# Speed of Light Unaffected by Moving Emitter

• B
• marcosdb
In summary, the conversation discusses the concept of special relativity and how observations and measurements must be considered in a specific frame of reference. The conversation also touches on the synchronization of clocks and how they can be used to measure the speed of light. It is important to note that measurements and observations will be different depending on the chosen frame of reference.
marcosdb
TL;DR Summary
Verifying my understanding of SR

Is my understanding correct that if we have a moving vehicle moving to the right at speed v, as above, with a light source in center going in both directions, that (upon emitting the light at time T), a detector at D1 & D2 would both detect light reaching it at T2? (even though in the time it takes for the light to travel, the distance to D1 has shortened & the distance to D2 has grown?)

According to which frame of reference?

Are the detectors attached to the cart? And which frame's definition of "at the same time" are you using? The answer to your question depends on your answers to both these questions.

Detectors are attached to the cart & moving with the cart

As for frame of reference, suppose we had an infinitely precise clock located at D1 & D2, synchronized; would it register the same value T2? (if I then walked over to them & looked at them?)

PeroK
[Corrected in view that first I have interchanged the labelling of ##D_1## and ##D_2## in the OP]

Make the distance between the light source and detectors ##D_1## and ##D_2##, as measured in the considered frame of reference (the "lab frame") ##L##. Let the signal, sent out at ##t=0## reach detector ##D_1## (##D_2##) at time ##t_1## (##t_2##). Since despite the moving of the source the speed of light is ##c## relative to the lab frame in both directions. Thus you have
$$c t_1=L-v t_1,$$
because in the time ##t_1## the right end of the vehicle has moved by the distance ##v t_1## against the direction of the light. Thus you have
$$t_1=\frac{L}{c+v}.$$
For the other detector it's
$$c t_2=L+v t_2,$$
because the left end of the vehicle moves in direction of the light signal, which thus needs to travel for a distance enlarged by ##v t_2##. So you have
$$t_2=\frac{L}{c+v}.$$
Of course, you must have ##v<c## for all this to be consistent.

Last edited:
marcosdb
marcosdb said:
As for frame of reference, suppose we had an infinitely precise clock located at D1 & D2,
So, also attached to the cart?
marcosdb said:
synchronized
But synchronised in which frame? Synchronised with clocks on the ground? Then no. Synchronised with clocks on the cart? Yes.

FactChecker, Dale and marcosdb
Ibix said:
So, also attached to the cart?
Precisely, both clocks are on the cart, moving with the cart.

Ibix said:
But synchronised in which frame
Synchronized to each other (both clocks on the train, moving), so I guess the answer as to which frame is the cart's frame

Given that, if we stop the clocks at the moment the light reaches them (i.e. take a reading on an LCD display), the expectation is that they would have the same number, correct?

And we could synchronize both of those clocks before we start moving (on the ground), correct, and they'd continue to be synchronized? (because both would have the same acceleration applied & would be moving at same velocity?)

PeroK
marcosdb said:
Synchronized to each other (both clocks on the train, moving), so I guess the answer as to which frame is the cart's frame

Given that, if we stop the clocks at the moment the light reaches them (i.e. take a reading on an LCD display), the expectation is that they would have the same number, correct?
Yes.

However, if you have clocks on the ground that are synchronised in the ground frame and just happen to be in the same places as your detectors when they trigger then those clocks will not show the same time, for the reason you gave in #1.

cianfa72, Dale, PeroK and 1 other person
Draw a position-vs-time diagram….
showing the endpoints and midpoint of the cart and the light signals.

Ibix
marcosdb said:
TL;DR Summary: Verifying my understanding of SR

Is my understanding correct that if we have a moving vehicle moving to the right at speed v, as above, with a light source in center going in both directions, that (upon emitting the light at time T), a detector at D1 & D2 would both detect light reaching it at T2? (even though in the time it takes for the light to travel, the distance to D1 has shortened & the distance to D2 has grown?)
This paragraph shows that you taken no consideration of the frame of reference in which any of you observations or measurements are made.

1) There is no such thing as a "moving" vehicle. The vehicle may be moving to the right at speed ##v## in the ground frame. In the vehicle's frame, it is not moving.

2) At time ##T## could be in the ground frame or the vehicle frame.

3) At time##T2## could be in the ground frame or the vehicle frame.

4) The time it takes for light to travel - again, in which frame.

5) The distance from the emission event to the detection at D1 is less than the distance from the emission event to the detection at D2, as measured in the ground frame.

malawi_glenn
marcosdb said:
As for frame of reference, suppose we had an infinitely precise clock located at D1 & D2, synchronized
In the ground frame, or in the cart frame?

Use position coordinates for the detectors instead (I assume they are attached to the cart):
##x_1## for the position of the left detector in the ground frame.
##x_1^\prime## for the position of the left detector in the ground frame.
##x_2## for the position of the right detector in the ground frame.
##x_2^\prime## for the position of the right detector in the ground frame.

In the cart frame, the light pulse will reach the left detector D1 at the time ##t_1^\prime## and the right detector D2 at the time ##t_2^\prime## and these two times are equal ##t_1^\prime = t_2^\prime##.
Assume that the light pulse is emitted from the center of the cart when the origins of the two reference frames are coinciding, and let ##t=t^\prime = 0## for that event.
Using the Lorentz-transformation, we can find the corresponding position and time coordinates in the ground frame (unprimed quantities).

Got it, that makes sense & confirms my understanding (apologies for not making clear that all references were to cart frame & not ground frame, and that the cart was moving at velocity v in relation to ground frame)

Why is it, then, that when we put this in a circle (Sagnac effect), we don't have the same expectation?

The Wikipedia article reasons that "the light has to travel further as detector is moving away/towards", which seems to only be true in relation to the lab frame; but the detector is in the disc frame, where it would seem, as a parallel to this:

PeroK said:
There is no such thing as a "moving" vehicle. The vehicle may be moving to the right at speed in the ground frame. In the vehicle's frame, it is not moving.

the disc would "not be moving" (in the disc frame, where the detector is), and as such we would expect the 2 beams of light to reach at the same time (in the disc frame)

marcosdb said:
Why is it, then, that when we put this in a circle (Sagnac effect), we don't have the same expectation?
The speed of light is the same in all inertial frames of reference. A rotating frame is not inertial. And there is no symmetry between frames here - one frame is rotating and objects at rest will feel centripetal forces while the lab frame will not. Furthermore your detectors are co-located so there's no room for questions of simultaneity. The light either arrives at the same time or it doesn't.

I would advise getting more strongly to grips with inertial frames before attempting to work with non-inertial ones. @robphy's suggestion of drawing a displacement-time graph is an excellent one - it's the first step to Minkowski diagrams, which IMO are the most powerful tool for understanding special relativity.

malawi_glenn
marcosdb said:
Got it, that makes sense & confirms my understanding (apologies for not making clear that all references were to cart frame & not ground frame
In the vehicle frame, there is simply a emitter and two detectors all at rest with respect to each other. There is no mystery.

I honestly don't think you understand that at all.
marcosdb said:
Why is it, then, that when we put this in a circle (Sagnac effect), we don't have the same expectation?
The Sagnac effect can wait until you have a grasp of the principle of relativity.
marcosdb said:
The Wikipedia article reasons that "the light has to travel further as detector is moving away/towards", which seems to only be true in relation to the lab frame; but the detector is in the disc frame, where it would seem, as a parallel to this:

the disc would "not be moving" (in the disc frame, where the detector is), and as such we would expect the 2 beams of light to reach at the same time (in the disc frame)
Exactly, you are totally confused!

PeroK said:
Exactly, you are totally confused!

So you can help un-confuse me, then?

Why is the spinning disc (with an emitter & 2 detectors, all at rest with respect to each other) not the same as the train (with an emitter & 2 detectors, all at rest with respect to each other)?

What is the difference?

marcosdb said:
Why is the spinning disc (with an emitter & 2 detectors, all at rest with respect to each other) not the same as the train (with an emitter & 2 detectors, all at rest with respect to each other)?

What is the difference?
The spinning disc is accelerating, for one thing.

PeroK said:
The spinning disc is accelerating, for one thing.

Does this mean, then, that if I have a fiberoptic cable running east-west from point A to B on the surface of Earth, that it actually takes longer for the signal to go from A -> B than B -> A?

As an excersice for the original problem.
Let the cart move w.r.t. the ground at constant speed ##0.60\cdot c## and let the lenght of the cart be (as measured in its own frame of reference) ##1000## m. The light pulses are emitted at the same time and at the center of the cart according to a stationary observerer in the cart's frame. The light pulses will hit detector D1 and D2 at the same time according to the cart's reference frame. What is the separation in distance and in time for these events as calculated in the ground frame?

marcosdb said:
Does this mean, then, that if I have a fiberoptic cable running east-west from point A to B on the surface of Earth, that it actually takes longer for the signal to go from A -> B than B -> A?
In which frame? This is not a closed circle so there isn't a unique answer.

marcosdb
Ibix said:
In which frame? This is not a closed circle so there isn't a unique answer.

So suppose we start with a fiberoptic cable that goes fully around earth, and a computer which has 2 ports so it can communicate with itself; when I send a signal out in the eastward direction, I measure the latency; then I send a signal out in the westward direction & also measure latency.

The expectation is that the latency would always be less in the westward direction?

And would that then not apply to all points on that circle (that anyone attached to that cable, sending a signal in the westward direction, would always have lower latency than the signal they receive back from the same point)?

Of course. But all of those people have only one clock - the one where they are. There's nothing to synchronise. If you just have a length of cable you need a clock at each end, and then you need to ask how you synchronised them. And that is equivalent to asking what frame you are using.

cianfa72 and marcosdb
Last part confusing me: wasn't the Michelson–Morley experiment using the rotation of the earth as a way to measure differences in the speed of light?

Would it not amount to the fiberoptic example I gave above, but with opposite results?

marcosdb said:
wasn't the Michelson–Morley experiment using the rotation of the earth as a way to measure differences in the speed of light?
It used out-and-back light, not a timed pulse one way and independently another timed one the other way. So it used one clock (effectively the interference effect gives you a difference in arrival times), not two as required by your experiment. So it's completely different to your short length of cable, without the clock synchronisation problem that has. It's also completely different to your ring-around-the-world because it isn't comparing light going in opposite directions in a rotating frame. Michelson and Morley were considering light travelling in orthogonal directions in what might as well be an inertial frame, just using the spin of the Earth to re-orient the apparatus between data runs.

PeroK and marcosdb
marcosdb said:
Why is the spinning disc (with an emitter & 2 detectors, all at rest with respect to each other) not the same as the train (with an emitter & 2 detectors, all at rest with respect to each other)?

What is the difference?
Proper acceleration. It is the acceleration measured by an accelerometer.

In the spinning disk the accelerometer attached to the disk measures the centripetal acceleration, while an unattached accelerometer measures no acceleration. The attached and unattached accelerometers are not equivalent.

For the train both accelerometers read zero. They are equivalent.

PeterDonis
marcosdb said:
Last part confusing me: wasn't the Michelson–Morley experiment using the rotation of the earth as a way to measure differences in the speed of light?
That was indeed the prediction based on the rigid aether model. The MMX ruled out a rigid aether.

SR does not predict a fringe shift for a rotating Michelson Morley interferometer, but it does predict a fringe shift for a rotating Sagnac interferometer. They are different designs and produce different results.

Michelson Morley interferometers rule out rigid aether theories. Sagnac interferometers rule out dragged aether theories. Both support special relativity.

In your recent posts you seem to be equating rotational motion with inertial motion, rigid aether predictions with special relativity predictions, and Michelson Morley interferometers with Sagnac interferometers. You need to be a little more careful not to mix up such distinct concepts. I sympathize: relativity starts to connect things that seem like they should be separate, so it can become confusing to know which previously distinct things go together and which don’t.

Ibix, PeroK and PeterDonis
marcosdb said:
Precisely, both clocks are on the cart, moving with the cart.Synchronized to each other (both clocks on the train, moving), so I guess the answer as to which frame is the cart's frame

Given that, if we stop the clocks at the moment the light reaches them (i.e. take a reading on an LCD display), the expectation is that they would have the same number, correct?

And we could synchronize both of those clocks before we start moving (on the ground), correct, and they'd continue to be synchronized? (because both would have the same acceleration applied & would be moving at same velocity?)
But then you must describe anything in the rest frame of the cart. For this you must do the Lorentz transformation with boost velocity ##v## of my posting #5. NOTE that I had made a mistake before and interchanged the labelling of the detectors. It's now corrected.

The events of the light signal arriving at detectors D1 and D2 are
$$x_1 = \begin{pmatrix} c t_1 \\ -L-v t_1 \end{pmatrix}=\begin{pmatrix} L/(1+\beta) \\ -L/(1+\beta) \end{pmatrix},$$
$$x_2=\begin{pmatrix} c t_2 \\ L-v t_2 \end{pmatrix}=\begin{pmatrix} L/(1-\beta) \\ L/(1-\beta) \end{pmatrix},$$
where ##\beta=v/c##.

$$\hat{\Lambda}=\begin{pmatrix}\gamma & -\beta \gamma \\ -\beta \gamma & \gamma \end{pmatrix},$$
where ##\gamma=1/\sqrt{1-\beta^2}##. Thus the events in the rest frame of the cart are
$$x_1'=\hat{\Lambda} x_1 = \gamma \frac{L}{1+\beta} \begin{pmatrix}1+\beta \\ -1-\beta \end{pmatrix} = \gamma L \begin{pmatrix}1 \\ -1 \end{pmatrix}=L' \begin{pmatrix}1 \\ -1 \end{pmatrix},$$
where ##L'=\gamma L## is the distance of the detector from the origin as measured in the rest frame of the cart (length contraction!).

Further
$$x_2'=\hat{\lambda} x_1 = \gamma \frac{L}{1-\beta} \begin{pmatrix} 1-\beta \\ 1-\beta \end{pmatrix} = \gamma L \begin{pmatrix} 1 \\ 1 \end{pmatrix} = L' \begin{pmatrix} 1 \\ 1 \end{pmatrix},$$
i.e., in this frame the light signal reaches both detectors at the same time, ##t'=L'/c## as it must be.

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