Godel metric in a cylindrical chart

In summary: Lol, it is like you read my mind. I was just looking at those differentials and noticed how with the line element that I just used, t and z seem to always be timelike, but r and θ seem to always be spacelike.In that case, would I need a different line element or something to see the case where the angular coordinate is...timelike?I don't think so, but I'm not sure.I don't think so, but I'm not sure.
  • #36
space-time said:
Here is a quote from you from another thread

Yes, but, as I noted before, this was assuming qualifiers that are not applicable in Godel spacetime.

space-time said:
Is it that a CTC is not future directed due to it's closed nature?

Sort of. In a spacetime with CTCs everywhere, like Godel spacetime, there is no single way to distinguish the "future" and "past" halves of the light cones that works everywhere. You can make that choice along a particular timelike curve, but you can't continuously extend that choice to all timelike curves the way you can in flat Minkowski spacetime.

space-time said:
I just need an outright clarification of what circumstances necessitate that (dxa/ds] be greater than 0.

There isn't one that works for all spacetimes. That's why I said in an earlier post that you can't take shortcuts. There simply is no way to state such a criterion in terms of coordinates. Coordinates are not physically meaningful. You need to look at invariants. Sometimes, if you have a well chosen coordinate chart, you can figure out meaningful invariants from it, but this is never guaranteed and there is no general rule for how you do it. You have to look at each individual case.
 
Physics news on Phys.org
  • #37
A useful tool for getting some physical insight into a metric or line element is the orthonormal basis. This can most easily be done with an algebraic manipulation of the line element. For the Godel metric, if we write

$$ \textbf{d}\omega0= \textbf{d}t + \sqrt{2}\sinh^2r \,\textbf{d}\varphi \quad \textbf{d}\omega1 = \textbf{d}r \quad \textbf{d}\omega2 = \sqrt{\sinh^4 r + \sinh^2 r} \, \textbf{d}\varphi \quad \textbf{d}\omega3 = \textbf{d}z$$

[add]. I've omitted a multiplicative factor from the line element, out of laziness.

Then the line element would be written as
$$-ds^2 = (d\omega0)^2 )+ (d\omega1)^2 + (d\omega2)^2 + (d\omega3)^2 $$

The metric would be written in this notation as

$$g = -\textbf{d}\omega0 \otimes \textbf{d}\omega0 + \textbf{d}\omega1 \otimes \textbf{d}\omega1 + \textbf{d}\omega2 \otimes \textbf{d}\omega2 + \textbf{d}\omega3 \otimes \textbf{d}\omega3$$

Hopefully , it's clear that this is a locally Lorentizn metric, similar to the line element ##-dt^2 + dx^2 + dy^2 + dz^2##. The similarity to the flat space metric allows for easy physical interpretation.

Note that ##\textbf{d}\omega0## is a linear functional of a vector, i.e. a scalar-valued function of a vector. It's a rank 1 tensor. Without the boldface, ##d\omega0## is a differental.

##\otimes## represents the tensor product. The expression for the metric gives the metric g, a rank 2 tensor, as a sum-of-tensor-products of rank 1 tensors.

Much as vectors form a basis, dual vectors form a cobasis. So we describe what we're doing as creating an orthonormal basis of one-forms, or an orthonormal cobasis.

Notationally , this has so far all been written in index free notatoin. ##\textbf{d}\omega0## might also be written in index notation as some one-form ##v_a##, with a lower index. Raising the index by multiplying ##v_a## with the inverse metric, i.e. ##g^{ab} v_a## would create the associated dual vector ##v^a##.
 
Last edited:
  • Like
Likes space-time
  • #38
PeterDonis said:
Yes, but, as I noted before, this was assuming qualifiers that are not applicable in Godel spacetime.
Sort of. In a spacetime with CTCs everywhere, like Godel spacetime, there is no single way to distinguish the "future" and "past" halves of the light cones that works everywhere. You can make that choice along a particular timelike curve, but you can't continuously extend that choice to all timelike curves the way you can in flat Minkowski spacetime.
There isn't one that works for all spacetimes. That's why I said in an earlier post that you can't take shortcuts. There simply is no way to state such a criterion in terms of coordinates. Coordinates are not physically meaningful. You need to look at invariants. Sometimes, if you have a well chosen coordinate chart, you can figure out meaningful invariants from it, but this is never guaranteed and there is no general rule for how you do it. You have to look at each individual case.
I see. Well, just one more question

Let's say that I do use the CTC
x(s) = [0, 1, s, 0]

Now, we know that the proper time experienced when traveling along a timelike curve between two events can be calculated by evaluating the following integral from s1 to s2:

$$\int \sqrt{-g_{ab}\dot x^a \dot x^b} \, ds$$

Now, in the case of this CTC, from s = 0 to s = 2π, this integral evaluates to:

(2π/ω)##\sqrt{2sinh^4 r - 2sinh^2 r}##

At r = 1, this evaluates to about 6.45/ω

Would this in turn mean that the amount of time covered by the CTC would be
(6.45/ω) seconds (or whatever unit of time you are using)? By "covered", I mean to say: Does the fact that an observer moving along this curve would experience a proper time of 6.45/ω seconds (from s = 0 to 2pi) mean that from this observer's perspective, time would progress "normally" for 6.45/ω seconds before reverting back to the original point in time that the observer began at, and that the observer would just keep reliving cycles of the same 6.45/ω seconds over and over again?

In short, I am basically asking if the proper time along this CTC from s = 0 to s = 2pi would dictate how much time would pass "normally" before the loop got reset.

If not, then how would you figure out how much time passes before the loop resets itself?
 
  • #39
space-time said:
Does the fact that an observer moving along this curve would experience a proper time of 6.45/ω seconds (from s = 0 to 2pi) mean that from this observer's perspective, time would progress "normally" for 6.45/ω seconds before reverting back to the original point in time that the observer began at, and that the observer would just keep reliving cycles of the same 6.45/ω seconds over and over again?

Basically, yes, but the proper time doesn't "revert" or "reset"; there is no sharp discontinuity anywhere along the curve. It is just that the events along the curve are such that they repeat in a cycle that is 6.45/ω seconds long.
 
  • Like
Likes space-time

Similar threads

Back
Top