

(Teresa) Thank you so much for your replies. I can't wait to share them with the students tomorrow. These are all questions that they generated. They are thinking about this material far more deeply than any former class.

Q: How, in terms of fundamental particles, does beta decay occur? We are specifically wondering how a neutron appearantly is converted into a proton and a beta particle.

A:The quark content of the neutron is u d d. In beta decay one of the d quarks transforms into a u quark and massive particle called a W- and a u quark. The uud then form a proton while the W- transforms into an electron and a anti-neutrino (actually, precisely an anti-electron type neutrino.) The neutrino is usually not observable since it doesn't interact much with matter. This kind of decay is called a "Weak" decay as it occurs by the Weak interaction. By the way the W- is a "virtual" particle in this decay; e. g. real W- particles have masses much much larger than the d quark, but since the transition occurs over a very very small time then it can happen actually as a consequence of the Heisenberg uncertainty principle. See http://particleadventure.org/frameless/npe.html for pictures

Q: Does anti- matter change simultaneously to the change in matter?

A:I am not sure I fully understand the question. When we now produce anti-matter in the laboratory, we change energy into matter + anti-matter. So we produce, for example, a proton and an anti-proton simultaneously. This was supposed to happy also in the early Universe, so there was, immediately after the big bang, an equal amount of matter and anti-matter. We now have found that there is difference between the decay rates of matter and anti-matter, so this contributes to an imbalance of matter and anti-matter but the effects we know about are not large enough to account for the fact that the Universe is mostly matter these days.

Q: Where is the anti-matter? Has it been isolated?

A:Its anti-baryons composed of three anti-quarks, or mesons with the opposite quark combinations than for matter. The anti-quarks have opposite charges to the quarks. Examples: a) Anti-proton: anti-u, anti-u, anti-d, so the charge is -1. b) Particle K- meson composed of s anti-u, while the anti-particle is a K+ composed of anti-s u. Yes anti-matter is easily produced in particle accelerators. It was first discovered in an experiment at Berkeley Ca. in 1955. The 1959 Nobel Prize was awarded to Emilio Segre and Owen Chamberlain who led the experiment. One of my friends Tom Ypsilantis was a graduate student who worked on the experiment. Even before that happened in 1932 Carl Anderson discovered the positron which is an electron with positive charge using cosmic rays. Electrons are fundamental particles just as we believe quarks are. See http://cerncourier.com/main/article/45/9/23 Recently (1995), atoms of anti-protons and positrons have been produced -anti-Hydrogen. See http://hussle.harvard.edu/~atrap/

Q: What are some examples of applications of particle physics in everyday life?

A: a) Pays my salary, or at least part of it.
b) More seriously, we are trying to understand the fundamental forces in nature and basic states of matter. This is also connected with observations from the cosmos, the "dark matter" problem and the even more mysterious "dark energy" problem. The dark matter problem has been known from the 1930's when Fritz Zwicky found that Galaxies didn't rotate as one would expect them too, unless they were embedded in a cloud of invisible matter, called dark matter. We still don't know what this is.
c) There are also concerns in our theory, called the "Standard Model," that we really don't understand how particles get their mass and how the forces could translate to higher energy scales.
d) There are also spin offs. Technologies that arise from particle physicists doing their research that are useful for others. Examples are new accelerators using protons that can much more effectively kill tumors than using x-rays (photons) and the construction of machines that produce intense bunches of light, called "light sources," which are electron accelerators used for many studies in biology and material science.

Q: Can you explain the charge nature of the anti-neutron. We understand that it is made op of oppositely charged anti-quarks, but if the effect is the same what is the overall difference between the neutron and the anti-neutron?

A: The anti-neturon is formed of anti-quarks, so in a world dominated by anti-matter it would behave the same as a neutron in our matter dominated world. On the other hand if an anti-neutron ever met a neutron it would annihilate into energy and there would be no matter left.

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Are there any good books at the elementary level for modern physics concepts?

Question by: Stan Jankowski, Faculty Respondent: Steve

I found you by simply typing "ask a physicist" in a search engine. I am a great fan of the late Dr. Isaac Asimov and have read his book "Atom" from cover to cover many times. The book was published by Penguin Group, N.Y., N.Y. in 1991. I would like to know if your are aware of a book that covers more recent developments and is written on a similar level. Although I have a healthy interest in the subject, I have no formal education. If your so inclined, pose my question to your colleagues.

A book which is quite readable is the book by Brian Greene, called "An Elegant Universe" http://www.amazon.com/Elegant-Universe-Superstrings-Dimensions-Ultimate/dp/0375708111
The first 2/3 of the book is very appropriate for anyone with a healthy interest in modern physics. The last 1/3 gets into more details about String theory, and is a bit more complicated.
Anyway, I highly recommend this book for a nice read. Also, NOVA made a special on this, which is a bit "watered down", but is also very good.
See: http://www.pbs.org/wgbh/nova/elegant/ ---> Click on "Watch the program"

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Tides & Cosmology questions

Question by: Martin Glover, Faculty Respondent: Steve

1. From earth the moon has the same apparent diameter as the sun. I assume that the sun is denser than the moon so why doesn’t the sun have at least as great an affect on earth's tides as the moon?

The sun does have an affect on the tides, but it turns out to be less because of a a (r/R)^4 dependence of the forces. Here, r is the radius of the earth, and R is the distance from the moon or sun. Note also that it is only the horizontal forces which create tides, not the total gravitational force. If you want to understand this better, I found a nice web page: http://www.du.edu/~jcalvert/geol/tides.htm

2. How many light years across is the universe? I've seen estimates that are far greater than the estimated age of the universe. How is that possible? If inflation only brought the universe to grapefruit size that doesn't seem to be much of a factor in bringing it to its present size.

Well this is a tricky one. The visible Universe is about 14.5 billion light years across. In a static Universe, this would mean that the furthest objects are 14.5 billion light years (BLY) away. Here, a light-year is the distance light travels in a year. However, the Universe id not static, in fact it's expanding. That is General Relativity and Cosmology indicate that space itself is stretching. A result of this stretching is that we can "see" things, in principle, which are well beyond the 14.5 BLY distance. Unfortunately, the light from many of these objects are too faint to see them, but in principle, their light can reach us.

3. The universe is expanding and the farther away it is the faster it's expanding. Are there parts of the universe that are so far away and moving away from us so fast that we never see their light?

Yes, absolutely, this should be the case. On the other hand, the only thing we can discuss is the "visible Universe". We don't know what's beyond that. Is the Universe infinite or finte, open or closed, flat or curved? We think it's flat, based on observations of the Cosmic Microwave background. Lots of great information is available at WMAP's web site. http://map.gsfc.nasa.gov/

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Variety of questions: Electroscope, Stationary States, Neutrons, more

Question by: Tom O' Kane, Faculty Respondent: Sheldon

why do the divergent leaves of a simple gold-leaf electroscope with a charge of -2 not vibrate as those two electrons move through the respective leaves? or are the two electrons (and hence their mutually repulsive force) stationary in space with respect to each other? and if so, how is this possible given that the other electrons in the leaves are moving with respect to their nuclei and hence also to the two electrons resulting in constantly changing forces on those two electrons? Or are there really no 'moving' 'particles' called electrons at all but rather a stationary wave function over the entire area of the leaves? if the latter is the spatial distribution of the charge constant over each leaf's wave function or does it increase toward the end of the leaf? and is the mass (inertial energy) of the 'electron' likewise distributed? and surprisingly, does the constancy of divergence of the simple gold-leaf electroscope therefore show that the classical model of 'electrons'-'moving'-around-a-nucleus is incorrect?

First of all we call the picture of an electron moving in a fixed orbit around a nucleus the "Bohr atom." We know that description is wrong as then the electrons would be accelerating to stay in a circular orbit (centripetal acceleration) and a charged object which acclerates radiates photons. Rather the electrons around a nucleus are in states of fixed energy and angular momentum. There wave function, when squared, describes the spatial probability density of where the electron is at any particular time. Now in the electroscope the free electrons are not bound to any particular atom and exist on the surface where they do move.

(Tom) Followup Q: So what you're saying is the 2 added electrons eventually stop moving and thereafter maintain a constant distance from each other (regardless of the Au atoms' electrons) as localized particles?
A: Yes
(Tom) Followup Q: Since you are a particle physicist perhaps you could clarify something basic for me. Classically an accelerating electron would radiate its energy away and spiral into the nucleus for an explanation of why see. Fig 3 http://www.arrl.org/tis/info/whyantradiates.html but precisely the same argument could be used for a gravitating mass (which pull could even be felt by nearby planets). So why do planets not spiral in to their respective focii by formally the same argument?
A: Neutrally charged objects don't radiate. (there might be some gravitational wave radiation however, which means that this is actually happening, but at a very very very slow rate.)
(Tom) Followup Q: When you mention "Neutrally charged objects"..... is there any Quantum Mechanical rule or Principle that would preclude one Neutron orbiting another Neutron in an otherwise gravitationally free space (say like a gravipause)? I don't mean closely packed like in a Neutron Star but orbiting, perhaps slowly at a, perhaps a few neutron-diameters, distance? Is this Theoretically possible?
A: Neutrons are attracted to neutrons by the strong force, so they could orbit one another.
(Tom) comment: That's interesting!
That would mean that if their resultant De Broglie wavelengths were ever less than 2*Pi*r there would exist discrete quantized orbitals just like in an atom. (Since they have no charge, I assume such neutrons would have no magnetic quantum number, but maybe they do). A substance made from such a material would surely have some very interesting properties: It would be a great electrical insulator for a start. If the bonding strength between such pairs were as strong as the strong force itself, it would be harder than diamond and since there would be no internal electrical repulsion it might be more flexible than rubber!
And now that I think of it, if such bonding were to occur, there would be no way to distinguish which neutron 'belonged' to which pair, the whole thing would be one giant 'molecule' of extreme density, prevented from collapsing in on itself by nothing more than the discreteness of its energy levels resulting from quantized De Broglie wavelengths! so it would be a neutron star anyway!
And goodness me! I think I've just understood for the first time what a neutron star actually IS and why it is the last stop (in terms of matter density) before a black hole!
That's very gratifying!

A comment: ( isn't science wonderful! so many lovely questions !)

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Development of a more fundamental theory of particles

Question by: Shantanu Thatte, India, Faculty Respondent: Steve

What I am developing is that why it is not possible that all the subatomic particles be made up of some very fundamental particles. Like in a picture tube only three colors make up all the colors why no in the universe.

Perhaps the most important issues to fundamental physics are:
(1) A theory which unifies the forces, and also solves many of the shortcomings of the Standard Model of Particle Physics

(2) Uncover a deeper symmetry, which explains the spectrum of fundamental particles (quarks, leptons and gauge bosons)

These are ambitious goals, and many, many sceintists are devotingh their lives to answering these two questions. Many theories, such as string theory, hope to address these issues. String theory would describe all paricles (fermions and bosons) as the vibrational modes of a string. This theory has evolved over the years, and one can actually have 2D "branes", as well as other objects. I encourage you to look at Brian Greene's String Theory special on NOVA http://www.pbs.org/wgbh/nova/elegant/ for a pedagogocal presentation on string theory.
Any theory that solves (1) must also do battle with solving the conflict between quantum mechanics (QM) and general relativity (GR). Basically, QM describes microscopic physics (atoms, quarks, etc) and GR describes the way energy (e.g, mass) and space interact with each other. Usually one deals with problems that are very small scale, such as the cosmos, in which case one applies GR or the very small (atoms, quarks, etc), in which case one applies QM. However, there are situations where both need to be applied, such as the center of a black hole, or the early Universe. However, these two theories are in conflict when applied to such situations. This is a major problem in fundamental physics, and a successful therory which combines GR and QM, must be developed. This is an exceedingly hard problem, and many, many scientists are workin on it.

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What are the smallest particles which exist, and what is their function?

Question by: Bob Smith, Lancaster, Pennsylvania, Faculty Respondent: Steve

How small of particles are know to exist? What are they called and what is their function? I found out about you from an internet search.

The smallest known particles to exist are quarks and leptons. The quarks are called: up, down, charm, strange, top, bottom, and the leptons are: electron, muon, tau, and electron-neutrino, muon-neutrino, and tau-neutrino. Each of these also has a corresponding antiparticle. (just add "anti-" in front of the name)
They are no larger than about 0.0000000000000000001 meters (10^-18 meters) in size, or about 1/1000th the size of a proton. The lightest two quarks (up and down) make up protons and neutrons. In addition, inside the proton is a "sea" of gluons, the force carriers responsible for keeping the quarks bound up into a volume of size 10-15 m. Also there are "virtual" pairs of up-antiup, down-antidown and strange-antistrange quarks inside the proton, which pop in & out of existence.
To learn more, I encourage you to have a look at the Particle Adventure web site http://www.particleadventure.org/

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