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Photon momentum and energy

  1. Feb 1, 2007 #1
    photon momentum and energy is a frequent topic on the forum. consider a tardyon (u<c) the momentum of which transforms as
    p=gp'(1+V/u') (1)
    E=gE'(1+Vu'/c^2) (2)
    state that special relativity theory ensures a smooth transition from the properties of the tardyon to the properties of a photon and make in (1) and (2) u=u'=c in order to obtain in its case
    p(c)=gp'(c)(1+V/c) (3)
    E(c)=gE'(c)(1+V/c) (4)
    Is there more to say?
    sine ira et studio
  2. jcsd
  3. Feb 2, 2007 #2

    Meir Achuz

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    Why use u at all? Why not just p=g(p+VE) and E=g(E+vp)?
  4. Feb 2, 2007 #3
    Doesn't look right. Here are the correct ones:

  5. Feb 2, 2007 #4
    photon (energy and momentum)

    Thanks. Because what I propose and what you propose represent one and the same thing (you consider c=1 as I see). Presenting as
    p=p'g(1+V/u') and E=E'g(1+Vu'/cc) in the case of a tardyon I emphsize the part played by its speed in the I' inertial reference frame. I think the two equatioins are correct.
    I state now (that is my right) that in special relativity we have a smooth transition from the equations which describe a tardyon to the equation which describe a photon i.e. if in the equations presented above i make u'=c I should recover the equations which account for the behaviour of a photon

    showing that in the case of a photon the transformation is performed in both cases by the same Doppler factor.
    Please let me know if there is a flow in my way of thinking.
    sine ira et studio:rofl:
  6. Feb 2, 2007 #5
    photon energy (mass??) momentum

    With all respect but I think your answers are stante pede)
    If you look with more attention to the equations I propose and to the equations to which you send me you will se that they represent one and the same thing. My presentation presents the advantage that it underlines the part played by the speed of the tardyon in the two inertial reference frames between which we perform the transformation. Thank you for your participation at my thread and let me know if there is some flow in my presentation.:rofl:
  7. Feb 3, 2007 #6
    Energy-momentum forms the coordinates of a 4-vector that transform together, i.e. each expresion contains both energy and momentum. Your equations transform momentum separate from the energy transformation. Can you prove that they mean the same thing as the standard transformations?
    Last edited: Feb 3, 2007
  8. Feb 3, 2007 #7
    from the tardyon to the photon

    Thanks for your help. IMHO special relativity does not start and does not end with the four vectors. Yes I can. Start with (I have in mind the OX(O'X') components of the momentum. Start with (W.G.V. Rosser, Classical Electromagnetism via Relativity, (Butherworth 1967) p.14)
    p=g(p'+VE'/cc)=gp'(1+V/u') (1)
    E=g(E'+Vp')=gE'(1+Vu'/cc) (2)
    What I have in mind is that the functions (1+V/u') and (1+Vu'/cc) have the same limit for u'=c and that is why I say that special relativity theory ensures a smooth transition from the equations that describe the behaviour of a tardyon to those that describe the behaviour of a photon.
    I trhink that we can express the right hand side of (1) as a function of E' only and the right side of (2) as a function of p' only. Special relativity is a flexible theory.
    Please let me know your oppinion.
    sine ira et studio
  9. Feb 3, 2007 #8
    The above formulas are not correct. The RHS needs to be in expression in BOTH p' AND E'. You must be doing something like assuming E'=p'c but this is true for photons only (massless particles), and not for tardyons as you are claiming in your opening sentence above.
  10. Feb 3, 2007 #9
    I am afriad that it is not ok.
    You are starting with
    p=g(V)*(p'+VE'/c^2) (1)
    E=g(v)*(E'+Vp') (2)

    This is ok (it is the standard 4-vector transformation). Then you are doing something strange to the above formulas that brings in u'. What is u'? Where is it coming from? What does all of this have to do with photons? How do you derive the second set of equalities?
  11. Feb 3, 2007 #10
    tardyon to photon transition

  12. Feb 3, 2007 #11

    Meir Achuz

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    I think your fight could be resolved by realizing that u=p/E and u'=p'/E',
    so BR is not wrong. I just think bringing in u, which has awkward transformation properties, leads to confusion (evidenced here) compared to just sticking with p and E, where it is trivial. You don't even have to consider any limit in treating photons. I long ago stopped using u (and c)because u is so close to 1 in HE physics that it is useless.
  13. Feb 3, 2007 #12


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    In one sentence, can someone define u' physically?
    Can u' be zero?
  14. Feb 3, 2007 #13
    Where is this one coming from? Especially, where did you get the c^2 factor?
  15. Feb 3, 2007 #14
    So far as I know
    p=mu=Eu/cc (relativistic mass exorcised?)
  16. Feb 3, 2007 #15
    Thanks! I think not everybody has the chance to work in HE (high energy I guess?) What are you using instead u and c? Have you published somewhere your point of view?
    sine ira et studio
  17. Feb 3, 2007 #16
    photon energy and momentum

    In a previous thread I asked you if taking a limit is associated with a phyhsical procedure? My question remained without being answered. As far as I know there are at least two kinds of velocities:
    1. u=proper length/coordinate time interval. The definition of the coordinate time interval involves clock synchronization, which involves the one way speed of light...Physicists also make a net difference between mean velocity (measurable?) and instantaneous velocity which as a result of a limit is not measurable. It is equal to zero when the measured proper length is zero (?)
    2. U=proper length/proper time interval which does not involve clock synchronization requiring only initialization of the moving clock which measures the proper time interval with the clock of the stationary reference frame with which it is instantly at the same point in space imposing the condition that both read zero at that very moiment. Physicists establish a relationship between u and U and so the concept of instantaneous velocity can be avoided. By definition U is zero when u is zero.
    I am convinced that I do not tell you anything new. I brought in discussion all that subject because I hope that I will receive competent points of view, concerning that fundamental problem and so the discussions could be usefull for the learners on the Forum as well.
    Please consider that I come on the Forum with the handicap that English is not my first language which makes my sentences not very clear,:rofl: .
    sine ira et studio
  18. Feb 3, 2007 #17

    Don't think so. The relationship between E and p for tardyons is:

  19. Feb 3, 2007 #18
    But I do think so. IMHO the equation you propose m(rest mass) only proves the invariance of the left side of your equation. I find in my old and new books that the momentum of a tardyon is
    whereas its energy is
    Am I wrong?
  20. Feb 4, 2007 #19


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    Yes, your equation [itex]p = Eu/c^2[/itex] (snipping out the [itex]mu[/itex] in the middle) is correct. I use it more often in the form

    [tex]\frac{u}{c} = \frac{pc}{E}[/itex]

    For example, an electron ([itex]mc^2[/itex] = 511 keV) with kinetic energy 500 keV has energy E = 511 + 500 = 1011 keV and [itex]pc = \sqrt{E^2 - (mc^2)^2} = \sqrt{1011^2 - 511^2}[/itex] = 872.35 keV, which gives u = (pc/E)c = (872.35/1011)c = 0.863c.
  21. Feb 4, 2007 #20

    So you start with :

    p=g(V)*(p'+VE'/c^2) (1)
    E=g(V)*(E'+Vp') (2)

    Then you use the fact that for TARDYONS ONLY (but NOT for photons) the following are true:
    E=g(u)mc^2 (I am using your notation).

    This also gives you:



    E'=pc^2/u' (3)
    p'=Eu'/c^2 (4)

    Substitute (3) into (1) and (4) into (2) and you get:

    p=g(V)p' (1+V/u') (5)
    E=g(V)E' (1+Vu'/c^2) (6)

    These are the relations you started with, now I understand how you got them. They are correct. Now what are you planning to do next? You cannot make u'=c as you claim in your initial post because (5) and (6) have been derived by using the tardyons equations. You have used :

    E=g(u)mc^2 remember? And they are applicable for massive particles only. You cannot turn around and apply a derivation that uses tardyon properties to photons.
    In other words (3)(4) are valid for tardyons ONLY, therefore (5)(6) are valid for tardyons ONLY, so you CANNOT make u'=c because u' is the speed of a tardyon.
    Last edited: Feb 4, 2007
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