I Conservation of momentum in a collision

[vanhees71]

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]

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.

 

[robphy]

That is explained in the video from time 30:53 minutes on.

This is essentially the strategy I mentioned in post 11.

 

[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.

 

[robphy]

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.

 

[bob012345]

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.

 

[Sagittarius A-Star]

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.

 

[bob012345]

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.

 

[Sagittarius A-Star]

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:

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

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

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

 

[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.

 

[Sagittarius A-Star]

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.

 

[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.

 

[Sagittarius A-Star]

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 $\vec p = p_t\frac{\vec v}{c}$.

 

[bob012345]

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!

 

[PeroK]

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.

 

[Grasshopper]

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.

 

[vanhees71]

This is obviously an interesting historical source but not a good text to learn the physics, because it uses the outdated concept of "relativistic mass".