Conservation of momentum in a collision

[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".