# Special Relativity help

1. Jan 28, 2010

### lockedup

1. The problem statement, all variables and given/known data
What is the energy of a photon whose momentum is the same as that of a proton with a kinetic energy of 10 MeV?

2. Relevant equations
K = mc$$^{2}$$($$\gamma$$-1)
p = $$\gamma$$mv
E = pc

3. The attempt at a solution
I figured I would go at it like this. I used the first equation listed to obtain a value for gamma and v. I put these into the second equation along with the mass of a proton. Then, I would put that value into the third equation. Is that even close?

2. Jan 29, 2010

### Lok

Seems ok if I understood it right. Your equations might help more.

3. Jan 29, 2010

### torquil

As far as I can understand, the procedure you describe is correct. As long as you have done the calculations properly, you should end up with the correct answer.

Torquil

4. Jan 29, 2010

### lockedup

Ok, so I have $$\gamma = 1 + 1.18 \times 10^{-19}$$ So, um, how am I supposed to square that (so I can find beta)? My calculator won't go that far... :*)

5. Jan 29, 2010

### dacruick

use an online calculator

6. Jan 29, 2010

### dacruick

or any computer program will do it

7. Jan 29, 2010

### dacruick

or you can just square the 1.18 and then double 19

8. Jan 29, 2010

### lockedup

They all give me 1

9. Jan 29, 2010

### dacruick

10. Jan 29, 2010

### dacruick

so for next time this is what you will do. you should square 1.18. then you should double the exponent. and voila

11. Jan 29, 2010

### lockedup

You're misreading the problem. If it was just 1.18e-19, I wouldn't have a problem. I need to square 1+1.18e-19. BTW, I found a calculator that will do it. Google can be tricksy sometimes...

12. Jan 29, 2010

### jdwood983

if $\gamma=1+1.18\times10^{-19}$ then you should just approximate $\gamma\simeq1$.

Also, use www.wolframalpha.com It's like having Mathematica available to you with 0 cost

13. Jan 29, 2010

### vela

Staff Emeritus
Your value for $\gamma$ is way too small. The proton's kinetic energy is about 1% of the rest energy, so $\gamma$ should be about 1.01.

14. Jan 30, 2010

### Altabeh

The speed of light = 299,792,458 m/s;
Proton mass: $$1.67262158*10^{-27} kg$$;
$$K= 1.602176487*10^{−15} J$$.

So with a simple mind-based calculation you at least can get that the Lorentz factor is around 1. The following shows its exact value up to 9 decimal digits:

$$\gamma=1.000010658$$.

AB

15. Jan 30, 2010

### chrispb

This is such a complicated way to do this problem. The energy of a proton is its kinetic energy plus its mass, and that squared is p^2 + m^2 (with c=1). Find p. E(photon) = p. Done.