# Speed of Light always remains constant ? think again.

1. Jun 7, 2005

### Anomalous

[SOLVED] Speed of Light always remains constant ? think again.

Take 3 points P1, P2 and D

Point D is at a fantastic distance from P1 P2.

Point P1 is very near to P2 and both are at equal distance from D.

A Rocket R2 is waiting at P2

Rocket R1 is approaching at tremendous velocity at point P1 and towards D.

At the instance when the rocket R1 reaches P1, both the rockets simultaneously fire a powerful laser beam ( may be a few meters in length ) towards D.

Few hours after firing the laser, R2 starts accelerating towards D and quickly showing its raw power, matches the speed and starts flying with R1.

Now, will the speed of already fired lasers that havent yet reached D with respect to R2 decrease ? now that it has come closer to the R1 hours after firing the laser, and while both the lasers together are approaching D.

2. Jun 7, 2005

### Meir Achuz

The speed of each laser pulse is c.
Whatever the rockets do after firing is irrelevant.
The frequency of the pulse from R1 will be increased by the relatilvistic Doppler effect.

3. Jun 7, 2005

### Staff: Mentor

Like you, Anamalous, most scientists before Einstein assumed speed was relative (including the speed of light) and time was constant. As it turns out, they were wrong: time is relative and speed (of light) is constant.

How do we know? We measure both the (relative) rate of the passage of time and the speed of light on a virtually constant basis.

Last edited: Jun 7, 2005
4. Jun 7, 2005

### Anomalous

1) LOL, obviously.

2) Why ?

3) But what about R2. And wont U care if R2 had achived 99% speed of light before catching up with R1 ?

Last edited by a moderator: Jun 7, 2005
5. Jun 7, 2005

### Anomalous

Its clear here that R2 covered the same distance that R1 did after firing the lasers yet laser fired by R2 is not going to over take laser fired by R1 and hence comapred to R1 speed of light for R2 was less when it was accelerating near R1.

I know that I am wrong here , but I dont know how, and until then for me anyone who doesnot explain that to me is wrong.

6. Jun 7, 2005

### Crosson

The reason that no one can explain the flaw in your reasoning is that you show a complete ignorance of the special theory of relativity. Read a book about SR, look at the reasons why the speed of light is a constant from all points of view at all times, then see if you still have this question.

7. Jun 7, 2005

### learningphysics

Yes, according to R2, the speed of light is not constant because R2 is accelerating. But when R2 stays in a particular inertial frame it will measure the same speed of light as anyone else in an inertial frame.

Speed of light is the same in all inertial frames of reference.

In an non inertial frame the speed of light can be different, even 0.

Here's a specific example from Rindler's Introduction to Special relativity.
Take the example of a spaceship undergoing constant proper acceleration a (in it's own instantaneous frame it is always accelerating at a). Suppose that when the ship starts off from rest, a photon is emitted at a distance $c^2/a$ behind the ship just as it leaves. Now from the point of view of the ship, the photon remains at that same distance behind it at all times. In other words that relative to the ship, the speed of the photon is 0.

The idea is that in a particular inertial frame, we will always measure the speed of light in vaccuum to be c.

8. Jun 7, 2005

### dextercioby

In its (sic!) "own instantaneous frame" it's (sic!) not moving...

Daniel.

P.S.Quote Rindler :grumpy: I'm sure that,if he's (sic!) that Rindler,he wouldn't have made that mistake.

9. Jun 7, 2005

### learningphysics

Yes it is not moving. velocity = 0. But acceleration does not = 0. So can't I say it is accelerating?

Hmmm... maybe I should have said that in the inertial frame where the ship's velocity is 0, its acceleration is a (proper acceleration).

Here's how Rindler's writes it:
"If we define the proper acceleration $\alpha$ of P as that which is measured in P's instantaneous rest frame"

10. Jun 7, 2005

### Integral

Staff Emeritus
This is a pretty ambiguous statement. I would assume the lasers are in the different ships so will therefore be traveling with the ship. What is the point of this? How does this have anything to do with the initial pulses fired by the different lasers?

I assume that the initial pulses were fired when the moving ship was by passing the stationary ship so, according to Maxwell's equations the pulses will be traveling in the same (or nearly so) direction at the same speed. Let us assume that they are traveling side by side. Now if we further assume that the lasers are of the same type (frequency) then when the pulse arrive at point D the observer there will be able to tell which ship was moving at the time the pulse was initialized as it will be blue shifted by an amount related to the speed of the ship. Since the pulses of light have traveled the same distance and since they were initiated at the same time they will arrive at D at the same time.

It is not clear to me what your proposed mechanism is that the speeding up of the stationary ship will magically effect the laser pulse it previously initiated?

11. Jun 7, 2005

### dextercioby

It's unbelieveble.In its rest frame the ship's not moving,which means it has 0 velocity and 0 acceleration.He must refer to a comoving frame.

Daniel.

12. Jun 7, 2005

### DaveC426913

I'm following your thought experiment all the way up to this last paragraph. It is nonsensical. Can you restate and elaborate on your thinking?

13. Jun 7, 2005

### robphy

If the accelerating observer carries an accelerometer, he will notice that it is not in its "center" position. His 4-acceleration is a nonzero spacelike vector.

14. Jun 7, 2005

### pervect

Staff Emeritus
My \$.02 on the accelerating spaceship. An accelerating spaceship can set up a coordinate system (which is not a frame) by using the coordinates (x,t) of an instantaneously co-moving observer.

Such a coordinate system is strictly limited in it's possible size. The coordinate system does not and cannot map every coordinate pair (xi, ti) of an inertial coordinate system to a coordinate pair (xa, ta) of the accelerated observer. This is covered in detail in MTW's section on "accelerated obsevers". There are two difficulties which prevent the coordinate system from covering all of space-time: the fact that geodesics allow multiple coordiantes for some events, and the existence of an event horizon.

The event horizon associated with the coordinate system of an accelerated observer is known as the "Rindler horizon". In the coordinate system of an accelerated obsever, a light beam exactly at the Rindler horizon will have a constant value for the distance coordinate, much as an outgoing light beam exactly at the event horizon of a Schwarzschild black hole will have a constant value for the 'R' coordiante.

The fact that the rate of change of the coordinate with respect to time is zero could be called a "zero velocity", I suppose. However, if you put any inertial observer moving at any rate at the same point in space-time, he will still find that the speed of light is equal to 'c' in his local coordinate system.

Last edited: Jun 8, 2005
15. Jun 8, 2005

### Anomalous

1) right.

2) It shows that moving of R1 wont change speed of light ( according to established evidence ) when the lasers were fired and hence both Lasers will always travell together.

3) That they were fired by rockets with a speed difference.

4) Bingo.

5) U tell me, I am a Rookee.

16. Jun 8, 2005

### Anomalous

Can U ask which part didnt make sense to U sir.

Last edited by a moderator: Jun 10, 2005
17. Jun 8, 2005

### arildno

Anomalous:
That the speed of light is constant for different observers is a postulate; the maths used in relativity uses this as one of its axioms, if you like (it is a necessarily true statement within relativity theory).

Thus, if you disagree to that, you've basically to complete one of two tasks:
1) Prove that relativity is mathematically inconsistent in some damaging way.
You've not done that.

2) Provide empirical results which must be interpreted as meaning that the speed of light cannot be regarded as constant.
You've not done that, either.

18. Jun 8, 2005

### Tom Mattson

Staff Emeritus
Well what exactly are you looking for? Are you looking for someone to do the calculation for you in both SR and Galilean relativity? Are you looking for someone to convince you that the SR prediction is correct? Are you looking for someone to do the experiment?

You asked a very cut-and-dry question, and you have been given an answer. You aren't going to get anywhere by snickering at the respondents. If you want something else, then ask for it.

19. Jun 8, 2005

### Crosson

I would advise you to read "A brief history of time" by stephen hawking, to understand more about the constancy of the speed of light.

In a nutshell, the distance traveled by light, and the time taken, are different from different points of view (called frames of reference). But, the speed is always the same from everyones point of view.

20. Jun 8, 2005

### learningphysics

I think it's an important distinction to make.... that the speed of light is constant in all inertial frames, but not from all points of view.