# I Will the velocity of light be the same in all reference frames?

1. Sep 26, 2016

### Aswin Sasikumar 1729

If yes then imagine what I am going to say...
From a source two photons are emmutted symultaniously.
If one of the photon had eyes to see what will 'he ' measures the velocity of the other photon which is moving with 'him'? Won't it be zero???!

2. Sep 26, 2016

### Demystifier

By "all" reference frames, one usually means only those which move slower than $c$. This condition is not satisfied in your case.

3. Sep 26, 2016

### Ibix

There isn't a reference frame in which light is at rest. The notion is contradictory, as you yourself noted, since light would have to be both at rest and moving c.

So the answer is that it isn't possible to describe what a light pulse would see. The question doesn't even makes sense in relativity.

4. Sep 26, 2016

### Staff: Mentor

Note that the correct statement is that the speed of light is the same in all inertial reference frames. It can be different in non inertial frames.

5. Sep 26, 2016

### robphy

One would have to specify the definition of "speed".

From a given event, light travels on the lightcone, regardless of the reference frame (inertial or non inertial).

[I should clarify that this is a statement about the tangent space at that event... a space of direction vectors in space-time from that event.]

Last edited: Sep 26, 2016
6. Sep 26, 2016

### Aswin Sasikumar 1729

7. Sep 26, 2016

### Aswin Sasikumar 1729

A law must be a law.So 'ALL' should mean all itself.

8. Sep 26, 2016

### DrGreg

The correct statement should be that the speed of light is the same in all inertial reference frames. A frame in which light is at rest is not inertial.

9. Sep 26, 2016

### Staff: Mentor

A typical example is a rotating reference frame where the speed of light can be substantially different from c at great distances from the axis. The restriction on c is for inertial frames.

10. Sep 27, 2016

### Demystifier

For all numbers $x$ there is a number $y$ such that $xy=1$. This is a mathematical law. Yet, there is an exception: $x=0$.

The reason why velocity of light is an exception has the same mathematical origin as the purely mathematical example above. In both cases there is a division by zero involved.

11. Sep 27, 2016

### Demystifier

Some people above suggested that there is no reference frame in which light is at rest, or that such a frame is not inertial. Both statements are wrong. There is such a frame and it is an inertial frame. But such a frame is not a Lorentz frame. It is a light-cone frame:
https://en.wikipedia.org/wiki/Light-cone_coordinates

12. Sep 27, 2016

### Staff: Mentor

Light cone coordinates are not an inertial frame. By the second postulate light travels at c in all inertial frames and light does not travel at c in light cone coordinates. Furthermore, the coordinate basis of light cone coordinates do not even define a valid reference frame since a reference frame, by definition, consists of three spacelike and one timelike orthonormal vector fields.

13. Sep 27, 2016

### Demystifier

Hm, it seems that you and me are using different definitions of the word "frame". In my "dictionary", the frame is the same as the system of coordinates. In your "dictionary", frames are a subclass of systems of coordinates.

Apart from this terminology difference, I think there is no any reason to think of light-cone coordinates as non-inertial coordinates. They are inertial, in the sense that lines of the coordinate mesh are geodesics. Indeed, light moves along one of such lines, and such a motion of light is inertial in the sense that no force (except gravity) influences it.

14. Sep 27, 2016

### Staff: Mentor

In standard GR terminology a reference frame is called a frame field, tetrad, or vierbein https://en.m.wikipedia.org/wiki/Frame_fields_in_general_relativity

It is a set of four vector fields, three of which are spacelike, one of which is timelike, and all of which are orthonormal to each other.

A coordinate system is not a set of four vector fields, but the distinction can be blurry because you can always use the coordinate basis to generate some vector fields known as the coordinate basis. However, a coordinate basis may not be orthonormal etc. so not all coordinate bases qualify as valid reference frames.

While I agree that the coordinate lines are geodesics, that is simply not how a reference frame is defined in relativity.

15. Sep 27, 2016

### vanhees71

A frame is something in the real world. You can take, e.g., a corner of your lab and the three edges of the room as the spatial reference frame and a clock to define a reference frame.

16. Sep 27, 2016

### Demystifier

OK, you convinced me, I will update my dictionary accordingly.

17. Sep 27, 2016

### Aswin Sasikumar 1729

Sir I am not having knowledge in what u people were discussing.But this above statement u gave is wrong.In fact incomplete .The correct law is :"for all non zero x there exist a y such that xy=1".

18. Sep 27, 2016

### Demystifier

Right. Likewise, the correct relativity law is: For all inertial frames with non zero $\gamma^{-1}$ the speed of light is the same, where
$$\gamma^{-1}=\sqrt{1-v^2/c^2}$$
and $v$ is the speed of the frame.

19. Sep 27, 2016

### Aswin Sasikumar 1729

I am convinced with this answer but there is no such word "non zero" in the Einstein's statement which is in our book.Thank u.

20. Sep 27, 2016

### Mister T

Let's start with the postulate that the speed of light is independent of the speed of the source. From there you can deduce that such a speed must be the fastest speed possible, and that no observer can travel that fast. Thus your "photon" traveling at the speed of light cannot be replaced with an observer, as observers cannot move that fast. If they could, the premise on which you based your conclusion would have to be false.