A thought experiment of the relativity of light

In summary, all of the observers will agree that the light source is expanding outwards from the point of emission in a perfect sphere with radius ##cT## where ##T## is their measurement of the time since emission. However, the speed with which the origins move will be different for different observers.
  • #1
CClyde
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A light source in uniform motion emits a flash of light.

A spherically symmetric wavefront propagates from a central point, the source, or the “origin” of emission.

The wave front remains at c relative to the origin as measured by all observers.
How fast do these origins move?

Am I missing something?
 
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  • #2
CClyde said:
How fast do these origins move?
How fast are your observers moving? Perhaps one of your observers is at rest relative to the device that emitted the flash of light, perhaps that device is moving relative to some of the other observers.

All of them will observe that the flash is expanding outwards from the point of emission in a perfect sphere with radius ##cT## where ##T## is their measurement of the time since emission. Only those observers who are at rest relative to the device will find that the device remains at the center of that sphere. All others will find that the device was at the center of the sphere at the moment of emission but has since moved away from that point.

Because of the relativity of simultaneity the set of points that one observer considers to be at a fixed distance from the point of emission at time ##T## (that is, the surface of a sphere of radius ##cT##) will not all be at that distance at the same time according to the other observers. Thus observers moving relative to one another will all agree about the spherical expansion of the light pulse, but they will be considering different events in spacetime to construct their spheres.

(And if this does no make sense to you, then you'll want to better understand the relativity of simultaneity. Any statement about where the light is at time ##T## is a statement about when the light reaches different places at the same time, and relativity of simultaneity tells us that "at the same time" is different in different frames)
 
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  • #3
CClyde said:
How fast do these origins move?
Any speed ##|\vec v|<c##.

CClyde said:
Am I missing something?
If you are new to relativity then undoubtedly you are missing something. You will need to express more of your concern if you want any help though.
 
  • #5
I am not new to relativity, but something I thought was self evident is being questioned.

If there are two of the events described in the thought experiment, and the light sources are in motion relative to each other during emission, are the origins of each emission at rest with each other?

If so, which I thought was the case, is this not the case for all origins?
 
  • #6
CClyde said:
the origins of each emission
Are single points (events) in spacetime, so it makes no sense to ask whether they are "at rest" or not relative to each other. That concept doesn't even apply to events.
 
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  • #7
CClyde said:
If there are two of the events described in the thought experiment, and the light sources are in motion relative to each other during emission, are the origins of each emission at rest with each other?
As @PeterDonis said, an event doesn’t have a velocity for the same reason that a point doesn’t have a direction.

Of course, the emitters are massive objects and they do not need to be at rest with respect to each other. But the emission events are still events.
 
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  • #8
CClyde said:
If there are two of the events described in the thought experiment, and the light sources are in motion relative to each other during emission, are the origins of each emission at rest with each other?
It depends what you mean by "the origins of each emission". If you mean the physical light sources themselves then they obviously remain in relative motion. In their own rest frames they each remain at the center of their expanding sphere of light, but the other source remains in motion and not at the center of their own sphere.

If, on the other hand, you mean "the place where the emission occurred" then this is by definition at the center of the sphere of light. The light source may or may not remain there. These positions are by definition at rest in your chosen frame, but different frames will not agree what this means. For example, one observer might place bouys at the locations of the emission events. In that observer's rest frame the bouys would remain at rest at the centers of the sphere of light, but other frames would say they were in motion and not at the center, albeit agreeing that they bouys are at rest with respect to one another.

Finally, you could mean "the emission events". These are frame independent, but "speed" and "rest" are not things that can be defined for them, and your query would make no sense.
 
  • #9
The motion of a light origin (coordinates of an emission event) is frame dependent to “inertial” observers.
The motion of light origins relative to each other is not a frame dependent measure (in flat spacetime).
For example:
The two sources of the events can remain at rest with each other and remain at the origin of their own light sphere(wavefront).
The the two sources can remain at rest with each other and move relative to the origin of their light spheres.

But as the two origins must remain at the center of a universal physical constant, there is no impetus by which we can claim they move relative to each other, without violating the constancy of c they hold with the wavefront.
 
  • #10
CClyde said:
The motion of a light origin (coordinates of an emission event) is frame dependent to “inertial” observers.
The motion of light origins relative to each other is not a frame dependent measure (in flat spacetime).
For example:
The two sources of the events can remain at rest with each other and remain at the origin of their own light sphere(wavefront).
The the two sources can remain at rest with each other and move relative to the origin of their light spheres.

But as the two origins must remain at the center of a universal physical constant, there is no impetus by which we can claim they move relative to each other, without violating the constancy of c they hold with the wavefront.
These are just words. Can you provide some mathematics to show what you are talking about.

The key is the Lorentz Transformation (between IRF's) - rather than the classical, Newtonian, Galilean transformation, which you may be assuming. If you are assuming this, then you may prove that the invariance of the speed of light is incompatible with classical physics. This is true and is the key point upon which classical mechanics and SR differ fundamentally.
 
  • #11
CClyde said:
The motion of a light origin (coordinates of an emission event) is frame dependent to “inertial” observers.
The motion of an event is undefined. It does not exist in any frame.
 
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  • #12
CClyde said:
If so, which I thought was the case, is this not the case for all origins?
PS I think @Nugatory gave the perfect answer to your question in post #2. I suggest you study what he said.
 
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  • #13
CClyde said:
The motion of a light origin (coordinates of an emission event) is frame dependent to “inertial” observers.
If you mean spatial coordinates here then I agree with what you write. If you mean spacetime coordinates then Dale's comment applies - events do not have motion.

I think all of the confusion here is around exactly what you mean by "light origin". I think what you mean is that when the light sources emit a flash they also drop bouys that immediately accelerate to rest in some predetermined frame, and those bouys remain at what you call the light origin in that frame. Assuming that's the case, then if there are two light sources that drop bouys pre-programmed to come to rest in the same frame then those bouys will remain at rest with respect to each other. Everyone will agree that those bouys are at rest with respect to one another. Only the frame that sees them at rest will see them at the center of the spherical wavefront of the light.
 
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  • #14
Ibix said:
I think all of the confusion here is around exactly what you mean by "light origin".
I agree. @CClyde is explicitly using the word event in their description, but then seems to switch to a different concept.

The event defining the spacetime origin of a flash of light would be some event ##\mathbf{R}=(t_0,x_0,y_0,z_0)##. The worldline defining the spatial origin of a flash of light would be ##\mathbf{r}(t)=(t,x_0,y_0,z_0)##.

I think that the OP is confusing ##\mathbf{r}(t)## with ##\mathbf{R}## in their communication with us and possibly in their own mind
 
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  • #15
I mean spatial coordinates of an event.
We’re talking about the founding premise of the theory of relativity.
The difference in the length of a light path traversing a rigid rod AB when it is at rest in a stationary system and moving in that system.
In both cases the length is determined by the “bouy” dropped at A and the length of the rigid measure to B and the length of light path from the bouy to B.
When AB is in motion the length is greater for the stationary observer.
How do you know this if you do not acknowledge the event coordinates of A when A is no longer at A? (the bouy, origin, point of emission etc.)

When the light path is considered in terms of the spherically symmetric wavefront, “A” does not do anything differently.
If we conducted a million tests of the relativity of simultaneity at different locations, directions, velocities and times we would find the same result.
It is the definition of light, its independence of the motion of the source. This does not mean the dropped buoys do their own thing, it means every one will be and will remain at rest with every other.
No you cannot go looking for these buoys, but you can find and measure their existence “during” the kinematics of light.
This is what I have understood to be the mechanics of light since I was old enough to understand it.
What am I missing? (don’t say I’m not old enough)
 
  • #16
CClyde said:
I mean spatial coordinates of an event.
We’re talking about the founding premise of the theory of relativity.
The difference in the length of a light path traversing a rigid rod AB when it is at rest in a stationary system and moving in that system.
In both cases the length is determined by the “bouy” dropped at A and the length of the rigid measure to B and the length of light path from the bouy to B.
When AB is in motion the length is greater for the stationary observer.
How do you know this if you do not acknowledge the event coordinates of A when A is no longer at A? (the bouy, origin, point of emission etc.)

When the light path is considered in terms of the spherically symmetric wavefront, “A” does not do anything differently.
If we conducted a million tests of the relativity of simultaneity at different locations, directions, velocities and times we would find the same result.
It is the definition of light, its independence of the motion of the source. This does not mean the dropped buoys do their own thing, it means every one will be and will remain at rest with every other.
No you cannot go looking for these buoys, but you can find and measure their existence “during” the kinematics of light.
This is what I have understood to be the mechanics of light since I was old enough to understand it.
What am I missing? (don’t say I’m not old enough)
This post is incomprehensible to me. SR is built on the Lorentz Transformation:$$t' = \gamma(t - \frac v {c^2} x), \ x' = \gamma(x - vt)$$This relates the coordinates of an event in one inertial reference frame to those in another inertial reference frame, where ##v## is the relative velocity between the origins of the frames, which coincide at ##t = t' = 0, x = x' = 0##.

From this, you can derive the velocity transformation formula:$$u' = \frac{u - v}{1 - \frac{uv}{c^2}}$$Where ##u, u'## are the velocity of a particle or wave in each reference frame.

Now, if we take ##u = c##, then (for any ##v##) we find that:$$u' = \frac{c - v}{1 - \frac v c} = c$$ And we see that a light wave has speed ##c## that is invariant across all inertial reference frames. Hence, light moves out in a sphere from its point of emission in any inertial reference frame. And, independent of the motion of the source. It does not move out in a sphere relative to the source in all inertial reference frames - but only in the reference frame where the source is at rest.

This is covered more comprehensively in one of the appendices in Special Relativity, by Helliwell.
 
  • #17
Thank you PeroK

I don’t think I said the speed of light varied for any frame, if I did, or it was implied it was not intended.

I am not questioning the Lorentz Transformation.

If it helps, I’ll rephrase my question with respect to the Lorentz Transformation:
Why would you even consider using the Lorentz Transformation unless you acknowledge the difference in measures between frames in motion in the first place?

The difference is the measure of the rigid rod length AB that light traverses (end A to end B) is different than the light path length measure A to B when AB is in motion.

When in motion, the spatial coordinates of the rigid rod end A at the time of emission from A, differ from the spatial coordinates of the end A at the time light reaches B. (c+v)This requires we recognize the spatial coordinates of A at the time of emission did not move with AB after emission and in doing so, we acknowledge there exists a spatial coordinate A after the light has left A.

This is not a difficult concept. In fact it is so fundamentally self evident that it appears it is not even considered by many.
 
  • #18
CClyde said:
If it helps, I’ll rephrase my question with respect to the Lorentz Transformation:
Why would you even consider using the Lorentz Transformation unless you acknowledge the difference in measures between frames in motion in the first place?
Let's focus on this question. If you have a single reference frame, then light moves at ##c## in all directions and there is nothing more to discuss. SR only becomes interesting when you consider the same events from a different reference frame. To do this, you need a transformation rule, and indeed you need the Lorentz transformation.

If you use the old Galilean transformation (##t' = t## and ##x' = x - vt##), then you cannot reconcile that with the invariance of the speed of light.

Once you have the Lorentz transformation, then you have the mathematical justification for the invariance of ##c## and the issue is resolved, IMO.

CClyde said:
The difference is the measure of the rigid rod length AB that light traverses (end A to end B) is different than the light path length measure A to B when AB is in motion.

When in motion, the spatial coordinates of the rigid rod end A at the time of emission from A, differ from the spatial coordinates of the end A at the time light reaches B. (c+v)This requires we recognize the spatial coordinates of A at the time of emission did not move with AB after emission and in doing so, we acknowledge there exists a spatial coordinate A after the light has left A.

This is not a difficult concept. In fact it is so fundamentally self evident that it appears it is not even considered by many.
Unless you can put this into mathematics, with events and coordinates, then there is nothing to discuss. What you write is not recognisable as physics to me.
 
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  • #19
That’s unfortunate, there is a lot to be learned in the principles of physics before the mathematics.

This guy was famous for avoiding the math until necessary.

“…We now inquire as to the length of the moving rod, and imagine its length to be ascertained by the

following two operations:—
(a) The observer moves together with the given measuring-rod and the rod

to be measured, and measures the length of the rod directly by superposing the measuring-rod, in just the same way as if all three were at rest.

(b) By means of stationary clocks set up in the stationary system and syn- chronizing in accordance with § 1, the observer ascertains at what points of the stationary system the two ends of the rod to be measured are located at a definite time. The distance between these two points, measured by the measuring-rod already employed, which in this case is at rest, is also a length which may be designated “the length of the rod.”

If you would like to read more, here is a link.

https://www.physics.umd.edu/courses...einstein_electrodynamics_of_moving_bodies.pdf
 
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  • #20
CClyde said:
That’s unfortunate, there is a lot to be learned in the principles of physics before the mathematics.
That statement would make more sense if you were trying to help me understand the basics of SR.

CClyde said:
I love that paper, but it is too wordy by modern standards. The mathemetics is there, nevertheless. Moreover, what he's saying is comprehensible.
 
  • #21
CClyde said:
I mean spatial coordinates of an event.
So that would be ##\mathbf{r}(t)=(t,x_0,y_0,z_0)##.

I really didn't understand the rest of your post. Especially the buoy stuff.

CClyde said:
Why would you even consider using the Lorentz Transformation unless you acknowledge the difference in measures between frames in motion in the first place?
Because experimental evidence requires it.

CClyde said:
This requires we recognize the spatial coordinates of A at the time of emission did not move with AB after emission and in doing so, we acknowledge there exists a spatial coordinate A after the light has left A.

This is not a difficult concept. In fact it is so fundamentally self evident that it appears it is not even considered by many.
This is clear. There is a difference between ##\mathbf{R}=(t_0,x_0,y_0,z_0)## and ##\mathbf{r}(t)##. You are incorrect that it is not considered, you just were not stating it in a way that anyone could understand what you were saying. I do recommend using math as much as possible so that what you want to say can be stated clearly.

CClyde said:
This guy was famous for avoiding the math until necessary.
We have already arrived at that point in this conversation. It is necessary for clear communication here.
 
  • #22
CClyde said:
there is a lot to be learned in the principles of physics before the mathematics.
Yes, but your question in this thread is not about the principles of physics. It's about a particular scenario. You can't solve a particular scenario just using principles. You have to use math.
 
  • #23
CClyde said:
This is not a difficult concept.
I agree. I just think you are expressing it in an extremely confusing way and I'm not certain that you said it all right.
 
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  • #24
CClyde said:
I mean spatial coordinates of an event.
And an event is a point in spacetime.

CClyde said:
We’re talking about the founding premise of the theory of relativity.
It's certainly a core concept, but you don't appear to be using it correctly.

CClyde said:
The difference in the length of a light path traversing a rigid rod AB when it is at rest in a stationary system and moving in that system.
None of this involves single events. "The length of a light path" needs to be defined using a curve, which is not a single event.

CClyde said:
In both cases the length is determined by the “bouy” dropped at A
The dropping is an event, but the buoy itself is not an event. It would be described by a worldline, which is a timelike curve.

CClyde said:
the length of the rigid measure to B and the length of light path from the bouy to B.
These would also need to be described by curves. So you need to specify which curves.

CClyde said:
When AB is in motion the length is greater for the stationary observer.
This is a statement about arc lengths along curves, not about events.

CClyde said:
How do you know this if you do not acknowledge the event coordinates of A when A is no longer at A?
This doesn't even make sense. The concept you are groping towards here is the worldline of A.

CClyde said:
What am I missing? (don’t say I’m not old enough)
You appear to be missing, at the very least, the standard terminology that is used to describe these things. See above.
 
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  • #25
CClyde said:
This is not a difficult concept. In fact it is so fundamentally self evident that it appears it is not even considered by many.
You are egregiously wrong. The concepts you are vaguely groping towards in your posts are used all the time in relativity. But the people who use them know the correct language to describe them and the correct math to calculate with them.
 
  • #26
CClyde said:
A light source in uniform motion emits a flash of light.
A spherically symmetric wavefront propagates from a central point, the source, or the “origin” of emission.
The wave front remains at c relative to the origin as measured by all observers.
You may find it easier to see how this works if we clarify the notion of "observer".
(The discussion below comes from Taylor and Wheeler's "Spacetime Physics", a book that I cannot recommend too highly as an introduction to Special Relativity).

Suppose we fill the entire universe with three-dimensional rigid lattice of meter sticks. At every junction we place a clock and a robot with a pencil and a notepad. Whenever something interesting happens at a point in space, the robot at that point writes it down along with the time on their clock when it happened. Eventually, long after whatever experiment we're doing is complete, we collect all these pieces of paper, use them to reconstruct what happened. This lattice of rods and clocks defines a reference frame.

In your hypothetical, the robot at the point (0,0,0) in the lattice records "Light-emitting device was here and emitted a flash of light at time ##t=0##". The robots at (1,0,0), (0,1,0), (0,0,), (-1,0,0), (0,-1,0), .... all record that the flash of light passed them at time ##t=1## (we're measuring time in units of light-meters, where one light-meter is the time it takes for light to move one meter), the robots at the (2,0,0), (-2,0,0), ... all record that the flash of light passed them at time ##t=2##, and so forth. When we combine all of these results, we find that the the flash of light expanded in a sphere around the point (0,0,0) - and the previous and subsequent movement of the light source is irrelevant, all that matters is that the the emission event took place at time zero at the (0,0,0) point in our lattice, as recorded by the robot there.

We can also imagine that have two such lattices moving relative to one another (there's an annoyance that physical lattices would collide - we can ignore that). For simplicity we will set them up so that their (0,0,0) points coincide when their clocks at that point both read ##t=0##. This gives us two reference frames, and the (x,y,z,t) readings recorded by the robots in one frame will be related to the readings recorded in the other frame by the Lorentz transformations.

Both frames will measure that the light is expanding in a sphere of radius ##ct##.
 
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  • #27
CClyde said:
That’s unfortunate, there is a lot to be learned in the principles of physics before the mathematics.
Can you name some principle of physics that are not related to math? Thanks.
 
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  • #28
I've not said any are unrelated to math.
 
  • #29
So I think so far, we are agreed:

1. The spatial coordinates of a light emission (t0, x0, y0, z0) remain at the center of the sphere that is the propagating wavefront.

2. The constancy of c holds the wavefront symmetrical for all observers.

3. All such light emission coordinates are, by the definition of lights independence of the motion of the source incapable of motion relative to each other.

I cannot think of a more clear way to state this.

If this is not true according to what you think are the kinematics of light, please explain.
 
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  • #30
CClyde said:
So I think so far, we are agreed:

1. The spatial coordinates of a light emission (t0, x0, y0, z0) remain at the center of the sphere that is the propagating wavefront.

2. The constancy of c holds the wavefront symmetrical for all observers.

3. All such light emission coordinates are, by the definition of lights independence of the motion of the source incapable of motion relative to each other.

I cannot think of a more clear way to state this.

If this is not true according to what you think are the kinematics of light, please explain.

Maybe the 2nd animation under the headline "The Doppler shift in two dimensions" helps:
https://www.einstein-online.info/en/spotlight/doppler/#The_Doppler_shift_in_two_dimensions
 
  • #31
CClyde said:
1. The spatial coordinates of a light emission (t0, x0, y0, z0) remain at the center of the sphere that is the propagating wavefront.
Those are spacetime coordinates, not spatial coordinates. For spatial coordinates you only want the x, y, and z. Adding the t coordinate makes that an event, for which you can't discuss motion because it makes no sense.
CClyde said:
3. All such light emission coordinates are, by the definition of lights independence of the motion of the source incapable of motion relative to each other.
Assuming you drop the time coordinate, yes.
 
  • #32
Right, thanks lbix, it keep doing that.
The spatial coordinates (x0, y0, z0) of the emission.
 
  • #33
Sagittarius A-Star, thank you, but I’m referring to the light itself not the source.
 
  • #34
CClyde said:
The spatial coordinates of a light emission (t0, x0, y0, z0)
Those aren't just spatial coordinates.

CClyde said:
remain at the center of the sphere that is the propagating wavefront.
In each inertial frame, yes, if you limit to just the spatial coordinates and if you properly interpret what those coordinates mean--see below.

CClyde said:
The constancy of c holds the wavefront symmetrical for all observers.
In the sense of spatial coordinates, yes, provided you interpret them correctly--see below.

CClyde said:
All such light emission coordinates are, by the definition of lights independence of the motion of the source incapable of motion relative to each other.
This makes no sense. Coordinates don't "move".

This would be a lot easier if, as has already been strongly hinted, you would learn the standard terminology for these things, which has already been described. In that standard terminology, what you are trying to describe would be described this way:

In flat spacetime, an emitter whose worldline is a particular inertial worldline emits a spherical wave front of light at some event O on its worldline. In any inertial frame whose spacetime origin is taken to be the event O, the wave front of light will appear spherical, and the worldline of the spatial origin of that frame (which, except in one particular frame, the emitter's rest frame, will not be the same as the worldline of the emitter) will be at the center of the sphere at all times.

However, the emitter's worldline, except in one particular inertial frame, the emitter's rest frame, will not be at the center of the sphere except at time 0 in that frame (the time of event O, when the wave front is emitted).
 
  • #35
CClyde said:
1. The spatial coordinates of a light emission (t0, x0, y0, z0) remain at the center of the sphere that is the propagating wavefront.
But the inferred position of that center will not correspond to the subsequent uniform motion of the source for all observers. Strange but true. So what are you trying to conclude?
 

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