# Energy in Special Relativity

1. Mar 12, 2009

### Master J

2 particles, A & B, travel towards eachother with speed S relative to the laboratory.

Show that the energy of A as measured by B is : Mc^2 (1 + S^2/c^2) y^2

y=gamma=1/SQRT(1 - S^2/c^2) M=rest mass

So if I consider the rest frame of B, I get the speed of A as 2S/[y(1 + S^2/c^2)], using the velocity transformation, and considering the frame of A also has speed S.

I am unsure where to go now, I can't see how I can end up with that ^^^^^.

Any pointers guys? Cheers

2. Mar 12, 2009

### LongLiveYorke

It's probably more complicated to first try to find the velocity of particle A in B's frame and then to use that velocity to calculate the energy. There's a more direct way that makes the problem a bit simpler.

Recall, $E = M c^2 \gamma$

So, to measure the energy of A in B's frame, we only need to know the gamma of A in B's frame. We know the gamma of A in the lab frame (or y, which is a part of the final answer so we can keep everything in terms of y). Do you know how to transform gamma (y) between frames? If you can figure that out, then you've basically solved the problem.

3. Mar 12, 2009

### turin

You can do this in a very elegant way.

Let pA denote the 4-momentum of A.
Let EA denote the energy of A.
Let pB denote the 4-momentum of B.
Let mB denote the (rest) mass of B.
Let . denote contraction of the Lorentz indices.

Relativity tells you two things:

1) pA.pB = EA mB
2) pA.pB is a Lorentz invariant.

EDIT: Do A and B have the same mass?

4. Mar 19, 2009

### Master J

turin, your method seems a little beyond me, I haven't covered that before.

As for thransforming gamma between frames, could you give me a pointer in the right direction? I find these transformations can easily get me lost at times!

5. Mar 22, 2009

### turin

If you plan on ever using relativity, you should really invest some time in the elegant methods that utilize Lorentz invariants. Furthermore, it provides the most popular example (beyond Euclidean symmetries) of the use of invariants of a physical symmetry.

As for your transformation of gamma, here is a simplified representation in 1 spatial dimension.

$$\left(\begin{array}{cc} \gamma_1&\gamma_1\beta_1\\\gamma_1\beta_1&\gamma_1 \end{array}\right) \left(\begin{array}{cc} \gamma_2&\gamma_2\beta_2\\\gamma_2\beta_2&\gamma_2 \end{array}\right) = \left(\begin{array}{cc} \gamma_3&\gamma_3\beta_3\\\gamma_3\beta_3&\gamma_3 \end{array}\right)$$

it is a straightforward matrix calculation to determine $\gamma_3$ from $\gamma_1$ and $\gamma_2$.