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Relationship between lightlike and spatial compactification

+ 2 like - 0 dislike
56 views

The compactification of a spatial dimension, say $x^1$ given by the identification $x \sim x^1 + 2\pi R$ is said to be related to the lightlike compactification by a Lorentz boost :

$$ \left( \begin{array}{c} x^0 \\ x^1 \\ \end{array} \right) \sim \left( \begin{array}{c} x^0 \\ x^1 \\ \end{array} \right) +2\pi \left( \begin{array}{c} -R \\ R \\ \end{array} \right) $$

What exactly is the relationship between them? How can I see this? And what does it mean from a physics point of view?

asked Jul 21, 2013 in Theoretical Physics by Dilaton (4,175 points) [ revision history ]
retagged Apr 19, 2014 by dimension10
What source are you reading from?

This post imported from StackExchange Physics at 2014-03-09 16:14 (UCT), posted by SE-user Physiks lover
I just did the Lorentz transform and realised that it's much simpler that way, . It's weird that this question's had just 68 views (and just 1 vote) whereas some questions get like 10000s of views and 100s of votes in like 4 hours .

This post imported from StackExchange Physics at 2014-03-09 16:14 (UCT), posted by SE-user Dimensio1n0
@Dimension10 yes, I have seen it thanks :-). Obviously such questions are not popular enough to get more votes (ambiguity of this sentence intented :-P ...). Yep it is mostly about the Lorentz transformations and the application of the issue in matrix/string theory is interesting too.

This post imported from StackExchange Physics at 2014-03-09 16:14 (UCT), posted by SE-user Dilaton
@Dilaton: Uh... for all I know, the application in string theory is just getting Type IIA from M-theory compactified on a light - like circle . . . Don't know if there are other applications.

This post imported from StackExchange Physics at 2014-03-09 16:14 (UCT), posted by SE-user Dimensio1n0

1 Answer

+ 2 like - 0 dislike

I hate this "Awww... Snap!" thing with SE on chrome.   

Basically, it's because a Lorentz Transformation maps between spatial coordinates and temporal ones.

I just realised that instead of writing this old (and "new") answer rubbish (in quotes now), I could have just done the Lorentz transform . So, in units where $\hbar=c_0=G=k_e= \ell_s=1 $ (who cares if other than $c_0$, any of them are actually involved),

$$\gamma \left[ {\begin{array}{*{20}{c}} 1&{ - v} \\ { - v}&1 \end{array}} \right]\gamma \left[ {\begin{array}{*{20}{c}} 1&{ - v} \\ { - v}&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{x_0} - 2\pi R} \\ {{x_1} + 2\pi R} \end{array}} \right] = \left[ \begin{gathered} {x_0} \\ {x_1} + 2\pi R \\ \end{gathered} \right]$$

That's what we want .

$$ \begin{gathered} \gamma \left[ \begin{gathered} {x_0} - 2\pi R - v{x_1} - 2\pi Rv \\ {x_1} + 2\pi R - v{x_0} + 2\pi Rv \\ \end{gathered} \right] = \left[ \begin{gathered} {x_0} \\ {x_1} + 2\pi R \\ \end{gathered} \right] \\ \left. \begin{gathered} \frac{{1 - \sqrt {1 - {v^2}} }}{{\sqrt {1 - {v^2}} }}{x_0} - 2\pi R\left( {1 + v} \right) - \frac{v}{{\sqrt {1 - {v^2}} }}{x_1} = 0 \\ \frac{{1 - \sqrt {1 - {v^2}} }}{{\sqrt {1 - {v^2}} }}{x_1} + 2\pi Rv - \frac{v}{{\sqrt {1 - {v^2}} }}{x_0} = 0 \\ \end{gathered} \right\} \\ \end{gathered} $$

And those are the equations you're asking for (right?) .

3 choices:

  1. Spend the rest of your life solving them .
  2. Use Wolfram Alpha .
  3. Consider it "Solvable in Principle". 

Choice (2) is recommended.

answered Jul 21, 2013 by dimension10 (1,950 points) [ revision history ]
edited May 11, 2014 by dimension10
Yeah thanks, that is how I generally imagine it too. They should be related by a Lorentz boost with a Lorentz factor that depends on the radius of the compactified direction, and this dependence I dont understand exactly.

This post imported from StackExchange Physics at 2014-03-09 16:14 (UCT), posted by SE-user Dilaton
@Dilaton: Oh, I see your question now...

This post imported from StackExchange Physics at 2014-03-09 16:14 (UCT), posted by SE-user Dimensio1n0

Lubos Motl briefly describes it here.

This post imported from StackExchange Physics at 2014-03-09 16:14 (UCT), posted by SE-user Dimensio1n0

@Dilaton: I have updated my answer.

This post imported from StackExchange Physics at 2014-03-09 16:14 (UCT), posted by SE-user Dimensio1n0

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