Physical meaning of zero time metric

[madsmh]

TL;DR

Meaning of zero coefficient on $dt^2$ in special relativity.

I am reading Wald's General Relativity and just did problem 2.8(b). The result I get is $\omega^2(x'^2+y'^2)-1$ as the coefficient for $dt^2$, and I am wondering about the physical significance of when $x'^2+y'^2=\frac{1}{\omega^2}$, what would this mean?

Mads

 

[Sagittarius A-Star]

As delimiters, you can use double $ or double #. See in below LaTex Guide link under "Delimiting your LaTeX code".

 

[Hill]

TL;DR Summary: Meaning of zero coefficient on $dt^2$ in special relativity.

I am reading Wald's General Relativity and just did problem 2.8(b). The result I get is $\omega^2(x'^2+y'^2)-1$ as the coefficient for $dt^2$, and I am wondering about the physical significance of when $x'^2+y'^2=frac{1}{\omega^2}$, what would this mean?

Mads

I don't know about this case, but in the Schwarzschild metric it happens on the event horizon. In that case it signifies a coordinate, not a physical, singularity.

 

[Ibix]

TL;DR Summary: Meaning of zero coefficient on $dt^2$ in special relativity.

I am reading Wald's General Relativity and just did problem 2.8(b). The result I get is $\omega^2(x'^2+y'^2)-1$ as the coefficient for $dt^2$, and I am wondering about the physical significance of when $x'^2+y'^2=\frac{1}{\omega^2}$, what would this mean?

Mads

Guessing that this is a rigid rotating frame, what is the linear speed of a point at rest in this frame at that radius? If you are using the worldline of such a point as your timelike coordinate, what does this mean for your coordinates at this radius?

 

[Sagittarius A-Star]

I am wondering about the physical significance of when $x'^2+y'^2=\frac{1}{\omega^2}$, what would this mean?

This value of $r^2$ is not allowed for the transformation:
https://en.wikipedia.org/wiki/Born_coordinates#Transforming_to_the_Born_chart

 

[Orodruin]

Apart from what has been said, just because a coordinate is called t does not make it timelike. If the coefficient in front of $dt^2$ is zero, it means that the t-coordinate is lightlike.

 

[PeterDonis]

I am wondering about the physical significance of when $x'^2+y'^2=\frac{1}{\omega^2}$

I have not checked the details of your solution, but in general in rotating coordinates there will be a radius from the center (note that $x'^2 + y'^2$ is the radius from the center) at which a curve at rest in the coordinates (i.e., with constant $x'$, $y'$, $z'$) will no longer be timelike (it will be null at that radius and spacelike outside it). Physically this means that you can no longer have objects at rest in the chart at or outside that radius, because, from the standpoint of a non-rotating frame, they would need to be moving at or faster than the speed of light.

 

[PeterDonis]

This value of $r^2$ is not allowed for the transformation:
https://en.wikipedia.org/wiki/Born_coordinates#Transforming_to_the_Born_chart

That's not quite true. You can do the transformation, but worldlines with constant spatial coordinates in the rotating chart will be null (instead of timelike) at that $r^2$, and spacelike outside it.

The Wikipedia page appears to be focusing on the case where you want the worldlines of Langevin observers to be at rest in the Born chart, in which case you can only cover the region inside that value of $r^2$, because Langevin observers can only exist there. But the transformation itself does not require that restriction.

 

[Bosko]

TL;DR Summary: Meaning of zero coefficient on $dt^2$ in special relativity.
I am reading Wald's General Relativity and just did problem 2.8(b).

You should have sent the text of the problem so that we don't look for it.
Anyway ... Wald's book ... special relativity ... 2.8 (b) here it is

The result I get is $\omega^2(x'^2+y'^2)-1$ as the coefficient for $dt^2$,

$x'^2+y'^2$ is the radius $r'$, in x',y' plane and the angular velocity $\omega$ is $\frac{v}{r'}$

$dt^2=\frac{v^2}{r'^2}r'^2-1$

$dt^2=v^2-1$

$dt^2=-(1-v^2)$
It is time dilation from in special relativity (time component of Minkowski metric tensor $g_{00}$)
$dt^2=(1-v^2)dt'^2$

$dt=\sqrt{1-v^2}dt'$

$dt'=\frac{1}{\sqrt{1-v^2}}dt$

and I am wondering about the physical significance of when $x'^2+y'^2=\frac{1}{\omega^2}$, what would this mean?

Mads

$r'^2=(\frac{v^2}{r'^2})^{-1}$

$r'^2=\frac{r'^2}{v^2}$

$r'^2 v^2=r'^2$

$v^2=1$

$v=1$ ... speed of light

 

[pervect]

TL;DR Summary: Meaning of zero coefficient on $dt^2$ in special relativity.

I am reading Wald's General Relativity and just did problem 2.8(b). The result I get is $\omega^2(x'^2+y'^2)-1$ as the coefficient for $dt^2$, and I am wondering about the physical significance of when $x'^2+y'^2=\frac{1}{\omega^2}$, what would this mean?

Mads

A zero coordinate for dt^2 often (but not always) a sign of a coordinate singularity. This does not imply any physical significance - it's an artifact of the coordinate choice, which is a human choice.

There is some chance a zero coefficient of dt represents an actual singularity, however.

Another example: $dr^2 + r^2 d\theta^2$ has a zero coefficient for $d\theta^2$ at r=0. These are polar coordinates on a plane. There is nothing physical going on at r=0, it's a removable coordinate singularity.

See also the wiki entry https://en.wikipedia.org/wiki/Coordinate_singularity

A coordinate singularity occurs when an apparent singularity or discontinuity occurs in one coordinate frame that can be removed by choosing a different frame.

....

Stephen Hawking aptly summed this up, when once asking the question, "What lies north of the North Pole?".

 

[Orodruin]

A zero coordinate for dt^2 often (but not always) a sign of a coordinate singularity. This does not imply any physical significance - it's an artifact of the coordinate choice, which is a human choice.

The sign of a coordinate singularity is that the metric eigenvalues go to zero or infinity. Just looking at a single metric component will not tell you this (unless you have already diagonalized the metric, but in that case you are already looking at the eigenvalues).

There is absolutely nothing wrong with having a zero in front of the dt^2. It just means t is a lightlike coordinate.