# I Why $ds^2=0$ in the Metric

1. Dec 31, 2017

### 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 theres another reason ?

2. Dec 31, 2017

### Orodruin

Staff Emeritus
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.

3. Dec 31, 2017

### 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 $ds^2=0$...

4. Dec 31, 2017

### Orodruin

Staff Emeritus
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.

Light is massless and moves along null geodesics.

5. Dec 31, 2017

### Arman777

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

I havent seen it yet so I was confused about the reason. I guess I can understand why we use $ds^2=0$.

6. Dec 31, 2017

### Staff: Mentor

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.

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

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.

7. Dec 31, 2017

### Orodruin

Staff Emeritus
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.

8. Dec 31, 2017

### Staff: Mentor

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

9. Dec 31, 2017

### Arman777

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 cant set $ds^2=0$ for light ?

I see, thanks
Also,
I understand it now

10. Dec 31, 2017

### Staff: Mentor

Yes, it is true for all metrics and all spacetimes

11. Dec 31, 2017

Thanks a lot