# Conservation of momentum in a collision

Gold Member
In any collision having multiple particles each will have a different ##\gamma## factor if they have different velocities and any transformation to another reference frame will involve yet another ##\gamma## factor. It seems to me the simplest answer to the OP question is that it is always and only the total relativistic momentum and energy that are conserved but in the low velocity limit all the ##\gamma## factors approach 1. It would be more instructive to do a relativistic momentum conservation problem properly and then see how it looks Newtonian in the limit as ##\large \frac{v}{c}## is very small.

Also, I disagree that the concept of relativistic mass is a bad idea. It seems very natural and is the reason for the difference in Newtonian momentum and relativistic momentum. In both cases we can write ##\vec p = m \vec v## but understand in the Newtonian case ##m## is the intrinsic rest mass ##m_o## and in the relativistic case it is ##m = \gamma m_o##

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weirdoguy and Grasshopper
Gold Member
Just found another one by Tipler (Modern Physics, 4th ed.) that does something similar, but then it takes it into a direction that seems ripe for confusing me: since each momentum component has its own gamma, it expresses γ in terms of the primed coordinates, so you get γ.

$$γ' = \frac{1}{ \sqrt{1 - \frac{u'^2}{c^2}}} = γ \frac{1 - \frac{vu_x}{c^2}}{\sqrt{1 - \frac{u^2}{c^2}}}$$

where in this case

##γ = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}##

which is part of the derivation of the Lorentz transformation for energy and momentum. This is page 73 (Chapter 2.3) in Tipler, 4th edition.

I'm sure all the physicists here have seen this, but it seems unwieldy to me. One of those roll up the sleeves situations. I do get why this is needed, but it seems easier to just say "Hey, look at the Lorentz transformation for space and time. Now just replace space and time with momentum and energy!"

But, on the other hand, when I look at that transformation, while the energy transformation looks like the space transformation ##E' = γ(E - vp_x)## , and the momentum looks like the time transformation ##{p'}_x = γ(p_x - \frac{vE}{c^2})## , Tipler says they transform exactly the opposite (and that makes more sense to me since time is always associated with energy, and momentum is always associated with space).

So what's the explanation for this? Tipler says the following:

"Note the striking similarity between Equations 2-16 and 2-17 and the Lorentz
transformation of the space and time coordinates, Equations 1-18 and 1-19. The mo-
mentum p ##(p_x, p_y, p_z)## transforms in relativity exactly like r ##(x,y,z)##, and the total energy E transforms like the time t."

This I do not understand. If I replaced E with x and p with t, then I'd have ##x' = γ(x - vt)##, which is exactly like the space transformation, not the time transformation; and ##t' = γ(t - \frac{vx}{c^2})##, which is exactly like the time transformation, not the space transformation.

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But, on the other hand, when I look at that transformation, while the energy transformation looks like the space transformation ##E' = γ(E - vp_x)## , and the momentum looks like the time transformation ##{p'}_x = γ(p - \frac{vp_x}{c^2})## , Tipler says they transform exactly the opposite (and that makes more sense to me since time is always associated with energy, and momentum is always associated with space).
Tipler is correct. Space- and time transformations look exactly equal in a unit system with ##c := 1##. Momentum is a 3-vector and space is a 3-vector. Momentum transforms like space. You see it better, if you write down the related 4-vectors.

PeroK
Gold Member
Tipler is correct. Space- and time transformations look exactly equal in a unit system with ##c := 1##. Momentum is a 3-vector and space is a 3-vector. Momentum transforms like space. You see it better, if you write down the related 4-vectors.
Ah. Now that makes some sense.

(from Hyperphysics via a google search)

Which means it's just a unit coincidence that when written as I written it looks the opposite.

Gold Member
Just found another one by Tipler (Modern Physics, 4th ed.) that does something similar, but then it takes it into a direction that seems ripe for confusing me: since each momentum component has its own gamma, it expresses γ in terms of the primed coordinates, so you get γ.

$$γ' = \frac{1}{ \sqrt{1 - \frac{u'^2}{c^2}}} = γ \frac{1 - \frac{vu_x}{c^2}}{\sqrt{1 - \frac{u^2}{c^2}}}$$

where in this case

##γ = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}##

which is part of the derivation of the Lorentz transformation for energy and momentum. This is page 73 (Chapter 2.3) in Tipler, 4th edition.

I'm sure all the physicists here have seen this, but it seems unwieldy to me. One of those roll up the sleeves situations. I do get why this is needed, but it seems easier to just say "Hey, look at the Lorentz transformation for space and time. Now just replace space and time with momentum and energy!"
Exactly, why limit confusion when you can compound confusion by mixing transformations with collisions! Either conserve momentum then transform it or transform It then conserve it. Don't try to conserve momentum components while transforming them at the same time or it will be a big confusing mess.

I encourage you to do the simplest case in both Newtonian and relativistic assumptions and see how they are the same in the low velocity limit but vary as the velocities get higher. You don't need a complex relativistic collision like high energy particle collisions at CERN, a basic classical like two balls colliding situation can be computed under relativistic physics assumptions because they are always true regardless of velocity.

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Summary:: I'm looking to figure out why the gamma factor is needed
An approach based on the 4-momentum.

##\mathbf P = \begin{pmatrix}
p_t \\
p_x \\
p_y \\
p_z
\end{pmatrix} = \begin{pmatrix}
p_t \\
\vec p
\end{pmatrix}##

It's squared (invariant) Minkowski-norm is:

## (mc)^2 := \mathbf P \cdot \mathbf P = p_t^2 - p^2##

The object moves with ##v## in spatial direction and "moves" with ##c## in ##ct##-direction => ##p/p_t = v/c##.

## (mc)^2 = p_t^2 (1-v^2/c^2)##

## p_t = (mc) \frac {1}{\sqrt{1-v^2/c^2}} \ \ \ \ \ \ (=E/c)##

## p = \frac{v}{c} mc \frac {1}{\sqrt{1-v^2/c^2}} = \gamma mv##

That's independently valid for movements in every spatial dimension (x, y, z). Therefore:

## \vec p = \gamma m \vec v##

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Mentor
p=mu is, IMHO, one of the most intuitive things in physics
If this is true for your intuition, that means you need to retrain your intuition. The simplest retraining would be to retrain your intuition to think of momentum as the integral of force with respect to time. Historically, this is how momentum was thought of (the term "impulse" was often used to convey this).

And, as I have already said, with this intuitive definition, the correct relativistic formula for momentum is easily found. In other words, this intuitive definition of momentum automatically tells you the mathematical formula for momentum once you know the invariance properties of mechanics; Galilean invariance leads to the Newtonian momentum formula; Lorentz invariance leads to the relativistic momentum formula.

Grasshopper
Mentor
the energy transformation looks like the space transformation ##E' = γ(E - vp_x)## , and the momentum looks like the time transformation ##{p'}_x = γ(p - \frac{vp_x}{c^2})##
First, you have the wrong transformation formula for ##p_x##: It should be ##{p'}_x = \gamma \left( p_x - \frac{v E}{c^2} \right)##.

Second, you are being misled by the fact that you have chosen unnatural units. In units where ##c = 1##, the two formulas are exactly symmetric, either in terms of ##t## and ##x## or in terms of ##E## and ##p_x##. In other words, the "time" and "space" transformations in these units look exactly the same for the case of one spatial dimension. To see any difference between them, you have to look at the full case of 3 spatial dimensions; then the "space" transformation looks different (because it is transforming a spatial 3-vector), and the "space" transformation then clearly corresponds to the momentum transformation (since that is the corresponnding spatial 3-vector).

Gold Member
First, you have the wrong transformation formula for ##p_x##: It should be ##{p'}_x = \gamma \left( p_x - \frac{v E}{c^2} \right)##.

Second, you are being misled by the fact that you have chosen unnatural units. In units where ##c = 1##, the two formulas are exactly symmetric, either in terms of ##t## and ##x## or in terms of ##E## and ##p_x##. In other words, the "time" and "space" transformations in these units look exactly the same for the case of one spatial dimension. To see any difference between them, you have to look at the full case of 3 spatial dimensions; then the "space" transformation looks different (because it is transforming a spatial 3-vector), and the "space" transformation then clearly corresponds to the momentum transformation (since that is the corresponnding spatial 3-vector).
Yeah that’s just a writing error. I don’t know if you can see all my edits but I had a nightmare of LaTex issues, which screwed up my proof reading. I was supposed to be copying Tipler and just messed it up.

What I’m curious about now, however, is why the choice of units gives a false impression (or rather, why glancing at it ignorantly may lead to a false impression) when looking at it as it is often presented in the opening chapters of an elementary textbook (that is, not as a four-vector and not in the units preferred by professionals in the subject).

Mentor
What I’m curious about now, however, is why the choice of units gives a false impression
Because it makes the ##1 / c^2## factor in the second term on the RHS inside the parentheses appear in the equation for ##p_x## in the energy-momentum transformation, whereas it appears in the equation for ##t## in the time-space transformation. That can make it seem like ##p_x## "corresponds" to ##t## instead of ##x##, which is not the case.

Grasshopper
What I’m curious about now, however, is why the choice of units gives a false impression (or rather, why glancing at it ignorantly may lead to a false impression) when looking at it as it is often presented in the opening chapters of an elementary textbook (that is, not as a four-vector and not in the units preferred by professionals in the subject).
If you don't want to use the unit system with ##c:=1##, then you can still see the symmetry between time- and (one) space-transformation, if you transform exactly components of the 4-vectors (for example transform ##ct## instead of ##t##), and if you use ##\beta = \frac{v}{c}##.

##x' = \gamma (x-\beta*ct)##
##ct' = \gamma (ct-\beta*x)##

##p_x' = \gamma (p_x-\beta*\frac{E}{c})##
##\frac{E'}{c} = \gamma (\frac{E}{c}-\beta*p_x)##

SiennaTheGr8, PeterDonis and PeroK
Another approach to get the relativistic momentum formula (##p_y=p_z=0##), based on the 4-momentum

##\mathbf P = \begin{pmatrix}
p_t \\
p_x \\
p_y \\
p_z
\end{pmatrix} ##

Inverse Lorentz transformation:
##p_x = \gamma (p{'}_x+\beta*p{'}_t)##
##p{}_t = \gamma (p{'}_t+\beta*p{'}_x)##

If the primed frame is the rest frame of the object (##p{'}_x =0##):
##p_x = \gamma (\beta*p{'}_{t0})##
##p{}_t = \gamma (p{'}_{t0})##

With squared (invariant) Minkowski-norm ##(mc)^2 := \mathbf P \cdot \mathbf P = p{'}_{t0}^2##

##p_x = \gamma (\beta*mc) = \gamma mv_x##
##p_t = \gamma mc##

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Gold Member
What I’m curious about now, however, is why the choice of units gives a false impression (or rather, why glancing at it ignorantly may lead to a false impression) when looking at it as it is often presented in the opening chapters of an elementary textbook (that is, not as a four-vector and not in the units preferred by professionals in the subject).
My understanding is that professional physicists who do a lot of calculations related to this find using units where ##c=1## convenient. But for others it is better to keep it in otherwise one risks losing track of proper dimensions and proportions.

Also, have you tried doing a simple collision problem with relativistic assumptions yet?

Homework Helper
Gold Member
2022 Award
Summary:: I'm looking to figure out why the gamma factor is needed?
A simple answer is to consider the inelastic decay of a particle into two particles of unequal mass. In the rest frame of the original particle there is zero momentum. After the decay, you cannot have both ##m_1v_1 = m_2v_2## and ##\gamma_1m_1v_1 = \gamma_2m_2v_2##. Unless ##m_1 = m_2##.

In any case, it's clearly one or the other for elastic collisions as well.

Grasshopper
Gold Member
My understanding is that professional physicists who do a lot of calculations related to this find using units where ##c=1## convenient. But for others it is better to keep it in otherwise one risks losing track of proper dimensions and proportions.

Also, have you tried doing a simple collision problem with relativistic assumptions yet?
I actually have not at this point, which is weird, given that I have messed around with some of the (non-tensor) math. But that's not physics.

I do plan on actually going through some of the problems in that Tipler book (I've actually bought it, so I won't be a total cheap-o and steal from the author). It seems to only have two chapters on relativity, but I think it will be a good introduction before I go through a higher level book I have (Special Relativity by Woodhouse).

Gold Member
2022 Award
Just found another one by Tipler (Modern Physics, 4th ed.) that does something similar, but then it takes it into a direction that seems ripe for confusing me: since each momentum component has its own gamma, it expresses γ in terms of the primed coordinates, so you get γ.

$$γ' = \frac{1}{ \sqrt{1 - \frac{u'^2}{c^2}}} = γ \frac{1 - \frac{vu_x}{c^2}}{\sqrt{1 - \frac{u^2}{c^2}}}$$

where in this case

##γ = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}##

which is part of the derivation of the Lorentz transformation for energy and momentum. This is page 73 (Chapter 2.3) in Tipler, 4th edition.
This makes use of the fact that the time component of the normalized four-velocity in any frame is ##\gamma##, and the time component is given by the Minkowski product of the corresponding time-like basis vector. If you have given ##U^{\mu}## in a frame ##\sigma## and the four-velocity of frame ##\Sigma'## is ##V^{\mu}## then ##\gamma'=U^{\prime 0}=V^{\mu} U_{\mu}##.

To translate this to the non-covariant three-velocities just use
$$U^{\mu}=\gamma_u (1,\vec{u}/c), \quad V^{\mu}=\gamma_v (1,\vec{v}/c)$$
to get
$$\gamma'=\gamma_u \gamma_v \left (1-\frac{\vec{v} \cdot \vec{u}}{c^2} \right).$$
So Tipler is indeed correct (as expected).

Sagittarius A-Star and PeroK
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.
That is explained in the video from time 30:53 minutes on.

Grasshopper
Homework Helper
Gold Member
Grasshopper and Sagittarius A-Star
This is essentially the strategy I mentioned in post 11.
Yes. In the video they also explain the advantage of using the 4-momentum, which can be easily Lorentz-transformed.

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Homework Helper
Gold Member
Yes. In the video they also explain the advantage of using the 4-momentum, which can be easily Lorentz-transformed.

Sure. If one starts with (or has gotten to) the 4-momentum (or 4-vectors in general), then there really isn’t anything left to explain about inclusion of the time-dilation factor.

But given the typical intro textbook introduction, the inputs will have to be conservation of momentum and velocity composition.

vanhees71
Gold Member
That is explained in the video from time 30:53 minutes on.
He never finished the problem by showing the relativistic momentum was conserved in the ##S'## frame. I tried but quickly got lost in the algebra.

He never finished the problem by showing the relativistic momentum was conserved in the ##S'## frame.
He does it in the video at time 39:27 minutes "Now comes the big reason ...". A Lorentz-transformation of a conserved 4-vector (here: 4-momentum of an isolated system of free particles) is also conserved.

I tried but quickly got lost in the algebra.
If the above formal argument does not satisfy you, then do a detailed calculation:

You can write down in frame ##S## the 4-momentum of the particle system before and after the explosion. You see that they are equal.

Then you can Lorentz-transform the 4-momenta of each particle before and after the explosion to frame ##S'##, add them up there (to get the system's 4-momentum) and see that the 4-momentum of the system in ##S'## is also equal before and after the explosion.

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Gold Member
He does it in the video at time 39:27 minutes "Now comes the big reason ...". A Lorentz-transformation of a conserved 4-vector (here: 4-momentum of an isolated system of free particles) is also conserved.

If the above formal argument does not satisfy you, then do a detailed calculation:

You can write down in frame ##S## the 4-momentum of the particle system before and after the explosion. You see that they are equal.

Then you can Lorentz-transform the 4-momenta of each particle before and after the explosion to frame ##S'##, add them up there (to get the system's 4-momentum) and see that the 4-momentum of the system in ##S'## is also equal before and after the explosion.
I was transforming the velocities to ##S'## using the relativistic vector addition relations like he did and then setting up the relativistic momentum in ##S'## and my ##\gamma## factors were complicated. Following his logic I was trying to show ##-\gamma_v(v)m v = \large \gamma_1(v_1) \frac{mv_1}{2} +\gamma_2(v_2) \frac{mv_2}{2}## with ## v_1 = \frac{u-v}{1 -\Large \frac{uv}{c^2}}## and ## v_2 = \frac{-u-v}{1 +\Large \frac{uv}{c^2}}##.

It became unworkable.

I tried but quickly got lost in the algebra.
It became unworkable.

Maybe you missed, that the mass of each paticle after the explosion is not ##\frac{1}{2}m##, but ##\frac{1}

{2\gamma_u}m## (mass defect).
##\mathbf P = \begin{pmatrix}
p_t \\
p_x \\
p_y \\
p_z
\end{pmatrix}##

##p{'}_t = \gamma (p{}_t - \beta * p{}_x)##
##p{'}_x = \gamma (p{}_x - \beta * p{}_t)##

Before the explosion:

##\mathbf P = \begin{pmatrix}
mc \\
0 \\
0 \\
0
\end{pmatrix} \ \ \ \ \ \mathbf P{'} = \begin{pmatrix}
\gamma mc \\
\gamma (-\beta * mc) \\
0 \\
0
\end{pmatrix} = \gamma m \begin{pmatrix}
c \\
-v \\
0 \\
0
\end{pmatrix}##

After the explosion:

##\mathbf P_{left} = \begin{pmatrix}
\gamma_u \frac{m}{2\gamma_u}c \\
\gamma_u \frac{m}{2\gamma_u}(-u) \\
0 \\
0
\end{pmatrix} \ \ \ \ \ \ \mathbf P{'}_{left} = \begin{pmatrix}
\gamma ( \frac{m}{2}c - \beta * \frac{m}{2}(-u))\\
\gamma ( \frac{m}{2}(-u) - \beta * \frac{m}{2}c) \\
0 \\
0
\end{pmatrix} = \gamma \frac{1}{2}m \begin{pmatrix}
c + vu/c\\
-u-v \\
0 \\
0
\end{pmatrix}##

##\mathbf P_{right} = \begin{pmatrix}
\gamma_u \frac{m}{2\gamma_u}c \\
\gamma_u \frac{m}{2\gamma_u}(+u) \\
0 \\
0
\end{pmatrix} \ \ \ \ \ \ \mathbf P{'}_{left} = \begin{pmatrix}
\gamma ( \frac{m}{2}c - \beta * \frac{m}{2}(+u))\\
\gamma ( \frac{m}{2}(+u) - \beta * \frac{m}{2}c) \\
0 \\
0
\end{pmatrix} = \gamma \frac{1}{2}m \begin{pmatrix}
c - vu/c\\
+u-v \\
0 \\
0
\end{pmatrix}##

##\mathbf P_{System} = \mathbf P_{left} + \mathbf P_{right} = \begin{pmatrix}
mc \\
0 \\
0 \\
0
\end{pmatrix} \ \ \ \ \ \mathbf P{'}_{System} = \mathbf P{'}_{left} + \mathbf P{'}_{right} = \gamma m \begin{pmatrix}
c \\
-v \\
0 \\
0
\end{pmatrix}##

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bob012345
RW137
Inelastic collision:

Consider two particles, 1 and 2, with the same mass m. Particle 1 has velocity +v and particle 2 has velocity –v. They collide and stick together after the collision. The velocity of this new particle 3 is zero. As the momentums add up to zero, the lab frame is also the center of mass (CM) frame.

Now we pass over to the frame of particle 2:

V2/CM = - V implies VCM/2 = +V.​

As the compound particle 3 is at rest in the CM frame, it has the velocity +v in the frame of particle 2. Any theory must come to this conclusion.

Case 1: v is not relativistic and you may apply Newton’s theory.

V1/2 = V1/CM + VCM/2 = V + V = 2V.​

Mass of the compound particle: M = 2m.

Total momentum before the collision: p = m·v1/2 = m·(2v).

Total momentum after the collision: p’ = M·v’ = 2m·v’.

Conservation of momentum: p’ = p implies v’ = v.

This is the correct conclusion.

Case 2: v is relativistic.

In this case you must apply the relativistic addition of velocities and Newton takes you to the wrong conclusion. Not only that you must modify the formula of momentum, but you also must get a new interpretation for the mass and the energy of the system (M different from 2m).

Hint: space-time.

Any mass m, even at rest in space, moves in the direction of the time axis. Let it move from A to B and get the unit vector u. Then look at the components of m·u. You will get momentum, energy and the relation of energy, momentum and mass. After this, solve the collision problems.

I was trying to show ##-\gamma_v(v)m v = \large \gamma_1(v_1) \frac{mv_1}{2} +\gamma_2(v_2) \frac{mv_2}{2}## with ## v_1 = \frac{u-v}{1 -\Large \frac{uv}{c^2}}## and ## v_2 = \frac{-u-v}{1 +\Large \frac{uv}{c^2}}##.
You missed, that the mass of each paticle after the explosion is not ##\frac{1}{2}m##, but ##\frac{1}{2\gamma_u}m## (mass defect). Correction of the right side of your equation:

##\gamma_1(v_1) \frac{mv_1}{2\gamma_u} +\gamma_2(v_2) \frac{mv_2}{2\gamma_u}##

Then I replace each ##\gamma_1(v_1) ## and ##\gamma_2(v_2) ## according to:
$$\gamma'=\gamma_u \gamma_v \left (1-\frac{\vec{v} \cdot \vec{u}}{c^2} \right).$$

##\gamma_v\gamma_u (1-\frac{uv}{c^2}) \frac{mv_1}{2\gamma_u} +\gamma_v\gamma_u (1+\frac{uv}{c^2}) \frac{mv_2}{2\gamma_u}##

Then I replace ##v_1## and ##v_2## according to:
## v_1 = \frac{u-v}{1 -\Large \frac{uv}{c^2}}## and ## v_2 = \frac{-u-v}{1 +\Large \frac{uv}{c^2}}##

##\gamma_v \frac{m(u-v)}{2} +\gamma_v \frac{m(-u-v)}{2} = -\gamma_v mv##

I think, the calculation in posting #59 (Lorentz-transformation of 4-momentum) is easier.

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bob012345 and PeroK
SiennaTheGr8
How about an appeal to empiricism? Particle accelerators demonstrate every day that the quantity ##E \vec v## is conserved even as ##v \rightarrow 1##, whereas ## m \vec v ## isn't.

vanhees71, Sagittarius A-Star and PeroK
How about an appeal to empiricism? Particle accelerators demonstrate every day that the quantity ##E \vec v## is conserved even as ##v \rightarrow 1##, whereas ## m \vec v ## isn't.
The appeal to empiricism is correct. Particle accelerators demonstrate, that the momentum of an isolated system is conserved, as the momentum of a particle is defined as ##\vec p = p_t\frac{\vec v}{c}##.

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Gold Member
You missed, that the mass of each paticle after the explosion is not ##\frac{1}{2}m##, but ##\frac{1}{2\gamma_u}m## (mass defect). Correction of the right side of your equation:

##\gamma_1(v_1) \frac{mv_1}{2\gamma_u} +\gamma_2(v_2) \frac{mv_2}{2\gamma_u}##

Then I replace each ##\gamma_1(v_1) ## and ##\gamma_2(v_2) ## according to:

##\gamma_v\gamma_u (1-\frac{uv}{c^2}) \frac{mv_1}{2\gamma_u} +\gamma_v\gamma_u (1+\frac{uv}{c^2}) \frac{mv_2}{2\gamma_u}##

Then I replace ##v_1## and ##v_2## according to:

##\gamma_v \frac{m(u-v)}{2} +\gamma_v \frac{m(-u-v)}{2} = -\gamma_v mv##

I think, the calculation in posting #59 (Lorentz-transformation of 4-momentum) is easier.
I sure did miss that! Thanks, that made it work out. I know that wasn't the best way to attempt the problem but I was just trying to follow the logic of the video. I also did it with 4-momentum notation and transformation and it was a whole lot easier!

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Sagittarius A-Star
Homework Helper
Gold Member
2022 Award
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?
Here's a simple example:

A mass of ##m##, moving with velocity ##v## collides with a stationary mass of ##3m##. Conservation of classical momentum results in the mass ##m## having velocity ##-\frac v 2## and the mass of ##3m## having velocity ##\frac v 2## after the collision.

And, if we transform to a different frame using the Galilean transformation, then we see that momentum is conserved in all other reference frames.

If, however, we transform to the initial rest frame of the mass ##m## and use the relativistic velocity transformation for the classical velocities, then we have velocities for the two masses of $$v_m = \frac{-3v/2}{1 + v^2/2c^2} \ \text{and} \ v_{3m} = \frac{-v/2}{1 - v^2/2c^2}$$
In this new frame, the total momentum before the collision was ##-3mv##; and, the total momentum after the collision is $$-\frac{3mv}{2}(\frac 1{1+ v^2/2c^2} + \frac 1{1- v^2/2c^2}) = -3mv(\frac{1}{1 - v^4/4c^4})$$ Which is not equal to the initial momentum in that frame. In other words, if classical momentum is conserved in the rest frame of one particle, then under the Lorentz Transformation (relativistic velocity transformation), momentum is not conserved in the rest frame of the other particle.

vanhees71 and Grasshopper
Gold Member
Excellent explanation.
Thank you for coming back this. It is interesting, as PeterDonis pointed out in post #42, that you only have two choices.

But seeing it this way helps me get a better a snapshot of all the balls that were being juggled by these people early in the 20th century trying to figure special relativity out. Perhaps it’s not useful for learning physics, but I see it as useful for getting a better grasp on the history of physics discoveries.

SiennaTheGr8
@Grasshopper

If you're interested in the history here, you'll want to read the "Non-Newtonian Mechanics" section of this paper by Lewis and Tolman from 1909.

Grasshopper