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This doesn't seem intuitive (like much of relativity) so I'll explain my reasoning below and hopefully someone can point out where I'm going wrong.

  1. Imagine we're a stationary observer watching a train go by at a 80% the speed of light. We see someone at the center of the train turn on a light.

  2. We observe the light hitting the back of the train before the front (because the back was travelling towards the light when it was turned on and the front travelling away from the light, hence the light traveled a longer distance to the front from our perspective).

  3. We note that for the person in the center the light is travelling at different speeds (relative to them). We conclude this because they are traveling in the same direction as the light going forwards and away from the light going backwards.

  4. We note that any way the person in the train tries to measure the speed of the light in a given direction they will always get the same answer (e.g. the speed of light). This is because they can only ever calculate the two-way speed of light. We conclude they can't tell that the light is traveling at different speeds relative to them.

  5. We summarise that the speed of light is different in different directions for moving frames (noting that some special directions may be equal (e.g. directly up and down in the train will exactly balance out)).

  6. We note that we are in a moving frame (e.g. the Earth is rotating around its axis and also around the sun, etc) therefore the speed of light is different in different directions for us (but we can't tell it).

A few points to clarify:

  • I'm talking about the speed of light in a different directions, a two-way measurement of the speed of light will cancel out any differences (so doesn't prove much).

  • Just looking to understand which step above is wrong (if any), would prefer minimal maths in answers.

UPDATE

So it seems I really should have added some context to why I'm asking this question :)

My intent is to understand the underlying set of rules that would generate the observables of Relativity. More specifically I am a programmer, so would like to consider how I could program the simplest set of rules to reproduce what we see.

From the comments and very clear answer from g s it seems there is an existing convention to take c as the average of the two way measurement of the speed of light and in the best case my logic is indistinguishable from the more common interpretation that c is equal in all directions, in worst case, it's just wrong.

So my intuition on this is that light being measured equally in all directions is something that would fall out of some simpler rules (e.g. something I can code) whereas it actually being equal from all frames of reference seems impossible to code.

So if there is a fault in my logic then great I can ditch that idea, but if its an equivalent approach to the existing convention, then that has a lot of meaning to me as I can try code it.

I may also try add an answer to my own question just to further show where my thinking is going (and hopefully people can point out any flaws in that).

Quantumplate
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    Can you clarify your point #3? The person in the centre of the train observes (or measures) the light hitting the front and back simultaneously, in their frame. – PM 2Ring Mar 04 '21 at 06:02
  • The stationary observer sees the light travelling at different speeds relative to the person in the center of the train. Because the person at the center can only measure the two-way speed of light, they will measure these different speeds as the same (so you're right they measure the light hitting the front and back at the same time). If the speeds are different and the light hits the back before the front, there is no measurement they can make which can tell this fact so they conclude the events are simultaneous (and the speed of light equal in all directions) even if its not. – Quantumplate Mar 04 '21 at 06:21
  • @Quantumplate that's not correct. It can be measured by clocks on board or you can reverse the situation and have the light originate from either end of the train and hit the middle. – Señor O Mar 04 '21 at 06:24
  • @SeñorO Two clocks at different locations can't be synchronised, see "Slow clock-transport" here, – Quantumplate Mar 04 '21 at 07:04
  • Well, they can be synchronised, but verifying that separated clocks are synchronised requires a round-trip signal (even for clocks at rest relative to each other). And as you mentioned earlier, we cannot measure the one-way speed of light. – PM 2Ring Mar 04 '21 at 07:10
  • BTW, we've had numerous questions (and some great answers) on this site regarding the one-way speed of light, especially since a certain Veritasium video on that topic was published late last year. Eg, https://physics.stackexchange.com/q/590942/123208 – PM 2Ring Mar 04 '21 at 07:15
  • @annav I'm not sure I understand, won't the stationary observer see the same distance traveled by light from middle-front-middle as middle-back-middle (and so no interference)? – Quantumplate Mar 04 '21 at 07:18
  • @Quantumplate theyre in the same frame, so yes they can be – Señor O Mar 04 '21 at 07:59
  • @annav isn't LIGO using a two-way measurement so any difference in speed of light in different directions will average out and always give no interference? Looking at it from the suns frame of reference, light will hit one mirror before the other but both beams will return to the middle at the same time (as the distance as seen by the sun is the same either way)? – Quantumplate Mar 04 '21 at 08:56
  • @Quantumplate Anyway, I may be wrong that the interference fringes would change. Only if the two way speed of light were (backwards and frontwards were affected by the fact that the earth rotates, which would go against the constant c. – anna v Mar 04 '21 at 09:20
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    @Quantumplate I will delete my comments – anna v Mar 04 '21 at 09:21
  • @SeñorO how would you synchronize them? If I have two light clocks at the center and move one back I can see the zigzag path shortens as it moves so this one gains a little time compared to the center. The one moving forward will lengthen the light path and so loose some time relative to the center. Its essentially small deceleration/acceleration so clock at front will be slower than clock at back? I'm assuming all clocks are effected equally by time dilation (so equally applies to atomic clock). Any getting in sync then requires light signals sent between clocks which doesn't work either? – Quantumplate Mar 04 '21 at 20:03
  • @PM2Ring thanks for the link. I've seen the questions, they all look like a method to measure the one way speed (and then answers debunking that method). I don't have any way to measure the one way speed, so subtle difference, just stating that speed in one direction <> speed in opposite direction (for an inertial frame and in most but not all directions). – Quantumplate Mar 05 '21 at 02:40
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    FWIW, I wasn't proposing that your question is a dupe of that one. But it's worthwhile to browse the existing posts on this topic. Sure, you can use a convention that assigns different one-way light speeds to different directions, but it's more convenient for calculations if you adopt the convention that the one-way speed is the same in all directions, as discussed in this answer by Dale. Also see what John Norton has to say on this topic. – PM 2Ring Mar 05 '21 at 04:56
  • @PM2Ring If it helps, I am a programmer, so my interest is from the point of view of programming rules to produce such a system, e.g. I would like to understand the underlying rules more than a convention for calculation for those within the frame (altho I appreciate the need for such a convention). – Quantumplate Mar 05 '21 at 06:47
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    @Quantumplate, for that case which allows anisotropic one – way speed of light you can explore this model https://arxiv.org/abs/1201.1828 The model considers Einstein synchronization of clocks in a moving frame as well as introduction of universal simultaneity for all observers. –  Mar 05 '21 at 07:31
  • @Albert that link is useful, thank you. Going to take me a while to get through it all tho :) – Quantumplate Mar 07 '21 at 22:55

3 Answers3

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Depending on how you phrase the question: either the question is meaningless, or the answer is no.

We can examine this by getting rid of the first pulse of light and instead starting our experiment as one observer sees simultaneous flashes from the walls to either side of her. This way, the light from each flash is only ever traveling in one direction (leftward for the rightward flash, rightward for the leftward flash).

enter image description here

Suppose a system with events L and R, equidistant from a central point, such that L is leftward and R is rightward. Suppose a set of inertial observers at that central point with various relative velocities. Note that L and R are instantaneous events and can therefore not have any such thing as velocity. Suppose for now that the speed of light is the same in all directions.

Note that there is no divine external reference frame that is really truly stationary, relative to which any of our inertial observers can be said to be moving at a particular objective velocity. We can pick any of our inertial observers, or invent any other inertial observer, and simply declare that observer to be our stationary reference frame.

Call our observers A, B, and C and suppose that they are traveling at different parallel velocities along the line between L and R. Let us arbitrarily declare that A is stationary, such that B is moving relative to A at velocity V, C is moving relative to A at velocity -V. Therefore C is moving relative to B at velocity -2V and B is moving relative to C at velocity 2V.

Note that we can now set B stationary by subtracting V from everything or C stationary by adding V to everything.

Suppose that, a short time after we start our experiment running, A observes L and R simultaneous.

In that case, because B is moving towards R, B observes R then L. Because C is moving towards L, C observes L then R.

Now add V to everything.

C is now our arbitrarily stationary frame with respect to which everything else is moving. However, the laws of physics haven't changed and the speed of light is the same in all reference frames, so the events L and R which were equidistant from the central point weren't propagating towards C from L at c+V and towards C from R at c-V, they were propagating towards C at c.

C still observes L then R, because for C, L really happened before R.

Likewise B still observes R then L and A still observes R and L simultaneous despite now being one of the "moving" frames.

Subtract V to get back to where we started and subtract V again. Now B is stationary. The laws of physics haven't changed and the speed of light is the same in all reference frames. Signals were never propagating towards B from L at c-V and towards B from R at c+V, they were propagating towards B at c.

So again, B sees R then L, A sees R and L simultaneous despite now being one of the "moving" frames, and C still sees L then R.

From this we can see that there is no divine observer frame that is really truly stationary and can therefore make an objective judgment about what is stationary, nor in which order events that are close together in time but distant in space really took place. Rather, for A, who is just as stationary as anyone, L and R are simultaneous; for B, who is just as stationary as anyone, R does precede L; and for C, who is just as stationary as anyone, L does precede R. For B and C, A just happens to be moving at just the right velocity to see the illusion of simultaneity.

However, they will all agree on the same order in which the events must have been observed, regardless of which one of them is stationary.

They all agree that first, C saw L and B saw R. Then A saw both. Then C saw R and B saw L.

Now suppose that the speed of light is different in the different directions by a fixed velocity K, our observers are traveling at the same velocities relative to one another, and A still saw L and R at the same time. We started out by treating the speed of light as being the same in both directions, and each signal is moving on just one direction towards our observers. (L's signal is always moving right, R's signal is always moving left.)

This is identical to adding K to the velocity of all three observers. But we already know what happens when we add a velocity to all three observers: they all still see the exact same things. If we picked the case K=V or K=-V, we could even keep two out of the three of our configuration diagrams. It doesn't matter what we set K to, as long as A still sees L and R simultaneous, everyone always sees the same things as in the case K=0.

Thus, the answer to "Is the speed of light different in different directions by a fixed velocity K?" is that the question is meaningless, since all possible permutations result in the same observables.

Now suppose that the speed of light is different in the different directions by a variable velocity J such that J is different for every observer and equals the opposite of their velocity plus some constant (possibly 0).

Then in every frame, all three observers must see the events as simultaneous. But since they are moving relative to one another, they all now need to disagree on in which order everyone else observed them - and in fact, since the light in their reference frame bounces off of other observers at whatever speed it's moving, they can watch everyone else observe them in different order. In A's frame the events are simultaneous and B moving towards R, so B's right side is illuminated before B's left side, and vice versa for C. But in B's frame the events are simultaneous and A is moving towards L, A's left side is illuminated before A's right side. So on for each combination of observers.

This is a logical impossibility, so the answer to "Is the speed of light different in different directions by a by a variable velocity J such that J is different for every observer and equals the opposite of their velocity plus some constant?" is emphatically no.

Suppose all of our observers have little photosensitive diodes attached to bombs and they explode if they get a flash from the right unless they got a flash from the left already. Let's shift everybody just a tiny bit to the left so that A sees L just a tiny bit before A sees R.

A sees L just a tiny bit before R, so it is fine. In A's frame, light reaches B from R before light from L, so in A's frame, B explodes. In A's frame, light from L reaches C before light from R, so C is fine.

B, on the other hand, has had the speed of light from R reduced and the speed of light from L increased by exactly the speed at which B is moving relative to A. As a result, B sees L just a tiny bit before R, so B does not explode.

B explodes and not-explodes.

Explodes and not-explodes is a contradiction, therefore the premise that led to the contradiction is false.

g s
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    Re: fixed velocity difference case: velocity is not speed. It has direction, so the velocity of light from L is K+c and the velocity of light from R is K-c. The speed is the absolute value. – g s Mar 04 '21 at 22:33
  • Can I confirm that you are adding V to everything, except the speed of light for L & R? This is because "the speed of light is the same in all reference frames"? e.g. all frames will measure it as c anyway? – Quantumplate Mar 05 '21 at 02:36
  • Yes, when I added and subtracted velocities, I was changing the velocities of A, B, and C, not the velocity of the light. – g s Mar 05 '21 at 04:21
  • I added a short addendum further explaining the reason why the answer to the meaningful question (the variable velocity difference case) is "no". – g s Mar 05 '21 at 05:01
  • When you add V to everything, if there is a (vertical) light clock in frame C are you adding V to the photons in the clock so that they stay vertically aligned with the clock? If so is there a reason you add V to some light (that in the frame) but not other light (that outside it)? If not, won't the photons continue left and leave the clock? – Quantumplate Mar 05 '21 at 06:26
  • If we take V as .9c, then add V to everything, this will give .9c for A, 1.8c for B and 0 for C. Is this A's perspective of frame C (as the 1.8c makes sense in that regard, but doesn't if it's C's perspective)? – Quantumplate Mar 06 '21 at 19:29
  • So I think I agree with the "fixed velocity K" outcome (altho still some clarifications about how we got there). The observation that its meaningless applies to people in the frame trying to calculate things. For me, I am a programmer and wish to understand the rules so I can reproduce them (in code), so this has some meaning for me. – Quantumplate Mar 06 '21 at 19:35
  • If we use a relativistic velocity for V, we have to add it relativisticly as $V+V = 2V / (1+ V^2/c^2)$. Importing the mathematical conclusions of relativity into our mathless argument for why we can trust the mathematical conclusions of relativity is probably a bad idea, since if we're going to do that anyway we could just skip right to the end and say $c+V = (c+V)/(1+V/c) = c(c+V)/(c+V)=c$, the speed of light is invariant with velocity, job done, nothing to see here. – g s Mar 07 '21 at 03:48
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Your point #3 asserting that the speed of light is different in different directions is based on an assumption, that speeds add in a particular way. But not all quantities in nature add up simply. If you take two raindrops of radius X and Y and merge them, the resulting drop does not have radius X+Y. So the question is, is speed like radius (not additive) or like volume (additive). Only experiments can tell. Experiments show that in fact relative speeds do not add simply.

Eric Smith
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  • Thank you for your answer, #3 is from the stationary observer's point of view, they are noting that light is traveling faster relative to the center of the train. The stationary observer notes the light travels at same speed in both directions (relative to them), but faster in the backwards direction (relative to the center of the train (from the stationary observers point of view)). – Quantumplate Mar 07 '21 at 20:10
  • No, in practice all the observers will agree that the speed of light is the same (the constant value c as predicted by Maxwell's equations). Velocities do not combine by addition, they combine in such a way that speed v combined with speed c produces speed c. This is the velocity addition formula https://en.wikipedia.org/wiki/Velocity-addition_formula. – Eric Smith Mar 07 '21 at 23:46
  • In your next reply you talk about a "true speed" of light being different in different directions but any attempt to measure it gives c. But if we cannot measure this "true speed" then what's the point of it? This was Einstein's insight: if something cannot even in principle be measured, we might as well dispense with it and work with what we can measure. – Eric Smith Mar 07 '21 at 23:49
  • So the point for me I guess, is that it feels like "light being measured as c in all frames" will fall out of some more fundamental rules (which I can maybe program) while I can't think of a way to program it as a fundamental rule. These discussions and different points of view help tho. If you look at the analogy in my self-answer, if I coded this, the programmer could freeze the program, get out a ruler and measure the distance and see the ship was not centered in the wave. The possibility that this absolute off-center could be interpreted from within the frame as centered intrigues me. – Quantumplate Mar 08 '21 at 02:52
  • No, I don't think it will fall out of a more fundamental rule. In classical physics the one way speed of light can be measured. The whole point of the Michelson-Morley experiment was that they tried to measure differences in the one way speed of light in different directions and failed to do so -- that was one of the major impetuses for the development of relativity. In your self-answer analogy the "water" serves the same role as the "aether" of Michelson and Morley, and people on the boats could detect the motion of the boats through the water. – Eric Smith Mar 09 '21 at 00:37
  • You may be confused by recent videos and discussions about how it is impossible to measure the one way speed of light. That's because to measure the one-way speed of light requires setting up synchronized clocks in different places, and that in turn requires assumptions about the one way speed of light being the same in all directions or (equivalently) that the time dilation factor is independent of direction so that you can carry the clocks to their final destinations and keep them in sync. – Eric Smith Mar 09 '21 at 00:51
  • The water in the self answer isn't like ether, it doesn't produce a "wind" or "friction" on anything. It's solely there to show the separation of light from an inertial frame, if you like you can remove it and instead of a wave have a mathematically expanding circle (this is what I will do in the program). – Quantumplate Mar 09 '21 at 09:55
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So I'm going to create an (incomplete) analogy to highlight my thinking, that the (one-way) speed of light can be different in different directions (relative to an inertial frame of reference) but any attempt to measure that will still give what we observe (e.g. c).

The purpose of this (for me) is to try identify the simplest set of programmable rules that reproduces what we observe in Relativity.

The Analogy

So I'm thinking of a 2m x 2m pool of water (to represent space-time).

I have some metal toys (say a spaceship, a train, anything used in Relativity thought experiments).

I can pick up a toy and move it over the water (let's assume I have a super steady hand and can maintain a constant speed and direction without shaking). This represents an inertial frame of reference. I will limit the speed of the toy to less than c (defined in next point).

I also have a pile of small metal balls which I can hold under the toy and drop into the water. We assume the speed of the ball and distance to the water is negligible such that the ball creates a ripple directly below where it was dropped. This is a single ripple which propagates outwards at speed c regardless of the speed of the frame of reference. Dropping a ball represents a light emitter and the wave represents a light wave.

For sake of simplicity we say we can detect a light wave (from one of our frames of reference) and when we do the wave disappears (maybe we freeze our movement until the wave dissipates?) and we infer that a photon traveled in a straight line from the center of the ripple to the point on the wave where it was detected (keeping in mind this is a rough analogy). This will represent a light detector.

We also need some way to represent mirrors so the toy (frame of reference) will be able to drop a barrier into the water that has the (magical) properties that as it moves through the water it does not displace any water but if a wave hits it then it reflects the wave. If it helps you can imagine a barrier just above the water which the wave will hit but won't touch the flat water. Both depictions have issues but for the sake of the analogy lets go with it and see where it leads us.

Observations

  1. So we can see straight away that the speed of the wave (the speed of light) is independent of the toy moving over it (the frame of reference). No matter the direction or speed of our toy, the wave will always propagate at c.

  2. If we place a toy at the center (a person standing) to act as our stationary frame of reference then they will always measure c as c, so far so good.

  3. If we now take a spaceship toy and move it towards the top left corner and drop a metal ball as we go we see that the spaceship is closer to the wave in the direction of travel and further away from the opposite side. When I say we, I'm saying us from our Godlike view above the pool.

  4. The stationary person in the center can see that the waves have different speeds for different directions relative to the spaceship. They conclude that c will have a different value on the spaceship in different directions.

  5. The spaceship thinks this is weird and decides to test this. They lower a mirror into the water at the front of the spaceship then drop a ball off the back where there is a also a light detector (essentially the wave is traveling from the back of the spaceship to the front, bouncing off a mirror then back to the detector at the back).

  6. The stationary person calculates that the wave and ship are traveling in the same direction so it will take a while to hit the mirror and reflect back, after that the wave is traveling in the opposite direction to the ship and it will be much quicker to do the return trip.

  7. The stationary observer concludes that the speed of the wave, relative to the ship, differs in both directions, but the time taken for the round trip is equal to that of a wave traveling the same trip at speed c. e.g. the two way measurement cancels out the difference in one-way speeds.

And this is as far as I have got. I need to verify if Step 7 is correct (it may be that I need to include time-dilation and calculate c from the perspective of the ship instead). I can probably just fake time dilation with clocks adjusted relative to the frame of references speed (relative to the stationary frame). The reason to fake it is that programming in matter would be a lot of work and the time dilation is better understood (e.g. as motion relative to a frames speed being limited by the speed of light).

Conclusions

  • I am hoping that if I keep going with this I can introduce time dilation and show that no matter the frame or perspective, they will always measure the speed of the wave as c regardless of whether it is actually that relative to the frame.

  • It also shows that by decoupling the speed of the wave (the speed of light) from the frame of reference we very intuitively get that the speed of the wave can never be anything but c and less intuitively (or still to be proven), any measurement of c from a moving frame (regardless of their speed or direction of measurement) will still measure c relative to them.

  • It seems that by taking the opposite approach (e.g. that of someone in the frame) it makes it easier to understand the physics in the frame but harder to derive the initial rules of the system (what I'm trying to do).

So we return to my question (for the purpose of creating a programmable set of rules).

Is there a flaw in assuming the one-way speed of light can differ (even tho we can't measure it)?

Quantumplate
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