# Please explain what is wrong with my relativistic momentum problem

1. May 18, 2014

### Virous

Hello!

First, please, zoom in the attached file and take a look on it. Image 1 demonstrates the situation: two identical balls collide, horizontal component of velocity of both remains unchanged and vertical component becomes opposite as a result of the collision. Table in image 3 shows all the velocities in terms of a (horizontal component) and b (vertical).

Now we change the frame of reference into the one, which moves to the right on the speed of a. It is demonstrated on Image 2. New velocities are calculated via the relativistic velocity addition formula and are shown in the table on image 4.

Finally I calculated total momentum before (image 5) and total momentum after (image 6). Results are not equal! What did I do wrong?

http://stg704.rusfolder.com/preview/20140518/7/40730757_3_1176191.png

Deep thanks!

2. May 18, 2014

### xox

Relativistic momentum is $\vec{p}=\gamma(v)m \vec{v}$. You seem to be missing the "gammas" in your formulas.

3. May 18, 2014

### Virous

No, I dont. On images 5-6 y - is gamma with respect to a.

Other gammas (with respect to other things) are already expanded.

4. May 18, 2014

### Staff: Mentor

The velocities of the balls won't be the same in the new reference frame, so the betas are different for the different particles.

It would help to see how you derived those expressions for the momenta.

5. May 18, 2014

### xox

OK, try writing by components:

$$\gamma(v_A) \vec{v_A}+\gamma(v_B) \vec{v_B}=\gamma(v'_A) \vec{v'_A}+\gamma(v'_B) \vec{v'_B}$$

It is difficult to decipher from your formulas.

6. May 18, 2014

### Virous

True, Im sorry. I have to expand a little bit more.

mfb, I did :)

So:

Hopefully the table on Image 4 is clear. It is just the velocity addition formula as follows:

http://stg660.rusfolder.com/preview/20140518/2/40731152_3_1176206.jpg

I used the "x" one for "a" calculations, since "a" is parallel to frame motion and "y" one for "b" calculations since it is perpendicular. Anyway, no doubt in that table, since it is given in the book :)

Now, to calculate momentum used this one:

http://stg741.rusfolder.com/preview/20140518/2/40731162_3_1176207.jpg

Then a simple substitution of what was in the table instead of velocity there :)

7. May 18, 2014

### Virous

And, of course, I summed up initial momenta of balls on image 5 and finals of image 6

8. May 18, 2014

### Meir Achuz

You have to use conservation of momentum to find the final velocities in either case.
The only way you can violate conservation of momentum is by making an error..

9. May 18, 2014

### Virous

No, Im doing an opposite thing. I took an experimental fact, which is the case, and wanna make sure, that my momentum is conserved. The book says, that it is so streight forward, so they let me to do this on my own. I spent a day, but it is not conserved :)

That is why Im here trying to figure out, what did I do wrong.

10. May 18, 2014

### dauto

what are you using for the γ and β in the expression for the velocities?

11. May 18, 2014

### xox

But you seem to be using the SAME $\gamma$ in all you formulas , this is not right. The " gammas" must be ALL DIFFERENT, see post 5. It is difficult to see what you are doing, I asked you two write the formulas component by component (first x, then y). Then we can see what is going wrong. Can you do it for x first?

12. May 18, 2014

### Virous

x has no problems. It is conserved. I do not include it at all. Working only with y.

dauto, Im not. Other gammas are expanded.

13. May 18, 2014

### xox

ok

...but you seem to have the SAME "gamma" in all the terms (it is difficult to decipher your writeup, so I might be wrong).

It is also not clear what you have as equation as conservation, there is no equation to speak of, where is your:

$$\gamma(v_A) \vec{v_A}+\gamma(v_B) \vec{v_B}=\gamma(v'_A) \vec{v'_A}+\gamma(v'_B) \vec{v'_B}$$

Last edited: May 18, 2014
14. May 18, 2014

### Virous

Yes, I do, but it comes from relativistic velocity addition, as I wrote before. It must be the same, since we are in the same frame. Again, it is given in the book. Then Im using the second formula (momentum one) and substitute data from the table, getting different gammas.

15. May 18, 2014

### xox

There are FIVE "gammas": $\gamma(v_A),\gamma(v_B), \gamma(v'_A),\gamma(v'_B), \gamma(V)$

I do not see them but this might be due to the way you write your formulas. Can you post them in LaTeX?

16. May 18, 2014

### Virous

Yes, Ill try to :) I have to leave now. In approximately 8 hours Ill be back and take one more try! Thank you for staying with me :)

17. May 18, 2014

### xox

I found the mistake, it was difficult due to the presentation. Looking at image 5:

You calculate $\gamma(v_A)$ correctly but your $\gamma(v_B)$ is incorrect, you need to use $v_B^2=(\frac{2a}{1+\beta^2})^2+(\frac{b}{\gamma(1+\beta^2})^2$

You have used only the y -component of $v_B$, you completely forgot to add the x component.

18. May 19, 2014

### Virous

Even earlier :) So, we start from the very beginning once again step by step.

Image 1 shows a situation: collision of two balls. This comes from an experimental fact, so it must be true and the momentum must be conserved. Initial and final velocities of two balls are given in the table on image 3. Here a represents horizontal and b - vertical components.

Now if one computes the momenta of each of the balls before and after, one will find, that it is conserved in both classical ($p = mu$) and relativistic ($p= \gamma mu$) mechanics. It is simple.

Now we move to another frame, which moves with a constant velocity a to the right with respect to the first frame. It moves in such a way, that in classical mechanics the Ball 1 will have 0 horizontal velocity, and the Ball 2 will have a horizontal velocity of -2a. Vertical velocities are the same.

In relativistic mechanics it is more complicated and we have to use relativistic velocity addition formula to get all the velocities listed in the table on image 4 (in the brackets the first equation is a horizontal and the second is a vertical components).

Calculation of momentum for horizontal components shows, that it is conserved. We are not interested in it. Now the vertical ones:

$p= \gamma mu$

Total momenta before the collision:

$\sum p_{before}=\gamma(u_{1}) mu_{1}+\gamma(u_{2}) mu_{2}=\frac{mu_{1}}{\sqrt{1-(\frac{u_{1}^2}{c^2})}}+\frac{mu_{2}}{\sqrt{1-(\frac{u_{2}^2}{c^2})}}$

That is how I got my equations. Then we just substitute the data from the table instead of u1 and u2 and do the same for total momenta after.

And then we find, that now the momentum is not conserved. Thats the problem!

19. May 19, 2014

### Virous

xox, firstly, Im not interested in horizontals, since they are conserved as they are (checked that). But I tried to add them to the equation as well. It makes no difference. Momentum is still not conserved.

20. May 19, 2014

### PAllen

This may be equivalent to what xox is saying, but you cannot use the co-linear velocity addition formula:

(u + v) / (1 + uv/c2)

for any non-colinear situation. Further, you cannot use it component wise in such a situation. There are several ways to state a more general velocity additions formula. My favorite is:

γ(relative) = γ(u)γ(v)(1-uv cos(θ))

then, from relative gamma you get relative speed.

21. May 19, 2014

### Virous

Offf...

I can, just a formula is a little bit different. Again, I added velocities correctly, because it is stated in the book.

In your formulas, PAllen, when θ=90, the formula turns into my one for verticals, and when it is 0 into my one for horizontals:

http://stg660.rusfolder.com/preview/20140518/2/40731152_3_1176206.jpg

22. May 19, 2014

### PAllen

And this is the problem. The velocity addition formula you use is wrong for this situation. It only applies to colinear velocities, and cannot be used component-wise.

Much simpler (for this problem), actually, is to arrive at values for S' by Lorentz transform. Then you don't need to worry about general velocity addition formulat - it is all contained in the Lorentz transform, from which all such auxiliary formulas are derived.

23. May 19, 2014

### Virous

If there is a mistake it is either in the nature (less probable) or in the last step. Velocity addition is correct definitely :)

24. May 19, 2014

### Virous

This is not :) I used two formulas: for col-linear and perpendicular velocities where appropriate.

25. May 19, 2014

### Virous

http://stg660.rusfolder.com/preview/20140518/2/40731152_3_1176206.jpg

Left one works for col-linear. Right one for perpendicular. They both can be derived from your one with θ.