Lightwatch/Lightclock (Lichtuhr)

[Martin68]

TL;DR Summary

Why does the light follow the clock?

Hi all,
a maybe silly question: In the SRT-experiment with the lightwatch (which is taken to show the time dilation for a moving clock), why does the light follow the clock? Light is indepedent of the emitting source. So I would expect that the clock is moving away and the light is moving straight away and could never hit the mirror (because it was moved away).
An example for that experiment is shown in http://www.einstein-online.info/vertiefung/LichtuhrZeitdilatation@set_language=de.html
(sorry, a German site)
Best regards
Martin

 

[A.T.]

In the SRT-experiment with the lightwatch (which is taken to show the time dilation for a moving clock), why does the light follow the clock?

If the light hits the mirror in the rest frame of the clock, then it must hit the mirror in every other frame.

 

[Ibix]

Welcome to PF.

The speed of light is independent of its source, true, but the direction it travels isn't. If that weren't the case, you'd end up with inconsistent results where using different frames (one where the source is moving and one where it isn't) disagree about whether a light pulse strikes a target, like the mirror.

 

[Martin68]

Hi A.T. and Ibix,
many thanks for the fast answers ... and for the moment it is clear to me. I will have to think about that, maybe there are more questions :#)
Best regards
Martin

 

[MikeeMiracle]

I think the simple answer is that this is just an illustration to help you visualise what is occurring, i.e that the light takes a longer path when the clock is moving before it hits the other plate than if it was stationary. i.e it appears to have slowed down.

 

[Nugatory]

So I would expect that the clock is moving away and the light is moving straight away and could never hit the mirror (because it was moved away).

It may make more sense if you think about the path the light is taking through space. Imagine that the light clock is moving through a cloud of dust. As the flash of light bounces back and forth between the mirrors, it will momentarily illuminate the dust particles.

Someone at rest relative to the clock will say that light is bouncing straight up and down; someone at rest relative to the dust will say that the light is following a zigzag path through space. But - crucially - the same dust particles are illuminated by the beam in both descriptions. That is, the path of the light is the same in both descriptions, no matter what the mirrors are doing.

If the dust particles are photosensitive (little grains of silver nitrate?) they will be permanently changed by exposure to light, and we can use this to determine the path of the light after the experiment is done. In both descriptions these light-affected particles will form a zigzag line through the cloud. Someone at rest relative to the clock wil say that the zigzag line is because the dust particles were moving sideways while the light moved straight up and down; someone at rest relative to the dust will say that the zigzag line is because the dust was not moving while the light was moving sideways as well as up and down. These apparently contradictory explanations are actually just two different ways of saying the same thing: the clock is moving relative to the dust.

 

[PeroK]

Summary: Why does the light follow the clock?

Hi all,
a maybe silly question: In the SRT-experiment with the lightwatch (which is taken to show the time dilation for a moving clock), why does the light follow the clock? Light is indepedent of the emitting source. So I would expect that the clock is moving away and the light is moving straight away and could never hit the mirror (because it was moved away).
An example for that experiment is shown in http://www.einstein-online.info/vertiefung/LichtuhrZeitdilatation@set_language=de.html
(sorry, a German site)
Best regards
Martin

Here's yet another way to look at it. Suppose you are playing with your new Lichtuhr. It's all looking good. The light is bouncing up and down between the mirrors.

But, then, a naughty physicist imagines a Gedankenexperiment, where you and your light clock are moving at nearly the speed of light in his reference frame.

Suddenly, your new Lichtuhr stops working. The light starts missing the mirror and will no longer bounce backwards and forwards. You can't figure it out.

After a few minutes, the physicist stops thinking of your Lichtuhr in the other reference frame and it starts to work again. Now that no one is imagining your Lichtuhr traveling at relativistic speed, the light bounces back and forwards "normally" again.

Personally, I don't think that the naughty physicist's trick would work. In other words, you can't change the physical reality of the impacts with the mirrors by considering, or observing the clock in a different frame of reference.

 

[timmdeeg]

So I would expect that the clock is moving away and the light is moving straight away and could never hit the mirror (because it was moved away).

The second animation shows the lower clock moving relativ to the upper clock and thereby the light bouncing up and down vertically between the two mirrors. That's because in the frame of the lower clock relativ to which this clock is at rest the mirror will be hit as shown. And this fact doesn't change from the perspective of the upper clock. Viewed from this perspective the mirror of the lower clock must be hit too but the light travel distance is larger as shown in the lower diagram.

 

[Martin68]

Many thanks for the further explanations
If we take as given that the sent out photons are taking over the directional vector of the moving clock, then everything is plausible. But the question still exists: WHY does the light is taking the directional vector?
Is there also a physical explanation for that, or may I not ask this question because it is only a model for the reality?
Is the sent out photon bound to the emitting source? What would happen, If we would hold the clock at one position after the photon was sent out? Would the photon move ahead in the diagonal direction, or would it also stop the side-movement?
Best regards
Martin

 

[PeterDonis]

the question still exists: WHY does the light is taking the directional vector?

Because that's how the clock is constructed: the light source points from one side of the clock to the other.

Describing the clock as "moving" is a misstatement; you choose a frame in which it is moving, or a frame in which it is at rest. Physically, the parts of the clock are all at rest relative to each other, so the light source just points from one side to the other of the clock in the obvious way. That's the only physical thing that is happening with the clock. What happens in a frame in which the clock is moving has nothing whatever to do with the clock: it only has to do with the frame.

What would happen, If we would hold the clock at one position after the photon was sent out?

Then you would be changing the physics of the clock: you would be changing the motion of one part (the mirror the light source is aimed at) after another part has operated (the light source has emitted its photon). But then it's no longer a light clock, because the definition of a light clock requires that all of its parts are at rest relative to each other and that no external influence operates on them.

 

[A.T.]

If we take as given that the sent out photons are taking over the directional vector of the moving clock, then everything is plausible.

All you have to take as given, is that all observers must agree if the mirror gets hit. Anything else would implausible.

Is there also a physical explanation for that,

Using the wave model, you can get it from relativity of simultaneity along a wave front emitting line.

What would happen, If we would hold the clock at one position after the photon was sent out?

If you stop a clock moving to the right, then you start moving it to the left its initial rest frame. The photon will miss, and both frames will agree on that.

 

[DrGreg]

Maybe this animation from an old post of mine will help

Maybe this diagram helps

Here's something (it could be a pulse of light, it could be a ball) bouncing up and down in a train.

An observer in the train (top) infers the thing is moving vertically up and down.

An observer on the ground (bottom) infers the thing is "sliding sideways" in a zig-zag path.

This is valid in both relativistic and Newtonian mechanics.

Source: https://www.physicsforums.com/threads/how-does-light-slide-sideways.804112/post-5048516

 

[PeroK]

Many thanks for the further explanations
If we take as given that the sent out photons are taking over the directional vector of the moving clock, then everything is plausible. But the question still exists: WHY does the light is taking the directional vector?

If you are sitting at a railway station and a train goes past. A passenger looks out the window and sees you fly past at $100m/s$. WHY are you moving? WHY do YOU take the directional vector? WHY do you not remain at rest relative to the person in the train?

 

[Dale]

But the question still exists: WHY does the light is taking the directional vector?

Because if it didn’t then our description of nature would not be self consistent. If it hits the mirror in one frame then it must hit the mirror in all frames since that is a measurable outcome. In order to do so it must “take the directional vector” as you say.

Is there also a physical explanation for that, or may I not ask this question because it is only a model for the reality?

There is always a detailed mechanistic explanation, but if you want that sort of explanation then you have to describe the mechanism in more detail.

Probably the easiest to understand is a spherical source with a collimator. The spherical source emits in all directions and the collimator blocks the light that does not go to the mirror. In the frame where the collimator is moving it is easy to see that the light must “take the directional vector” to pass the collimator.

 

[Martin68]

Wow, so many people helping me understand ... fantastic :-))

Ok, in the example with the ball in the train it is clear: The ball has got the side-movement from the train.
Until now I thought, a photon does not have that. I thought it would be like a stone that is thrown into the water. Image two scenarios:
a) A stone simply falls perpendicular into the water: We get a wavefront annular to the point where the stone hits the water.
b) A stone is thrown with an additional horizontal velocity: Then we see the same, the middle point of the wavefront stands still at the starting point in the water, it will not move to the side.

So a photon will behave in another way: The starting-point of the wavefront will drift sideways with the same velocity like the emitting light-source.

 

[PeroK]

Wow, so many people helping me understand ... fantastic :-))

Ok, in the example with the ball in the train it is clear: The ball has got the side-movement from the train.
Until now I thought, a photon does not have that.

This is the problem. Somehow you got a crazy idea about photons. Now that you have that idea it's really difficult to change your mind. There have quite a few people before you on this forum who got this idea. I can sort of understand it. You read:

"The speed of light is independent of the source, or the motion of the observer."

And, you interpreted this as:

"The motion of light is - somehow, fantastically - independent of the motion of the observer."

And, now, because you misinterpreted this idea, it's really difficult to change your mind. No matter how crazy it is.

On the positive side, if you are intelligent, you must see that the idea makes no physical sense. Perhaps we can't help you anymore. You either stay with this crazy idea or accept it was all a misunderstanding on your part.

There is nothing special about the motion of light, as opposed to the motion of particles, except that the speed is invariant.

Can you accept that statement? If not, why not?

 

[Martin68]

There is nothing special about the motion of light, as opposed to the motion of particles, except that the speed is invariant.

Can you accept that statement? If not, why not?

Yes, I have never seen this in that way. But I'm fine with that . Many thanks!

 

[Mister T]

Ok, in the example with the ball in the train it is clear: The ball has got the side-movement from the train.
Until now I thought, a photon does not have that.

Replace "photon" with "pulse of light" and you essentially have the same thought as most physicists did in the late 1800's. It would be a mechanism that could be used to distinguish between a state of rest and state of steady motion. Despite elaborate experimental apparatus designed to detect the effect, it can't be detected.

 

[jbriggs444]

Using the wave model, you can get it from relativity of simultaneity along a wave front emitting line.

Such a beautiful and pithy statement.

You have a horizontal surface emitting a wave front simultaneously in the frame of the emitting surface. The resulting wave front propagates vertically in the frame of the emitting surface.

From the viewpoint of a rest frame compared to which the surface is moving, the emission is not simultaneous. Instead, the emission event sweeps across the moving surface and the emitted wave front emerges at an angle and propagates at an angle.

Of course it is the same wave front viewed from two different perspectives.

 

[Martin68]

[If I don't react to all statements: My English is not so good to transform all described models from English to german :#)]
... but I try to understand everything

 

[vanhees71]

If you need a German introduction to relativity, maybe my lecture notes are of use for you:

https://itp.uni-frankfurt.de/~hees/publ/theo3-l3.pdf

 

[Martin68]

If you need a German introduction to relativity, maybe my lecture notes are of use for you:

https://itp.uni-frankfurt.de/~hees/publ/theo3-l3.pdf

Many thanks ... but: I am only an interested 'normalo'. For that, I would have to repeat all stuff of my maths-lecture (my course of studies over 20 years ago was software-engeneering) and additionally eat some very thick maths-books :#)))

 

[timmdeeg]

Ok, in the example with the ball in the train it is clear: The ball has got the side-movement from the train.

Do you think that in this example the ball feels a physical sideways blow?

 

[Martin68]

Do you think that in this example the ball feels a physical sideways blow?

No, for the ball this is an inertial system (I don't know if this is physically and in english correct, you call it 'frame'?)

 

[timmdeeg]

No, for the ball this is an inertial system

That's fine, from your wording I understood you otherwise. The "sliding sideway" is due to the relative motion. It's enforced because the ball has to hit the floor.

 

[Martin68]

from your wording I understood you otherwise

... oh oh, my english and my ability to write what I want to express :-))

 

[PeroK]

No, for the ball this is an inertial system (I don't know if this is physically and in english correct, you call it 'frame'?)

This might be a good point to introduce the relativistic velocity addition formula. This is simplest in one dimension.

We have a particle traveling at velocity $u'$ in a one inertial reference frame, S'. This reference frame is moving at speed $v$ relative to another reference frame, S.

You could think of S as the reference frame of a planet and S' as the reference frame of a spaceship, moving relative to the planet with velocity $v$. The spaceship fires a projecile at speed $u'$ (relative to the ship itself).

Now, in classical physics, the speed of the projecile relative to the planet, $u$, would simply be:

$u = v + u'$

Note that $v, u'$ can be positive or negative, but let's consider both positive for this example. The ship is moving away from the planet and fires the projecile away from the planet as well.

The relativistic version of this formula is:

$u = \frac{v + u'}{1 + vu'/c^2}$

There are three things worth noting:

1) This is equivalent to the invariance of the speed of light. If you wanted you could replace the postulate about light with a postulate about this formula and develop SR from this formula.

2) If $v, u'$ are small compared to $c$, then:

$u \approx \frac{v + u'}{1 + 0} = v + u'$

And you get the well-known classical case.

3) If $u' = c$. That is to say, the ship fires a photon or light beam, then:

$u = \frac{v + c}{1 + vc/c^2} = c\frac{v/c + 1}{1 + v/c} = c$

And we recover the invariance of the speed of light. I.e. the light beam has the same speed, $c$, in the frame of the planet.

In this context, perhaps it is clearer that the key property about light is its invariant speed. Nothing more.

The two-dimensional case, which would apply in the case of light clock, is mathematically slightly more complicated, but you can look it up under "relativistic velocity addition". Here, for example:

https://en.wikipedia.org/wiki/Velocity-addition_formula#Special_relativity

 

[timmdeeg]

... oh oh, my english and my ability to write what I want to express :-))

Yes it isn't easy sometimes, I know, I'm German too. But people around here are helpful and they understand.

 

[PeroK]

Yes it isn't easy sometimes, I know, I'm German too. But people around here are helpful and they understand.

Most of the time we don't even notice! Seriously, your English is so good. That applies to @Martin68 as well.

 

[bahamagreen]

Some translations of German language relativity books used the English word "velocity" for the German word for "speed".

 

[A.T.]

You have a horizontal surface emitting a wave front simultaneously in the frame of the emitting surface. The resulting wave front propagates vertically in the frame of the emitting surface.

From the viewpoint of a rest frame compared to which the surface is moving, the emission is not simultaneous. Instead, the emission event sweeps across the moving surface and the emitted wave front emerges at an angle and propagates at an angle.

Of course it is the same wave front viewed from two different perspectives.

Here is a paper that has some images visualizing this:
https://arxiv.org/pdf/physics/0409013.pdf

 

[vanhees71]

Some translations of German language relativity books used the English word "velocity" for the German word for "speed".

The problem is that in German we don't have enough words to be as precise as in English. The most severe example that in German we have only "Wissenschaft", in English you have the very important difference between science ("Naturwissenschaft") and humanities ("Geisteswissenschaft"), which are two distinct incompatible worlds...

 

[A.T.]

The problem is that in German we don't have enough words to be as precise as in English. The most severe example that in German we have only "Wissenschaft", in English you have the very important difference between science ("Naturwissenschaft") and humanities ("Geisteswissenschaft"), which are two distinct incompatible worlds...

Seems like it's not a matter of "not enough words", but how words are formed: invent completely new ones vs. combine existing ones.

 

[Ibix]

Seems like it's not a matter of "not enough words", but how words are formed: invent completely new ones vs. combine existing ones.

Actually, English being a bit of a mix of both germanic and latin derived languages, we often have two words for more or less the same thing. We just seem to have repurposed the germanic-rooted and latin-rooted words for "how fast is something going" to mean different things in scientific language.

 

[A.T.]

Actually, English being a bit of a mix of both germanic and latin derived languages, we often have two words for more or less the same thing.

Yeah, "invent" wasn't the right term here.