I The Einstein Clock aka Light Clock

[DmitryS]

TL;DR

apparent problem of Einstein clock

Hello, everyone, hope someone will resolve my doubts.
I have posted here some two years ago asking for an explanation of the Lorentz transforms derivation found in the Einstein 1905 paper. The answer I got seemed quite satisfactory. Two years after I revisit this derivation and this is what I see.

In the Einstein original paper, the Lorentz transforms derivation included as a premise that light is always propagated along the direction perpendicular to the line of motion when viewed from the stationary system with the velocity sqrt(c^2 − v^2). This expression can be considered the first mentioning of what was to become the model of the so-called light clock, where the light travelling vertically for one observer is moving at the angle arccos (v/c) for the other.

There is also a different derivation of the Lorentz transforms with the help of the spherical flash of light originating from the origins of the two inertial frames of reference when those coincide. The transforms are derived based on the assumption that the expression for the sphere of light must be identical for either observer.

I summarized my doubts as follows.

1. The derivation of the Lorentz transforms through the invariance of the light sphere suggests that postulating constancy of the speed of light should be equal to: ‘Any point of space where light originates is considered motionless relative to the inertial observer’. That is to say, if a spaceship generates a pulse spherical light wave at a certain point, the ship will continue moving staying at its centre, while the on-land observer will see the spaceship moving through this sphere of light while seeing the centre of the sphere static (whence relative simultaneity). Thus, the two spheres, one with the centre at O, the other at O’, are identical in every way. Those rays that go up will go up, and those that are in line with motion will be in line with motion. The flux of rays over the entire surface of the sphere will be the same for both observers.

2. On the other hand, the light clock model suggests an entirely different interpretation of the speed of light’s constancy, namely: ‘Any photon will have the absolute value of the speed equal to the speed of light relative to an inertial observer along the trajectory resulting from the composition of motions of that photon and of the observer moving relative to any other inertial observer’. In short, it suggests that the ray going straight up for Observer O’ and the ray going at the angle arccos(v/c) to the line of motion for Observer O are the same physical object.

3. And that means (if we marry this model to the one with the two spheres of light) that the flux of rays limited by the cone with the angle 2arccos(v/c) with the axis coinciding with the line of motion for Observer O must be the same as the flux of the hemisphere towards the same direction for Observer O’. The angle of 2arccos(v/c) will open up and become 2π in the O’ system. If we consider the flux to be constant over the sphere for Observer O, it won’t be constant for Observer O’: the hemisphere on the right will be not so dense with rays as the hemisphere on the left. That, of course, will mean that there’s an objective difference between the two observers and the relativity principle does not hold.

4. The least we can say is that these two models can’t go together. One of them should be false, and since the light clock model violates the relativity principle, the invariance of the two spheres must be correct.

This is it in a nutshell. I never saw this problem mentioned in literature. Possibly someone here has an answer.

 

[PAllen]

I only have time for a quick reply now, just focusing on some issues with your first bullet:

1. The derivation of the Lorentz transforms through the invariance of the light sphere suggests that postulating constancy of the speed of light should be equal to: ‘Any point of space where light originates is considered motionless relative to the inertial observer’. That is to say, if a spaceship generates a pulse spherical light wave at a certain point, the ship will continue moving staying at its centre

.. in the ship's inertial frame (assuming the ship moving inertially)

, while the on-land observer will see the spaceship moving through this sphere of light while seeing the centre of the sphere static (whence relative simultaneity). Thus, the two spheres, one with the centre at O, the other at O’, are identical in every way

No. Though each a circle in their respective frame, each circle represents a different set of events.

Those rays that go up will go up, and those that are in line with motion will be in line with motion. The flux of rays over the entire surface of the sphere will be the same for both observers.

This is fundamentally wrong. Look up aberration of light rays. If the flux is even per angle in one frame, it will not be even in the other frame. The angle of all light rays other than directly forward or back will be different between the two frames.

 

[Dale]

TL;DR: apparent problem of Einstein clock

Thus, the two spheres, one with the centre at O, the other at O’, are identical in every way. Those rays that go up will go up

This is not correct and it is not an implication of the light cone (sphere).

The light cone says that if there are rays in every direction in one frame (a sphere) then there will also be rays in every direction in the other (also a sphere). It does not map specific rays in one frame to another. That is done with the Lorentz transform.

 

[Ibix]

Any point of space where light originates is considered motionless relative to the inertial observe

Any point is stationary relative to the observer - this is effectively the definition of a point in this context.

Thus, the two spheres, one with the centre at O, the other at O’, are identical in every way. T

No they aren't. They are different slices through the light cone, and consequently have different angular distributions of light. To see this, draw a Minkowski diagram in x-t coordinates showing a light cone, a plane of constant t, and a plane of constant t'. Since the latter is slanted in this coordinate system you will see that one end of the primed coordinates' sphere is further from the origin than the other in the unprimed coordinates. Thus if the light intensity is isotropic in one set it must be anisotropic in the other.

On the other hand, the light clock model suggests an entirely different interpretation of the speed of light’s constancy, namely: ‘Any photon will have the absolute value of the speed equal to the speed of light relative to an inertial observer along the trajectory resulting from the composition of motions of that photon and of the observer moving relative to any other inertial observer’.

This isn't different. You will find that the position of any pulse of light lies on a sphere centered on its emission point.

the flux of rays limited by the cone with the angle 2arccos(v/c) with the axis coinciding with the line of motion for Observer O must be the same as the flux of the hemisphere towards the same direction for Observer O’.

Without checking your maths, yes, you seem to be describing relativistic aberration.

That, of course, will mean that there’s an objective difference between the two observers and the relativity principle does not hold.

No. The principle of relativity doesn't say all observables are the same, just that the laws of physics have the same form. Measuring a different distribution of light does not require different laws of physics - in fact, you just used those laws to derive different intensities.

 

[Sagittarius A-Star]

The angle of 2arccos(v/c) will open up and become 2π in the O’ system.

Do you mean the case $v=0$? You get $2\arccos{} 0= \pi$, but not $2\pi$.

It seems that you want to describe the headlight effect.
https://demonstrations.wolfram.com/TheHeadlightEffect/

 

[FactChecker]

That is to say, if a spaceship generates a pulse spherical light wave at a certain point, the ship will continue moving staying at its centre, while the on-land observer will see the spaceship moving through this sphere of light while seeing the centre of the sphere static (whence relative simultaneity). Thus, the two spheres, one with the centre at O, the other at O’, are identical in every way.

If they record identical measurements of position and time for the sphere of light, but their measurement systems of position and time are not identical, would you call the two spheres "identical"?
That is the situation. Both are equally valid, but their systems of measurement are not the same.

 

[DmitryS]

Friends, thank you for all the inputs, because they allow me to make my position clearer.
I don't think the aberration will apply here. As Ibix said, "Any point is stationary relative to the observer - this is effectively the definition of a point in this context". Right, so, wherever the flash is, we consider its source motionless. How do we observe aberration from a motionless source?

To put an end to the assumption that aberration cures the matter, let's forget about the spaceship. Let's consider only the flash coming from the same point of space at the moment of coinciding O and O'. All of you said that the flux will not be the same for both observers.

I don't think it would be right to suggest that a mirror placed somewhere ahead the motion of O' in the line of motion will reflect more light for Observer O than for Observer O'.

My question is not about small details. I expect there's something graver going on here.

 

[Ibix]

"Any point is stationary relative to the observer - this is effectively the definition of a point in this context". Right, so, wherever the flash is, we consider its source motionless. How do we observe aberration from a motionless source?

You don't. But you can have a moving source, one which is not always at the same point.

I don't think it would be right to suggest that a mirror placed somewhere ahead the motion of O' in the line of motion will reflect more light for Observer O than for Observer O'.

The mirror will reflect the same fraction of the emitted light according to both observers (edit: hmm, not sure, see my next post). But they will have different measures of the intensity distribution and different measures of the distance from emission at which reflection occurs, and hence different measures of the angular subtense of the mirror. The combination of different intensity distribution and different angular subtense of the mirror will lead to the fraction of emitted light striking the mirror being invariant.

My question is not about small details. I expect there's something graver going on here.

It is entirely about understanding aberration and the "headlight effect" mentioned by @Sagittarius A-Star, I'm afraid.

 

[Dale]

I don't think the aberration will apply here.

Aberration always applies. Aberration is a disagreement between two different reference frames about the direction of light. If you have different frames and each frame is describing a direction of travel for light, then there is aberration.

How do we observe aberration from a motionless source?

By having different frames describe waves moving away from the source. It isn’t about the source, it is about the waves produced by the source.

Consider a phased array source where the phases are set such that the emitted light is isotropic in a frame where the source is moving. The aberration will be with respect to that frame, the frame where the waves are isotropic, not the frame of the source.

Similarly here. If your single-event source emits light that is isotropic in one frame then it will be anisotropic in other frames due to aberration. It is not avoidable.

My question is not about small details.

Well, your question is about relativistic aberration. So I guess that means aberration is not a small detail. I agree.

 

[Ibix]

The mirror will reflect the same fraction of the emitted light according to both observers

On reflection (sorry) I'm not sure this is correct, and need to do a bit of maths.

It is definitely true that if the source emits a finite number of rays (perhaps it's a lamp surrounded by a wire mesh) then the number of rays reflecting off the mirror is an invariant, although the frames will not agree about the distribution of the rays. However, the total energy emitted by the source is frame dependant, as is the distribution of that energy between the rays. The amount of energy striking and reflected from the mirror is also frame dependant. So, although I am sure that the fraction of rays striking the mirror is invariant, and I'm sure that the energy carried to it is not invariant, I'm not sure about the fraction of the energy emitted that strikes the mirror.

The maths looks messier than I can do without pen and paper (on the train at the moment, and no table) and a cursory search didn't turn up a textbook formula for the energy distribution. I suspect one can combine my ray model with each ray's energy-momentum four vector to figure things out, but I also suspect there's a nasty integral in there.

However, I already mentioned a Minkowski diagram, which the @DmitryS (and anyone else) should really draw any time he thinks he's found a paradox, because they're great at showing you where your misunderstanding was. Here's a light cone in yellow:

Time is on the vertical axis, as usual, with horizontal spatial axes. The emission event is at the origin. Now we can add a blue plane showing what this frame calls "now" a short while after the emission:

Taking a slightly higher perspective we can see that the intersection of the light cone and "now" is a circle:

Assuming that the light intensity is isotropic in this frame then it is clear that the light will have the same intensity at any angle because it's the same distance from the source.

But we can instead draw a plane of constant t' (red):

Notice that the intersection is elliptical, slanted, and its center is offset from the z axis. This shows you immediately that the light cannot be isotropic in the primed frame, because you can see that some parts of the ellipse are (in the unprimed frame where light is isotropic) closer to the emission event's location than others, and will therefore see brighter light by the inverse square law.

The relativity of simultaneity strikes again.

 

[cianfa72]

Assuming that the light is isotropic in this frame then it is clear that the light will have the same intensity at any angle because it's the same distance from the source.

Why assume the isotropy of light in this frame (i.e. in the frame where the horizontal blue plane has equation t= const). The light cone (yellow) is invariant and in that frame is isotropic, isn't it ?

 

[Ibix]

The light cone (yellow) is invariant and in that frame is isotropic, isn't it ?

Only if I specify it to be so. You can draw the same lightcone and planes in any frame. I want to do it in the frame where the light intensity is isotropic so that I can conclude it is not osotropic in other frames without explicitly doing any maths.

 

[Dale]

The light cone (yellow) is invariant and in that frame is isotropic, isn't it ?

Consider a dipole antenna that emits a brief pulse. The emitted radiation is not isotropic in any frame. But the shape of the wavefront is spherical in every frame.

 

[cianfa72]

Consider a dipole antenna that emits a brief pulse. The emitted radiation is not isotropic in any frame. But the shape of the wavefront is spherical in every frame.

Yes sorry, only later I realized the point was about isotropy of light intensity and not light propagation process (speed).

 

[PeterDonis]

wherever the flash is, we consider its source motionless.

No, you don't. The light source is not a point. It's a worldline--the path through spacetime of an object (whatever object emits the light). The event of emission of the light takes place at a point on that worldline, but the motion of the source is the motion of the worldline--which is different in every frame. There is only one frame in which the source is motionless.

 

[Dale]

As Ibix said, "Any point is stationary relative to the observer - this is effectively the definition of a point in this context". Right, so, wherever the flash is, we consider its source motionless. How do we observe aberration from a motionless source?

Also, note that @Ibix said that the point is motionless, not that the source is motionless when it is at that point.

 

[DmitryS]

That's what I call a discussion.

Right, I want to summarize now what I see, because I can't really reply to all the minor things that matter in all your comments, but I thank you all anyway.

1. Let's return to the very beginning. What matters here, as I think, is the relativity principle, which in my mind does not reduce to simply the invariance of the physical laws. Still, the foundation of it is that we cannot distinguish between the situation of any inertial observer, i.e. we must be unable to say which one 'really' has moved.

That's why I prefer to go on an assumption (withot insisting it be correct for now) that the spherical flash of light taking place at the origins of the coordinate systems the moment they coincide must be the same for both Observer O and Observer O' as they both perceive it as a flash coming from a motionless source in their respective frames. If one should insist that this is not so, that the uniform flux of rays for Observer O shouldn't be such for Observer O', we could always remind them that the situation is symmetrical, or otherwise there is an observer in absolute motion and an observer at absolute rest. Each observer has had that flash in their frame considering its source motionless, and we might as well start the argument from the point of view of Observer O'. The outcome of this argument must not depend on the starting point (that's another possible formulation of the relativity principle). If we start it from the point of view of Observer O', we'll get the thinning of flux on the opposite side, and I don't know how to proceed from here.

The arguments mentioning aberration imply continuous radiation, but what we deal with here is a singular momentary burst of photons at the same moment of time for both observers, and it will be very difficult to substantiate the divergence of their trajectories in either frame, unless we postulate - not quite clear on what grounds - that for one observer the quality of flux was this, ergo, for the other one it should be that (again, dependence on the starting point).

I'm saying it again I will not insist that the above is absolutely correct. My purpose for now is that the participants see my problem as I see it.

2. I'm not quite convinced about the outcomes of the mirror thought experiment. I understand @Ibix couldn't go into detail for the lack of convenience and paper, but I guess I must insist we talk specifically about the flux of photons, physical objects, and not the characteristics of radiation, since we deal with a limited number of oscillators and at the end of the day it's all that matters. The number of photons flying back past Observer O' must be the same as that flying back past Observer O.

3. For the Minkowski diagram, let me start from afar. Not so long ago I watched a YouTube video where the presenter claimed the invariance of the Einsteinian equations (ct)^2 = x^2 + y^2 + z^2 and (ct')^2 = x'^2 + y'^2 + z'^2 was all wrong, and to prove his point he showed some drawings plotted by maths software where there were ellipses not circles. I knew he was wrong. I saw his mistake immediately, because some time before I had fooled around with a maths soft myself and made it draw ellipses from the Lorentz transforms. This mistake was representing not the view of the moving observer, but instead the view of the moving observer as seen by the stationary observer.

I'm afraid you can't prove anything by slicing the light cone belonging to the one observer by the plane perpendicular to a diffent time axis. The equations (ct)^2 = x^2 + y^2 + z^2 and (ct')^2 = x'^2 + y'^2 + z'^2 are not different slices of the same light cone - they are two different light cones. Indeed, those are equations of 4-cone with ct or ct' for the axis, and the sphere of light is its 3D projection. So slicing the light cone with ct for the axis by the plane orthogonal to ct' is the error of the same nature as the afore-mentioned YouTuber made, representation of the view of the moving observer not as it is but as seen by the stationary observer.

@Ibix wrote: "You can draw the same lightcone and planes in any frame. I want to do it in the frame where the light intensity is isotropic so that I can conclude it is not osotropic in other frames without explicitly doing any maths". - I'm afraid not. You draw a non-synchronized plane, and that testifies to nothing but to the view of O' by O, not the view by O' themselves.

Before we go any further and speak about what aberration in my mind really has to do with all that (a somewhat complicated topic), we need to agree that there is a problem with the light clock.

 

[Sagittarius A-Star]

Before we go any further and speak about what aberration in my mind really has to do with all that (a somewhat complicated topic), we need to agree that there is a problem with the light clock.

The explanation of aberration in the light-clock scenario does not only work with classical particles, but also with EM waves.

The reason is relativity of simultaneity.

Assume, a horizontal electromagnetic wavefront segment of vertically moving light crosses the horizontal x-axis. In the primed rest frame of the light clock, the left and right side of the wavefront segment cross the x'-axis simultaneously, that means
$\Delta t' = 0$.

With inverse LT and length contraction follows with reference to the unprimed frame, that the left and right side of the wavefront cross the x-axis with the following time difference:

$\Delta t = \gamma (\Delta t' + {v \over c^2} \Delta x') = \gamma (0 + {v \over c^2} \Delta x') = {v \over c^2}\Delta x$.

For small angles: When the right side of the wavefront crosses the x-axis, the left side has already moved further across the x-axis by

$\Delta d \approx c \Delta t = c {v \over c^2}\Delta x = {v \over c} \Delta x$.

That means, the small wavefront segment is tilted by

$\Delta d / \Delta x \approx v/c$.

 

[PeterDonis]

What matters here, as I think, is the relativity principle, which in my mind does not reduce to simply the invariance of the physical laws.

What else would be involved?

they both perceive it as a flash coming from a motionless source in their respective frames.

No, they don't. The source can only be motionless with respect to one of them.

Until you fix this fundamental error, everything else you think will be wrong.

 

[PAllen]

That's what I call a discussion.

Right, I want to summarize now what I see, because I can't really reply to all the minor things that matter in all your comments, but I thank you all anyway.

1. Let's return to the very beginning. What matters here, as I think, is the relativity principle, which in my mind does not reduce to simply the invariance of the physical laws. Still, the foundation of it is that we cannot distinguish between the situation of any inertial observer, i.e. we must be unable to say which one 'really' has moved.

That's why I prefer to go on an assumption (withot insisting it be correct for now) that the spherical flash of light taking place at the origins of the coordinate systems the moment they coincide must be the same for both Observer O and Observer O' as they both perceive it as a flash coming from a motionless source in their respective frames. If one should insist that this is not so, that the uniform flux of rays for Observer O shouldn't be such for Observer O', we could always remind them that the situation is symmetrical, or otherwise there is an observer in absolute motion and an observer at absolute rest.

Let's stop right here. We will not get far by insisting that "we all must agree that 2 + 2 = 5". A given flash bulb (for example) has one world line, and at any point in its history, it is at rest in the origin of one family of frames connected by rotation. In all other frames, it is in motion. In a frame where it is at rest, its emission is isotropic in frequency spectrum and intensity (for an ideal flash bulb). In all other frames, its emission is affected by Doppler and aberration and is very much NOT isotropic in either frequency spectrum or intensity. Let's agree on these facts before proceeding.

 

[PeroK]

Friends, thank you for all the inputs, because they allow me to make my position clearer.
I don't think the aberration will apply here. As Ibix said, "Any point is stationary relative to the observer - this is effectively the definition of a point in this context". Right, so, wherever the flash is, we consider its source motionless. How do we observe aberration from a motionless source?

To put an end to the assumption that aberration cures the matter, let's forget about the spaceship. Let's consider only the flash coming from the same point of space at the moment of coinciding O and O'. All of you said that the flux will not be the same for both observers.

I don't think it would be right to suggest that a mirror placed somewhere ahead the motion of O' in the line of motion will reflect more light for Observer O than for Observer O'.

My question is not about small details. I expect there's something graver going on here.

The light clock is really very simple. If you accept the invariance of the speed of light, then time dilation, length contraction and the relativity of simultaneity follow from a few simple thought experiments.

That said, any simple matter can be unnecessarily complicated by confused thinking.

 

[PAllen]

Let's stop right here. We will not get far by insisting that "we all must agree that 2 + 2 = 5". A given flash bulb (for example) has one world line, and at any point in its history, it is at rest in the origin of one family of frames connected by rotation. In all other frames, it is in motion. In a frame where it is at rest, its emission is isotropic in frequency spectrum and intensity (for an ideal flash bulb). In all other frames, its emission is affected by Doppler and aberration and is very much NOT isotropic in either frequency spectrum or intensity. Let's agree on these facts before proceeding.

Maybe it would help to imagine two flash bulbs, one in motion relative to the other. They both flash when one just passes the other. In all frames, the light from the two flashes is different, even though the flash bulbs are each identical in behavior in their rest frames. And the difference is exactly Doppler and aberration.