Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,082 questions , 2,232 unanswered
5,355 answers , 22,793 comments
1,470 users with positive rep
820 active unimported users
More ...

  Why does the the stationary and axially symmetric vacuum spacetime become flat when the line density of an infinite line source equals 1/2?

+ 4 like - 0 dislike
1165 views

When an infinite line source with line density $\sigma$ is located on the $z$ axis, the stationary, axially symmetric vacuum space-time is $$ds^2 = - r^{4\sigma}dt^2+ r^{8\sigma^2-4\sigma}(dr^2+dz^2)+r^{2-4\sigma}d\phi^2$$ When the line density equals $1/2$, the metric is $$ds^2 = - r^{2}dt^2+ dr^2+dz^2+d\phi^2$$ ,which is flat and is essentially the Rindler coordinate of Minkovski spacetime. So why does the metric become flat, when the line density becomes so large and equals to 1/2 ?

asked Dec 31, 2014 in Theoretical Physics by Alienware (185 points) [ revision history ]
edited Mar 14, 2015 by Arnold Neumaier

1 Answer

+ 4 like - 0 dislike

This is because cosmic strings are not gravitating matter in the usual sense but space-time defects. You will in fact obtain a flat space-time for every $\sigma = k/4, k \in \mathbb{Z}$ but in the coordinates you use this will be every time Minkowski but in a different set of coordinates.

You can see this by the canonical construction of cosmic strings as can be found e.g. in Griffiths & Podolský: Consider Minkowski in cylindrical coordinates: $$ds^2 = - dt^2 + d\rho^2 + dz^2 + \rho^2 d \varphi^2$$ Now introduce a defect by glueing $\varphi=0$ to $\varphi=2 \pi(1- \delta)$  instead of $2 \pi$. Now rescale $\phi = (1-\delta) \varphi$ and your metric will be $$ds^2 = - dt^2 + d\rho^2 + dz^2 + (1-\delta)^2 \rho^2 d \varphi^2$$ The parameter $\delta$ can also be linked with the matter density $\lambda$ of the defect interpreted as a massive string, $\lambda=\delta/4$.

However, once you circulate from $\delta=0$ to $\delta=1$, you should start again identifying $0$ to $2 \pi$ and get Minkowski again, which is not reflected in the construction of the coordinate $\phi$ given above. On the other hand, the metric in Weyl coordinates as you introduce it seems to be satisfactory in this respect, albeit at the cost of a re-covering of Minkowski at every loop.

(Do note that if you obtained this formal solution by putting a $\delta$-peak of matter on the $\rho=0$ axis in Weyl coordinates, you have righteously unleashed the exact-solution horrors upon yourself as the metric singularity you produce can erase the matter itself! *Evil laughter with that reverb suggesting it is coming from hell.*)

answered Mar 13, 2015 by Void (1,645 points) [ no revision ]

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsOv$\varnothing$rflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...