# The Equivalence of Mass and Energy

1. Apr 20, 2006

### BoogieL80

I was working on the following problem:

An electron and a positron each have a mass of 9.11 x 10-31 kg. They collide and both vanish, with only electromagnetic radiation appearing after the collision. If each particle is moving at a speed of 0.30c relative to the laboratory before the collision, determine the energy of the electromagnetic radiation.

I feel like I'm missing something basic here. I know the formula is E=mc2. However my webassign says my answer is wrong. Am I wrong to assume that the mass is 2*9.11e-31? Also don't I just multiply 0.30*c2?

2. Apr 20, 2006

### nrqed

No. The total energy is $E = \gamma m c^2$. Only if a particle is at rest does E= mc^2.

3. Apr 20, 2006

### BoogieL80

I'm sorry. Why are you using gamma in that equatin?

4. Apr 20, 2006

### Hootenanny

Staff Emeritus
$$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$

So the full equation then becomes;

$$E = \frac{m_{0}c^2}{\sqrt{1-\frac{v^2}{c^2}}}$$

The gamma is used as a 'shortend' verson of the equation.

~H

5. Apr 20, 2006

### BoogieL80

Thank you.

6. Apr 21, 2006

### dfx

Incidentally we were studying pair production and my physics teacher claimed u simply use e = 2mc^2 regardless of the velocity.

7. Apr 21, 2006

### Hootenanny

Staff Emeritus
This is valid since the minimum amount of energy required to produce a pair of particles is the amount of energy equivilant to their rest mass. Any additional energy will be in the form of kinetic energy. If two particles are produced with zero kinetic energy then the formula becomes;

$$E = \frac{m_{0}c^2}{\sqrt{1-\frac{v^2}{c^2}}}$$

$$v = 0 \Rightarrow E = \frac{m_{0}c^2}{\sqrt{1-\frac{0^2}{c^2}}}$$

$$E = \frac{m_{0}c^2}{\sqrt{1}} = E = m_{0}c^2$$

~H

8. Apr 21, 2006

### Reshma

Yes, but you would be dealing with moving masses here. So, $E = 2\gamma m_0 c^2$ would be right.

9. Apr 21, 2006

### Hootenanny

Staff Emeritus
Not if you were calculating the minimum energy required to produce a pair of particles.

~H

10. Apr 21, 2006

### nrqed

That can't be right. If you shoot an electron and a positron at 0.99c, the gamma rays emitted will have more energy than if they were initially at rest.

11. Apr 21, 2006

### Hootenanny

Staff Emeritus
Apologise people, I didn't read the "regardless of velcoity" bit. My mistake.

~H

12. Apr 21, 2006

### Reshma

Oh, but how does $\gamma mc^2$ take care of the two particles here(electron and positron)?

13. Apr 21, 2006

### nrqed

You have to use this for each particle. So total energy = twice that if the two are moving at the same speed.

14. Apr 21, 2006

### Hootenanny

Staff Emeritus
As I said above, I misread part of the question. But I do not understand what you are asking here?

~H