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Question about a general way to prove E^2 = (cp)^2 + (mc^2)^2

  1. Aug 24, 2013 #1
    For particles with non zero mass m we know that:

    E = mc2γ (1)
    P = mvγ (2)

    with the usual meaning of the symbols.
    From those two equations is easy to prove the equation:

    E2 = (cp)2 + (mc2)2 (3)

    which doesn't contain the factor γ any longer and so it can even be used for particles of speed = c as photons. But is this really correct? We started from equations (1) and (2) which *are not* applicable to v = c particles.

    Is there a more general way to prove (3)?

    --
    lightarrow
     
  2. jcsd
  3. Aug 24, 2013 #2

    mfb

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    You have to know something about massless particles in order to derive an equation for massless particles.
    E=pc, for example, is sufficient - formula (3) follows as a direct consequence for m=0.
    E=hf and ##p=\frac{hf}{c}=\frac{h}{\lambda}## can be used, too.
     
  4. Aug 24, 2013 #3

    Dale

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    I think that the most general way to prove it is to show that P=(E,p) forms a four-vector. Then define m=|P| and (3) follows automatically.
     
  5. Aug 24, 2013 #4

    Vanadium 50

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    I would say the best thing to do is to start with (3) and use that to prove (1) and (2).
     
  6. Aug 24, 2013 #5
    Right, but then would it be correct to prove that a photon's mass is zero substituting E = cp in E2 = c2p2 + m2c4?
    Since the last equation, strictly speaking, was derived only for m ≠ 0 particles. What do you think?
     
  7. Aug 24, 2013 #6
    Good. How I prove that (E,p) is a four-vector?
     
  8. Aug 24, 2013 #7

    Bill_K

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    I think both equations are true, and it's an idle exercise to worry about which one is derived from what. Similarly, in electromagnetism does ∇·B = 0 imply the existence of A, or is it the other way around??
     
  9. Aug 24, 2013 #8
    I know but mine it's not a strictly mathematical problem: I'm trying to follow a "physical path", starting from experimental results and SR.
     
  10. Aug 24, 2013 #9

    mfb

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    SR, formulated with 4-vectors in a Minkowski space, is what DaleSpam suggested. You get all equations out of this formalism.
     
  11. Aug 24, 2013 #10

    WannabeNewton

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    ##p\cdot p = 0## for a photon so ##-E^2 + |\vec{p}|^2 = 0##. For massive particles, ##p\cdot p = -m^2## so ##E^2 = |\vec{p}|^2 + m^2##.
     
  12. Aug 24, 2013 #11

    pervect

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    As I recall, there's a technique to get the relativistic momentum relationships by considering the kinematics of the collision of billiard balls - i.e. their velocities must transform according to the usual relativistic formulae, and this constraint is sufficient to give an expression for the momentum as a function of velocity. I think Goldstein has this.

    Photons aren't billiard balls, but I suspect that by considering a pair of photons producicing a particle pair, you could derive a similar expression. I haven't tried to work out the detials though.

    [add]
    A more direct route than pair production might involve two photons bouncing off a small lightweight mirror, and insisting that the mirror's trajectory isn't modified, as an analogy to the "billiard balls".
     
    Last edited: Aug 24, 2013
  13. Aug 24, 2013 #12

    Dale

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    Simply show that it transforms correctly. I.e. show that ##E'=\gamma (E-p_x v)## and ##p_x'=\gamma (p_x -vE)##
     
  14. Aug 24, 2013 #13

    Bill_K

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    Can I use (1) and (2)? :tongue2:
     
  15. Aug 24, 2013 #14

    robphy

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    In physics, sometimes when studying particular phenomena (e.g., interactions of massive particles) we get a glimpse (e.g., your (1) and (2) and your (3)-applied-only-to-massive-particles) of the full story. Then, one may be led to consider theoretical-extensions ( (3) to the zero-mass case)... and checking it with experiment.


    Now... thinking about it a little more... ( seeking a "physical path" to (3) )...
    Here could be another alternative view incorporating (1)-and-(2) and (3).
    At face value,
    (1) is a plot of E as a function of m and v
    (2) is a plot of p as a function of m and v
    and you notice that (3) algebraically results from (1) and (2) and the definition [and elimination] of gamma.
    Apparently, (1) and (2) don't apply to the v=c case because gamma will be infinite,
    which forces m=0 in order to somehow make E and p finite for a typical photon. ( The applicability of (3) as a constraint isn't so obvious.)

    So, instead of (1) and (2) consider
    pc/E=v/c ( by taking (2)/(1) )
    .... in rapidity terms, this says (mc^2 sinhQ)/(mc^2 cosh Q)=tanh Q.
    The Euclidean analogue is that the slope is the ratio of the components of a vector.
    In particular, if the slope is 1 or -1, the magnitudes of the components are equal.

    Practically speaking, consider experiments for various choices of m and v and plot their data on a pc vs E graph... [assuming you know how to get p and E from the experimental data]*.
    You'll notice that variable-m constant-v experiments will trace out [a certain subset of] radial lines on the pE-plane.
    You'll notice that constant-m variable-v experiments will trace out [a certain subset of] hyperbolas on the pE-plane (3)... including the asymptotic hyperbola (lines with slopes 1 and -1) corresponding to the zero-mass case.

    *I think the tricky point for motivation is getting the expression for the relativistic momentum.
     
    Last edited: Aug 24, 2013
  16. Aug 24, 2013 #15

    jtbell

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    Also note that starting from Maxwell's equations, one can show that the energy and (magnitude of) momentum densities in an electromagnetic wave are related by E = pc.
     
  17. Aug 25, 2013 #16
    Thanks to everyone for the answers.

    --
    lightarrow
     
  18. Aug 25, 2013 #17

    rbj

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    it's how i learned that [itex] m = \gamma m_0 [/itex] (the now frowned-upon anachronism). but it required keeping the [itex]p=mv[/itex] relation.
     
  19. Aug 26, 2013 #18
    The proof depends on the fact that
    [tex]\gamma^2=\frac{1}{1-\frac{v^2}{c^2}}[/tex]
    which in this context is true. Implicitly we are assuming all the caveats of SR and (for this proof)
    [tex]m\ne 0[/tex]

    It is possible to relax the requirement that
    [tex]m\ne 0[/tex]
    by proceeding from a topological argument using the invariance of distances in space-time. You would still have the caveats of SR and so the same [tex]\gamma[/tex].

    Try to substitute the first expression for [tex]\gamma[/tex] into (1) and (2) and then formulate the right-hand side of (3). Simplify and see what you get.

    For the more general way, using the topological invariance of distances in space-time we start with
    [tex]p_\mu p^\mu=p_0 p^0+\vec{p}\cdot\vec{p}=-m^2 c^2[/tex]
    and get
    [tex]-(p^0)^2+|\vec{p}|^2=-m^2 c^2[/tex]
    or
    [tex]p^2+m^2 c^2=(p^0)^2[/tex]
    This only leaves the interpretation of [tex]p^0[/tex]. Based on units and consistency with situations where
    [tex]m\ne 0[/tex]
    we take
    [tex]p^0=E/c[/tex].
     
    Last edited: Aug 26, 2013
  20. Aug 26, 2013 #19
    This follows from a math theorem that says for any vector A the following is true:
    [tex]\mathbf{\nabla} \cdot (\mathbf{\nabla} \times \mathbf{A})=0[/tex]
    So if [tex]\mathbf{\nabla} \cdot \mathbf{B}=0[/tex] then there must exist an A such that
    [tex]\mathbf{B}=\mathbf{\nabla} \times \mathbf{A}[/tex]
     
    Last edited: Aug 26, 2013
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