# Homework Help: Relativistic collision

Tags:
1. Oct 6, 2014

### Rubber Ducky

1. The problem statement, all variables and given/known data

Two relativistic particles "L" and "R", each of rest mass $m_0$, are moving at speed $v$ towards each other (in the frame of an observer). They collide squarely and are stationary afterwards.

(a) From the perspective of one particle, what is the oncoming speed of the other?
(b) What is the rest mass, relativistic mass, total energy, kinetic energy, and momentum of each particle before the collision in the frame of the observer?
(c) Repeat (b) in the rest frame of particle "L"
(d) What is the total momentum and energy before and just after the collision, in the observer's frame?

2. Relevant equations

$u'=\frac{u-v}{1+\frac{v^2}{c^2}}$

$m=\frac{m_0}{\sqrt{1-\frac{u^2}{c^2}}}$
$E=\frac{m_0c^2}{\sqrt{1-\frac{u^2}{c^2}}}=K+m_0c^2=\sqrt{(pc)^2+(m_0c)^2}$
$p=\frac{m_0u}{{\sqrt{1-\frac{u^2}{c^2}}}}$

3. The attempt at a solution

(a) If the observer is in frame $S$, and the frame of L is $S'$, then $\left |u'_R\right |=\frac{v+v}{1+\frac{v^2}{c^2}}=\frac{2v}{1+\frac{v^2}{c^2}}$

(b) $m_L=\frac{m_0}{\sqrt{1-\frac{u^2}{c^2}}}=m_R$

The rest of this part is relatively (heh) straight forward, I think. I'm not sure how to apply the same formulas to part (c), though. For example, for the mass of R with respect to L, would I copy the formula from part (b), but replace $u$ by what I found in (a)?

Also, in the frame of L, shouldn't its rest mass be equal to its relativistic mass, since particle L is at rest in this frame, and rest mass is defined as the mass measured by a frame in which the object is at rest?

2. Oct 7, 2014

### Orodruin

Staff Emeritus
Many people on this forum will cringe from seeing this which is in the formulation of your problem. Relativistic mass is not really used among scientists and with good reason. Instead, total energy (which is exactly the same thing up to a factor of $c^2$, but we tend to use units where $c = 1$) will typically be used.

That being said, what you have done so far looks fine. For (c), you can use the same argumentation, but you now have to consider the actual velocity each particle has in the rest frame of L. The answer to:
would therefore be "yes".
Yes. Again, note that we would typically talk only about one mass, called the rest mass, invariant mass, or just mass).

3. Oct 7, 2014

### Rubber Ducky

Thanks for your reply. My prof feels the same way as you, and my textbook says as much; "we feel that relativistic mass does not lead to any deeper physical understanding, so we will not treat it in this text". So it's weird / frustrating to see a question about it.

For (c), in computing the relativistic mass, I get $m_R=\frac{m_0}{\sqrt{1-\frac{(u'_R)^2}{c^2}}}$. There's something I'm unsure of - in (a) I needed to find the speed only, but now I believe I'm looking at velocity?

$u'_R=\frac{-v-v}{1+\frac{v^2}{c^2}}=\frac{-2v}{1+\frac{v^2}{c^2}}$ since particle L perceives particle R as moving in the negative direction. The simplification for $m_R$ is fine - just wanted to know if I'm using the correct reasoning.

Now for total energy, in (c), isn't it true that $E_L=\frac{m_0c^2}{\sqrt{1-\frac{u'_Lv}{c^2}}}=m_0c^2$, since particle L perceives itself as being stationary? And $E_R=m_Rc^2$?

4. Oct 7, 2014

### Orodruin

Staff Emeritus
What you used was the expression for relativistic velocity addition (in one dimension, the one in which the Lorentz transformation boosts) and it is sign dependent so for one of the transformations in (a) you technically have to take the absolute value. For the total energy, what appears in the gamma factor should be $v^2/c^2$ (where v is the speed of an object (but the velocity squared is clearly the same as speed squared) and thus only speed is relevant for the gamma factor. This brings me to:

Yes, the total energy of any object in its rest frame is the rest energy. However, watch out with the gamma factor, it should be containing $u'_L{}^2$, not $u'_L v$.

5. Oct 7, 2014

### Rubber Ducky

Oops, I always make that mistake. I've come across something puzzling in (c) when finding the kinetic energy of particle R, $K_R$. I know $E_R=K_R+m_Rc^2$, so $K_R=E_R-m_Rc^2=0$, since $E_R=m_Rc^2$. But from L's perspective, particle R is certainly moving, so how can its kinetic energy vanish? I'm guessing $E_R=m_Rc^2$ is incorrect?

6. Oct 8, 2014

### Orodruin

Staff Emeritus
This is just one of the drawbacks of relativistic mass, confusion.

Kinetic energy is by definition total energy minus rest energy. The rest energy is proportional to the rest mass (invariant mass), not to the relativistic mass (total energy).