# An elastic collision

1. Feb 28, 2010

### Altabeh

Hi

I checked this problem many times but I didn't end up with the result wanted.

Assume that a particle of rest mass $$m_0$$, (relativistic) energy $$e_0$$ and (relativistic) momentum $$p_0$$ is moving in a straight line. This particle suddenly hits a stationary particle with rest mass $$M_0$$ ahead and they both get involved in an elastic collision. As a result of the collision, the second particle gains momentum $$P$$ and energy $$E$$ and the first one keeps moving with a new momentum, $$p$$, while its energy is now $$e.$$

In the Newtonian limit, using the conservation laws of energy and momentum we can get

$$P=\frac{2p_0M_0}{M_0+m_0},$$
$$p=\frac{p_0(m_0-M_0)}{M_0+m_0}.$$

But in the relativistic case, the two conservation laws get really sloppy and complicated, though it is claimed that, for example,

$$P=\frac{2p_0M_0(e_0+M_0c^2)}{2M_0e_0+m^2_0c^2+M^{2}_0c^2}.$$

How can we obtain this expression? Is this even correct?

AB

Last edited: Feb 28, 2010
2. Mar 1, 2010

### George Jones

Staff Emeritus
After mucking about with 4-vectors and Lorentz inner products, I think I get

$$P=\frac{2p_0M_0(e_0+M_0c^2)}{2M_0e_0+m^2_0c^2+M^{2}_0c^2 - e_0^2}.$$

I probably made a mistake; tomorrow I'll try to check my calculations.

3. Mar 1, 2010

### George Jones

Staff Emeritus
I found my silly mistake. Yes, I get

$$P=\frac{2p_0M_0(e_0+M_0c^2)}{2M_0e_0+m^2_0c^2+M^{2}_0c^2}.$$

4. Mar 1, 2010

### Altabeh

So that is a good progress that I'm now sure the question is right! How did you get this? Give me a clue, please!

AB

5. Mar 2, 2010

### George Jones

Staff Emeritus
The method I used, using 4-velocity to split spacetime into time and space, is not that well known. Introduction to Spacetime: A First Course on Relativity by Bertel Laurent is the only book that I have seen apply this method to collision problems. If you have access this book, the stuff in section 9.2, appropriately applied, gives the desired result.

I used this method for other things here

and here (Doppler effect)

If you don't have access to Laurent, I can try and outline the method.

6. Mar 2, 2010

### Altabeh

Hi

Thank you for introducing Laurent's book to me, though I went through the Internet by google and unfortunately didn't find the book! But looking at the method you use to derive the Doppler effect, it sounds a little bit neat but yet complicated enough to make me feel like I won't understand or guess what is behind all those equations! Not finding the book required to get what the procedure of derivation is, compelled me to do all calculations through the old algebraic\calculus methods of SR and finally I got the answer. A very interesting point is that my first try at solving the problem led to an expression for $$P$$ which involved two extra terms in the denominator. But later on, it came to my mind that using the relativistic energy formula these two would reduce to $$m_0^2c^2$$. I think you hit the same hedge halting the easygoing derivation for some time.

AB

7. Mar 2, 2010

### George Jones

Staff Emeritus
No, my mistake was very, very silly.

I wrote (setting $c=1$)

$$\left( M_0 + e_0 \right)^2 - p_0^2 = M_0^2 + 2M_0 e_0 + m_0^2 - e_0^2.$$

I used

$$-p_0^2 = m_0^2 - e_0^2 ,$$

but I forgot to square the second term in the binomial.

8. Mar 2, 2010

### Ben Niehoff

The method I would use to calculate something like this is the following:

Use four-vectors; i.e.

$$p = (\gamma m, \gamma m \vec \beta)$$

where

$$\vec \beta = \frac{\vec v}{c}$$

Then

1. Lorentz transform to the center-of-momentum frame.

2. In this frame, the results of the collision are easy to calculate: the velocity 3-vectors simply reverse themselves.

3. Finally, Lorentz transform back to the original frame.

To find the center-of-momentum frame, you just find the total four-momentum

$$p_{tot} = p + P = \left( \gamma m + M}, \gamma m \vec \beta \right)$$

The gamma factor for the CM frame is whatever is multiplying (m + M) in the 0 component of $p_{tot}$:

$$p_{tot} = \left( \frac{\gamma m + M}{m + M} (m+M), (m+M) \frac{\gamma m}{m + M} \vec \beta) \right)$$

so

$$\gamma_{CM} = \frac{\gamma m + M}{m + M}$$

and

$$\gamma_{CM} \beta_{CM} = \frac{\gamma m}{m + M} \beta$$

Note that in the last case, you only need to find the product [itex]\gamma_{CM} \beta_{CM}[/tex], because the Lorentz transform takes the form

$$\left( \begin{array}{cccc} \gamma & -\gamma \beta & 0 & 0 \\ -\gamma \beta & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right)$$

9. Mar 2, 2010

### Altabeh

Looks like a nice method, but unfortunately it's not so detailed to answer some questions huddled in my mind about it! It'd be awesome if you could show me a good source (book or articles ,...) that discusses it in a fairly detailed way.

Thanks.
AB