# Sears, Brehme and momentum

1. May 7, 2010

### snoopies622

In section 8-2 of Sears and Brehme's "Introduction to the Theory of Relativity", the authors derive the formula for the "relativistic mass" of a particle in motion by analyzing a completely inelastic collision between identical objects from more than one frame of reference. I'm having a little trouble following it because their analysis relies heavily on a Brehme space-time diagram, which is a kind I'm not used to.

My question however is, is this even possible to do? In an inelastic collision the rest masses of the objects change, and nowhere in their analysis do I see this mentioned or factored into the equations. Also, when I try to analyze the collision myself - without using their diagram but with the same overall approach - I get a different (wrong) answer. I don't know if I'm making a mistake or if authors are somehow cheating.

I won't go into any of the mathematical details here in this first entry just in case someone out there already knows the answer to this question based on the information I've given, or already has the text.

Thanks.

2. May 7, 2010

### starthaus

Normally this subject is treated through conservation of momentum in elastic collision. See a very good treatment here

3. May 7, 2010

### snoopies622

Thanks, starthaus. Yes, I've seen that approach. What attracts me to the Sears and Brehme one is that it takes place in only one dimension of space instead of two. That's why I'm struggling to understand it.

Here's a little of the math.

They start by saying, suppose an object of mass B moving at velocity v strikes another object of mass A that is at rest. If the two objects stick together and then move with velocity V, then we shall define the ratio of the two masses in this way:

$$\frac {m_A}{m_B} = \frac {v}{V} - 1$$

I see how this would work in classical mechanics, but I've never seen in used in relativistic mechanics, and I could be wrong but it does smell a little fishy to me.

Any thoughts?

Last edited: May 7, 2010
4. May 7, 2010

### starthaus

It is difficult for me to decide whether the above condition is justified or not. They seem to use as a starting point:

$$(m_A+m_B)V=m_Bv$$

This doesn't seem right since the masses seem to be the "relativistic" (ugly) masses. Obviously, the $$m_B$$ in LHS is not the same as the $$m_B$$ on the RHS since they encapsulate different $$\gamma$$ values. To wit:

$$m_{B-LHS}=m0_B*\gamma(V)$$
$$m_{B-RHS}=m0_B*\gamma(v)$$

So, it would seem that the proof has started on the wrong foot.

On the other hand, the proof on wiki is very clean, there are no errors.

Last edited: May 7, 2010
5. May 8, 2010

### snoopies622

OK, I finally got the right answer using their approach, but I agree with you starthaus, that mass-ratio equation doesn't really make sense. I wonder why it works in this case then. Coincidence, perhaps. I suppose a deeper analysis would reveal the answer.

6. May 8, 2010

### snoopies622

Of course this raises the question, what is a good experimental way to define the ratio of two masses?

7. May 8, 2010

### JustinLevy

Hmm... maybe something along the lines of "resistance to change in inertia". So apply a constant force to the masses, and see how their velocities change. Maybe that isn't good since if one is taking mass to no longer be intrinsic to the particle, then how could you know if the "intrinsic" quantities the force depend on aren't changing with velocity as well.

That is pretty close to the original idea of how they came up with the notion of "relativistic mass" in the first place. Since instead of F = (d/dt) mv, we have F = (d/dt) gamma m v.

Out of curiosity, in the book, what do they do with relativistic mass after bringing it up? Or are they just bringing it up for historical reasons?

8. May 9, 2010

### snoopies622

I haven't read ahead yet but just by skimming it looks like they use it to find the relativistic kinetic energy of a particle. Actually it looks like there's a lot of interesting stuff ahead, but those damned Brehme diagrams are on every page!