Question

Taylor series expansion of the equation for energy in Special Relativity

Answer 1

I have to make the Taylor expansion of the formula for energy according to special relativity by substituting \[x = v^2 \] and expand it to the order of \[v^4 \]

Answer 2

I'm attaching my calculations, because I'm pretty sure they are wrong.

Answer 3

First of all I noticed that your second derivative is incorrect the parenthesis should be to the 5/2 power.

Second you need to carry out the expansion one more time to get some more v^4 and v^2 terms

I think a better approach is not to make the change of variable ( x=v^2) and just keep v as you variable. that way you know that after the fourth term you can stop

Answer 4

Also for convergence purposes the expansion should be about 0 not an arbitrary velocity.

Answer 5

Hi - do you have to derive the whole thing from first principles ? You can just take the denominator and expand it using the binomial theorem for \[(1+x)^n\] where\[x=-\frac{ v^2 }{ c^2 }\] and \[n=-\frac{ 1 }{ 2 }\]
Three terms will give you the v^4 correction to the newtonian formula for kinetic energy

Answer 6

Thank you @gleem and @ProfBrainstorm ! Sorry for my late reply. Okay, I did what you suggested and expanded the expression to the powers of v 0, 2 and 4. I saw that the first expression is Einstein's formula and that the second is half the kinetic energy. But how about the terms at the power 1 and 3? Don't I miss taking them into account after such a substitution?

The problem states that I have to use Taylor expansion, so I can't apply the binomial expression.

Answer 7

If you're expanding in powers of v^2 then you do indeed have all powers of v^2, but because v is squared, you only end up with even powers of v.
The binomial expansion is what you end up with when you apply taylor's theorem to f(x) = (1+x)^n.

Answer 8

Okay, thank you!

Answer 9

I glanced at your working - remember there's no V0 and V, the expansion is around x=0, which is equivalent to V being zero.
Get all the calculus worked out using x, and only then substitute Vsquared back into the expressions to avoid confusion

Answer 10

Right. Okay.