# Homework Help: Lightlike compactification

1. Aug 13, 2007

### ehrenfest

I am trying to do problem 5 at the following website:

I got for (b)

$$x' ~ \gamma \left( \gamma (1 -\beta^2) x'^0 + 2 \pi ( 1- \beta) R \right)$$

and I got a similar identification for ct'.

For part (c), he wants us to find an identification in which the space coordinate is identified but the time coordinate is not. Does that mean is he asking to find x' ~ 0? x' ~ x' ? By using B_s as a variable and holding everything else constant?

Last edited by a moderator: May 3, 2017
2. Aug 14, 2007

### George Jones

Staff Emeritus
I think that you have a mistake in your latex, but, from what I can see, I think that I get something quite different.

3. Aug 17, 2007

### ehrenfest

I left out an identification sign.

$$x' tilde \gamma \left( \gamma (1 -\beta^2) x' - 2 \pi ( 1- \beta) R \right)$$

That does not make it quite different though. But your right I think that is wrong.

OK. So, I can find the non-primed coordinates in terms of the primed coordinates and plug that into the identification. Is it possible to isolate to then isolate the each primed coordinate on one side? Are operations to both sides allowed on identifications?

BTW what is the tilde syntax in latex

Last edited: Aug 17, 2007
4. Aug 17, 2007

### George Jones

Staff Emeritus
Use \sim. (Boy, was I ever wrong at first)

Do you mean

$$x'^0 \sim \gamma \left( \gamma (1 -\beta^2) x'^0 - 2 \pi ( 1- \beta) R \right)?$$

If so then, your result, after simplification, is much closer to my result than I realized.

Shortly, after Zwiebach came out, I started working through it systematically, with the intention of finishing the whole thing. I only made it through the first six chapters before life intervened. I did, however, stuff my hand-written solutions to many of the exercises and problems for the first five chapters into a file folder that, despite moving moving between cities a number of times in the last few years, I somehow have not managed to lose.

I compared your result to the last line of my solution without actually thinking. :uhh: We differ by only one sign - I got

$$-2 \pi ( 1+ \beta) R \right)$$

To simplify your expression, multiply the outside $\gamma$ through, and use

$$\gamma = \frac{1}{\sqrt{1 - \beta^2}} = \frac{1}{\sqrt{\left( 1 - \beta \right) \left( 1 + \beta \right)}}$$

I'm not sure whose sign is correct, as both give nice "Doppler-shift" results. I'll have another look.

Last edited: Aug 17, 2007
5. Aug 18, 2007

### ehrenfest

So, the complete answer for (a) is

$$x'^0 \sim \gamma \left( \gamma (1 -\beta^2) x'^0 - 2 \pi ( 1+\beta) R \right)?$$

$$x'^1 \sim \gamma \left( \gamma (1 -\beta^2) x'^1 + 2 \pi ( 1-\beta) R \right)?$$

Simplified, this is

$$x'^0 \sim \left( x'^0 - 2 \pi ( 1-\beta)^{-1} R \right)?$$

$$x'^1 \sim \left( x'^1 + 2 \pi ( 1+\beta)^{-1} R \right)?$$

So, for part (c), I get R_s^2 = R^2/gamma^2. How about you?

Last edited: Aug 18, 2007
6. Aug 20, 2007

### George Jones

Staff Emeritus
Not quite; careful with the square roots.

7. Aug 23, 2007

### ehrenfest

I double-checked everything and I cannot find any problems with my expression for $$x'^0$$ (= ct). So, if you're sure its wrong I'll post all my work. Just let me know.

8. Aug 23, 2007

### George Jones

Staff Emeritus
OK, maybe you should show how you got the term I referenced in my last post.

9. Aug 24, 2007

### ehrenfest

We know that $$x'^0 = \gamma(x^0 - \beta x')$$ and that
$$x'^1 = \gamma ( -\beta x^0 + x')$$

Do you agree that

$$x ^0 = (x'^0 +\beta x' \gamma)/\gamma$$ and that

$$x^0 = (x'' - \gamma x')/(-\gamma \beta)$$

just by rearranging the first two equations in my post.

This allows us to solve for x^0 and x^1 in terms of primed coordinates.

I get x^0 = (\beta x'^1 + x'^0)/(\gamma - \gamma \beta ^2) and

x^1 = (x'^1 + \beta x'^0)/(\gamma - \gamma \beta ^2)

We can then rewrite the identification

$$x'^0 = \gamma(x^0 - \beta x') \sim \gamma(x^0-2\pi R - \beta (x^1 +2\pi R))$$ by replacing the unprimed coordinates on the rhs of above with the expressions I found for them in the previous two equations

10. Aug 27, 2007

### George Jones

Staff Emeritus
Sorry, I posted the wrong term, thus misleading you.

I think there is a mistake going from

$$x'^0 \sim \gamma \left( \gamma (1 -\beta^2) x'^0 - 2 \pi ( 1+\beta) R \right)$$

to

$$x'^0 \sim \left( x'^0 - 2 \pi ( 1-\beta)^{-1} R \right).$$

I think there is a problem with the term

$$- 2 \pi ( 1-\beta)^{-1} R .$$

Last edited: Aug 27, 2007
11. Aug 27, 2007

### ehrenfest

You're right. It should be

$$x'^0 \sim \left( x'^0 - 2 \pi \sqrt{ \frac{1+\beta}{1-\beta}} } R \right).$$

and

$$x'^1 \sim \left( x'^0 + 2 \pi \sqrt{ \frac{1-\beta}{1+\beta}} } R \right).$$

12. Aug 27, 2007

### George Jones

Staff Emeritus
Good.

Now, I really do have a small problem

with the

$$2 \pi ( 1-\beta) R$$

term in

$$x'^1 \sim \gamma \left( \gamma (1 -\beta^2) x'^1 + 2 \pi ( 1-\beta) R \right).$$

13. Aug 27, 2007

### ehrenfest

I see. It should be

$$x'^1 \sim \gamma \left( \gamma (1 -\beta^2) x'^1 + 2 \pi ( 1+\beta) R \right).$$

So

$$x'^1 \sim \left( x'^1 + 2 \pi \sqrt{ \frac{1+\beta}{1-\beta}} } R \right).$$

14. Aug 28, 2007

### George Jones

Staff Emeritus
Yes.

15. Aug 28, 2007

### ehrenfest

For c then, I think that we get the desired S' with

$$\gamma R = \beta R_s$$

16. Aug 29, 2007

### George Jones

Staff Emeritus
I didn't get this, but I did get something very close to this.

17. Aug 29, 2007

### ehrenfest

$$R = \beta R_s \gamma$$ ?

18. Aug 30, 2007

### George Jones

Staff Emeritus
I get a negative sign.

19. Aug 30, 2007

### ehrenfest

OK. Well it should really be +/- because we have $$R^2 = B^2 R_s^2 \gamma^2$$ and then we take the square root of both sides, right?