# Conservation of momentum in a collision

Gold Member
Summary:
I'm looking to figure out why the gamma factor is needed
Now, deriving relativistic momentum isn't terribly difficult, but that's not the same as understanding it. I'm trying to figure out why conservation of momentum in special relativity requires the gamma factor.

When I looked at conservation of momentum in elementary physics, we basically just took two masses and two speeds, worked out a collision, and showed that the total momentum was the same before and after the collision. For example, suppose there are two equal masses moving at speeds equal in magnitude but opposite in direction along the same axis, and they collide in a perfectly inelastic collision. For example, ##\frac{1}{2}mu + \frac{1}{2}m(-u)= m(0) = 0##. Both before and after collision total momentum equals 0, so momentum is conserved.

What I would like to know is, how would I go about showing that in special relativity, momentum is not conserved unless I include the gamma factor coefficient for each object? I'm thinking that a simple collision in which I can make the objects approaching along an axis isn't going to do the job. Is it more related to something like an observer passing by wouldn't agree that the momentum is conserved in the collision? I can see why this might be the case, since the transformation equation includes both energy and momentum, and since velocities add using relativistic velocity addition. Or can it be shown in a single reference frame?

.

I'm just looking for some general direction about how I can prove to myself that the gamma factor in the momentum equation is necessary, without just someone telling me the answer, if that's possible.

Anything helps. Thanks.

## Answers and Replies

PeterDonis
Mentor
I'm trying to figure out why conservation of momentum in special relativity requires the gamma factor.
Um, because momentum itself does?

I'm not sure why this is even a question. In relativity, the definition of momentum includes the gamma factor.

LBoy and robphy
A system of free, non-interacting particles has the Lagrangian\begin{align*}
L = -\sum_a m_a \sqrt{-g_{\mu \nu} \dot{x}_a^{\mu} \dot{x}_a^{\nu}}
\end{align*}where ##\dot{x}_a = dx_a/d\lambda_a##. Given that the infinitesimal Poincaré transformation ##\tilde{x}^{\mu} = x^{\mu} + \epsilon \delta^{\mu}_{\mu_0}(x^{\nu})## applied to all the particles leaves ##L## invariant, by Noether's theorem the quantity\begin{align*}
p_{\mu_0} = \sum_a \dfrac{\partial L}{\partial \dot{x}^{\mu}_a } \delta^{\mu}_{\mu_0} = \sum_a \dfrac{m_a g_{\rho \mu} \dot{x}^{\rho}}{\sqrt{-g_{\rho \sigma} \dot{x}_a^{\rho} \dot{x}_a^{\sigma}}} \delta^{\mu}_{\mu_0} = \sum_a \dfrac{m_a \dot{x}_{\mu_0}}{\sqrt{-g_{\rho \sigma} \dot{x}_a^{\rho} \dot{x}_a^{\sigma}}}
\end{align*}is conserved (conventionally ##\sqrt{-g_{\rho \sigma} \dot{x}^{\rho} \dot{x}^{\sigma}}## is written as ##\gamma^{-1}##).

Abhishek11235, vanhees71 and Dale
robphy
Homework Helper
Gold Member
The short answer: you are dealing with vectors and these vectors are not generally parallel.

In an energy-momentum diagram,
a collision is describable by a polygon (analogous to an ordinary collision problem in PHY 101).
One can solve for unknown components or magnitudes or angles/rapidities
by using fancy 4-vector methods
or by using components as would be done in a statics problem.
Because not all of the 4-momenta are parallel, one needs the particle-##\gamma##'s (i.e. ##\cosh\theta_i##'s, dot-products) to find components.

(Non-example: imagine trying to do a general static problem involving force vectors,
but only using the magnitudes of the vectors. Components (dot-products) are needed. Components involve ##\cos\phi##.)

Grasshopper
Gold Member
Um, because momentum itself does?

I'm not sure why this is even a question. In relativity, the definition of momentum includes the gamma factor.
I think a more concise way to ask the question is this:

If we use the Newtonian definition of momentum, it is not conserved in special relativity. Assuming the last sentence is true (I have heard it around in various physics discussions... I know, that's not an acceptable source. Apologies), why is that? Or more specifically, how do I go about showing myself that on paper?

My gut is telling me it's related to connection between energy and momentum — assuming anything I said about Newtonian momentum is true. But gut feelings mean nothing to those without very strong physics backgrounds.

Gold Member
A system of free, non-interacting particles has the Lagrangian\begin{align*}
L = -\sum_a m_a \sqrt{-g_{\mu \nu} \dot{x}_a^{\mu} \dot{x}_a^{\nu}}
\end{align*}where ##\dot{x}_a = dx_a/d\lambda_a##. Given that the infinitesimal Poincaré transformation ##\tilde{x}^{\mu} = x^{\mu} + \epsilon \delta^{\mu}_{\mu_0}(x^{\nu})## applied to all the particles leaves ##L## invariant, by Noether's theorem the quantity\begin{align*}
p_{\mu_0} = \sum_a \dfrac{\partial L}{\partial \dot{x}^{\mu}_a } \delta^{\mu}_{\mu_0} = \sum_a \dfrac{m_a g_{\rho \mu} \dot{x}^{\rho}}{\sqrt{-g_{\rho \sigma} \dot{x}_a^{\rho} \dot{x}_a^{\sigma}}} \delta^{\mu}_{\mu_0} = \sum_a \dfrac{m_a \dot{x}_{\mu_0}}{\sqrt{-g_{\rho \sigma} \dot{x}_a^{\rho} \dot{x}_a^{\sigma}}}
\end{align*}is conserved (conventionally ##\sqrt{-g_{\rho \sigma} \dot{x}^{\rho} \dot{x}^{\sigma}}## is written as ##\gamma^{-1}##).
I've seen an easy use of the Lagrangian to show that time translation invariance implies energy conservation. But how do you get that particular Lagrangian? (never mind the fact that I barely know what a tensor is, or how to decipher Einstein summation convention, of course ;) )

Following along, it seems you just basically did something similar and ended with the correct momentum equation.

EDIT — What happens if you use a different transformation, or start with a different equation representing the system? (since you're using what appears to be Einstein Summation Convention, I'm assuming that's some sort of tensor equation, and further assuming there's a four-vector somewhere in there)

But how do you get that particular Lagrangian?
The action of a single, free particle must be Lorentz invariant and therefore proportional to ##ds##, that is, ##S = k\displaystyle{\int} ds##. In order for ##S## to have the dimension of an action, ##k## must have energy units. Furthermore, because timelike geodesics achieve maximum spacetime distance between two given events, it is required that ##k<0## such that these correspond to a minimum of the action. The simplest possible choice is ##k=-m## and therefore ##S = -m \displaystyle{\int} ds##.

For a system of non-interacting particles, the overall action is the sum of the individual actions. (If the particles interact, the general form is given by the Tetrode-Focker action).

I've seen an easy use of the Lagrangian to show that time translation invariance implies energy conservation.
The Lie algebra of the Poincaré group includes ten generators ##f##: four translations (##f^{\mu} = \delta^{\mu}_{\mu_0}##) three Lorentz boosts (##f^{\mu} = {(K_{\mu_0})^{\mu}}_{\nu} x^{\nu}##) and three spatial rotations (##f^{\mu} = {(J_{\mu_0})^{\mu}}_{\nu} x^{\nu}##). If the Lagrangian is invariant under an infinitesimal coordinate transformation ##x^{\mu} \mapsto \tilde{x}^{\mu} = x^{\mu} + \epsilon f^{\mu}##, then\begin{align*}
\delta L &= L(x^{\mu} + \epsilon f^{\mu}, \dot{x}^{\mu} + \epsilon \dot{f}^{\mu}) - L(x^{\mu}, \dot{x}^{\mu}) = 0 \\
\delta L &= \epsilon f^{\mu} \dfrac{\partial L}{\partial x^{\mu}} + \epsilon \dot{f}^{\mu} \dfrac{\partial L}{\partial \dot{x}^{\mu}} = 0 \\
\end{align*}Finally since ##\dfrac{\partial L}{\partial x^{\mu}} = \dfrac{d}{d\lambda} \dfrac{\partial L}{\partial \dot{x}^{\mu}}## we have upon dividing by ##\epsilon##,\begin{align*}
f^{\mu} \dfrac{d}{d\lambda} \dfrac{\partial L}{\partial \dot{x}^{\mu}} + \dot{f}^{\mu} \dfrac{\partial L}{\partial \dot{x}^{\mu}} = \dfrac{d}{d\lambda} \left( \dfrac{\partial L}{\partial \dot{x}^{\mu}} f^{\mu}\right) = 0\end{align*}therefore the quantity ##\dfrac{\partial L}{\partial \dot{x}^{\mu}} f^{\mu}## is conserved. Invariance of ##L## under time translations is expressed by ##f^{\mu} = \delta^{\mu}_0## and gives ##E = mu_0## as the conserved quantity. Invariance of ##L## under spatial translations is expressed by ##f^{\mu} = \delta^{\mu}_i## with ##i \in \{1,2,3\}## and gives ##p_i = mu_i## as the conserved quantity. (Recall that ##u^i = dx^i/ds##).

Last edited:
Abhishek11235, vanhees71 and Grasshopper
Dale
Mentor
2021 Award
Summary:: I'm looking to figure out why the gamma factor is needed

I'm trying to figure out why conservation of momentum in special relativity requires the gamma factor.
Well, it should be obvious that it requires some function that goes to infinity as v approaches c. Otherwise you could take an object, bring it close to c, split it in half with one half stopping and the momentum going all into the other half, and its speed would exceed c.

SiennaTheGr8 and Grasshopper
PeterDonis
Mentor
If we use the Newtonian definition of momentum, it is not conserved in special relativity..
If you are using the Newtonian definition of momentum, then what you are doing is inconsistent with special relativity. So it doesn't even make sense to ask why that definition of momentum isn't conserved in special relativity.

pervect
Staff Emeritus
As I recall, Goldstein has a couple of different discussions of the derivation of relativistic momentum in "Classical Mechanics".

If you want an informal, off-the-cuff answer, though, consider how odd it would be to have the energy of a particle increase without bound as it approaches the speed of light, but have the momentum limited to mv.

This assumes that you are aware that in special relativity, the velocity of a particle can never exceed c, but the energy of a particle has no upper bound, thus we can create very high energy particles in particle accelerators, but they never exceed the speed of light.

Specifically, consider on of Hamilton's equations:

##\dot{q} = v = \partial H / \partial p##

In the context of a free particle, we can identify the hamiltonian H with the energy E and the position coordinate q with position x, so that ##\dot{q}## is equal to velocity. Then we can write:

##v = \partial E / \partial p##

So we'd have v limited to c, E increasing without limit, but p approaching a finite value. It just wouldn't work.

robphy
Homework Helper
Gold Member
• Naively assume that newtonian momentum is conserved in a collision.
• Apply a velocity transformation.
• Observe that newtonian momentum is not conserved in the boosted frame.
• Observe that using the relativistic momenta in place of the newtonian momenta allows conservation of [relativistic] momentum in the boosted frame. (It might useful to follow what happens to the particle-##\gamma##'s to see that they are needed, and what goes wrong if those time-dilation factors are one.)

Ibix
Sagittarius A-Star
Summary:: I'm looking to figure out why the gamma factor is needed

With proper acceleration ##\alpha=\gamma^3 a## (follows from relativistic velocity addition) and if the force is in direction of movement, then the magnitude of relativistic 3-momentum =

##p = \int F \cdot dt = \int m\alpha \cdot dt = m \int \gamma^3 a \cdot dt = m \int \gamma^3 \frac{dv}{dt} \cdot dt = m\int_0^v \gamma^3 \cdot d v = m\gamma v ##

The integral can be calculated via:
https://www.integral-calculator.com/

Dale and Grasshopper
Gold Member
If you are using the Newtonian definition of momentum, then what you are doing is inconsistent with special relativity. So it doesn't even make sense to ask why that definition of momentum isn't conserved in special relativity.
I know. What I’m trying to do here is follow the reasoning that leads to the correct momentum representation.

vanhees71
Gold Member
2021 Award
I've seen an easy use of the Lagrangian to show that time translation invariance implies energy conservation. But how do you get that particular Lagrangian? (never mind the fact that I barely know what a tensor is, or how to decipher Einstein summation convention, of course ;) )

Following along, it seems you just basically did something similar and ended with the correct momentum equation.

EDIT — What happens if you use a different transformation, or start with a different equation representing the system? (since you're using what appears to be Einstein Summation Convention, I'm assuming that's some sort of tensor equation, and further assuming there's a four-vector somewhere in there)
For a free particle the Lagrangian should be a function of ##\dot{\vec{x}}## only, because otherwise you'd have a force (when using Cartesian coordinates for space), where the dot here means derivatives with respect to time. From the point of view of manifest covariance that's a pretty disgusting object, because it's not even the spatial part of four-vector components, but now comes the magic of the action principle: All you need to get a relativistic equation of motion is to have an action that is a Lorentz scalar (note that this is a sufficient but not necessary condition, which becomes important when doing the more mathematical path via Noether's theorem). So we look for a Lorentz-invariant action we can build with ##\dot{\vec{x}}##. There are also the same symmetries of space (for any inertial observer, and of course we work in an inertial reference frame here) as Newtonian physics, because it's the same Euclidean 3D space in special relativity as in Newtonian physics. So the Lagrangian should be a function of ##\dot{\vec{x}}^2##, because that's the only scalar quantity under spatial rotations you can form from this single vector. Now the only scalar action you can build from this is involving ##\mathrm{d} \tau=\sqrt{1-\dot{\vec{x}}^2/c^2} \mathrm{d}t##. So the action obviously is defined by the Lagrangian
$$L=C \sqrt{1-\dot{\vec{x}}^2/c^2}.$$
To determine also the constant ##C## we assume that for ##|\dot{\vec{x}}|\ll c## we should get a Lagrangian that is equivalent to the free-particle Lagrangian in Newtonian physics, namely
$$L_{\text{Newton}}=\frac{m}{2} \dot{\vec{x}}^2.$$
So we expand ##L## in powers of ##\dot{\vec{x}}^2##, and we can stop already after the first-order contribution:
$$L=C \left [1-\frac{1}{2} \frac{\dot{\vec{x}}^2}{c^2} + \mathcal{O}[(\dot{\vec{x}}^2/c^2)^2 \right].$$
From this we read off
$$C=-m c^2,$$
because then
$$L=-m c^2 + \frac{m}{2} \dot{\vec{x}}^2+ \mathcal{O}[(\dot{\vec{x}}^2/c^2)^2].$$
The constant term doesn't do any harm, because it doesn't contribute to the equations of motion at all. So we conclude that a valid Lagrangian for a relativistic free particle is
$$L=-mc^2 \sqrt{1-\dot{\vec{x}}^2/c^2}.$$
Now from Noether's theorem we know, how to define energy and momentum. Momentum is the conserved quantity due to spatial translation invariance, and it is given by
$$\vec{p}=\frac{\partial L}{\partial \dot{\vec{x}}}=\frac{m \dot{\vec{x}}}{\sqrt{1-\dot{\vec{x}}^2/c^2}}=m\gamma \dot{\vec{x}}.$$
Energy is the conserved quantity due to time-translation invariance, and is given by the Hamiltonian, i.e.,
$$E=H=\dot{\vec{x}} \cdot \vec{p}-L=m c^2 \gamma.$$

Einstein44, Grasshopper and Dale
PeterDonis
Mentor
What I’m trying to do here is follow the reasoning that leads to the correct momentum representation.
Then it seems like you are not asking why Newtonian momentum is not conserved in relativity. You are asking why relativistic momentum has the gamma factor in its formula. That's a different question (and many of the responses you are getting actually seem to be directed at that question).

vanhees71
PeroK
Homework Helper
Gold Member
2021 Award
Summary:: I'm looking to figure out why the gamma factor is needed

Now, deriving relativistic momentum isn't terribly difficult, but that's not the same as understanding it. I'm trying to figure out why conservation of momentum in special relativity requires the gamma factor.

When I looked at conservation of momentum in elementary physics, we basically just took two masses and two speeds, worked out a collision, and showed that the total momentum was the same before and after the collision. For example, suppose there are two equal masses moving at speeds equal in magnitude but opposite in direction along the same axis, and they collide in a perfectly inelastic collision. For example, ##\frac{1}{2}mu + \frac{1}{2}m(-u)= m(0) = 0##. Both before and after collision total momentum equals 0, so momentum is conserved.

What I would like to know is, how would I go about showing that in special relativity, momentum is not conserved unless I include the gamma factor coefficient for each object? I'm thinking that a simple collision in which I can make the objects approaching along an axis isn't going to do the job. Is it more related to something like an observer passing by wouldn't agree that the momentum is conserved in the collision? I can see why this might be the case, since the transformation equation includes both energy and momentum, and since velocities add using relativistic velocity addition. Or can it be shown in a single reference frame?

.

I'm just looking for some general direction about how I can prove to myself that the gamma factor in the momentum equation is necessary, without just someone telling me the answer, if that's possible.

Anything helps. Thanks.
At a basic level you can show that if relativistic momentum is conserved in one IRF it is conserved in them.all. Under the Lorentz transformation, of course. That's not true of Newtonian momentum under the Lorentz transformation.

You could then conduct experiments to determine whether relativistic momentum is conserved or not- the lab frame would do.

If those experiments failed to show conservation of momentum, then it seems to me that all the fancy stuff with Lagrangians wouldn't be valid in the first place!

PS this is probably another good test of whether we fundamentally have SR or Newton. I.e. determine experimentally which version of momentum is conserved.

Gold Member
At a basic level you can show that if relativistic momentum is conserved in one IRF it is conserved in them.all. Under the Lorentz transformation, of course. That's not true of Newtonian momentum under the Lorentz transformation.

You could then conduct experiments to determine whether relativistic momentum is conserved or not- the lab frame would do.

If those experiments failed to show conservation of momentum, then it seems to me that all the fancy stuff with Lagrangians wouldn't be valid in the first place!

PS this is probably another good test of whether we fundamentally have SR or Newton. I.e. determine experimentally which version of momentum is conserved.
Experimental data of course confirms that Newton's formula for momentum is inaccurate. But as far as I can tell, all you have to do to confirm SR is demonstrate that the speed of a beam of light is independent of the speed of its source. Honestly, it feels like to me that Einstein's simple example at the start of his "On the Electrodynamics" paper is convincing enough, since it suggests that electromagnetic phenomena obey the principle of relativity, and since you can derive a wave function from Maxwell's equations that has speed c.

Referring to this paragraph:

"It is known that Maxwell's electrodynamics—as usually understood at the present time—when applied to moving bodies, leads to asymmetries which do not appear to be inherent in the phenomena. Take, for example, the reciprocal electrodynamic action of a magnet and a conductor. The observable phenomenon here depends only on the relative motion of the conductor and the magnet, whereas the customary view draws a sharp distinction between the two cases in which either the one or the other of these bodies is in motion. For if the magnet is in motion and the conductor at rest, there arises in the neighbourhood of the magnet an electric field with a certain definite energy, producing a current at the places where parts of the conductor are situated. But if the magnet is stationary and the conductor in motion, no electric field arises in the neighbourhood of the magnet. In the conductor, however, we find an electromotive force, to which in itself there is no corresponding energy, but which gives rise—assuming equality of relative motion in the two cases discussed—to electric currents of the same path and intensity as those produced by the electric forces in the former case."

https://www.fourmilab.ch/etexts/einstein/specrel/www/

To me that clearly indicates that the principle of relativity applies to electromagnetic phenomena. Maybe that's not rigorous enough, logically speaking. But it's convincing to me.

PeroK
Gold Member
Then it seems like you are not asking why Newtonian momentum is not conserved in relativity. You are asking why relativistic momentum has the gamma factor in its formula. That's a different question (and many of the responses you are getting actually seem to be directed at that question).
That's true. But I feel like the answer to the gamma factor question will necessarily answer the other one.

How about I look at it from a different approach:

What if I wanted to derive the relativistic momentum equation, but NOT doing it the usual easy way (starting with a position coordinate, dividing by proper time, multiplying by mass, and taking the time derivative)? How about deriving it from a physical collision, using the Lorentz transformation and/or the velocity addition formula where needed? How would I go about doing that?

Gold Member
For a free particle the Lagrangian should be a function of ##\dot{\vec{x}}## only, because otherwise you'd have a force (when using Cartesian coordinates for space), where the dot here means derivatives with respect to time. From the point of view of manifest covariance that's a pretty disgusting object, because it's not even the spatial part of four-vector components, but now comes the magic of the action principle: All you need to get a relativistic equation of motion is to have an action that is a Lorentz scalar (note that this is a sufficient but not necessary condition, which becomes important when doing the more mathematical path via Noether's theorem). So we look for a Lorentz-invariant action we can build with ##\dot{\vec{x}}##. There are also the same symmetries of space (for any inertial observer, and of course we work in an inertial reference frame here) as Newtonian physics, because it's the same Euclidean 3D space in special relativity as in Newtonian physics. So the Lagrangian should be a function of ##\dot{\vec{x}}^2##, because that's the only scalar quantity under spatial rotations you can form from this single vector. Now the only scalar action you can build from this is involving ##\mathrm{d} \tau=\sqrt{1-\dot{\vec{x}}^2/c^2} \mathrm{d}t##. So the action obviously is defined by the Lagrangian
$$L=C \sqrt{1-\dot{\vec{x}}^2/c^2}.$$
To determine also the constant ##C## we assume that for ##|\dot{\vec{x}}|\ll c## we should get a Lagrangian that is equivalent to the free-particle Lagrangian in Newtonian physics, namely
$$L_{\text{Newton}}=\frac{m}{2} \dot{\vec{x}}^2.$$
So we expand ##L## in powers of ##\dot{\vec{x}}^2##, and we can stop already after the first-order contribution:
$$L=C \left [1-\frac{1}{2} \frac{\dot{\vec{x}}^2}{c^2} + \mathcal{O}[(\dot{\vec{x}}^2/c^2)^2 \right].$$
From this we read off
$$C=-m c^2,$$
because then
$$L=-m c^2 + \frac{m}{2} \dot{\vec{x}}^2+ \mathcal{O}[(\dot{\vec{x}}^2/c^2)^2].$$
The constant term doesn't do any harm, because it doesn't contribute to the equations of motion at all. So we conclude that a valid Lagrangian for a relativistic free particle is
$$L=-mc^2 \sqrt{1-\dot{\vec{x}}^2/c^2}.$$
Now from Noether's theorem we know, how to define energy and momentum. Momentum is the conserved quantity due to spatial translation invariance, and it is given by
$$\vec{p}=\frac{\partial L}{\partial \dot{\vec{x}}}=\frac{m \dot{\vec{x}}}{\sqrt{1-\dot{\vec{x}}^2/c^2}}=m\gamma \dot{\vec{x}}.$$
Energy is the conserved quantity due to time-translation invariance, and is given by the Hamiltonian, i.e.,
$$E=H=\dot{\vec{x}} \cdot \vec{p}-L=m c^2 \gamma.$$
Thanks. This is a little closer to my math level than the tensor stuff. I followed all of it, the expansion, and so on (I'm assuming the O means all members of the expansion of higher order) , but my only curiosity is the constant ##-mc^2##. I have the power of hindsight to know that that is correct, and units speaking, it needs to be mass times velocity squared. But I have to figure out why it would be that in particular — if you approach it from this angle.

Although on the other hand, I feel it doesn't matter too much, since the particular value of c is a choice of units. But obviously it's energy, so m, and there's a ##c^2## term in the radical. So binomial expansion etc.

EDIT — Never mind on that. It's pretty obvious, since it needs to, as you said above, be the same as Newtonian kinetic energy when v<<c.

It's just because in the limit ##c \rightarrow \infty## the Lagrangian must reduce to the classical expression (up to the addition of a total time derivative, in this case a constant ##-mc^2##).

Gold Member
It's just because in the limit ##c \rightarrow \infty## the Lagrangian must reduce to the classical expression (up to the addition of a total time derivative, in this case a constant ##-mc^2##).
Thanks. That dawned upon me just a second ago.

ergospherical
PeterDonis
Mentor
I feel like the answer to the gamma factor question will necessarily answer the other one.
A question that doesn't even make sense in the first place can't have an answer.

What if I wanted to derive the relativistic momentum equation, but NOT doing it the usual easy way (starting with a position coordinate, dividing by proper time, multiplying by mass, and taking the time derivative)?
I'm not sure this is an accurate description of any way of deriving the relativistic momentum formula. You are mixing coordinate-dependent and invariant quantities, and that's never a good idea. (Also, your last "time derivative" is ambiguous; do you mean with respect to coordinate time or proper time?)

The "usual way" of deriving the relativistic momentum formula is to integrate force with respect to time to get total impulse, just as the relativistic energy formula is derived by integrating force with respect to distance to get total work.

How about deriving it from a physical collision, using the Lorentz transformation and/or the velocity addition formula where needed? How would I go about doing that?
In other words, you want to derive the relativistic formula for momentum from the assumption that momentum must be conserved in collisions?

PeroK
Homework Helper
Gold Member
2021 Award
Experimental data of course confirms that Newton's formula for momentum is inaccurate. But as far as I can tell, all you have to do to confirm SR is demonstrate that the speed of a beam of light is independent of the speed of its source. Honestly, it feels like to me that Einstein's simple example at the start of his "On the Electrodynamics" paper is convincing enough, since it suggests that electromagnetic phenomena obey the principle of relativity, and since you can derive a wave function from Maxwell's equations that has speed c.

Referring to this paragraph:

"It is known that Maxwell's electrodynamics—as usually understood at the present time—when applied to moving bodies, leads to asymmetries which do not appear to be inherent in the phenomena. Take, for example, the reciprocal electrodynamic action of a magnet and a conductor. The observable phenomenon here depends only on the relative motion of the conductor and the magnet, whereas the customary view draws a sharp distinction between the two cases in which either the one or the other of these bodies is in motion. For if the magnet is in motion and the conductor at rest, there arises in the neighbourhood of the magnet an electric field with a certain definite energy, producing a current at the places where parts of the conductor are situated. But if the magnet is stationary and the conductor in motion, no electric field arises in the neighbourhood of the magnet. In the conductor, however, we find an electromotive force, to which in itself there is no corresponding energy, but which gives rise—assuming equality of relative motion in the two cases discussed—to electric currents of the same path and intensity as those produced by the electric forces in the former case."

https://www.fourmilab.ch/etexts/einstein/specrel/www/

To me that clearly indicates that the principle of relativity applies to electromagnetic phenomena. Maybe that's not rigorous enough, logically speaking. But it's convincing to me.
Well, you asked about relativistic momentum. I fail to see the relevance of this.

Gold Member
Thanks for your patience and effort, everyone. It is appreciated.

I think I figured out a reasonable solution to my problem. But first, let me give the question a final phrasing that is as clear as I can get it: How do I show that classical momentum is not conserved in a universe where coordinate transformations are Lorentz transformations?

As for answering that, I didn't want to get very abstract, because to my untrained eye it's just math. What I mean is, I can see that ##p=γmu## approaches classical momentum when ##u << c##, but I am not able to see why ##mu## wouldn't be conserved (even if I can clearly see that ##p = mu## has nothing that stops ##u## from going to infinity). So, I tried to look at a more concrete physics example. If you're still interested in helping, how far off is this?

Imagine two observers, S and S', with S' moving at speed v in the x-direction with respect to S. Each observer has an identical baseball (call them balls A and B). Each of them throws their ball along the y-axis at speed ##u_0##, as measured in their own reference frame, so that the balls collide. The collision is perfectly elastic.

Since each observer will see his ball rebound with the same speed he threw it, if momentum is conserved, the sum of those y components of momentum must be zero, because the momentum is simply reversed after collision.

But because of relativistic velocity addition, momentum ##mu_y## is not the same for each ball, which means it can't sum to zero. And this is true according to both observers due to symmetry of the set up, so the resolution can't simply hide in changing reference frames.

For example, looking at it from the perspective of observer S (in this example, subscripts of x or y will indicate the component along an axis, and subscripts of A or B will indicate which ball; e.g., ##u_{y_B}## means the y component of ball B according to S):

According to observer S, ball A moves along the y-axis with speed ##u_0##. The ball thrown by the other observer, ball B, has an x component of velocity equal to v, and a y component of $$u_{y_B} = \frac{u'_{y_B}}{γ(1 - \frac{v u'_{x_B}}{c^2})}$$

but ##u'_{x_B} = 0## and ##u'_{y_B} = -u_0## , so the y component of ball B's velocity is
$$u_{y_B} = \frac{-u_0}{γ}$$

And it happens that for these magnitudes,
$$u_{y_B} = \frac{u_0}{γ} < u_{y_A} = u_0$$

So the magnitude of the velocity of ball B is smaller than the magnitude of the velocity of ball A. But in a perfectly elastic collision, you'd expect the y components of the momentum to cancel out, and they can't here, because the balls are exactly the same, and they are thrown with the same speed according to each observer, but the velocity addition requires that additional gamma factor for the y component of velocity.

But, if you multiply ##mu_y## by gamma, it seems to me that it neatly takes care of the problem. That definition of momentum would be conserved, because
$$γ \frac{u_0}{γ} = u_0$$

and so the summation of the momentum in the y direction would be zero, as it should be. Now, I'm fairly confident about the first part, but not so much this part. I haven't worked it out in my head yet. But it does seem pretty clear to me that the vector sum of the y components of velocity alone cannot simply be zero, because the relativistic addition of velocity formula is not ##u_y = u'_y##. And if they have the same mass, but the y components do not cancel out, then the y momentum components do not cancel out, and there is some sort of contradiction here, because that must mean momentum isn't conserved.

So the definition of momentum has to be adjusted.

Obviously we don't have this problem in Galilean relativity, since
##u_x = u'_x + v##
##u_y = u'y##
##u_z = u'z##

But if you introduce the Lorentz transformation, and the velocity addition formulas that follow,
p = Σmu cannot be conserved.

Last edited: