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Could someone help show that in special relativity, conservation of momentum is independent of inertial frame by applying Lorentz transform.Or better, can you derive the formula for relativistic momentum under the requirement of conservation of momentum for inertial

Here’s what I’m hoping you can help me with. Your naive to SR, so you try and define momentum just as in high school. You suppose conservation of momentum in frame S and apply Lorentz transform And realize an observer in frame S’ does not observed conservation of momentum. So you need to adjust the definition of momentum so that conservation of momentum is in all inertial frames. As a mathematician, how do you go about finding the new formula for momentum? This should not require high power tools

Adam
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    Welcome New contributor Adam! Have you taken any time to search this site for Q & A that may answer (or help to answer) your question? For example, is there anything at the Q & A that you would find helpful? How can one motivate the relativistic momentum? – Alfred Centauri May 26 '20 at 17:07
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    Are you happy that momentum is a four-vector? – jacob1729 May 26 '20 at 17:10
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    @jacob1729 That'd be a peculiar source of one's happiness. :) –  May 26 '20 at 17:13
  • Which axioms are we allowed to assume? – probably_someone May 26 '20 at 17:14
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    The question in the title is different from the questions in the body. Invariance and conservation are two different things. Please ask one question. BTW, four-momentum isn’t invariant. – G. Smith May 26 '20 at 17:19
  • The the metric in SR is Lorenz invariant. This can be verified directly applying the Lorenz transform And doing plug and chug. I’m looking for a similar computation showing If you have conservation of momentum in one inertial Frame the making the Lorenz transform you get conservation in the other inertial frame. – Adam May 26 '20 at 17:39
  • What is conserved is $p^\mu p_\mu$. If you change the reference frame, the momentum do change to $p'=\Lambda p$, but what is kept invariant is $p'^\mu p'\mu = p^\mu p\mu$. – user2820579 May 26 '20 at 18:48
  • A question after reading your latest edit (as of the time I write this): why the emphasis on "as a mathematician"? For example, assume you consider the totally inelastic collision of two identical particles. In the unprimed (inertial) coordinates, the particles have pre-collision velocity of $u$ and $-u$ along a line through each particle and the final particle velocity is zero. In the primed frame, relatively moving with velocity $u$ along the same line, the pre-collision particle velocities are $0$ and $-\frac{2u}{1+u^2/c^2}$ and the final particle velocity is $-u$. (cont.) – Alfred Centauri May 26 '20 at 19:50
  • Clearly, the 'ordinary' momentum $p=mv$ is not conserved in the relatively moving frame. Having said all that, how would a mathematician approach finding a notion of momentum that is conserved that is different from the approach a physicist would use? – Alfred Centauri May 26 '20 at 19:53
  • Hello Alfred, I am looking for a derivation that does not use thought experiments or relays on much physics intuition. For example, we know that ordinary momentum fails. So, Maybe we argue we need a correction factor. Let’s try p=F(v)mv. – Adam May 26 '20 at 20:04
  • Ok, now demanding conservation of momentum in all frames, applying Lorentz transform, let’s see if we can find F. This seems so natural to ask and do – Adam May 26 '20 at 20:06
  • Adam, it does seem natural to ask and to do so why the "as a mathematician" 'filter'? Do you think that this approach is unique to mathematicians? – Alfred Centauri May 26 '20 at 20:18
  • I agree it’s bad use. As a math grad student, I read some solutions and they answer the problem using high powered tools in physics or complicate the problem because they don’t want to get hands dirty doing computations. – Adam May 26 '20 at 20:26
  • @Adam Given the accepted answer with the Lagrangian, maybe "This should not require high power tools" should be removed from the question... unless these are standard tools to you, starting with momentum "from high school". – robphy May 26 '20 at 20:59
  • Adam, just a helpful hint: if you would like for me to be notified when you reply to one of my comments, just put @AlfredCentauri somewhere in the comment. On the other hand, if you directing a comment to the author of a post, it's not needed. Also, I think the Lagrangian / Noether charge approach to this is the most elegant approach. But, to be sure, it's an approach physicists use extensively. Are you interested in an answer using the $p=F(v)mv$ approach? – Alfred Centauri May 26 '20 at 22:19

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Conservation of energy-momentum is a fundamental principle in relativity; it is "built in" to Einstein's equation $$G^{ab}=8\pi G T^{ab}.$$ It can be proven for interactions in quantum field theory, but that is a heavy duty proof. Otherwise, it is best to take it as a fundamental principle (it can also be proven from Noether's theorem, but that depends on an equivalent reformulation of Newton's laws and the argument can be seen as circular).

As for the other part of your question, $4$-momentum, or energy-momentum $(E,\mathbf p)$, is a $4$-vector. So there is nothing to prove. $4$-vectors are same in all frames. Perhaps it may help if you define the velocity $4$-vector for a body at rest $v=(1,0,0,0)$, and obtain its form after Lorentz transformation. Then you can define $4$-momentum in the usual way $$p = mv.$$

Charles Francis
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  • Here’s what I’m hoping you can help me with. Your naive to SR, so you try and define momentum just as in high school. You suppose conservation of momentum in frame S and apply Lorentz transform And realize an observer in frame S’ does not observed conservation of momentum. So you just adjust the definition of momentum so that conservation of momentum is in all inertial frames. As a mathematician, how do you go about finding the new formula for momentum? This should not require high power tools. – Adam May 26 '20 at 18:19
  • It doesn't. As I said, you just define the 4-velocity, $v$ as a vector, and define momentum $mv$ in the usual way. I think your problem is that sometimes the most trivial things are difficult to see, just because you don't expect them to be quite so trivial. – Charles Francis May 26 '20 at 18:25
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    @Adam In my opinion, that kind of clarification is what belongs in the question, not in the various comments. As it stands, it seems (I'm guessing) that your question is being considered for closing because it maybe doesn't contain enough details, is too broad, or doesn't show enough work. Maybe someone else can comment. – robphy May 26 '20 at 18:26
  • $\gamma$ is absorbed into the definition of $4$-velocity. Since one wants to use vector (and tensor) equations, one has to replace 3-velocities, which are non relativistic, with $4$-velocity vectors, as I described. I almost never write an equation in which $\gamma$ is explicit. That is for low level (and often somewhat unmathematical) treatments. Imv those treatments create a great deal of confusion. – Charles Francis May 26 '20 at 18:49
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Since the OP asks to find the, and I quote here,

formula for relativistic momentum under the requirement of conservation of momentum for inertial frames

(the last word being my guess which makes most sense), we do the following.

We first define the particle orbits as functions $x^\mu(\tau)$ in spacetime, where $\tau$ is an arbitrary Lorentz-invariant parameter. The action is $$A=\int d\tau\ L(x^\mu(\tau),\dot x^\mu(\tau),\tau)$$ where $\dot x^\mu(\tau)$ denotes the derivative with respect to the parameter $\tau$. If the Lagrangian depends only on invariant scalar products of the form $x^\mu x_\mu,x^\mu\dot x_\mu,\dot x^\mu \dot x_\mu$, then it is invariant under Lorentz transformations $$x^\mu\to \dot x^\mu=\Lambda^\mu_\nu x^\nu$$ where $\Lambda$ satisfies $\Lambda g\Lambda^T=g$ with $g_{\mu\nu}=(1,-1,-1,-1)$.

For a free massive point particle in spacetime, the Lagrangian is $$L=-mc\sqrt{g_{\mu\nu}\dot x^\mu\dot x^\nu}.$$ It is invariant under $\tau\to f(\tau)$ for arbitrary and sufficiently smooth $f$. Under translations like $$\delta_sx^\mu(\tau)=x^\mu(\tau)-\epsilon^\mu(tau)$$ the Lagrangian is invariant, satisfying $\delta_sL=0$. Thus applying Euler-Lagrange to calculate the variance, we get $$0=\int_{\tau_\mu}^{\tau_\nu}d\tau\left(\frac{\partial L}{\partial x^\mu}\delta_sx^\mu+\frac{\partial L}{\partial \dot x^\mu}\delta_s\dot x^\mu\right)=-\epsilon^\mu\int_{\tau_\mu}^{\tau_\nu}d\tau\frac{d}{d\tau}\left(\frac{\partial L}{\partial \dot x^\mu}\right).$$

Thus the Noether charges are $$-\frac{\partial L}{\partial\dot x^\mu}=mc\frac{\dot x^\mu(\tau)}{\sqrt{g_{\mu\nu}\dot x^\mu\dot x^\nu}}=mcu^\mu\equiv p^\mu$$

and satisfies $$ \frac{d}{d\tau}p^\mu(\tau)=0$$

This is the conservation of 4-momentum, once we note that $p^\mu$ is indeed the 4-momentum which can be noted by defining $\tau$ to be the physical time $t=x^0/c$. Also note that $u^\mu$ is the dimensionless 4-velocity of the particle, and hence the 4-momentum retains it's look from Newtonian mechanics.

Thus if one can agree on the Lagrangian, then the definition of 4-momentum as the conserved Noether charge of it falls from the definitions and Euler-Lagrange.

Yuzuriha Inori
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Conservation of momentum is mingled with the conservation of energy when you go to relativistic speed. $$E^{'} = \gamma (E - v p)$$ $$p^{'} = \gamma (p - \frac{v E}{c^2})$$ Now if you have conservation $E_1+E_2 = E_3+E_4$ and $p_1 + p_2 = p_3 + p_4$ then because Lorentz transformation is linear, they will just transform into $E^{'}_1 + E^{'}_2 = E^{'}_3 + E^{'}_4$ and $p^{'}_1 + p^{'}_2 = p^{'}_3 + p^{'}_4$ in the new frame.

stackoverblown
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In a collision, the conservation of 4-momentum can be described by a polygon (just like a free-body diagram of forces on an object in static equilibrium): $$\sum_i \tilde P_{i,\rm before} - \sum_j \tilde P_{j,\rm after}=\tilde 0.$$

Then, as @stackoverblown says, Lorentz Transformations are linear transformations (just like Euclidean rotations and Galilean Transformations are). So, this polygon transforms into another polygon (as determined by the Lorentz Transformation).

robphy
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  • I agree that’s the idea for the proof. Could you fill in the computations. – Adam May 26 '20 at 17:53
  • @Adam How would you fill in the computations if we were dealing with (say) a triangle [sum of two vectors equals a third] and applying a rotation (a plain old euclidean rotation) to the space? [No differential geometry needed... just linear algebra... possibly matrix algebra.] – robphy May 26 '20 at 17:55
  • What I’m looking for is something like the computations when verifying the metic in SR is Lorentz invariant.

    Suppose conservation of momentum in a particular inertial frame S, so P_Final = P_ initial.

     – Adam May 26 '20 at 18:01
  • Apply Lorenz transform and show in frame S’ P’_final = P’_inital. – Adam May 26 '20 at 18:07
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Momentum is only conserved if there is no external force on your system. Since force is the time derivative of momentum, momentum is conserved if the external force on a particle is zero. Let me make this more mathematical: $0 = \vec f = d\vec p / dt$. This holds for the non-relativistic and the relativistic case.

Of course one can write down a lagrangian and then apply Noether's theorem, as this is more math. But basically the answer is as simple as I put it.

my2cts
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