# Relativistic energies

1. Apr 29, 2007

### Brewer

1. The problem statement, all variables and given/known data
Two neutrons A and B are approaching each other along a common straight line. Each has a constant velocity v as measured in the laboratory. Show that the total energy of neutron B as observed in the rest frame of neutron A is given by,

$$(1+\frac{v^2}{c^2})(1-\frac{v^2}{c^2})^{-1}m_pc^2$$

2. Relevant equations
Either:
$$E=\gamma mc^2$$

or

$$E^2 = m^2c^4 + p^2c^2$$

3. The attempt at a solution
I'm completely stuck on this. I think that the best way to get to the required answer is via the first of the two equations that I wrote down. However I can't manipulate the fraction into what is required. I get close I think, but its never quite what is required. I think the closest I get is by doing the following:

$$E = \gamma mc^2 = \frac{mc^2}{(1-\frac{v^2}{c^2})^{\frac{1}{2}}} * \frac{(1-\frac{v^2}{c^2})^2}{(1-\frac{v^2}{c^2})^2} = \frac{mc^2(1-\frac{v^2}{c^2})^2}{(1-\frac{v^2}{c^2})} = mc^2(1-\frac{v^2}{c^2})$$

From here I'm almost tempted to say that the bracket on the top is the difference of two squares, which would give me 2 brackets with the correct signs, but the v/c part wouldn't have the correct power, and the second bracket wouldn't be raised to the correct power either.

Another thought I had is that the velocity of B from A's reference frame is 2v (although I'm not 100% convinced about this - it involved some odd hand waving on my part to get to this result!), but i assume that once I have the answer in the correct form I should be able to insert the correct value and the answer will fall out properly.

If you should decide to help me I would appreciate an early step to help me on the way - I'd quite like the challenge of getting to the result by myself as much as possible. Its just I feel I may have gone wrong somewhere, or that I'm using the incorrect equations.

Thanks for reading this - I hope you can follow my above working/train of thought!

2. Apr 29, 2007

### neutrino

You are dealing with relativity. Therefore, you should abandon your Galilean ways of doing things.

3. Apr 29, 2007

### Brewer

But Galilean is so much easier!!

4. Apr 29, 2007

### neutrino

Try saying that to your teacher.

5. Apr 29, 2007

### Brewer

From that do you mean that the relativistic velocity is $$\frac{2v}{1+\frac{v^2}{c^2}}$$?

6. Apr 29, 2007

### neutrino

Yup!

10char

7. Apr 29, 2007

### Brewer

Apart from that was I on the right lines?

8. Apr 29, 2007

### neutrino

Yes you were.

9. Apr 29, 2007

### Brewer

So its just a case of algebra - its looking messy so far, I don't seem to able to get things to cancel out just yet.

10. Apr 29, 2007

### robphy

With rapidities and some knowledge of hyperbolic trig identities, the method of solution becomes more obvious and the answer falls out nicely.

Here's a starting point:
$$v_B=\tanh\theta$$

$$E_B=m\cosh\theta$$ in the lab frame. ($$E_A=m\cosh(-\theta)$$ in the lab frame.)
$$E'_B=m\cosh (\theta +\theta)$$ in the A-frame. ($$E'_A=m\cosh(-\theta+\theta)=m\cosh(0)$$ in the A-frame.)
Now write $$\cosh(2\theta)$$ in terms of $$\theta$$, then in terms of $$\tanh\theta$$.

In SR, the velocity is not additive, but the rapidity is additive.

Last edited: Apr 29, 2007
11. Apr 29, 2007

### Brewer

Whats a rapidity? I can't seem to find anything on the web - I'm not likely to have used them before under a different name am I?

Anyway, I seem to be having a few problems with my derivation - could someone take a look for me please?

$$u^2 = (\frac{2v}{1+\frac{v^2}{c^2}})(\frac{2v}{1+\frac{v^2}{c^2}}) = \frac{4v^2}{1+ \frac{2v^2}{c^2} + {v^4}{c^4}}$$

right?

Using this I can get to $$E = \frac{mc^2}{1-\frac{v^2}{c^2}}$$

but I can't see where the $$1+\frac{v^2}{c^2}$$ comes from. Apart from that I think I'm there.

I realised that when I was doing it before I wasn't getting rid of the square root on the bottom of the fraction by multiplying top and bottom by the bracket squared. Instead this time I multiplied top and bottom by the denominator. Still doesn't explain that other mystery term - its just mystifying!

Thanks for the info of rapidities - I think its beyond the scope of my course thus far, but I will look into it for my own personal information.

Last edited: Apr 29, 2007