Please explain what is wrong with my relativistic momentum problem

[Virous]

If you get back to the very first post of mine, I have all the velocities in the primed frame in the table on Image 4. They are all calculated via relativistic addition formulae. Then, images I have in my previous post, just quoted by you are for primed frame. Unprimed frame is OK with me.

 

[xox]

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!

OK,

In the unprimed frame:

1. Before collision

\vec{u_A}=(a,b)
\vec{u_B}=(-a,-b)

2. After collision


\vec{u'_A}=(a,-b)
\vec{u'_B}=(-a,b)



Obviously the momentum before collision is equal to the momentum after collision:

p_{Ax}+p_{Bx}=p'_{Ax}+p'_{Bx}=0
p_{Ay}+p_{By}=p'_{Ay}+p'_{By}=0

If you get back to the very first post of mine, I have all the velocities in the primed frame in the table on Image 4. They are all calculated via relativistic addition formulae. Then, images I have in my previous post, just quoted by you are for primed frame. Unprimed frame is OK with me.

Yes, but we found out that your "gammas" were incorrect. Using the corrected "gammas", try writing the speeds and the equations of momentum conservation (as I did above) as measured from the boosted frame. Please do not mix the components, this makes error detection more difficult. BTW, I found another error, your Image 5,6 do not represent the "total momentum", you are missing one term in both "before" and "after". There should be 3 terms, not 2.

 

[Virous]

I don`t get where are they incorrect. So what are the correct equations?

 

[xox]

I don`t get where are they incorrect. So what are the correct equations?

You need to learn how to find out the errors, work, do not expect us to do all the work for you.

 

[Virous]

I do not expect anyone to do anything for me. I spent a weekend trying to solve this. Everything I reached I already posted. I have no idea, where my errors are. That is why I`m here.

 

[Virous]

POST #56 has my calculations. Just, please, point out where the mistake is.

 

[Virous]

Again, the book says, that it is entirely straight forward. As we see, it is so straight forward, that 6 people can`t solve it on 4 pages of the forum :)

 

[xox]

I do not expect anyone to do anything for me. I spent a weekend trying to solve this. Everything I reached I already posted. I have no idea, where my errors are. That is why I`m here.

You need to redo some of the calculations and post them. I pointed out the calculations that need to be redone and how to redo them.

 

[xox]

Again, the book says, that it is entirely straight forward. As we see, it is so straight forward, that 6 people can`t solve it on 4 pages of the forum :)

Actually, Peter Donis solved it all for you. And I spent a lot of time pointing out your mistakes. We are not supposed to do your homework, you need to learn how to do it right.

 

[Virous]

It is not a homework. It is my own attempt to get into mathematics of special relativity. If it was a homework, I would ask the person who gave it to me :)

Yes, I saw the solution of Peter Donis. However, I need to understand the mistake in my one, since I don`t believe, it`s a good idea to take different approach, until you understand, what`s the problem with the first one.

All of my calculations are already posted. I`ll go through them once more...

So, what we have initially is this:

This table shows the velocities of two balls before and after collision in terms of their vertical and horizontal components. That is how it happens in frams S.

Now we move to another frame S', which moves to the right with velocity of a. Recalculating of velocities gives:

This table was taken directly from the book (however, I checked it for errors several times and from my point of view it is fully correct). In this table gamma stands for \frac{1}{\sqrt{1-(\frac{a}{c})^2}} and beta stands for a/c.

Now I tried to calculate momenta before and after. And what I did (by substitution to xox`s formulas) I have on these two images:

Before:

After:

It is clearly visible, that momenta are not conserved. I do not understand why.

 

[Virous]

I`m very sorry for taking your time. You said, that I`m omitting x components. I included them in the way you showed me. If I did it incorrectly, please, tell me where.

 

[Virous]

Without x components of momenta:

 

[Virous]

After:

 

[xox]

I`m very sorry for taking your time. You said, that I`m omitting x components. I included them in the way you showed me. If I did it incorrectly, please, tell me where.

Yes, you corrected that error. You also corrected the fact that you were missing one term in your first post. Now, stop being so stubborn and separate the equation of conservation into the x and y components, as I showed you. You have a sign error someplace but it is very difficult to debug since you insist on adding the x and y components of the momentum.

 

[Virous]

Just forget about x-components. They are fine. Y components are just the two posts above your last one.

 

[xox]

Just forget about x-components. They are fine. Y components are just the two posts above your last one.

This is better, now please put the "before" and "after" in one post, so I can compare the results.

 

[Virous]

Sure!

Before:

After:

 

[PeterDonis]

Virious, your $u_{by}$ values are wrong; the denominator should have $\left( 1 - (a / c )^2 \right)$, not $\left( 1 + (a / c)^2 \right)$.

Also, you really need to learn to use the forum LaTeX feature to write equations, instead of posting images. Check out this guide:

https://www.physicsforums.com/showpost.php?p=3977517&postcount=3

 

[Virous]

PeterDonis, Ball B moves in the negative direction in frame S and frame S' moves in the positive. Hence we have (-a)*(a) which gives -a^2. And - times - gives +

 

[Virous]

Yes, yes, sorry about LaTeX. I`m using it whenever I can. But because I have all of these equations in Wolfram Mathematica, I just screenshot them to save time. They are quite huge and it will take a lot to copy them. Sorry again!

 

[Virous]

To derive these equations (in the table) I used velocity addition formula:

u'=\frac{u}{\gamma (1-\frac{u_{x}v}{c^2})}

v here is a velocity of S' in S. It is a.
ux is -a, since Ball B moves to the left.
uy is -b.

If you`ll substitute, you`ll get the denominator with "+"

 

[Virous]

Wow, your previous post just disappeared :)

 

[xox]

Sure!

Before:

After:

OK, it is getting close, we are down to an apparent sign inversion between "before" and "after" total momentum.

Now, I will give you a hint: look at the first term in the "before" and work the expression under the radical

(1+\frac{a^2}{c^2})^2-4\frac{a^2}{c^2}-\frac{b^2}{c^2}(1-\frac{a^2}{c^2})=(1-\frac{a^2}{c^2})(1-\frac{a^2}{c^2}-\frac{b^2}{c^2})

 

[PeterDonis]

PeterDonis, Ball B moves in the negative direction in frame S and frame S' moves in the positive. Hence we have (-a)*(a) which gives -a^2. And - times - gives +

Actually, looking back at my post #37, I was wrong. The correct values for $u_{ay}$ and $u_{by}$ before (the only change in the after values is that the signs are swapped) are, as I noted in an earlier post, the ratios $u^y / u^t$ of the values from my post. Those are (again, I'm using units where $c = 1$):

$$
u_{ay} = \frac{b}{\sqrt{1 - a^2}}
$$

$$
u_{by} = \frac{- b \sqrt{1 - a^2}}{1 + a^2}
$$

So the corresponding momentum values are (note that I'm using your values for $u_{ax}$ and $u_{bx}$, which are correct):

$$
p_{ay} = \frac{m u_{ay}}{\sqrt{1 - u_{ax}^2 - u_{ay}^2}} = \frac{m b}{\sqrt{1 - a^2} \sqrt{1 - b^2 / \left( 1 - a^2 \right)}} = \frac{m b}{\sqrt{1 - a^2 - b^2}}
$$

$$
p_{by} = \frac{m u_{by}}{\sqrt{1 - u_{bx}^2 - u_{by}^2}} = \frac{- m b \sqrt{1 - a^2}}{\left( 1 + a^2 \right) \sqrt{1 - \left( 2 a \right)^2 / \left( 1 + a^2 \right)^2 - b^2 \left( 1 - a^2 \right) / \left( 1 + a^2 \right)^2}} = \frac{- m b}{\sqrt{\left[ \left( 1 + a^2 \right)^2 - 4 a^2 \right] / \left( 1 - a^2 \right) - b^2}} = \frac{- m b}{\sqrt{1 - a^2 - b^2}}
$$

So the $y$ momentum does cancel, as desired.

 

[Virous]

It was like this even before all the corrections :) The point is, that if we use the classical momentum formula (p=mu), we get this error. Introduction of gamma according to the book must solve this problem, but it doesn`t.

 

[PeterDonis]

Introduction of gamma according to the book must solve this problem, but it doesn`t.

Check my post #84, your $y$ velocity components were indeed wrong, I just gave a mistaken correction previously.

 

[xox]

It was like this even before all the corrections :) The point is, that if we use the classical momentum formula (p=mu), we get this error. Introduction of gamma according to the book must solve this problem, but it doesn`t.

This is not the issue, you haven't finished your calculations, use the hint I gave you, some very nice simplification will happen.

 

[Virous]

Your hint didn`t exist, when I saw your post the first time. Sorry :(

 

[xox]

Your hint didn`t exist, when I saw your post the first time. Sorry :(

post 83, you are very close, if you do the calculations correctly you should get the total momentum to be null.

 

[Virous]

Oh, yeeeeees! I found it! It simplifies to zero. My problem was, that I fully relied on my Math software, and it for some reason didn`t simplified it completely, because of possiblity of imaginary roots in case of overcoming the speed of light!

Thanks to everyone! So, summarizing everything, my initial mistake was in the fact, that I had to add velocity components geometrically via pyphagora`s theorem!

Thanks to all!