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I Different methods of deriving the energy-momentum equation

  1. Feb 5, 2017 #1
    Hello,
    In deriving the energy-momentum equation:
    [tex] E^2 = (pc)^2 + (mc^2)^2 [/tex]
    the following equations are used:
    [tex] p = ymv [/tex]
    [tex] E = ymc^2 [/tex]

    But both equations are equations that depend on mass, while the final result does not and applies to massless particles. Besides the energy-momentum equation is often cited as the equation relating mass and relativity rather than the one from which it was derived. This leads me to think that this equation is more fundamental than either of the ones used to derive it, so there should be a way to derive it without these two. However, I cannot find any such method. Does anyone knows of how I could derive it without using those two equations?

    Thank you.
     
  2. jcsd
  3. Feb 5, 2017 #2
    This came up in a thread here just a few days ago.

    Peter Donis and some other users indicated that it is indeed possible to show that ##\sqrt{E^2 - (pc)^2}## is Lorentz-invariant without recourse to the mass concept. Maybe someone would be kind enough to chime in here and show us how this would be done. My guess is that you'd start with Minkowski spacetime and the translational symmetries of space and time, then use Noether's Theorem to define energy and momentum as the associated conserved quantities, and then do some analysis to show that those conserved quantities constitute the components of a four-vector. One would then define "mass" as the vector's magnitude.

    But I'm not sure really.
     
  4. Feb 5, 2017 #3
    Sorry for the reposting.
    Would you happen to have a link to the thread? Or the title?
     
  5. Feb 5, 2017 #4

    PeterDonis

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    You could do that, but it would only apply to inertial motion (since Noether's theorem only gives you constants of the motion for inertial motion). But you can always define a 4-momentum vector for any object, even one moving non-inertially; it's the tangent vector to the object's worldline at a given event, normalized such that its components in any inertial frame are the energy and momentum of the object at that event as measured by an observer at rest in that frame. This vector is a 4-vector by definition.

    Yes, after defining the vector as above.
     
  6. Feb 5, 2017 #5
    Thanks, PD.
     
  7. Feb 5, 2017 #6
  8. Feb 6, 2017 #7
    Thank you for the input and the link
     
  9. Feb 6, 2017 #8

    Ibix

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    The thing I don't get (you can detect when I realised I didn't completely understand this from the date of my last post in the thread Sienna linked) is how one gets to the four-momentum in the first place.

    I can see how to get to the four velocity - it's the tangent vector to the particle's worldline. But the four-momentum (for a massive particle) is that times the mass. And as far as I'm aware the Lagrangian is ##g_{\mu\nu}P^\mu U^\nu## (give or take a square root), which implies knowledge of the four momentum.

    Do you just observe that the sum of four-velocities is conserved in certain types of collision (spoiler: between particles of equal mass), and observe that there is some scalar property of particles (which I will call m for no reason at all) that can multiply the four velocity such that the sum of ##m U^\mu## is conserved for all collisions? And then take the Newtonian limit and show that m is the Newtonian mass? Then study the interaction between particles on null worldlines and massive particles to derive ##P^\mu## for null paths?
     
  10. Feb 6, 2017 #9

    PeterDonis

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    So is the 4-momentum. Both are tangent to the particle's worldline (as is any scalar multiple of either). To choose a particular tangent vector you have to pick a normalization. The 4-velocity is normalized to be a unit vector--physically that means its components in a particular inertial frame are ##\gamma## and ##\gamma v##, where ##v## is the object's ordinary velocity in that frame and ##\gamma## is the gamma factor associated with ##v##. The 4-momentum is normalized so that its components in a given inertial frame are the particle's energy and momentum in that frame. Both normalizations are perfectly valid physically, and neither one is logically prior to the other.
     
  11. Feb 6, 2017 #10
    I'm still confused too, Ibix.

    The step I'm missing is what comes after defining energy and momentum in a general, non-mass-dependent way (using Noether's theorem?). How would you then demonstrate that these conserved quantities do indeed constitute the components of a 4-vector? Equivalently, how would you determine that ##E^2 - (pc)^2## is Lorentz-invariant?
     
  12. Feb 6, 2017 #11

    PeterDonis

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    To apply Noether's theorem, we use the time and space translation symmetries of Minkowski spacetime to define four conserved quantities for inertial motion. So we have four Killing vectors, which I'll call ##t^\mu##, ##x^\mu##, ##y^\mu##, and ##z^\mu##, and they define four conserved quantities by contraction with ##p_\mu##, which is a vector tangent to the particle's worldline. (Note that this works for any tangent vector, so we still have to choose a normalization as I described in my previous post. I'll assume that we've chosen the normalization so that the values of the conserved quantities are in fact the energy and momentum that would be measured in an inertial frame.)

    So we have four conserved quantities of the form ##X = \xi^\mu p_\mu##, where ##\xi^\mu## is one of the above Killing vectors. But each one of these conserved quantities is a Lorentz scalar--it doesn't change when we change frames! What is going on here?

    What is going on is that we forgot that the translation symmetries of Minkowski spacetime are not just four Killing vector fields; they are a four-parameter group of Killing vector fields. Picking four mutually orthogonal Killing vectors ##t^\mu##, ##x^\mu##, ##y^\mu##, ##z^\mu## at an event is equivalent to picking a particular inertial frame (and using these Killing vectors as its coordinate basis vectors). So transforming between frames means transforming from one set of four Killing vectors to another. Given a fixed tangent vector ##p_\mu##, this transformation will change the four conserved quantities derived by contracting ##p_\mu## with the four Killing vectors; but it should be obvious that the four conserved quantities must transform under this change as the components of a 4-vector, because the Killing vectors are just the coordinate basis vectors, and that's how the components of a 4-vector are defined--by contracting a chosen vector with the basis vectors.
     
    Last edited: Feb 6, 2017
  13. Feb 6, 2017 #12
    Thank you. I'll have to put in some legwork to understand all that, but that's what I was looking for.
     
  14. Feb 6, 2017 #13
    The relations ##E=\gamma mc^2## and ##p=\gamma mv## are valid only for massive particles, so it's no surprise that you can't use them to derive a relation that applies to massless particles.

    Note that if you start with ##mc^2 = \sqrt{E^2-(pc)^2}## you can show that it's valid for both massive and massless particles.

    You can do the same with the relation ##p=(\frac{E}{c^2})v##.
     
  15. Feb 6, 2017 #14
    Okay, so let's see if I understand this correctly. When you normalize you are, in essence, defining ##m## as the scalar magnitude?
     
  16. Feb 6, 2017 #15

    PeterDonis

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    No. Normalizing is picking a scalar magnitude (out of the infinite number of possible ones) that meets whatever requirement you are trying to meet.
     
  17. Feb 6, 2017 #16
    But does that choice not serve as the definition of ##m##?
     
  18. Feb 6, 2017 #17

    PeterDonis

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    I'm not sure why, unless you want to describe "choosing a normalization that makes the components of the 4-vector match the energy and momentum measured in a given inertial frame" as a definition.
     
  19. Feb 6, 2017 #18

    Orodruin

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    I would give the 4-momentum a slight advantage since it also works for light-like world lines whereas the 4-velocity does not.
     
  20. Feb 7, 2017 #19

    Ibix

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    So, what you are doing is, for some particle:
    1. Defining a quantity called energy, which is related to the heat released bringing the particle to a stop relative to me
    2. Defining a quantity called momentum, which is related to how much a standard target, initially at rest relative to me, recoils when struck by the particle.
    3. Noting that we can make these definitions even if "the particle" is a light pulse (although my wording was sloppy for the energy in this case).
    4. Noting that there exists an affine parameter ##\lambda## along the worldline of the particle such that ##dx^0/d\lambda## behaves like energy and the three ##dx^i/d\lambda## behave like the components of the momentum. The modulus of that vector is naturally a Lorentz scalar, and can easily be shown to be the Newtonian mass in the appropriate limit.
    Right? Sorry for extreme pedantry.

    Except for the last step we could, in fact, make the same case in Newtonian physics. Then Newton would have defined mass as the constant of proportionality between the recoil velocity of our standard target and the initial velocity of the particle. But it's not so natural because we don't need to consider massless things in Newtonian physics - their status isn't clear due to electromagnetism being a relativistic theory.
     
  21. Feb 7, 2017 #20

    PeterDonis

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    This won't do because it requires you to already know the rest mass (since otherwise all you will know from the above experiment is the kinetic energy relative to you). That's why SiennaTheGr8's method of defining energy and momentum using Noether's theorem is a better choice.
     
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