Reconciling Constant speed of light with Relativity

1. Jul 7, 2015

Sturk200

If the speed of light is constant regardless of the state of motion of the source, then doesn't this imply that it is possible to calculate the velocity of a reference frame by measuring the time it takes for light to traverse some known distance in that frame. For instance if our frame is moving in the positive x direction with velocity v and a pulse of light is emitted from the origin, shouldn't it be possible to calculate v by considering the time it takes for the pulse to reach some position x? The actual distance travelled by the light beam will be the measured distance x plus the amount our reference frame has travelled in the requisite time it took for the light to reach x. In maths, ct=x+vt, and therefore v=c-x/t, where x is the apparent distance travelled by the light as measured in our reference frame, t is the amount of time it took for the light to make that trip, and v is the velocity of our reference frame. In sum, since the speed of light is constant, the amount of time it takes for light to travel some fixed distance will depend on one variable only, and that is the state of motion of the frame in which the time is measured, and thus it should be possible to deduce the state of motion of a reference frame, simply by measuring how long it takes for light to travel a fixed distance. My question is, doesn't something have to be wrong here, since according to relativity it should be impossible for an inertial observer to deduce his own state of motion by observing local phenomena?

2. Jul 7, 2015

axmls

Our frame is moving with velocity v relative to what?

3. Jul 7, 2015

mathman

There is no absolute velocity, so the phrase has no meaning.

4. Jul 7, 2015

Sturk200

Let's define a reference frame such that ct=x. That is, so that when we emit a light pulse from the origin and measure the time it takes for that pulse to reach a position x, the product of the known speed of light and the time will be precisely equal to x. Can we do that and then define v w/r/t this newly defined frame?

5. Jul 7, 2015

PAllen

This will be true in every frame, whatever their relative state of motion, and whatever the speed of sources in each frame. That was the key finding leading to SR (starting well before Einstein) - that the speed of light is the same independent of emitter speed in a frame, and is the same for frames moving relative to each other.

6. Jul 7, 2015

Sturk200

Maybe I'm not being clear. Take a ruler and set it in motion in the positive x direction. Imagine that we are in the rest frame of the ruler. Now emit a pulse of light from the left end of the ruler and time how long it takes for the pulse to reach the ten inches mark. If the ruler is not moving at all relative to the ideal frame I tried to set up, then the time it takes for the light to hit the mark will be exactly 10 inches/c, because the light will actually have had to travel only ten inches. But if the ruler is moving, as we supposed, and if it's moving awfully fast too, then it will take a longer time for the light to reach the ten inches mark, because (1) it has farther to travel and (2) the speed of light is independent of the state of motion of the source. Now the difficulty I am having is this, if it takes longer for light to travel ten inches in my frame when my frame is moving in the direction in which light is emitted, then can't I deduce my state of motion from this extra time?

To illustrate more clearly, consider the comparison to a tennis ball in classical relativity. If I am traveling at constant velocity and fire a tennis ball at ten mph, it will always take the same of time to travel ten inches, no matter how fast I'm going or what direction I'm going in -- and this is because my state of motion imparts a velocity to the tennis ball, providing it with the means of traversing any extra distance created by my motion with no added time. Of course, when we are dealing with light there is no such effect, and my velocity will change how long it takes for light to travel ten inches in my frame.

7. Jul 7, 2015

PAllen

This is false, pure and simple. The rest of your difficulty follows from this. Pick 10 such rulers moving at different speeds. In frame in which each ruler is at rest, it will still take the same time to reach the 10 inch mark. Pick any combination of emitter motions and ruler motions, and every ruler will measure light from every emitter to be going at the same speed.

8. Jul 7, 2015

Sturk200

Interesting. I guess my confusion is more basic than I thought. So then would the following be correct?

We place a light detector in the center of a ruler and two light sources, one on either end of the ruler, each facing the detector in the center. When we place the ruler at rest, on a table, the light detector registers simultaneous pulses every five seconds. Now we give the ruler a large constant velocity along the axis of its length. Will the light pulses still be registered as simultaneous?

9. Jul 7, 2015

Staff: Mentor

So the basis of relativity (in traditional derivations) are the two postulates. The first is the principle of relativity, which implies that you cannot measure an absolute velocity, and the second is the invariance of c.

Relativity starts from these two postulates and then proceeds to derive the Lorentz transformation which describes the relationships between distances and times as measured relative to different inertial reference frames.

It turns out that, in order to avoid the type of measurement you suggest, different frames will disagree on the distances and times involved.

10. Jul 7, 2015

Staff: Mentor

Don't forget relativity of simultaneity - it's important.

If the two sources flash simultaneously in a frame in which the ruler is at rest, they will not flash simultaneously (that's relativity of simultaneity at work - don't forget it) in a frame in which the ruler is not at rest. The difference (along with length contraction) will be just enough to make up for the different distances traveled from source to detector so that the flashes reach the detector together no matter which frame you choose.

Did I mention the importance of rememembering relativity of simultaneity?

11. Jul 7, 2015

Staff: Mentor

No, it doesn't. The ruler is length contracted. Once you take that into account, you find, as PAllen said, that $ct = x$ holds in every inertial frame.

12. Jul 7, 2015

1977ub

Yes if you are traveling with the ruler, but not to us who remained behind.

13. Jul 8, 2015

harrylin

That is exactly what led to "special relativity":

"We will raise [..] the “Principle of Relativity” to the status of a postulate, and also introduce another postulate, which is only apparently irreconcilable with the former, namely, that light is always propagated in empty space with a definite velocity c which is independent of the state of motion of the emitting body. These two postulates suffice for the attainment of a simple and consistent theory [..]" (emphasis mine).
- https://www.fourmilab.ch/etexts/einstein/specrel/www/

Others here already mentioned the main elements of the solution.

14. Jul 8, 2015

Janus

Staff Emeritus
A bit of clarification here. The pulses will arrive simultaneously at the detector according to both frames, They will not however be emitted simultaneously in both frames.

Last edited: Jul 8, 2015