Particle energy and momentum in different reference frames

[warfreak131]

Homework Statement

A particle is observed in a certain reference frame (i.e. observer A) has a total energy of 5GeV and a momentum of 3GeV/c.
(a) What is the energy of the frame in which its momentum is equal to 4GeV/c.

Homework Equations

$$E'={\gamma}E-{\gamma}{\beta}(cp)$$
$$cp'=-{\beta}{\gamma}E+{\gamma}(cp)$$

The Attempt at a Solution

I plugged in cp'=4GeV, cp=3GeV, and E=5GeV, and solved the second equation for v. Then I plugged in v to the applicable variables in the first equation, and I got 4GeV, but the answer is 5.62GeV. And a different part of the question asks for the relative velocity of the two frames, and for that I get .6 c, as opposed to .19 c.

 

[vela]

You might find the invariant m² = E²-p² useful.

Also, the Lorentz transformations can be written in terms of the hyperbolic trig functions:

$$\begin{align*}<br /> E' &= E\cosh \theta + p\sinh \theta \\<br /> p' &= E\sinh \theta + p\cosh \theta<br /> \end{align*}$$

The velocity is then given by $\beta = \tanh \theta$.

 

[warfreak131]

Is the invariant some other form of the equation

$$E^{2}=(cp^{2})+(mc^{2})^{2}$$?

 

[vela]

Yeah, it's the same equation with c=1.

 

[warfreak131]

okay, i managed to get the velocity correct, but i still can't manage to get E' to come out correct

 

[warfreak131]

when the question says "total energy", does it mean E^2=(cp)^2+(mc^2)^2 or does it mean E=ymc^2?

 

[vela]

Either way should give you the same answer. Show us your calculations so we can spot your error.

 

[fzero]

when the question says "total energy", does it mean E^2=(cp)^2+(mc^2)^2 or does it mean E=ymc^2?

Both E's are the same since

$$\sqrt{ (pc)^2 + (mc^2)^2} = \gamma m c^2.$$

I was actually able to get E' to work out, but found $$\beta = 0.92 c$$ which disagrees with the relative velocity you quoted.

 

[vela]

I got β=0.186, so you should be able to get it to work out.

 

[warfreak131]

I got β=0.186, so you should be able to get it to work out.

i managed to get that for beta as well. but when i plug it into the transformation equations to find E', it doesn't come out to be correct