# Speed of light measured the same by all inertial observers?

1. Apr 7, 2005

### metrictensor

What is the speed of light measured the same by all inertial observers?

2. Apr 7, 2005

### DaveC426913

$$3*10^8 \frac{m}{s}$$
or
186,262 miles per second

A Google search would have found this faster.

(Or did you mean to ask why??)

3. Apr 7, 2005

### kleinwolf

Is this reasoning correct : suppose we have a non-inertial observer (towards another one)....then considering the local (in time) observer fitted at every time to the non-inertial one...then the speed of light is c for it, so it has to be c for the non-inertial observer at any time...so the speed of light is c for any observer ???

4. Apr 7, 2005

### HallsofIvy

Staff Emeritus
That sounds like it's awfully close to Zeno! If I understand what you are saying: "A non-inertial observer is accelerating but at each instant he is going at exactly the same speed as some inertial observer. For that inertial observer, the speed of light is c, therefore for the non-inertial observer, the speed of light is always c also!"

I'm not at all sure that is valid. The fact that, at each instant, the non-inertial observer is going at some speed doesn't mean that he observes what an inertial at the same speed would observe.

5. Apr 7, 2005

### jcsd

No, beacsue the coordinates tha he measures the speed of light to be c at any instant, whilst being stationary to him at thta isnatnt change from instant to instant as they arel accelarting relative to our non-inertial obsrebre at thta instant.

6. Apr 7, 2005

### pervect

Staff Emeritus
This is true for the specific case of an observer accelerating in a constant direction at a constant rate (a constant proper acceleration), with some important limitations.

The main limitation is that the speed of light in such an accelerating frame is equal to 'c' only at the origin of the frame. "Above" and "below" the origin of the frame clocks appear to run at different rates- one can use the usual formula for gravitational time dilation to calculate the apparent rates at which clocks run, one can also derive the same result from the Lorentz transform without using the gratatioanl time dilation formulas.

7. Apr 7, 2005

### dextercioby

A Google search would have found

$$c= 299,792,458 \ \frac{\mbox{m}}{\mbox{s}}$$

Daniel.