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Photon mass

  1. Apr 3, 2006 #1
    Is the student not correctly taught if the intructor says:
    the photon has no rest mass but it has a m=E/cc dynamic mass? Could such a statement have missleading consequences?
     
  2. jcsd
  3. Apr 3, 2006 #2

    robphy

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    IMHO, to properly answer this, one needs the [precise] definition of "dynamic mass" [and "rest mass"] that is being given to the student. (To address the misleading issue, one should also consider the student's ability to reason with that precision.)
     
  4. Apr 3, 2006 #3

    Meir Achuz

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    Yes! Many of the questions and answers in this forum show that confusion.
     
  5. Apr 3, 2006 #4

    rbj

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    to the contrary. the confusion (here in this forum) results when blanket statements are made that photons have no mass, without any qualification.

    you end up having to explain why momentum [itex] p = mv [/itex] exists with photons, yet they have no mass. you end up having to explain why photons have energy [itex] E = h \nu = m c^2 [/itex], yet have no mass.

    Photons have mass, but no rest mass.
     
  6. Apr 3, 2006 #5
    I'd rather make the blanket statement that photons have no mass, and end up having to explain that [itex]p=m-0v[/itex] is no longer a helpful relation. And, in all fairness, [itex]E=m_0c^2[/itex] is not too helpful either. Of course, we all know the proper relation between energy, mass and momentum:

    [tex]E^2 = m_0^2 c^4 + p^2 c^2[/tex]

    Furthermore, in more advanced physics, "dynamic mass" or "relativistic mass" hardly ever enters into consideration, so it is more or less a useless concept. Ironically the concept of changing mass is a manifestation of the conservative human need to resist change from the classical relations of [itex]p = Mv[/itex] where [itex]M= \gamma m_0[/itex].
     
    Last edited: Apr 3, 2006
  7. Apr 3, 2006 #6

    pervect

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    I would avoid making up a new name for yet another sort of mass, if a perfectly servicable name already exists.

    I'm not aware of the terms "dynamic mass" being used in textbooks or the literature.

    It's confusing enough that we already have:

    invariant mass, relativistic mass, Bondi mass, ADM mass, Komar mass, (and I think someone mentioned Dixon mass, which I want to learn more about someday).

    There is no need at all to add another synonym for "relativistic mass" to this stew.

    If you mean to ask "is the term relativistic mass outmoded", I would say, yes, it is. It's not incorrect, though - though it is frequently used incorrectly, one reason that it has become outmoded (students are too likely to misunderstand it).
     
    Last edited: Apr 3, 2006
  8. Apr 3, 2006 #7

    rbj

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    but, as long as we're being fair, i didn't say [itex]E=m_0c^2[/itex]. i said [itex]E=m c^2[/itex]



    and we can show that the above is true from

    [tex] E = m c^2 [/tex]

    [tex] m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}} [/tex]

    and

    [tex] p = m v [/tex]



    is that relation wrong ([itex]p = mv[/itex] where [itex]m = \gamma m_0[/itex])? it comes from my human need to have to commit to memory the fewest fundamental equations as possible. all i have to add to the list is [itex]m = \gamma m_0[/itex].
     
  9. Apr 4, 2006 #8

    Meir Achuz

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    Q.E.D. I rest my case.
     
  10. Apr 4, 2006 #9

    rbj

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    do you realize how anemic the "case" you're resting on is?

    (non-existent is a more accurate term.)
     
  11. Apr 4, 2006 #10

    Meir Achuz

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    Your confusion arises because you are trying to discuss a correct theory using equations and concepts from an incorrect theory.
    Now, you can argue among yourselves.
     
  12. Apr 4, 2006 #11

    rbj

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    your argument remains vapid. you make no defense of it and expect it to be taken for granted. do you also get paid for doing no meaningful work?

    must be nice, if you do.
     
    Last edited: Apr 5, 2006
  13. Apr 4, 2006 #12
    as I see the discussion is far from what I have asked. Consider an atom located on a sensitive balance and equlibrated by a mass located on the other pan. the atom emits two photons of the same frequency in horizontal and opposite directions. if the balance goes out from the state of equilibrium then you explain the fact using which mass of the photon?
     
  14. Apr 5, 2006 #13

    rbj

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    someday they will redefine the kilogram in such a way to fix Planck's constant to a defined value (sorta like the meter is defined today to fix the speed of light to a defined value 299792458 m/s):

    my hypothetical (and rhetorical) question is, if you had a box of negligible mass with perfectly mirrored internal surfaces with such a collection of photons in it and put it on a scale, how much would it weigh? at least 1 kg?
     
  15. Apr 5, 2006 #14
    whom are you quoting?
     
  16. Apr 5, 2006 #15

    ZapperZ

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    I would also add "effective mass" to that list you have there. And I think you have made a terrific point.

    This whole thread is going WAY beyond what is called for. If someone who doesn't undertand physics that much would ask for the mass of something, I would seriously doubt that he/she is asking for "effective mass", "relativistic mass", etc.. etc. These are concepts that he/she does NOT understand, or maybe not even aware of. So if we start dealing out all of these things, we do nothing but add to the confusion.

    Photons have no mass. Done! We all can safely assume that answers the question in the simplest manner. We can THEN, address the issue of "momentum". I have zero problem in addressing that, because even in classical E&M where the concept of photons does not exist, you can still find radiation "pressure" of an EM wave. We had no need to invoke any exotic explanation for such a thing. There is no issue whatsoever in explananing a wave having a momentum (no mass concept is involved here either last time I checked Jackson's text). And there is also no mass in defining the crystal momentum in solid state physics. So defining a momentum of something with no mass concept is no big deal. It cannot be used to argue for the usage of a "relativistic mass".

    If people are so uptight about making sure we define and distinguish between "rest mass" and "relativistic mass" each time a simple question like this comes up, then I would insist that you also distinguish between "bare mass" and "effective mass" of whatever particle you are using. I can easily make the case that MOST of the masses that you measure are "effective masses" and not the particle's bare mass, and that such effective masses can CHANGE! So I can play this annoying game too till the cows come home.

    Zz.
     
    Last edited: Apr 5, 2006
  17. Apr 5, 2006 #16
    photon

    would you say all that in front of students you teach?
     
  18. Apr 5, 2006 #17

    rbj

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    it was originally from a paper by Peter Mohr and Barry Taylor that i thought was the one that can be found at

    http://ejde.math.unt.edu/conf-proc/04/m1/mohr.pdf

    but, it appears (from a slight rewording and change of numbers) that i got it from somewhere else (but from Mohr and Taylor, if you Google "kilogram" and "definition", you get a lot of hits with their names on it - some of the papers are not free) and i cannot find the exact quote at present. but the paper above has an equivalent quote.

    Mohr and Taylor are physicists at NIST ( http://physics.nist.gov/cuu/ ) and when i hear someone at NIST saying "the kilogram should be redefined to this:", i tend to think that someday soon (within years) it will be.

    if they did this (which would have the effect of defining Planck's constant to h = 6.626069311 × 10-34 J/Hz) and, if someday, we could get a really good measure of the universal gravitational constant, G (we likely will never measure G to 9 digits), then they could also redefine the second so that G is also a defined value, and then the meter, kilogram, and second would effectively be defined as constant multiples of the Planck length, Planck mass, and Planck time. for experimental purposes, it's probably best to leave the definition of the second to what it is. these Cesium clocks are getting cheaper and cheaper to duplicate for reference use in modern physical experiments.
     
  19. Apr 5, 2006 #18

    ZapperZ

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    Yes I would, because they won't come back with a retort "But is that rest mass or relativistic mass?"

    If I'm teaching a higher level physics course, then I would tell you that such a question doesn't come up. Having taken so many advanced classes throughout my academic years, I have never once heard a student asking such a question. And trust me, we were known to ask the instructors a lot of questions.

    Zz.
     
  20. Apr 5, 2006 #19

    rbj

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    but if i was your student, Z, there would be other questions that would bounce back at you. (i have identified them previously.)
     
  21. Apr 5, 2006 #20

    ZapperZ

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    And I have answered them, especially on your reason for using the "relativistic mass" term JUST to avoid explaning the existence of momentum without a mass.

    And guess what, if you were in an intro physics class where you have zero clue on what special relativity is, why would you even know the existence of a relativistic mass, much less understand what it is? When asked for the mass of an electron, you do also go on in a lengthy treatise on it's bare mass and effective mass without caring if you are adding to the confusion BEYOND what that person can comprehend?

    This has NOTHING to do with the physics. It has everything to do with finding the appropriate answer to right audience. I have to deal with such a thing regularly since I am actively involved in many outreach programs - I just finished participating in a week-long activity for high school girls during the Women in Science week here at the lab. I have been asked this VERY question many times. Do you think this type of audience would know the difference between such a thing? Do you think when they think about the concept of "mass" that they have a clear idea of the significance of "relativistic mass", "effective mass", "inertial mass", mass renormalization, etc, etc? Do you go into a lengthy spiel on this regardless of who asked?

    Zz.
     
  22. Apr 6, 2006 #21

    pervect

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    It's very easy. The mass (invariant mass!) of a system is not the sum of the masses of its parts. That's all there is to it. This rather trivial observation makes the whole thing a total non-issue.

    The energy of an isolated system is the sum of the energies of its parts.
    The momentum of an isolated system is the sum of the momenta of its parts.

    The mass of an isolated system is given by the relation E^2 - p^2 = m^2 (in geometric units, add factors of 'c' if you desire.

    This is all there is to it. The only remaining complexity is for non-isolated systems, which students rarely have to deal with.
     
  23. Apr 6, 2006 #22

    Doc Al

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    Latex testing...

    [tex]E = m c^2[/tex]

    (Seems to work fine here too.)
     
  24. Apr 6, 2006 #23

    krab

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  25. Apr 6, 2006 #24

    rbj

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    i like Lev Okun. i've had a few email conversations with him (along with Michael Duff and Gabriele Veneziano) about their "Trialogue on Fundamental Constants" paper (i think Duff might call me a "4-constants" partisan, even though i completely agree with Duff's main thesis).

    but i just do not see his point regarding "The pedagogical virus of 'relativistic mass' ". it seems to me, pedagogically, that it's precisely the other way around:

    1. after time-dilation, length-contraction, and Lorentz transformation is derived, then, given the same axioms, relativistic mass is derived. it is equivalent to saying that the momentum of an inertial body as viewed by an observer it is moving past is

    [tex] p = \frac{m_0 v}{\sqrt{1 - \frac{v^2}{c^2}}} [/tex]

    and we all agree with that statement. even the relativistic mass deniers. it's just that i continue to recognize that as


    [tex] p =m v [/tex]

    where

    [tex] m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}} [/tex]


    2. then it might be asked, what is the kinetic energy, [itex]T[/itex], of a body with mass at rest [itex]m_0[/itex] and velocity [itex]v[/itex]? what is the amount of work required to accelerate such a body to velocity [itex]v[/itex]? and we might hope that, when velocity [itex]v \ll c[/itex] that [itex]T = \frac{1}{2} m_0 v^2[/itex].

    when the work (in one dimension) is

    [tex] T = \int F dx [/tex]

    [tex] v = \frac{dx}{dt} [/tex]

    and force is

    [tex] F = \frac{dp}{dt} [/tex]

    and

    [tex] p = \frac{m_0 v}{\sqrt{1 - \frac{v^2}{c^2}}} [/tex]

    and after you do a little fanagling (substitution of variable and integration by parts), you get this result.

    [tex] T = m_0 c^2 \left( \left(1 - \frac{v^2}{c^2} \right)^{-\frac{1}{2}} - 1 \right) [/tex]

    which is the same as

    [tex] T = m c^2 - m_0 c^2 [/tex]

    or, as interpretation:

    [tex] T = E - E_0 [/tex]

    where the total energy of the body is:

    [tex] E = m c^2 [/tex]

    and the energy the body has when it isn't even moving is

    [tex] E_0 = m_0 c^2 [/tex].

    If commonly agree symbols are used (i would ask you to use [itex]m_0[/itex] for "invariant mass"), we all agree with this last equation. but, given the expression for "relativistic mass" (that you guys want to dispute the existence of):

    [tex] m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}} [/tex]

    then even

    [tex] E = m c^2 [/tex]

    is also true and is perfectly consistent with

    [tex] E^2 = E_0^2 + (p c)^2 = \left( m_0 c^2 \right)^2 + (p c)^2 [/tex]

    and we all agree with that equation.

    3. The pedagogical problem that i just can't understand from the deniers or relativistic mass is that

    [tex] E = m c^2 = E_0 + T [/tex]

    compactly (and accurately) relates total energy of a body as the sum of "rest energy" and "kinetic energy" and the kinetic energy becomes the classical expression

    [tex]T = \frac{1}{2} m_0 v^2[/tex]

    in the limit where the velocity of the body is at nonrelativistic speeds.

    How is it that anyone disputes this???

    4. Then, for photons, we start with the result of the photoelectric effect:

    [tex] E = h \nu [/tex]

    which we interpret as total energy (since the photon was exchanged for added kinetic energy for an electron after paying for the "work function" of the surface):

    [tex] E = m c^2 [/tex]

    so we equate the two and get an inherent (relativistic) mass for the photon as

    [tex] m = \frac{h \nu}{c^2} [/tex]

    which you guys don't seem to like, but it's the simplest pedagogical path to get the photon momentum of

    [tex] p = m v = m c = \frac{h \nu}{c} [/tex]

    since the velocity of the photon is, by definition, the speed of light.

    And the fact that the rest mass of the photon is zero comes out simply from

    [tex] m_0 = m \sqrt{1 - \frac{v^2}{c^2}} [/tex]

    when [itex] v = c [/itex].

    Now, none of you guys disagree with this momentum expression:

    [tex] p = \frac{h \nu}{c} [/tex]

    What is your pedagogically simpler means to get to that expression? why should the momentum of a photon be that expression (from pedagogically fundamental principles)? if, in your derivation, you use

    [tex] E^2 = E_0^2 + (p c)^2 = \left( m_0 c^2 \right)^2 + (p c)^2 [/tex]

    then you need to pedagogically justify that.
     
  26. Apr 6, 2006 #25

    robphy

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    Who disputes its existence? You've clearly written it down, you have used it in some computations, and it can be given a physical interpretation.
    The complaint is that it is probably not pedagogically good as the invariant rest-mass.
     
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