Calculating Neutron B's Energy in Neutron A's Rest Frame | Relativistic Energies

[Brewer]

Homework Statement

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,

[tex](1+\frac{v^2}{c^2})(1-\frac{v^2}{c^2})^{-1}m_pc^2[/tex]

Homework Equations

Either:
[tex]E=\gamma mc^2[/tex]

or

[tex]E^2 = m^2c^4 + p^2c^2[/tex]

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:

[tex]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})[/tex]

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!

 

[neutrino]

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!),

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

 

[Brewer]

But Galilean is so much easier!

 

[neutrino]

But Galilean is so much easier!

Try saying that to your teacher.

 

[Brewer]

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

 

[neutrino]

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

Yup!

10char

 

[Brewer]

Apart from that was I on the right lines?

 

[neutrino]

Yes you were.

 

[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.

Thank you for your help.

 

[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:
[tex]v_B=\tanh\theta[/tex]

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

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

 

[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?

[tex]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}}[/tex]

right?

Using this I can get to [tex]E = \frac{mc^2}{1-\frac{v^2}{c^2}}[/tex]

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

I realized 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.