# The principle of the constancy of the velocity of light

1. Sep 17, 2006

### myoho.renge.kyo

in this thread i have quoted A Einstein. i give him all the credit for all the definitions and ideas (if I have interpreted them correctly). if I have not interpreted his ideas correctly, it is my fault, and i am sorry. i need help to interpret his ideas. that is why i join this forum. thanks!

all our judgments in which time plays a part are always judgments of simultaneous events. for example, if I say that at 11:00 pm, 9/16/2006, in Burbank, California, a train arrived, I am saying that the pointing of the small hand of my watch to 11:00 pm and the arrival of the train are simultaneous events.

A = a point of space where there is a clock and an observer who determines the time value of an event in the immediate proximity of A by finding the position of the hands (of the clock) which are simultaneous with this event.

B = a point of space where there is another clock (in all respects resembling the one at A) and another observer who determines the time value of an event in the immediate proximity of B by finding the position of the hands (of the clock) which are simultaneous with this event.

A time = the time value of the event in the immediate proximity of A (determined by the observer with the clock at A)

B time = the time value of the event in the immediate proximity of B (determined by the observer with the clock at B)

common A and B time = the time value of the event in the immediate proximity of either A or B (either one determined by either the obsever with the clock at A or the observer with the clock at B). a common A and B time cannot be defined unless we establish that the time required by light to travel from A to B equals the time required by light to travel from B to A. thus to say that there is now a common A and B time is to say that the clock at A and the clock at B synchronize.

stationary system (K) = a system of coordinates in which the equations of Newtonian mechanics hold good.

stationary rigid rod (k) = a stationary rigid rod lying along the axis of x of K. the length of k is l as measured by a measuring-rod which is also stationary. a uniform motion of parallel translation with velocity v along the axis of x of K is now imparted to k.

the length of k in the moving system (the moving system in this case is k itself) = l = the length of k ascertain by the following operation: an observer moving together with the given measuring-rod and k measures the length of k directly by superposing the measuring-rod, in just the same way as if all three were at rest.

the length of the moving system k in the stationary system K = rAB = the length of k ascertain by the following operation: by means of stationary clocks set up in the stationary system K and synchronizing, an observer ascertains at what points (x1 and x2) of the stationary system K the two ends (A and B) of the moving system k are located respectively at a definite time (tA). the observer then measures the distance between these two points by the measuring-rod already employed, which in this case is at rest

at the end A of the moving system k there is a clock, and at the end B there is another clock. these two clocks and the stationary clocks in the stationary system K synchronize. with the clock at A there is an observer, and with the clock at B there is another observer.

at tA (the time value determined by the observer with the clock at A) a ray of light departs from A towards B. at tB (the time value determined by the observer with the clock at B) the ray of light is reflected from B. and at t'A (the time value determined again by the observer with the clock at A) the ray of light reaches A again.

the principle of the constancy of the velocity of light states that "the ray of light moves in the stationary system K with the determined velocity c, whether the ray be emitted by the stationary system k or by the moving system k."

A. Einstein states in the Principle of Relativity, p 45, that

l = c * (tB - tA), where tA and tB = the time values measured by the observer with the clock at A and the observer with the clock at B. for these observers, the ray of light moves with the determined velocity c.

and that

rAB / (c - v) = (tB - tA), where tA and tB = the time values measured by the observers with the clock at x1 in the stationary system K and the observer at x2. for these observers, the ray of light moves with the determined velocity c - v.

so in essence he is also saying that the principle of the constancy of the velocity of light states that "the ray of light moves in the stationary system K with the determined velocity c if the ray is emitted by the stationary system k, but if the ray is emitted by the moving system k, the ray of light moves in the stationary system K with the determined velocity c - v."

why? or why not? thanks! (11:00 pm, 9/16/2006, thru 12:00 am, 9/17/2006, in Burbank, California)

Last edited: Sep 17, 2006
2. Sep 17, 2006

### michael879

this is wrong, the velocity of a ray of light is viewed as c from any reference frame no matter where it is emitted.

*edit*
according to the second postulate of special relativity, the speed of light is c in all inertial reference frames.

3. Sep 17, 2006

### Zeit

Hello,

As michael879 told you, the second postulate of SR says that the speed of light is the same in all inertial reference frame. This is logical ; why, for someone thinking he is at rest, the speed of light should be different when he is in his car or in front of his computer? This is the idea of SR, there is no difference between someone really at rest or someone moving, for someone else, at a constant speed and linear movement. The only difference is the point of view.

You said "the ray of light moves in the stationary system K with the determined velocity c - v." This is not true, light is light, no matter where. In K, even if k is moving, light has the same speed. The difference is that the ray of light travels along the x-axis, as the rod does. So when, in K, the light goes in the same direction than the rod, the ray takes more time to arrive to the end of the rod. So, the observer in K would say "I expect that for k, this ray of light takes also more time to travel the lenght of the rod, so it is like if the light ray was slower". K would make a mistake, because he applies the newtonian transformation laws from his reference frame to the other, which is not what he should do. But, K could say that for him, it is like if the ray goes at the speed c-v compared to the end of the rod.

Do you understand the difference? In classical physics, K would have say "because, for me, it takes more time for the ray to arrive to the end of the rod, it is like if the rod was motionless and that the ray travels at the speed c-v. Where does the rod is motionless? In his reference frame!!". In SR, because Nature seems to make no difference between reference frames, K would stop at "... the speed c-v.".

Is it clear what I tried to say? I know this is a mess...

4. Sep 17, 2006

### myoho.renge.kyo

A. Einstein says on p 42 in The Principle of Relativity the following:

(tB - tA) = rAB / (c - v)

(t'A - tB) = rAB / (c + v)

[when measured in the stationary system K]

and on p 45 the following:

For a ray of light emitted at the time t = 0 in the direction of the increasing x [when measured in the moving system k]

x = ct

But the ray moves relatively to the initial point of [the moving system k], when measured in the stationary system [K], with the velocity c - v, so that

x / (c - v) = t

why? thanks! (7:00 am thru 8:00 am)

since i was a kid i was facinated by Einstein's ideas. i always wanted to understand them. i am not here to refute his ideas. i am only here to understand them, and in the process of understanding, i might ask hard questions. thanks again.

5. Sep 17, 2006

### Zeit

Hello,

Imagine that you are near a road. There are two cars, a blue one (B) and a red one (R). Note the lenght of R as L. The experience is to mesure the time needed by B (which always travels at a constant speed) to traverse L.

First, imagine that R is motionless. At t=0, the front of B is aligned with the back of R. You can find that t' = L/v_B .

Second, imagine that R is moving at the speed v_R so v_R < v_B. At t=0, the front of B is aligned with the back of R. But, this time, R is moving, so it would take longer to B to traverse L. You can find the duration t'' needed by B to traverse R doing t'' = L/(v_B - v_R).

In CM, you apply your result as the result of R reference frame, which means that if B was rather a ray of light, for R, this ray would have a speed slower than c. But, there is no reason that your reference frame should be different than R's reference frame.

I would like to help you more, but I'm handicapped by my little knowledge in English. Sorry.

Zeit

6. Sep 17, 2006

### michael879

Ive never read Einstein's writing, I learned relativity from a textbook but this is wrong and your probably reading it out of context. What you said is the classical mechanics view of light, not einstein's relativistic view.

7. Sep 17, 2006

### turbo

Read Chapter 22. Einstein said that the constancy of the speed of light in a vacuum is only applicable to the Special Theory of Relativity, and that in the General Theory, the speed of light in a vacuum must be variable, because refraction can only occur when the speed of light varies with location.

http://www.bartleby.com/173/

8. Sep 17, 2006

### Zeit

Sorry, but what myoho.renge.kyo quotes is true. As I said, if you do not directly apply this result to another reference frame (in which case we use Galileo's transformations, so classical mechanics). But, the fact is that Galileo was right... for an only reference frame...

Look at my example with the cars. I never apply my results as what R sees. But it is true that if, for me, a ray light traverses the lenght of a red car moving, it would take more time for the ray of light to traverse it than if the car was motionless. If you look at two photons travelling in the same direction, never the second photon would come closer of the first that it is now.

What myoho.renge.kyo quotes is Einstein's explanation of simultanneity concept, the idea that time (and we could find after that lenght also) is relative to each other.

I do not understand. Refraction occurs when a ray of light travels through two different transparent "locations" (sorry, I have not find a better word...). The slow down of speed of light is caused by the absorption-emmision of the photons. What does it have to do with GR? Thanks.

9. Sep 17, 2006

### JesseM

Einstein's derivation in his original 1905 paper (available online here) is a bit confusing, but when he talks about the light moving at velocity c - v, he's talking about how it would move in a coordinate system that was related to your first coordinate system K by a coordinate transformation which leaves y,z, and t unchanged, while replacing x with x-vt. This is not a valid inertial coordinate system in special relativity, it doesn't represent the measurements of any observer using rulers and synchronized clocks of the type Einstein introduced to define inertial coordinate systems.

If it helps, I tried to follow along Einstein's derivation in a post on a usenet physics group a while ago, I'll repost it here:

10. Sep 17, 2006

### michael879

yea, this is true from the frame of reference that sees the car as moving. The earth's frame sees the light go at c-v compared to the car. However the light never appears to go at any velocity other than c from any inertial reference frame. The earth frame still sees the light as going at c as does the car's reference frame.

11. Sep 17, 2006

### Staff: Mentor

This is often called the "separation velocity" of two objects which are moving in some inertial reference frame. It is not the same sort of thing as the velocity of a single object in some inertial reference frame, and the v < c restriction doesn't apply to it. Note that it does not equal the velocity of one object in the inertial reference frame in which the other one is stationary. To find that, you have to use the relativistic "velocity addition" formula.

12. Sep 18, 2006

### alpaolo

Excuse for the question.
There is a person wich can say to me an answer about speed of light?
The maximum speed of light is c, the result of michelson & morley experiment says: there isn't ether in space.
Is possible to say:
The photons generated by a light emitter was indipendent by a generator and then, there isn't anything that can "drag" the photons with sufficent force to measure the variation ( With exclusion of curvature of the rays by the high mass gravity)?

13. Sep 18, 2006

### michael879

Ive seen your drag theory on another thread. c isnt just the maximum speed for light in an inertial frame, its the minimum too. your drag theory doesnt explain how when someone tries to "chase" a photon it still always goes at c relative to them.

14. Sep 18, 2006

### JesseM

No, the fact that Michelson-Morley expected the light to move at different speeds relative to the earth has nothing to do with "drag". By analogy, suppose you see two cars driving by you, and they are moving at 80 mph and 50 mph. In Newtonian physics, in the frame of the car moving at 50 mph, the speed of the second car would be 80-50 = 30 mph. Nothing to do with drag, just a question of observers moving at different speeds measuring other objects to have different speeds relative to themselves. But in relativity this is no longer true; if I see a light beam go by me followed by a ship, and I measure the light beam to be moving at 1 c and the ship to be moving at 0.7 c relative to me, then the ship will not measure the light beam to be moving at 1 c - 0.7 c = 0.3 c relative to itself, instead it will measure the light beam to still be traveling at 1 c relative to itself. This has to do with the fact that in relativity, each observer uses their own rulers and clocks at rest relative to themselves to measure distances and time (and speed = distance/time), but each observer measures the rulers of other observers (which are moving along with those other observers in the first observer's frame) to be shrunk, and their clocks to be slowed-down and out-of-sync with each other.

15. Oct 4, 2006

### baryon

You mean Gallilean, not Newtonian.:surprised

16. Oct 4, 2006

### JesseM

Newtonian physics uses the Galilei transform to describes measurements made in different inertial reference frames, so I guess either could be correct, although I don't usually see people using the phrase "Galilean physics", maybe since Galileo didn't really have any unified framework for describing physical laws like Newton's three laws of motion.

17. Oct 5, 2006

### pess5

Einstein's special theory says ...

Light's speed wrt the stationary measuring observer is c.
Light's speed wrt any moving body is c-v, per the stationary measuring observer.​

The reason light's speed is c wrt the stationary measuring observer is because v=0 since all inertial measurers assume themselves the stationary per their own vantage. So c-v = c-0 = c.

Some will cite that the medium supports only a speed c change. Some will suggest that light travels orthogonal to 3-space, and so we can never record a change in its speed as a function of own own material motion thru the 3-space. Light cannot be held in a state of rest, and so we can assume that light's dynamics are mutually linked to the intrinsic properties of the fused spacetime itself.

18. Oct 5, 2006

### JesseM

Actually, that is not the meaning of Einstein's equation there--see my first post on the thread where I went over Einstein's method, he was actually introducing a new coordinate system which was related to an actual physical coordinate system by a Galilei transform, and thus this new coordinate system would not correspond to the actual measurements of any inertial observer in relativity. The v was not an observer's measurement of his own velocity, but rather his velocity as measured in the "stationary frame" which was based on actual rulers and clocks. The coordinates that would actually be measured by this observer if he used his own rulers and clocks were denoted by the greek symbols tau, xi, eta, and zeta (ie. $$\tau, \xi, \eta, \zeta$$).

19. Oct 5, 2006

### pess5

Yes, I realize all that. I was just being simplistic, that's all. What I said is correct, but was not intended to replace the Lorentz Transforms, nor the Fitzgerald Contraction (which was the posted eqn actually). If you are considering light wrt yourself, or wrt another of your frame, v=0 and so c-v or c+v equals c. Just another simplistic way of looking at it, since c-v was in the derivation at hand. So I just responded in the context of the stated post.

Yes, I'll take a look at your post. I have myself been through the LT derivation a gazillion times over many years. It's clear that you have as well. Einstein relates the perspectives of frames K and k in his kinematic model, setting the outbound and reflection legs equal between perspectives, takes the partial derivatives, integrates, substitutes for x'=x-vt and x=ct, deduces no length change along orthogonal axes, deduces that direction of v has not effect on the length contraction, which all leads to the Lorentz Transformations. My nickel description here