# Relativity and Momentum

1. Aug 7, 2008

### UAPhys03

1. The problem statement, all variables and given/known data
What is the momentum of a proton with a kinetic energy of 750MeV?

2. Relevant equations
E = m0c2 + K
E2 = p2c2 + m0c2
Mp = 1.67 x 10-21

3. The attempt at a solution

I think whats confusing me is K = 750MeV, but I tried to solve for p

First, I converted 750MeV to 750x106 eV

Then solved for E in the first equation
E = 750x106 eV

Then used that to solve for p in the second equation
But i get a wrong answer.

Last edited: Aug 7, 2008
2. Aug 7, 2008

### hage567

In your second equation, $$m_oc^2$$ needs to be squared. So it should be $$E^2 = p^2c^2 + (m_oc^2)^2$$

You also need to express the proton mass in electronvolts. Are you doing that?

3. Aug 8, 2008

### Redbelly98

Staff Emeritus
In other words, you get that E = K. But that's only true when m0=0, which is not the case for a proton.

You did mean to say proton, and not photon?

4. Aug 8, 2008

### UAPhys03

Hey,

So I fixed the equation and figured it out

E = m0c2 + KE

m0c2 = 1.5x10-10j = 939.4MeV

E = 939 + 750 = 1689MeV

E2 = p2c2 + m02c2

16892 = p2c2 + 9392

p2c2 = 191000

pc = 1404MeV

p = 1404MeV/c

Thanks!

5. Aug 9, 2008

### apunisheriii

is it okay to solve like this??

1/2(Mp)v2=K
1/2(p2)/Mp=K
p=square root of 2K(Mp)
p=6.34x10^-19 kg m/s

6. Aug 9, 2008

### Andrew Mason

Your arithmetic is not correct. It should be: p2c2 = 1,970,000

AM

7. Aug 9, 2008

### Andrew Mason

Not for relativistic speeds. A proton with 750 Mev of kinetic energy is moving at relativistic speed. At non-relativistic speed you can use:

$$\text{KE} = \frac{1}{2}mv^2 = \frac{p^2}{2m}$$

$$p = \sqrt{2m\text{KE}}$$

You had it right if you were using Mp for proton mass and p for momentum, but it is confusing.

AM

8. Aug 9, 2008

### apunisheriii

How can i know whether it is in non-relativistic or relativistic speed???

9. Aug 9, 2008

### Redbelly98

Staff Emeritus
For non-relativistic speeds, KE has to be a lot less than mc2. Equivalently, this means the speed is a lot less than c.

In this problem, KE is 750 MeV. That is comparable to mc2 = 940 MeV, so it is relativistic in this case.

10. Aug 9, 2008

### UAPhys03

So for relativistic speed wouldn't you just substitute that equation,

K = 1/2((lorenz factor) - 1)mv2

substitute for p

K/((lorenz factor) - 1) = p2/2m

and you could figure out how fast the proton is moving given the kinetic energy?

11. Aug 9, 2008

### Redbelly98

Staff Emeritus
No, the relation is

K = ( (Lorenz factor) - 1 ) m c2

Given K, solve for the Lorenz factor, from which you can get v.

12. Aug 10, 2008

### apunisheriii

So can i use E=hf to calculate the momentum as it's relativistic??

13. Aug 10, 2008

### Redbelly98

Staff Emeritus
Why would you think that? Momentum does not appear anywhere in your equation ???

14. Aug 11, 2008

### apunisheriii

i thought the wavelength in the equation E=hc/lambda is the same as the de Broglie's wavelength......
is the de Broglie's wavelength of a electron with non-relativistic speed couldnt be sub into the E=hf equation??

15. Aug 11, 2008

### Redbelly98

Staff Emeritus
Sometimes, not always. There actually 3 "speed regimes", not 2, to think about.

1. Non-relativistic: K is much less than mc2
2. Moderately relativistic: K is comparable to mc2
3. Extremely relativistic: K is much larger than mc2

That equation, E = hf = hc/lambda = pc , is valid in the extremely relativistic case. Another way of seeing this is from two equations that are true in all 3 cases:

E2 = (pc)2 + (mc2)2
and
E = K + mc2

In case 3 (and only then) we can neglect the mc2 terms because it is much smaller than the other quantities, and we get

E = K = pc (and also E = hc/lambda = hf)

Hope that helps.

16. Aug 12, 2008

### apunisheriii

o,
finally i understood!!
thnx ya!!