What is the relative velocity of the two reference frames?

[byerly100]

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

 

[OlderDan]

A particle as observed in a certain reference frame has a total energy of 5 GeV and a momentum of 3 GeV/c (i.e., cp, which has the dimension of energy, is equal to 3 GeV).

a) What is its energy in a frame in which its momentum is equal to 4 GeV/c^2?
b) What is its rest mass in amu?
c) What is the relative velocity of the two reference frames?

I tried using E^2-(cp)^2=(E.)^2 but got 5.83 GeV. The answer to a is 5.66 GeV.

1 amu=931.49 MeV/c^2
The answer to b is 4.3 amu.
The answer to c is 0.187c.

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

 

[byerly100]

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

 

[OlderDan]

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

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

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

From your orignal data,
(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

 

[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.

 

[OlderDan]

Yes, I was using the dot for 0 subscript.

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

I could also use help on the rest of the problem.

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

 

[OlderDan]

I was trying b. I'm still working on it.

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

 

[byerly100]

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

931.49 MeV/c^2 = 1 amu

 

[OlderDan]

It is (m.c^2)^2.
4 Gev= (m.c^2)

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

 

[OlderDan]

It is (m.c^2)^2.
4 GeV= (m.c^2)
4.44x10^-8 kg(?)=m.

1 amu = 1.66x10^-27 kg

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.

 

[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...

 

[byerly100]

v= c^2p/E

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

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

 

[OlderDan]

v= c^2p/E

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

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

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 -

 

[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.

 

[OlderDan]

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.

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.