# Relativistic Energy

1. Oct 11, 2016

### Kunhee

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

Two particles of rest mass m0 approach each other with equal and opposite velocity v, in a laboratory frame. What is the total energy of one particle as measured in the rest frame of the other?

But the question gives a clue which reads "if (v/c)^2 = .5, then E = 3m0c^2."

3. The attempt at a solution

My current understanding of this question is that the total energy is simply m0c^2 multiplied by the Lorentz factor. Could you explain to me how the clue is relevant?

2. Oct 11, 2016

### TSny

Maybe the clue is to be used to check your answer for the special case when (v/c)2 = .5

3. Oct 11, 2016

### Kunhee

When (v/c)^2 = .5 then the equation is just that E = (1.414)(m0c^2). Where did the 3 come from?

4. Oct 11, 2016

### TSny

Yes

5. Oct 11, 2016

### TSny

You are asked to find the energy of one of the particles according to a reference frame moving with the other particle. The statement (v/c)^2 = .5 specifies v for each particle in the lab frame. But you need to work in the frame of one of the particles.

6. Oct 11, 2016

### Fightfish

It's not a hint per se - it's just providing you with a particular test case to enable you to check if your obtained answer is correct.

What would be a hint is: what is the velocity of the other particle as measured in the rest frame of one of the particles?

7. Oct 11, 2016

### Kunhee

Oh I see.
Could you help me set up the equation in the particle's frame of reference?

Before I can plug into E = y m0 c^2
I need to find the m0 when the Lorentz factor is applied first because we are
observing it from the frame of the particle?

8. Oct 11, 2016

### Fightfish

$m_0$ is unchanged - it is the rest mass of the particle. What is changed is the Lorentz factor - because the particles move with different velocities in different frames.

9. Oct 11, 2016

### Kunhee

Does that mean I need to first do velocity transformation
u = (u' + v) / (1 + u'v/c^2)

And use the resulting velocity to plug into
E = y m0 c^2 ?

10. Oct 11, 2016

### Fightfish

Yes, that is correct - but it would be good if you try not to view it as doing some stuff and then plugging into a formula, because that often hinders with the understanding of the concepts.

11. Oct 11, 2016

### Kunhee

I see. Thank you!

12. Oct 11, 2016

### Kunhee

-

Last edited: Oct 11, 2016
13. Oct 11, 2016

### Fightfish

Perhaps it would help if you show your final expression for the energy. The transformed velocity $v'$ is a function of $v$, so your final answer will / should be in terms of $v$, because well, that is your given initial parameter.

14. Oct 12, 2016

### Kunhee

u' = [u - (-u)] / [1 - (u)(-u)/c^2] = 2u / [1+(u/c)^2

Plugging the u' into Lorentz factor for the Energy equation:

E = y m0 c^2 = m0 c^2 / [1 - ( [2u / [1+(u/c)^2] / c )^2] ^ -1/2

And with further simplification I get m0 c^2 / (1/3) = 3

Got it. Thanks!