# What is the relative velocity of the two reference frames?

1. Oct 8, 2006

### byerly100

c) What is the relative velocity of the two reference frames?

Last edited: Oct 9, 2006
2. Oct 8, 2006

### OlderDan

Maybe just try part a) again, but make sure you account for the total energy and the momentum in both frames.

Not sure where you went wrong. I got 5.66

3. Oct 8, 2006

### byerly100

I probably didn't account for both frames. How did you use the 4 GeV/c?

4. Oct 8, 2006

### OlderDan

E^2-(cp)^2 = (E.)^2 = (m.c^2)^2 is invariant

(I assume you are using the dot for the zero subscript)

(E.)^2 = 25GeV^2 - 9GeV^2 = 16GeV^2

4GeV/c is the p in the second frame. Substitute that into your energy equation and solve for E

5. Oct 8, 2006

### byerly100

Yes, I was using the dot for 0 subscript.

E^2-(16 GeV^2)=16 GeV^2
E= 5.66 GeV

I could also use help on the rest of the problem, b and c.

Last edited: Oct 8, 2006
6. Oct 8, 2006

### OlderDan

If you look back at the equations used, you should get part b) quite easily. Give it a try.

7. Oct 8, 2006

### OlderDan

What is the 16Gev^2 on the right hand side of your earlier equation?

8. Oct 8, 2006

### byerly100

16Gev^2=E.^2
4 GeV= E.

931.49 MeV/c^2 = 1 amu

Last edited: Oct 8, 2006
9. Oct 8, 2006

### OlderDan

So.. solve it for the rest mass and you've got it.

10. Oct 8, 2006

### OlderDan

Check your original post for conversion from MeV/c^2 to amu

Part c) is the hardest part. Do you know how to find the velocity of the particle in each frame from the known energy and momentum? Once you have that, you need to use the relitivistic velocity addition formula. I'll let you work on that a bit.

Last edited: Oct 8, 2006
11. Oct 8, 2006

### byerly100

4 x 10^9 eV / 931.49 x 10^6 eV = 4.3 amu (should there not be a /c^2 in the conversion information?)

I'm trying c now.

v/c = cp/E
v=c^2p/E

v. = c^2(3 GeV/c) / 4 GeV = 2.25x10^8 m/sec
v=2.12x10^8 m/sec

I got u= 2.86x10^8 m/sec...

Last edited: Oct 8, 2006
12. Oct 9, 2006

### byerly100

v= c^2p/E

ux=ux'+v/(1+vux'/c^2)

I got something for part c but it was off (wrong).

Last edited: Oct 9, 2006
13. Oct 9, 2006

### OlderDan

You left out parentheses in your numerator, but I expect you used them. You just have a sign problem. You know the velocity of the object in two frames. You are looking for the relativistic difference between the two. Try replacing the + in your equation with -

14. Oct 9, 2006

### byerly100

ux'=(ux-v)/(1-vux/c^2)

I got 0.185c. I used 1.8x10^8 m/sec for one v and 2.12x10^8 m/sec for another v.

Last edited: Oct 9, 2006
15. Oct 9, 2006

### OlderDan

I see you just changed it.. I was trying to figure out where you got the first v

I also got the .185c. I think you got it.