Inserting limits into an integrated term (Quick question)

[rwooduk]

Homework Statement

Derive the momentum for a charged particle going through matter.

Homework Equations

None.

The Attempt at a Solution

I understand the derivation but there's one step I am not clear about, and I'm probably being really stupid but this:

if the -infinity term is squared then doesn't it become positive and the sum equal zero?

Thanks for any advice.

 

[Dick]

Homework Statement

Derive the momentum for a charged particle going through matter.

Homework Equations

None.

The Attempt at a Solution

I understand the derivation but there's one step I am not clear about, and I'm probably being really stupid but this:

if the -infinity term is squared then doesn't it become positive and the sum equal zero?

Thanks for any advice.

Yes, you are correct and that's a really sloppy derivation. What really happens is that when you factor the x^2 out of the square root you get |x|. And x/|x| is 1 at +infinity and -1 at -infinity. Does that help?

 

[rwooduk]

Yes, you are correct and that's a really sloppy derivation. What really happens is that when you factor the x^2 out of the square root you get |x|. And x/|x| is 1 at +infinity and -1 at -infinity. Does that help?

Thanks for the reply, It's hard to visualise what you mean, is there any chance you could Latex it? If not, it's fine, I'll just remember that it goes to (2/b^2) and not zero.

thanks again!

 

[Dick]

Thanks for the reply, It's hard to visualise what you mean, is there any chance you could Latex it? If not, it's fine, I'll just remember that it goes to (2/b^2) and not zero.

thanks again!

What would be the point of memorizing that something like that? $\sqrt{x^2+b^2}=|x| \sqrt{1+\frac{b^2}{x^2}}$. Notice the factor in front is not $x$, it's $|x|$!

 

[rwooduk]

What would be the point of memorizing that something like that? $\sqrt{x^2+b^2}=|x| \sqrt{1+\frac{b^2}{x^2}}$. Notice the factor in front is not $x$, it's $|x|$!

Ahh, i see now, yes that makes sense! Thanks very much for this!