Why Does ds^2 Equal 0 in Metrics?

[Arman777]

In FLWR metric or in Minkowski metric or in any general metric can we say that $ds^2=0$ cause speed of light should be constant to all observers ?

Or there's another reason ?

 

[Orodruin]

What do you mean? What $ds^2$ is depends on the world-line you are considering. $ds^2 = 0$ corresponds to a light-like curve. This is a coordinate independent statement and it will be true for the light-like curve regardless of the coordinate system you use.

 

[Arman777]

In FLRW metric when we measure the redshift we assume $ds^2=0$. Like also in minkowski metric $ds^2=0$ cause only in that case we can get c=dx/dt.

Probably I should add to the question why for a light $ds^2=0$...

 

[Orodruin]

In FLRW metric when we measure the redshift we assume $ds^2=0$.

This is not a measurement of redshift. It is a computation of the redshift based on the FLRW universe. Studying light, it is quite clear that we must use a light-like geodesics.

Like also in minkowski metric $ds^2=0$ cause only in that case we can get c=dx/dt.

Probably I should add to the question why for a light $ds^2=0$...

Light is massless and moves along null geodesics.

 

[Arman777]

This is not a measurement of redshift. It is a computation of the redshift based on the FLRW universe. Studying light, it is quite clear that we must use a light-like geodesics.

Hmm I see, I ll do more research on light-like geodesic case.

I haven't seen it yet so I was confused about the reason. I guess I can understand why we use $ds^2=0$.

 

[Dale]

In FLRW metric when we measure the redshift we assume ds2=0ds2=0ds^2=0.

The redshift you are talking about is for light. Light travels on null geodesics. Therefore $ds^2=0$. It is not a general statement about the metric, it is a specific statement about light.

Like also in minkowski metric ds2=0ds2=0ds^2=0 cause only in that case we can get c=dx/dt.

Same thing here. For light $ds^2=0$ for the reason you gave. But for massive objects $ds^2<0$ and for hypothetical tachyons $ds^2>0$.

Probably I should add to the question why for a light ds2=0ds2=0ds^2=0...

If you write down the metric, set $ds^2=0$, then what are you left with? The equation of a sphere of radius $ct$. This is something traveling at c in all directions, which is the second postulate. Therefore, $ds^2=0$ for light is the mathematical statement of the second postulate.

 

[Orodruin]

Same thing here. For light $ds^2=0$ for the reason you gave. But for massive objects $ds^2<0$ and for hypothetical tachyons $ds^2>0$.

I think it should be qualified that whether $ds^2 > 0$ or $ds^2 < 0$ for time-like world lines depends on the sign convention for the metric. Mathematicians and GR people generally prefer $ds^2 < 0$ while particle physicists prefer $ds^2 > 0$. Always check which convention is being used in the particular text. Of course, this does not affect $ds^2 = 0$ for null world lines.

 

[Dale]

I think it should be qualified that whether ds2>0ds2>0ds^2 > 0 or ds2<0ds2<0ds^2 < 0 for time-like world lines depends on the sign convention for the metric.

Yes, good point. My preferred convention is to write $ds^2$ when I am using the (-+++) convention and to write $d\tau^2$ when I am using the (+---) convention

 

[Arman777]

It is not a general statement about the metric, it is a specific statement about light.

Yes I tried to mean that, as I understood from your post for light $ds^2=0$. But Is this true for all of the metrics ?
In example For Minkowski metric we are working on light rays then $ds^2=0$, in FLRW metric , we are working on light then $ds^2=0$ etc. ?

Or there could be a metric where we can't set $ds^2=0$ for light ?

If you write down the metric, set ds2=0ds2=0ds^2=0, then what are you left with? The equation of a sphere of radius ctctct. This is something traveling at c in all directions, which is the second postulate. Therefore, ds2=0ds2=0ds^2=0 for light is the mathematical statement of the second postulate.

I see, thanks
Also,

Light is massless and moves along null geodesics.

I understand it now

 

[Dale]

But Is this true for all of the metrics ?

Yes, it is true for all metrics and all spacetimes

 

[Arman777]

Yes, it is true for all metrics and all spacetimes

Thanks a lot