# Derivation of Relativistic Momentum

1. May 17, 2010

### rrrright

Hi. I was wondering if anyone has a simple derivation of relativistic momentum from lorentz transformation or the relativistic velocity addition formulas. I have attempted to understand this example:

http://en.wikibooks.org/wiki/Special_Relativity/Dynamics#Momentum

but I have been having some difficulties understanding some of it. If anyone could help it would be much appreciated.

2. May 17, 2010

### Fredrik

Staff Emeritus

Which definition you should use depends on what method you are using to include particles and their interactions in your theory. If you do it just by writing down a force, then (four-)momentum is defined simply as mass times (four-)velocity. If you do it by writing down a Lagrangian L, the momentum corresponding to the ith position coordinate is defined by $$p^i=\frac{\partial L}{\partial\dot x^i}$$. If you do it by writing down a Hamiltonian, you don't define momentum, it's a primitive, just like position in the other two pictures I mentioned.

You seem to be talking about proving that momentum is conserved in particle collision, not about defining momentum. That's another story, and I'll let someone else answer that.

3. May 17, 2010

### starthaus

This is a very good derivation (the whole page is very good). If you need any help, LMK.

4. May 18, 2010

### rrrright

You will have to bear with me since I am just a high school student attempting to learn some of this on my own.

When they say:

$$u'_{yR}=-u'_{yB}$$

What do they mean and how do they reach this conclusion? I am not seeing it from the diagram.

Later in the explanation it states that

$$u'_{xR}=v$$

How do they come to this conclusion?

5. May 18, 2010

### starthaus

The author (R.C. Tolman, in a 1917 paper) chose the speeds such that $$u'_{yR}=-u'_{yB}$$. This is not a conclusion, it is a choice that facilitates the rest of the calculations.

This is also a choice that allows the determination of $$v$$ from the equation:

$$v=u'_{xR}=\frac{u_{xR}-v}{1-v*u_{xR}/c^2}$$

6. May 18, 2010

### rrrright

Okay that makes sense.

Am I correct in my assumption that in the second frame of reference the observer is moving at $$u_{xB}$$ but not $$u_{yB}$$?

7. May 18, 2010

### starthaus

Yes, the observer is moving along the x axis, with the same speed as the x component of the blue ball speed.