I Can a distant object behind a cosmic string appear as a double image?

[etotheipi]

Please see the end of question 2 of this document: https://www.damtp.cam.ac.uk/user/examples/3R1d.pdf
What does Tong mean by "Show intuitively how a distant object behind a cosmic string may appear as a double image"?

[For reference, the cosmic string in question is described by $(T_{\mu \nu}) = \mu \delta(x) \delta(y) \mathrm{diag}(1,0,0,-1)$]

 

[PeterDonis]

What does Tong mean by "Show intuitively how a distant object behind a cosmic string may appear as a double image"?

He means that, in this spacetime, you can see two images of the object, one coming from each side of the cosmic string--something like gravitational lensing.

 

[etotheipi]

He means that, in this spacetime, you can see two images of the object, one coming from each side of the cosmic string--something like gravitational lensing.

What's the argument he's looking for? That last equation $ds^2 = -dt^2 + dz^2 + d\tilde{r}^2 + \tilde{r}^2 d\tilde{\phi}^2$ is Minkowski spacetime in cylindrical coordinates but I can't at all see how to take that and then deduce anything about double images...

 

[PeterDonis]

That last equation $ds^2 = -dt^2 + dz^2 + d\tilde{r}^2 + \tilde{r}^2 d\tilde{\phi}^2$ is Minkowski spacetime in cylindrical coordinates

It looks like it is, but the question at the end, "Is this Minkowski spacetime?", should lead you to stop and think. Look carefully at the original perturbed metric, in terms of $\lambda$. Is it valid for all values of $r$? (Look at the formula for $\lambda$.) How about the coordinate transformations to get from that metric to the apparently Minkowski one? Are they valid for all radial coordinates?

 

[PeterDonis]

What's the argument he's looking for?

The last coordinate transformation changes the angular coordinate $\phi$. What change is made? What does it mean?

 

[etotheipi]

Right, I see $\lambda$ is undefined at the origin $r=0$, so I guess the new spacetime is Minkowski \ $\mathbf{0}$. As for the angular co-ordinate it transforms as $(1-8\mu G)d\phi^2 = d\tilde{\phi}^2$ but I can't see any motivation other than to make the metric into a nicer form...

 

[PeterDonis]

I see $\lambda$ is undefined at the origin $r=0$

That's true, but it's not the only interesting property it has. Look at the perturbed metric in terms of $\lambda$, and consider intervals for which $dt = dz = 0$.

As for the angular co-ordinate it transforms as $(1-8\mu G)d\phi^2 = d\tilde{\phi}^2$

Yes, which means the angular coordinate is being rescaled. But what does it even mean to rescale an angular coordinate? What happens to $\phi = 2 \pi$ when $\phi$ gets rescaled?

 

[etotheipi]

That's true, but it's not the only interesting property it has. Look at the perturbed metric in terms of $\lambda$, and consider intervals for which $dt = dz = 0$.

Ah, okay! Since $ds^2 \geq 0$, for these spacetime displacements we must have $\lambda \leq 1$ which is equivalent to $r \leq r_0 \mathrm{exp}\left( \dfrac{1}{8\mu G} \right)$, i.e. an upper bound on $r$.

Yes, which means the angular coordinate is being rescaled. But what does it even mean to rescale an angular coordinate? What happens to $\phi = 2 \pi$ when $\phi$ gets rescaled?

I'd say a point at angle $\phi$ in the old spacetime is rotated around the $z$ axis to angle $\tilde{\phi} = \phi\sqrt{1-8\mu G}$ in the new spacetime.

It's still not jumping out at me how to tie all these things together. Maybe I need to stare at it for longer...

 

[PeterDonis]

Since $ds^2 \geq 0$, for these spacetime displacements we must have $\lambda \leq 1$ which is equivalent to $r \leq r_0 \mathrm{exp}\left( \dfrac{1}{8\mu G} \right)$, i.e. an upper bound on $r$.

Not necessarily; $\lambda = 1$ is still a valid part of the spacetime. But at the value of $r$ where $\lambda = 1$, one can have a closed null curve with constant $t$, $z$, $r$. And at larger values of $r$, one can have closed timelike curves with constant $t$, $z$, $r$. So this is clearly not Minkowski spacetime.

I'd say a point at angle $\phi$ in the old spacetime is rotated around the $z$ axis to angle $\tilde{\phi} = \phi\sqrt{1-8\mu G}$ in the new spacetime.

Not just one point. Since $\phi = 2 \pi$ in the old coordinates (not "old spacetime", it's the same spacetime, just changing coordinates) goes to $\phi < 2 \pi$ in the new coordinates, the metric that "looks" Minkowski does not cover the full range of angular coordinate $\phi$ that actual Minkowski spacetime does. Instead it's like Minkowski spacetime with a pie-slice shaped portion cut out and the edges of the cut "sewn" back together (since there is no discontinuity in the spacetime there). So in the Minkowski-looking coordinates, we have that the locus $\phi = 2 \pi - \Delta$ (where $\Delta$ is the offset created by the rescaling of $\phi$) is the same as the locus $\phi = 2 \pi$.

Now consider what that means if we have a distant object at $\phi = 2 \pi - \Delta$, which is the same as $\phi = 2 \pi$, and we have an observer looking towards the cosmic string from direction $\phi = \pi - \Delta / 2$. How many images will he see? What would your intuitive response be? (Remember that the metric looks Minkowski in these coordinates apart from the missing pie slice.)

 

[etotheipi]

Since $\phi = 2 \pi$ in the old coordinates (not "old spacetime", it's the same spacetime, just changing coordinates)

Yes sorry, by new and old spacetime I meant to say in the new and old co-ordinate spaces, $\tilde{x}^{\mu}(M)$ and $x^{\mu}(M)$.

Now consider what that means if we have a distant object at $\phi = 2 \pi - \Delta$, which is the same as $\phi = 2 \pi$, and we have an observer looking towards the cosmic string from direction $\phi = \pi - \Delta / 2$. How many images will he see? What would your intuitive response be? (Remember that the metric looks Minkowski in these coordinates apart from the missing pie slice.)

I think I see what you're getting at; in the new "Minkowski-ish" co-ordinates, the angles $\tilde{\phi} = 0$ and $\tilde{\phi} = 2\pi - \Delta$ corresponding to the two sides of the wedge are both equivalent to $\phi = 0$ in the original co-ordinates [since $2\pi = 0 \, (\mathrm{mod} \, 2\pi)$].

Consequently a single image at $\phi = 0$ would become two images at $\tilde{\phi} = 0$ and $\tilde{\phi} = 2\pi - \Delta$. [And a camera positioned at $\pi - \Delta/2$, looking toward the origin, would see two images at $\pm \Delta /2$]. Is that right?

 

[PeterDonis]

I think I see what you're getting at; in the new "Minkowski-ish" co-ordinates, the angles $\tilde{\phi} = 0$ and $\tilde{\phi} = 2\pi - \Delta$ corresponding to the two sides of the wedge are both equivalent to $\phi = 0$ in the original co-ordinates [since $2\pi = 0 \, (\mathrm{mod} \, 2\pi)$].

Consequently a single image at $\phi = 0$ would become two images at $\tilde{\phi} = 0$ and $\tilde{\phi} = 2\pi - \Delta$. [And a camera positioned at $\pi - \Delta/2$, looking toward the origin, would see two images at $\pm \Delta /2$]. Is that right?

That's what I was thinking, yes.

 

[etotheipi]

That's what I was thinking, yes.

Thanks for the help! This is quite an interesting and surprising effect.

 

[Orodruin]

To me, ”show intuitively” does not indicate a need to actually show it mathematically, but rather to provide a heuristic argument.

 

[PeterDonis]

To me, ”show intuitively” does not indicate a need to actually show it mathematically, but rather to provide a heuristic argument.

In this case I think it more or less amounts to the same thing; we could state the "Minkowski spacetime with a pie slice cut out" property, which is the key feature, in words, but it's just as easy to state it mathematically.