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Four-momentum invariance between frames

  1. May 25, 2013 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations

    3. The attempt at a solution

    E2 - p2c2 = E02

    I know that this is true. But how do i relate p1 to p1'? and same for energy as well.

    I expanded the LHS= E0,12 + E0,22 + 2E1E2 - 2(p1c)(p2c)

    for the RHS =

    E0,12 + E0,22 + 2E'1E'2 - 2(p'1c)(p'2c)

    Obviously the rest mass energies equate on both LHS and RHS but how do i show that the 2E1E2 - 2(p1c)(p2c) terms are the same?
  2. jcsd
  3. May 25, 2013 #2


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    If you're allowed to use that ##(E,\mathbf p)## is a 4-vector, the problem is pretty easy. (You know that the "square" of a 4-vector is invariant, right?)

    Do you think you're allowed to use that, or are you supposed to use something else entirely?
    Last edited: May 25, 2013
  4. May 25, 2013 #3

    In the non primed frame,
    $$ E_1=\frac{E_{0,1}}{\sqrt{1-u_1^2}} $$
    $$ E_2=\frac{E_{0,2}}{\sqrt{1-u_2^2}} $$
    $$ p_1=\frac{E_{0,1}u_1}{c\sqrt{1-u_1^2}} $$
    $$ p_2=\frac{E_{0,2}u_2}{c\sqrt{1-u_2^2}} $$

    $$ E_1E_2-cp_1cp_2=\frac{E_{0,1}}{\sqrt{1-u_1^2}}\frac{E_{0,2}}{\sqrt{1-u_2^2}} -\frac{E_{0,1}u_1}

    {\sqrt{1-u_1^2}}\frac{E_{0,2}u_2}{\sqrt{1-u_2^2}} =
    E_{0,1}E_{0,2}\frac{1-u_1u_2}{\sqrt{1-u_1^2}\sqrt{1-u_2^2}} $$------(1)

    where u s are velocity divided by speed of light c. You get the similar results in the primed frame. In the frame primed,
    $$ E_1'=\frac{E_{0,1}}{\sqrt{1-u_1'^2}} $$
    $$ E_2'=\frac{E_{0,2}}{\sqrt{1-u_2'^2}} $$
    $$ p_1'=\frac{E_{0,1}u'_1}{c\sqrt{1-u_1'^2}} $$
    $$ p_2'=\frac{E_{0,2}u'_2}{c\sqrt{1-u_2'^2}} $$
    $$ E_1'E_2'-cp_1'cp_2'=\frac{E_{0,1}}{\sqrt{1-u_1'^2}}\frac{E_{0,2}}{\sqrt{1-u_2'^2}} -\frac{E_{0,1}u_1'}{\sqrt{1-u_1'^2}}\frac{E_{0,2}u_2}{\sqrt{1-u_2'^2}}
    = E_{0,1}E_{0,2}\frac{1-u_1'u_2'}{\sqrt{1-u_1'^2}\sqrt{1-u_2'^2}} $$ ----------(2)

    By applying Lorenz transformation from one system to another you may be able to check whether (1)=(2) .

    You know

    $$ E_1E_2-cp_1cp_2=
    E_{0,1}E_{0,2} g_{\mu\nu} U^\mu_1 U^\nu_2=g_{\mu\nu} p^\mu_1 p^\nu_2 $$

    ,where U is four-voloecity, is scalar product of two four vectors which is invariant under change of frame. Please be suggested to check invariance through Lorenz transformation of velocities.
    Last edited: May 26, 2013
  5. May 25, 2013 #4
    Hint: The scalar product of two four-vectors is invariant.
  6. May 26, 2013 #5
    The only question that was bugging me was this:

    We know that when a particle is at rest in a frame, the measured energy is E0 = m0c2. When the particle starts to move with speed v in that frame, the energy measured in that frame is γE0.

    Now consider it with a twist. Suppose a particle has measured energy E, momentum p in frame 1. Frame 2 is moving with respect to Frame 1 at speed v. Would the energy and momentum measured in frame 2 be γE and γp?

    We've only done basic special relativity (first year's work) so the last line it appears to be unfamiliar..
  7. May 26, 2013 #6
    I think the question here wants us to prove it...(By lorentz transformation I think)
  8. May 26, 2013 #7
    I think the question wants us to prove it. Also, we can't just assume that the product results in a '-' sign instead of the usual + sign in dot product. We have to show it in this case.
  9. May 26, 2013 #8


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    How do you know that? If we have defined E as the 0 component of the four-momentum, then it's easy to see. But if we're not allowed to use that E is the 0 component of the four-momentum, then some other definition of E must be used. Does your book contain another definition?

    You need to look at how your book defines terms like "momentum", "four-momentum" and "energy".
  10. May 26, 2013 #9
    You know the formula of Lorentz transformation between (ct,x,y,z) and (ct',x',y',z').
    Replace ct with E/c and x with p_x in the formula. This is what you want.
  11. May 27, 2013 #10
    I'm not sure how you can simply do this. The lorentz transformation is a transformation between 2 sets of coordinates: (x,y,z, t) and (x', y', z', t').
  12. May 27, 2013 #11


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    You can apply the Lorentz transformation to any four-vector. That's actually one possible way to define the term "four-vector". If we use this definition, then a good way to define the four-momentum p is to say that it's the four-vector with components (1,0,0,0) in the comoving inertial coordinate system. (I'm using the convention to put the time coordinate first, and units such that c=1).

    But I think this explanation may be more enlightening: The world line of a particle is the graph of a curve ##x:\mathbb R\to\mathbb R^4##. Now, there are many such functions that have the same graph, so we can make a clever choice of the curve x. We choose x such that for all ##t\in\mathbb R##, t is the proper time of the curve segment from x(0) to x(t). Such a curve is said to be parametrized by proper time.

    I'm not going to do it here, but it's possible to show that this implies that ##x'(t)^T\eta x'(t)=-1##. This is going to be very useful in a minute.

    The four-velocity is defined as the normalized tangent vector of the curve x. For all ##t\in\mathbb R##, the derivative x'(t) is a tangent vector of the curve at x(t). If we denote the four-velocity by u, we have
    $$u(t)=\frac{x'(t)}{\sqrt{-x'(t)^T\eta x'(t)}}=x'(t).$$ This is why we chose x to be parametrized by proper time.

    The four-momentum p is defined by ##p=mu##. So we have
    $$p=mu=mx'=m\frac{d}{dt}x.$$ Now we can see what happens to p if we do a Lorentz transformation of x. As x changes to ##\Lambda x##, p changes to
    $$m\frac{d}{dt}\left(\Lambda x\right).$$ If we denote this by p', we have
    $$p'=m\frac{d}{dt}\left(\Lambda x\right) =\Lambda\left(m\frac{d}{dt} x\right)=\Lambda p.$$
  13. May 27, 2013 #12


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    I really think that the only thing you're supposed to do in this problem is to show that if the 0 component (or 4 component, depending on your book's convention) of the four-momentum p is denoted by E, then that equality holds. You are supposed to know somehow that four-momentum is a four-vector and that the 0 (or 4) component is energy. How exactly you're supposed to know that I don't know. It's probably in your book.

    The problem is very easy if you have some familiarity with four-vectors. You should begin by proving that the "square" of every four-vector is invariant. The "square" is defined using the metric g,
    $$p^2=g(p,p)=p^T\eta p.$$ This is how I would write it anyway. I don't know anything about the notational conventions used by the book you're reading.
  14. May 27, 2013 #13
    You really almost had it proven in your original posting when you concluded that you had to prove that
    Actually, this equation should more properly have been written as
    [tex]E_1E_2-(\vec{p_1c})\centerdot \vec{(p_2c)}=E'_1E'_2-(\vec{p'_1c})\centerdot \vec{(p'_2c)}[/tex]
    where [itex](\vec{p_1c})\centerdot \vec{(p_2c)}[/itex] is the dot product of the spatial (3D) vector [itex](\vec{p_1c})[/itex] with the spatial (3D) vector [itex](\vec{p_2c})[/itex], as reckoned from the unprimed frame of reference, and with a similar interpretation for the analogous vectors as reckoned from the primed frame of reference.

    The left-hand side of the equation can be recognized as the scalar product of the 4 momentum of particle 1 with the 4 momentum of particle 2 (for +--- sign signature) as reckoned from the unprimed frame of reference, and the right-hand side of the equation can be recognized as the scalar product of the 4 momentum of particle 1 with the 4 momentum of particle 2 as reckoned from the primed frame of reference. But, the scalar product of any two 4 vectors is frame invariant. Therefore, QED.

  15. May 27, 2013 #14
    Yes, but i'm not sure if we're allowed to assume that in this question..
  16. May 27, 2013 #15
    Well, when you're sure, please let us know.

  17. May 27, 2013 #16


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    This is not the equality we're supposed to prove. And you need to be careful not to give away too much information. If this had been the correct equation, then you would have been giving away too much.
  18. May 27, 2013 #17


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    If you can't assume it, prove it.
  19. May 27, 2013 #18
    As the OP indicated, it is what is left over after terms from both sides of the equation that are known to be equal to one another have been cancelled.

    In my own defense, I might point out that, in my earlier post in this thread, I was less explicit when I merely hinted that "The scalar product of two four-vectors is invariant."

  20. May 28, 2013 #19


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    unscientific, I hope you understand that we can't tell you anything more if you don't make an effort of your own. You need to study the most basic stuff about four-vectors in your book. In particular, make sure that you understand the "scalar product" of four-vectors. (It's not actually an inner product, so it may be called something else in your book. What I'm talking about is a function that takes two four-vectors to a number, that's defined in all relativity texts).

    If you don't think you're allowed to use stuff about four-vectors (I think you are, and that this makes this problem trivial), then you need to study your book to see what you are allowed to use. If you want help here, you must tell us what you find.
  21. May 28, 2013 #20


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    The problem is really easy if you write everything in terms of rapidities. The rapidity ##\theta## that corresponds to velocity ##v## is given by ##\theta = \tanh^{-1} (v/c)##, where ##c## is the speed of light. You should be able to show for a particle of mass ##m## moving with velocity ##v## that
    E &= mc^2 \cosh \theta \\
    p &= mc \sinh \theta
    \end{align*} You'll also need to use the velocity-addition formula to find a relationship between the rapidities in frame A to the corresponding rapidities in frame B.

    Then just evaluate the two sides of the equations, use a few hyperbolic trig identities to simplify, and you'll be done.
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