Please explain what is wrong with my relativistic momentum problem

[xox]

Sure there is. You want to see if they simplify. If it didn't check the terms under the square root then you would complain that it didn't simplify things under square roots.

Do me a favor, please do the calculation in post 100 by hand, it is not very complicated. To make things easier, calculate ONLY the y component of the momentum, no need to do the x component. This halves the number of terms. Post the LaTeX. Please point out where you needed to "check the terms under the square roots".

 

[Dale]

If you don't look under the square roots then you get:

$$-\frac{2 a m}{\left(B^2+1\right) g}+\frac{b m}{\left(1-B^2\right) g y}-\frac{b
m}{\left(B^2+1\right) h y}$$

Where y and B are defined as in post 100 and $$g=\sqrt{1-\frac{b^2}{\left(1-B^2\right)^2 c^2 y^2}}$$ and $$ h= \sqrt{1-\frac{\frac{4 a^2}{\left(B^2+1\right)^2}+\frac{b^2}{\left(B^2+1\right)^2
y^2}}{c^2}}$$

So you need to check the terms under the square roots pretty quick.

 

[xox]

If you don't look under the square roots then you get:

$$-\frac{2 a m}{\left(B^2+1\right) g}+\frac{b m}{\left(1-B^2\right) g y}-\frac{b
m}{\left(B^2+1\right) h y}$$

Where y and B are defined as in post 100 and $$g=\sqrt{1-\frac{b^2}{\left(1-B^2\right)^2 c^2 y^2}}$$ and $$ h= \sqrt{1-\frac{\frac{4 a^2}{\left(B^2+1\right)^2}+\frac{b^2}{\left(B^2+1\right)^2
y^2}}{c^2}}$$

So you need to check the terms under the square roots pretty quick.

Please prove that I need "to check the terms under the square root", this is all I asked. You did not finish the calculations, I did and I did not have to check any radicand.

 

[Dale]

Prove what? I thought we were looking at the results of FullSimplify.

 

[xox]

Prove what?

that one needs to check any of the radicands for being positive.

 

[Dale]

They don't simplify otherwise.

 

[xox]

They don't simplify otherwise.

Continue the calculations, show that this is the case.

 

[Dale]

Continue what calculations? I finished and posted it. That is as simple as the expression goes without additional information.

 

[Dale]

I do have some simple Mathematica code for doing special relativity exercises. It is based on 4-vectors, so it is rather convenient to use for momentum and other similar problems.

Here is a copy of the code. Use at your own risk