Lightlike compactification

In summary, the conversation revolves around finding an identification for problem 5 on the given website. The identification involves identifying the non-primed coordinates in terms of the primed coordinates and plugging them into the identification. There is a difference in opinions on one of the terms in the identification, but it is eventually resolved and the correct answer is given.
  • #1
ehrenfest
2,020
1
I am trying to do problem 5 at the following website:

http://www.myoops.org/twocw/mit/NR/rdonlyres/Physics/8-251String-Theory-for-UndergraduatesSpring2003/F4BA42A3-4DD9-402F-BDA8-6D5CB14B0FAF/0/ps1.pdf [Broken]

I got for (b)

[tex] x' ~ \gamma \left( \gamma (1 -\beta^2) x'^0 + 2 \pi ( 1- \beta) R \right)[/tex]

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?
 
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  • #2
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
George Jones said:
I think that you have a mistake in your latex, but, from what I can see, I think that I get something quite different.

I left out an identification sign.

[tex] x' tilde \gamma \left( \gamma (1 -\beta^2) x' - 2 \pi ( 1- \beta) R \right)[/tex]

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
 
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  • #4
Use \sim. (Boy, was I ever wrong at first)

Do you mean

[tex]x'^0 \sim \gamma \left( \gamma (1 -\beta^2) x'^0 - 2 \pi ( 1- \beta) R \right)?[/tex]

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

[tex]-2 \pi ( 1+ \beta) R \right)[/tex]

To simplify your expression, multiply the outside [itex]\gamma[/itex] through, and use

[tex]\gamma = \frac{1}{\sqrt{1 - \beta^2}} = \frac{1}{\sqrt{\left( 1 - \beta \right) \left( 1 + \beta \right)}}[/tex]

I'm not sure whose sign is correct, as both give nice "Doppler-shift" results. I'll have another look.
 
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  • #5
George Jones said:
Use \sim. (Boy, was I ever wrong at first)

Do you mean

[tex]x'^0 \sim \gamma \left( \gamma (1 -\beta^2) x'^0 - 2 \pi ( 1- \beta) R \right)?[/tex]

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

[tex]-2 \pi ( 1+ \beta) R \right)[/tex]

To simplify your expression, multiply the outside [itex]\gamma[/itex] through, and use

[tex]\gamma = \frac{1}{\sqrt{1 - \beta^2}} = \frac{1}{\sqrt{\left( 1 - \beta \right) \left( 1 + \beta \right)}}[/tex]

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

Your sign is right.

So, the complete answer for (a) is

[tex]x'^0 \sim \gamma \left( \gamma (1 -\beta^2) x'^0 - 2 \pi ( 1+\beta) R \right)?[/tex]

[tex]x'^1 \sim \gamma \left( \gamma (1 -\beta^2) x'^1 + 2 \pi ( 1-\beta) R \right)?[/tex]

Simplified, this is

[tex]x'^0 \sim \left( x'^0 - 2 \pi ( 1-\beta)^{-1} R \right)?[/tex]

[tex]x'^1 \sim \left( x'^1 + 2 \pi ( 1+\beta)^{-1} R \right)?[/tex]

So, for part (c), I get R_s^2 = R^2/gamma^2. How about you?
 
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  • #6
ehrenfest said:
[tex]- 2 \pi ( 1+\beta) R \right)?[/tex]

Not quite; careful with the square roots.
 
  • #7
I double-checked everything and I cannot find any problems with my expression for [tex] x'^0 [/tex] (= ct). So, if you're sure its wrong I'll post all my work. Just let me know.
 
  • #8
ehrenfest said:
I double-checked everything and I cannot find any problems with my expression for [tex] x'^0 [/tex] (= ct). So, if you're sure its wrong I'll post all my work. Just let me know.

OK, maybe you should show how you got the term I referenced in my last post.
 
  • #9
We know that [tex] x'^0 = \gamma(x^0 - \beta x') [/tex] and that
[tex] x'^1 = \gamma ( -\beta x^0 + x') [/tex]

Do you agree that

[tex] x ^0 = (x'^0 +\beta x' \gamma)/\gamma [/tex] and that

[tex] x^0 = (x'' - \gamma x')/(-\gamma \beta) [/tex]

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

[tex] x'^0 = \gamma(x^0 - \beta x') \sim \gamma(x^0-2\pi R - \beta (x^1 +2\pi R)) [/tex] by replacing the unprimed coordinates on the rhs of above with the expressions I found for them in the previous two equations
 
  • #10
Sorry, I posted the wrong term, thus misleading you.

I think there is a mistake going from

[tex]x'^0 \sim \gamma \left( \gamma (1 -\beta^2) x'^0 - 2 \pi ( 1+\beta) R \right)[/tex]

to

[tex]x'^0 \sim \left( x'^0 - 2 \pi ( 1-\beta)^{-1} R \right).[/tex]

I think there is a problem with the term

[tex]- 2 \pi ( 1-\beta)^{-1} R .[/tex]
 
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  • #11
George Jones said:
Sorry, I posted the wrong term, thus misleading you.

I think there is a mistake going from

[tex]x'^0 \sim \gamma \left( \gamma (1 -\beta^2) x'^0 - 2 \pi ( 1+\beta) R \right)[/tex]

to

[tex]x'^0 \sim \left( x'^0 - 2 \pi ( 1-\beta)^{-1} R \right).[/tex]

I think there is a problem with the term

[tex]- 2 \pi ( 1-\beta)^{-1} R .[/tex]

You're right. It should be

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

and


[tex]x'^1 \sim \left( x'^0 + 2 \pi \sqrt{ \frac{1-\beta}{1+\beta}} } R \right).[/tex]
 
  • #12
ehrenfest said:
You're right. It should be

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

Good.

Now, I really do have a small problem

with the

[tex]2 \pi ( 1-\beta) R[/tex]

term in

[tex]x'^1 \sim \gamma \left( \gamma (1 -\beta^2) x'^1 + 2 \pi ( 1-\beta) R \right).[/tex]
 
  • #13
George Jones said:
Good.

Now, I really do have a small problem

with the

[tex]2 \pi ( 1-\beta) R[/tex]

term in

[tex]x'^1 \sim \gamma \left( \gamma (1 -\beta^2) x'^1 + 2 \pi ( 1-\beta) R \right).[/tex]

I see. It should be

[tex]x'^1 \sim \gamma \left( \gamma (1 -\beta^2) x'^1 + 2 \pi ( 1+\beta) R \right).[/tex]

So

[tex]x'^1 \sim \left( x'^1 + 2 \pi \sqrt{ \frac{1+\beta}{1-\beta}} } R \right).[/tex]
 
  • #14
Yes.
 
  • #15
For c then, I think that we get the desired S' with

[tex]\gamma R = \beta R_s [/tex]
 
  • #16
ehrenfest said:
For c then, I think that we get the desired S' with

[tex]\gamma R = \beta R_s [/tex]

I didn't get this, but I did get something very close to this.
 
  • #17
[tex] R = \beta R_s \gamma [/tex] ?
 
  • #18
ehrenfest said:
[tex] R = \beta R_s \gamma [/tex] ?

I get a negative sign.
 
  • #19
OK. Well it should really be +/- because we have [tex]R^2 = B^2 R_s^2 \gamma^2[/tex] and then we take the square root of both sides, right?
 

1. What is lightlike compactification?

Lightlike compactification is a mathematical concept used in theoretical physics to describe the behavior of spacetime at its boundary. It involves the idea of "light cones," which are structures that represent the paths that light can travel in a given spacetime.

2. How does lightlike compactification relate to general relativity?

Lightlike compactification is closely related to general relativity, which is a theory of gravity that describes how matter and energy affect the curvature of spacetime. In general relativity, light cones play a crucial role in understanding the behavior of objects in a curved spacetime.

3. What is the purpose of using lightlike compactification?

The purpose of using lightlike compactification is to provide a more complete and unified understanding of the behavior of spacetime, especially near its boundary. It allows for the inclusion of lightlike paths and other important structures that are not accounted for in traditional compactification methods.

4. How does lightlike compactification differ from other compactification methods?

Lightlike compactification differs from other compactification methods in that it includes the behavior of lightlike paths, which are essential in understanding the curvature and topology of spacetime. It also allows for a more complete and unified description of spacetime at its boundary.

5. Are there any real-world applications of lightlike compactification?

While lightlike compactification is primarily a theoretical concept, it has potential applications in the study of black holes and other extreme astrophysical phenomena. It may also have implications for the understanding of quantum gravity and other fundamental theories of physics.

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