# Schutz, page 226 again

1. Oct 18, 2005

### Jimmy Snyder

Here is a question that requires you to have a copy of the book (A First Course in GR) to answer it.

On page 226, near the bottom of the page is this sentence:

The second assumption is called the slow-motion assumption, since it implies that the typical velocity inside the source region, WHICH is $\Omega$ times the size of THAT REGION, should be much less than 1.

Emphasis mine. Can someone tell me what THAT REGION is? If you tell me that THAT REGION is the source region, then I will ask what WHICH refers to.

2. Oct 18, 2005

### pervect

Staff Emeritus
Don't have the book, but the way I read it, "WHICH" refers to the velocity, and THAT REGION is "the source region".

... since it implies that the typical velocity inside the source region, (said velocity) being omega times the size of the source region, should be much less than 1.

3. Oct 19, 2005

### Jimmy Snyder

You are right. I can be so blind at times. In this case, I was concentrating on the phrase "WHICH is $\Omega$ times the size of THAT REGION", and thinking that he was comparing one region to another. Now I just have to figure out why he can compare a velocity to (the length of?) a region.

4. Oct 19, 2005

### George Jones

Staff Emeritus
I'm not sure what "typical" means, but here's a the RMS speed for a toy example.

Consider a 1-dimensional simple harmonic oscillator that moves in a region of length L = 2A.

x(t) = A sin(Omega t)

v(t)^2 = (A Omega cos(Omega t))^2

The average of cos^2 is 1/2, so

<v(t)^2> = 1/2 A^2 Omega^2

RMS v = sqrt(<v^2>) = A Omega/sqrt(2) = (L Omega)/(2 sqrt(2))

I'm not sure that the factor of 2*sqrt(2) is important in the type of estimate that Schutz does.

Regards,
George

5. Oct 19, 2005

### Jimmy Snyder

Not important. He was just giving a justification for the term "slow motion assumption". If he cared about the factor, he could absorb it into the w in slow. I just had time to make a quick response to pervect's post before I ran off to work this morning, but I knew that when I got back home I would have to carry out the calculation that you did for me in your post. Thanks.