Einstein Light Clock Conundrum

[Bufofrog]

8 minutes ago, Otto Nomicus said:

Yup, I'm very sad about that. Now why don't you also do as I invited Eise to do? Maybe one of you can provide a logical explanation.

It has already been explained to you in this thread and it doesn't seem to have helped you understand.  I doubt I can explain it to you any better, so I am afraid you may just have to remain sad.

[studiot]

22 minutes ago, Otto Nomicus said:

No, actually I would just keep my eyes focused on the moving light clock and see the beam going straight up and down like always. I wouldn't see any zig-zag. You could imagine a zig-zag but the light beam obviously wouldn't be propagating along the slanted lines, because the laser was never aimed on a slant and neither was the mirror. The cycle would take the exact same time to complete and there would be no reason to think you needed to imagine time going slower than normal on the train, and if you did, then light would be seen to be traveling at slower than normal speed, so why would you want to do that?

Nobody here, including you, have explained how a laser beam directed vertically could have its beam emitted at an angle of about 40 degrees from vertical, in the case of the diagram I showed, and the mirror also magically adjust itself to the same angle. Why don't you do that now, I'm interested to hear the tale of the magic angled beam laser and the self adjusting mirror. Of course, the clock would have to catch up to the angled beam in time for it to hit the mirror. What if the clock stopped moving right after emitting the beam, the beam would miss the mirror, right? Or would it stop halfway through the trip to the mirror and adjust itself to going straight down? 

The problem is that you original sketch is fixed in your mind, but it is a flawed depiction of what happens.
Thus your text description is also flawed.

I will see if I have the time to produce an improved one.

By the way you don't need a laser to show this, in fact it is can be less than helpful.
Einstein didn't know about lasers back then so he constructed his light clocks using ordinary sources and light pulses. by which he meant the leading edge of the light 'signal'.

[Eise]

41 minutes ago, Otto Nomicus said:

Why don't you do that now, I'm interested to hear the tale of the magic angled beam laser and the self adjusting mirror.

No magic needed. Observing and seeing are not exactly the same. If I see that the moon is smaller than my hand, it doesn't mean that the moon really is smaller: I have to take into account that the moon is much farther away.

So in the situation of the light clock I am not asking what you see (hey, you could move your head in the other direction, and you probably would 'see' that some velocities are above c. I am asking what according to you, in the ground frame,  is the real distance the light beam travels in your frame of reference. (Hint: if you follow a moving car with your head, does that mean it has no speed? No, you must compensate for your own subjective view.) In fact, by turning your head, you are identifying yourself with the train frame. Turning a head is not part of what a frame of reference is.

Maybe you should go back to Swansont's example, I hope at least that you now understand how he meant it. Must the person bouncing the ball on the train compensate for the velocity of the train, so that you see a zig-zag trajectory?

And a final example: if somebody on a train throws a ball with a moderate velocity of, say, 20 km/h, and the train is moving 100 km/h, how fast is the ball flying according to you? 120 km/h? Or 20 km/h, because you moved your head to follow the train? And if he throws the ball out of the window, in the same direction as the train is moving, with which velocity will it hit a signal pole? 20 km/h, because you followed the train with your head? Or is it 120 km/h? (Don't try this at home, eh, in a real train. You know why...)

PS

Here is a link, that explains the light clock. Including a bouncing ball...

Edited January 27, 2023 by Eise

[Genady]

Sometimes exaggeration helps. So, in addition to wonderful explanations and examples above by Eise and others, consider this. Instead of a train, let's do it in a plane, a very fast one.

The light source is installed in the plane directly above seat 17A facing straight down to the seat. The plane is flying. When it is above NYC, the flash of light goes out. After very short time it hits the seat 17A. But the plane is really fast. By the time the light hits the seat, the plane is above Boston. The seat 17A moved 200 miles, from NYC to Boston. However, the light hits it anyway. Thus, the light went 0.5 m down and 200 miles NE.

[Otto Nomicus]

2 hours ago, Eise said:

No magic needed. Observing and seeing are not exactly the same. If I see that the moon is smaller than my hand, it doesn't mean that the moon really is smaller: I have to take into account that the moon is much farther away.

So in the situation of the light clock I am not asking what you see (hey, you could move your head in the other direction, and you probably would 'see' that some velocities are above c. I am asking what according to you, in the ground frame,  is the real distance the light beam travels in your frame of reference. (Hint: if you follow a moving car with your head, does that mean it has no speed? No, you must compensate for your own subjective view.) In fact, by turning your head, you are identifying yourself with the train frame. Turning a head is not part of what a frame of reference is.

Maybe you should go back to Swansont's example, I hope at least that you now understand how he meant it. Must the person bouncing the ball on the train compensate for the velocity of the train, so that you see a zig-zag trajectory?

And a final example: if somebody on a train throws a ball with a moderate velocity of, say, 20 km/h, and the train is moving 100 km/h, how fast is the ball flying according to you? 120 km/h? Or 20 km/h, because you moved your head to follow the train? And if he throws the ball out of the window, in the same direction as the train is moving, with which velocity will it hit a signal pole? 20 km/h, because you followed the train with your head? Or is it 120 km/h? (Don't try this at home, eh, in a real train. You know why...)

PS

Here is a link, that explains the light clock. Including a bouncing ball...

Valid point about it not originally being a laser, which wasn't invented until 1958, which may have been a prime reason why Einstein was susceptible to misconception, and why he could pass off his theory as plausible in those day, but since that time it was easy to see that his thought experiment couldn't work if it's a laser in the clock. Einstein's logic would be valid if photons were physical objects and the clock was analogous to a river. In that case the velocity would be the square root of the sum of the squares of the vertical velocity of the photons and the horizontal velocity of the clock/river. Can a moving clock apparatus continuously impart its horizontal velocity to photons after they leave the emitter and after they bounce off a mirror? Obviously not, because photons are not physical objects and therefore nothing physical can impart its velocity to it as it travels across it. If that were the case, shining a laser crosswise through moving water would deflect it in the direction of the moving water, and that has never been shown experimentally to occur.

The only other thing that one could postulate could cause the photons to move horizontally with the clock is inertia, and obviously photons are not subject to inertia. Why would you and Einstein think photons would behave exactly the same as physical objects when they're not physical objects? There simply is no horizontal velocity being imparted on the photons because there is nothing known in physics that could cause it. So how is it that light can bounce up and down in a moving frame and not appear to bend in the rearward direction? Because a frame moving at uniform straight line velocity is exactly the same as the frame not moving at all, that's why. In fact, light doesn't even have velocity, it only has a rate of electric and magnetic field propagation, we just perceive that as velocity but it's not the same thing because no force is involved and no mass is involved, therefore no velocity is involved, it's a misconception.

24 minutes ago, Genady said:

Sometimes exaggeration helps. So, in addition to wonderful explanations and examples above by Eise and others, consider this. Instead of a train, let's do it in a plane, a very fast one.

The light source is installed in the plane directly above seat 17A facing straight down to the seat. The plane is flying. When it is above NYC, the flash of light goes out. After very short time it hits the seat 17A. But the plane is really fast. By the time the light hits the seat, the plane is above Boston. The seat 17A moved 200 miles, from NYC to Boston. However, the light hits it anyway. Thus, the light went 0.5 m down and 200 miles NE.

Maybe you just perceived it as a vertical beam moving 200 miles NE but in fact it was a series of photons being produced, each slightly NE of the previous. None of the photons actually moved 200 miles NE at all. You can't just have a self propagating light beam bouncing back and forth for 200 miles with the emitter turned off. Why does a beam not just keep bouncing back and forth for a long time after an emitter is turned off until it dissipates from energy loss? Nobody knows but it doesn't.

Edited January 27, 2023 by Otto Nomicus

[Genady]

8 minutes ago, Otto Nomicus said:

in fact it was a series of photons being produced, each slightly NE of the previous. None of the photons actually moved 200 miles NE at all. You can't just have a self propagating light beam bouncing back and forth for 200 miles with the emitter turned off.

Nope. Only one photon left the light source and it did not bounce - there are no mirrors in this experiment. It just left the source and travelled to the seat. Let's to be specific, have a photosensor on the seat that received and registered one photon.

[Otto Nomicus]

59 minutes ago, Genady said:

Nope. Only one photon left the light source and it did not bounce - there are no mirrors in this experiment. It just left the source and travelled to the seat. Let's to be specific, have a photosensor on the seat that received and registered one photon.

It's just the way things work, uniform motion is indistinguishable from no motion. Photons don't move sideways so that's the proof that uniform motion is equivalent to no motion. The way another frame may seem to you from your own frame is irrelevant. You can't clock the speed of light in another frame anyway, how could you? You can't even clock the one-way speed of light in your own frame, much less another one a distance from yours.

[swansont]

5 hours ago, Otto Nomicus said:

No, actually I would just keep my eyes focused on the moving light clock and see the beam going straight up and down like always. I wouldn't see any zig-zag. You could imagine a zig-zag but the light beam obviously wouldn't be propagating along the slanted lines, because the laser was never aimed on a slant and neither was the mirror. The cycle would take the exact same time to complete and there would be no reason to think you needed to imagine time going slower than normal on the train, and if you did, then light would be seen to be traveling at slower than normal speed, so why would you want to do that?

There are a number of videos that show if you toss a ball vertically on a moving platform, it comes back down to the point of origin on the cart. 

such as

 

So insisting that the ball would come straight down when the cart is moving is a non-starter; we can see that the ball doesn’t do this. The ball travels a greater distance in the lab frame than in the cart’s frame.

The wrinkle for the light clock is the restriction that the speed of light is the same in both frames. d = ct. Since d is greater and c is the same, t must be greater

 

[studiot]

@Otto Nomicus

I'm sorry you chose not to respond to my offer so I will leave you with these thoughts as some dubious statements have been made on both sides, perhaps you are confusing each other.

 

Firstly if you are going to accept that a light path that measures 1metre  between two points in all in the same frame but say 1.3 metres in another frame then you must also accept that the time of transit of that different distance must also be differnt to maintain the principle of constacy of the speed of light in all frames.

Secondly the light in a light clock should not be thought of a constant stream of light.

There is no way you can observe the time of transit of such a light stream.
Light clocks use light pulses which are measured at the source and receiver but not in between.

 

[Genady]

1 hour ago, Otto Nomicus said:

It's just the way things work, uniform motion is indistinguishable from no motion. Photons don't move sideways so that's the proof that uniform motion is equivalent to no motion. The way another frame may seem to you from your own frame is irrelevant. You can't clock the speed of light in another frame anyway, how could you? You can't even clock the one-way speed of light in your own frame, much less another one a distance from yours.

To be specific, let's say you are in the plane. You clock the time it takes for photon to get from the source to the seat, 0.5 m, and calculate the speed of light, c.

I am on the ground. I clock the time it takes for the same photon to get from the same source to the same seat, 200+ miles, and calculate speed of light, c.

[swansont]

4 hours ago, Otto Nomicus said:

In fact, light doesn't even have velocity, it only has a rate of electric and magnetic field propagation, we just perceive that as velocity but it's not the same thing because no force is involved and no mass is involved, therefore no velocity is involved, it's a misconception.

Light definitely has a velocity (we can measure the propagation time over some distance) and can impart an impulse since it has momentum. So saying no force is involved is incorrect.

[Otto Nomicus]

10 hours ago, swansont said:

There are a number of videos that show if you toss a ball vertically on a moving platform, it comes back down to the point of origin on the cart. 

such as

 

So insisting that the ball would come straight down when the cart is moving is a non-starter; we can see that the ball doesn’t do this. The ball travels a greater distance in the lab frame than in the cart’s frame.

The wrinkle for the light clock is the restriction that the speed of light is the same in both frames. d = ct. Since d is greater and c is the same, t must be greater

 

I didn't dispute the obvious or say what you apparently think I said somewhere at some time, being that I'm not 5 years old. What I disputed was the idea that the clock moving horizontally meant that photons had to travel a longer path in the same amount of time. In Einstein's day, lasers hadn't been invented yet so apparently he could actually get people to believe that light emitted vertically could travel on a slant. What's YOUR excuse for believing it long after lasers were invented? Do you actually think that's logical, that a laser beam could change its direction of wave propagation from vertical to 40 degrees from vertical just because the emitter is moving sideways? You think that makes perfect sense huh? Interesting. 

[Eise]

16 hours ago, Otto Nomicus said:

Valid point about it not originally being a laser, which wasn't invented until 1958, which may have been...

Your whole argument has nothing to do with what I wrote. Please check what I wrote, and react on that.

Genady's example however brought me to a better illustration than my first one.

1. So again we have the train with its light clock. In the train's frame it works perfectly, because the clock and the observer are at rest with respect to each other. Agree so far?

2. Now you are looking at the moving time clock from the ground referential frame. Would that affect if the light clock works? Of course not. Nothing changes in the train's inertial frame. Being observed or not from an observer in another frame of reference does not change anything. So the light clock still works fine. The light beam will not miss the small mirror just because it is observed from another referential frame. Agree so far?

3. What you in the ground frame can do now is mark the point on the railway where the light beam leaves the laser first. Just look, and e.g. it is exactly at a point where a tree stands. This is your first mark. Then mark the point where the light hits the mirror, e.g. where a signal pole is standing. And then again mark the point on the railway where you see that the light hits the laser again, e.g. a railway crossing. (Of course you have to turn your head to follow the train, and to mark the points on the railway, so you do not see a zig-zag line). Do you agree that such a procedure would be possible?

4. Now you measure the distances between the marked tree, signal pole, and railway crossing. So this is the real distance the train has traveled. And plotting the vertical distance of the light beam against these marked points, you will get a zig-zag line, this time independent on how you move your head, based on real distances. So no change of direction of the laser is necessary: just plot the position of the light against the railway track.

And to extend on point 2: first you have to accept point 1: The light clock works perfectly from the frame of reference of the train. 

2a. As an example, assume the train has a speed of 0.8c relative to the ground frame.  Now we introduce a second observer, which travels at 0.4c in the same direction as the train. Both you and the second observer look at the light clock. Question for you: in which direction should the experimenter on the train point the laser so that:

• the light clock works for him
• the light clock works for you
• the light clock works for the second observer.

If you say that this is impossible, then logically, you are saying that being observed changes the working of the light clock.

Now you can choose to learn something by thinking my example through. Or you can bend in all kind of silly, wriggled arguments to show why you are right. Or you can just ignore my posting.

The problem is that special relativity is used to design technologies, like GPS or particle accelerators, and explains things like the relation between electrical and magnetic fields, the colour of gold, the liquidness of mercury, and a lot more. Special relativity, which of course includes time dilation, is daily practice for many physicists, and so proven to the bone. So are you just discussing if the light clock example is a correct illustration of time dilation, or are you arguing against special relativity?

Edited January 28, 2023 by Eise

[Otto Nomicus]

59 minutes ago, Eise said:

Your hole argument has nothing to do with what I wrote. Please check what I wrote, and react on that.

Genady's example however brought me to a better illustration than my first one.

1. So again we have the train with its light clock. In the train's frame it works perfectly, because the clock and the observer are at rest with respect to each other. Agree so far?

2. Now you are looking at the moving time clock from the ground referential frame. Would that affect if the light clock works? Of course not. Nothing changes in the train's inertial frame. Being observed or not from an observer in another frame of reference does not change anything. So the light clock still works fine. The light beam will not miss the small mirror just because it is observed from another referential frame. Agree so far?

3. What you in the ground frame can do now is mark the point on the railway where the light beam leaves the laser first. Just look, and e.g. it is exactly at a point where a tree stands. This is your first mark. Then mark the point where the light hits the mirror, e.g. where a signal pole is standing. And then again mark the point on the railway where you see that the light hits the laser again, e.g. a railway crossing. (Of course you have to turn your head to follow the train, and to mark the points on the railway, so you do not see a zig-zag line). Do you agree that such a procedure would be possible?

4. Now you measure the distances between the marked tree, signal pole, and railway crossing. So this is the real distance the train has traveled. And plotting the vertical distance of the light beam against these marked points, you will get a zig-zag line, this time independent on how you move your head, based on real distances. So no change of direction of the laser is necessary: just plot the position of the light against the railway track.

And to extend on point 2: first you have to accept point 1: The light clock works perfectly from the frame of reference of the train. 

2a. As an example, assume the train has a speed of 0.8c relative to the ground frame.  Now we introduce a second observer, which travels at 0.4c in the same direction as the train. Both you and the second observer look at the light clock. Question for you: in which direction should the experimenter on the train point the laser so that:

• the light clock works for him
• the light clock works for you
• the light clock works for the second observer.

If you say that this is impossible, then logically, you are saying that being observed changes the working of the light clock.

Now you can choose to learn something by thinking my example through. Or you can bend in all kind of silly, wriggled arguments to show why you are right. Or you can just ignore my posting.

The problem is that special relativity is used to design technologies, like GPS or particle accelerators, and explains things like the relation between electrical and magnetic fields, the colour of gold, the liquidness of mercury, and a lot more. Special relativity, which of course includes time dilation, is daily practice for many physicists, and so proven to the bone. So are you just discussing if the light clock example is a correct illustration of time dilation, or are you arguing against special relativity?

Did the light beam propagate on the slanted line just because you could draw a slanted line between the points where it left the top of the clock and where it hit the bottom of the clock? That was you drawing a slanted line, not photons propagating in the direction of the slanted line. The light clock cycle takes the same amount of time no matter who is looking at it from where, which is my point. 

Now about relativity explaining the relationship between electrical and magnetic fields. I'm sure you've seen the animations where the person will suggest that if we were moving with a positively charged particle along the outside of a wire with electric current in it at the same speed as the electrons, which is about 3 m per hour, that we and the particle would see the protons in the wire, and the spaces between them, contracted so the protons appear to be denser than the electrons, thus making the field seem to be more positive rather than negative, thus causing the particle to move away from the wire. That's probably what you're talking about.

Was it still the same number of protons in the wire? If so, is there extra space at one or both ends of the wire after the protons contracted together? Now you'll probably say no, the entire wire contracted, contracting the protons with it, so there didn't need to be either empty space left somewhere in the wire or there suddenly be more protons in the wire than when we were looking at it from the standpoint of the protons and the wire being stationary. Well, if the whole wire contracted and the protons contracted with it then did the spaces between the electrons also become contracted? If not, how did they avoid it? That would be the same as the spaces between the electrons expanding in the wire, would it not? In which case. fewer electrons would be inside the wire, right? How did they vanish just because you decided to take the viewpoint of the moving positively charged particle outside the wire?

There's obviously no way that the wire could contract and not also leave less room for the same number of electrons. Did the wire contract or did it not? Either way you have a conundrum. So it is rather obvious that relativity did not in fact explain that at all, yet another misconception just like the light clock thought experiment was based on a misconception. In both cases, nobody actually thought it through very much at all apparently.

Edited January 28, 2023 by Otto Nomicus

[Eise]

@Otto Nomicus: please go through my points one by one. So we can see better where our arguments deviate.

[Otto Nomicus]

1 hour ago, Eise said:

Your whole argument has nothing to do with what I wrote. Please check what I wrote, and react on that.

Genady's example however brought me to a better illustration than my first one.

1. So again we have the train with its light clock. In the train's frame it works perfectly, because the clock and the observer are at rest with respect to each other. Agree so far?

REPLY: Yup

2. Now you are looking at the moving time clock from the ground referential frame. Would that affect if the light clock works? Of course not. Nothing changes in the train's inertial frame. Being observed or not from an observer in another frame of reference does not change anything. So the light clock still works fine. The light beam will not miss the small mirror just because it is observed from another referential frame. Agree so far?

REPLY: Yup

3. What you in the ground frame can do now is mark the point on the railway where the light beam leaves the laser first. Just look, and e.g. it is exactly at a point where a tree stands. This is your first mark. Then mark the point where the light hits the mirror, e.g. where a signal pole is standing. And then again mark the point on the railway where you see that the light hits the laser again, e.g. a railway crossing. (Of course you have to turn your head to follow the train, and to mark the points on the railway, so you do not see a zig-zag line). Do you agree that such a procedure would be possible?

REPLY: Yup

4. Now you measure the distances between the marked tree, signal pole, and railway crossing. So this is the real distance the train has traveled. And plotting the vertical distance of the light beam against these marked points, you will get a zig-zag line, this time independent on how you move your head, based on real distances. So no change of direction of the laser is necessary: just plot the position of the light against the railway track.

And to extend on point 2: first you have to accept point 1: The light clock works perfectly from the frame of reference of the train. 

REPLY: So your point is that the photons propagated in the direction of the slanted line you drew? I beg to differ, I maintain that they still propagated in the same vertical direction as when the clock was not moving and propagated over the exact same vertical distance and therefore did not need to exceed 299,792,457 m/s to travel a longer path unless we pretend that time was slowed down in the clock/train frame compared to ours.

2a. As an example, assume the train has a speed of 0.8c relative to the ground frame.  Now we introduce a second observer, which travels at 0.4c in the same direction as the train. Both you and the second observer look at the light clock. Question for you: in which direction should the experimenter on the train point the laser so that:

• the light clock works for him
• the light clock works for you
• the light clock works for the second observer.

REPLY: No adjustment of the laser is required, because in all of those cases the photons would be propagating vertically, traveling the same vertical distance and taking the same amount of time.

If you say that this is impossible, then logically, you are saying that being observed changes the working of the light clock.

REPLY: Never said it was impossible.

Now you can choose to learn something by thinking my example through. Or you can bend in all kind of silly, wriggled arguments to show why you are right. Or you can just ignore my posting.

REPLY: The only thing I learned from all that is the fact that you have no comprehension of direction of photon propagation and apparently think it changes whenever the light clock moves horizontally. I guess that means your powers of logic are rather wanting.

The problem is that special relativity is used to design technologies, like GPS or particle accelerators, and explains things like the relation between electrical and magnetic fields, the colour of gold, the liquidness of mercury, and a lot more. Special relativity, which of course includes time dilation, is daily practice for many physicists, and so proven to the bone. So are you just discussing if the light clock example is a correct illustration of time dilation, or are you arguing against special relativity?

REPLY: I direct you to the last part of my prior reply, which you seem to have ignored rather than try to refute it.

 

[Eise]

48 minutes ago, Otto Nomicus said:

REPLY: So your point is that the photons propagated in the direction of the slanted line you drew? I beg to differ, I maintain that they still propagated in the same vertical direction as when the clock was not moving and propagated over the exact same vertical distance and therefore did not need to exceed 299,792,457 m/s to travel a longer path unless we pretend that time was slowed down in the clock/train frame compared to ours.

So here lies your problem. If the photons propagated also vertical in the ground frame, they would stick to the tree for you. The clock would not work anymore in the train frame. Of course now you will say that this can only be solved by pointing the laser in another, slanted direction. But why would the observer/experimenter do this? For him the photons return at the laser at the moment he passes the railroad crossing. That is my point 3 above, and you agreed with that. 

Furthermore the observer/experimenter on the train cannot do it for you in the ground frame and for the observer with a speed of 0.4c in the same direction as the train at the same moment.  Which slant should he choose so you both see the light clock still working, because, as you agreed with, observing doesn't change anything (point 2).

[studiot]

I still don't see any reply to my comments, which is a pity because you are travelling down the wrong path, pulling everybody with you.

 

The outline scheme of though in analysing the relativity of the light clock runs as follow.

1. The light clock is a self contained unit the output of which is not tracking the light path but a 1 second 'tick'.
2. This tick measures 1 second when compared against the tick used by the stationary observer's ie the two clocks stationary are side by side.
3. When the clock (and it is the whole clock not part of it) is in motion relative to the stationary observer, the observer receives signals of the number of ticks on the moving clock.
4. When he times those ticks using his own clock he finds that they are longer than those of his own clock.
5. This difference is calculated by relativity using distance information derived from the moving clock's motion. 
6. This is why others have said the primary observation is time dilation here.

Incidentally you have been stridently claiming that the angle of incidence within the clock does not vary.

This is not observationally true.
The correct analysis, backed up by experiment,  is not normally taught in elementary Special Relativity and leads to a formula for transforming between the angles of incidence observed by a stationary observer, θ, and by a moving observer θ', which are different, just as with the time ticks.

The invariant in this case is the number of ticks.

[math]\cos \theta  = \frac{{\cos \theta ' + \frac{v}{c}}}{{1 + \frac{v}{c}\cos \theta '}}[/math]

This formula is also associated with observed frequency changes to the light, known as the radia and the transverse doppler effects.

Edited January 28, 2023 by studiot

[Otto Nomicus]

33 minutes ago, studiot said:

I still don't see any reply to my comments, which is a pity because you are travelling down the wrong path, pulling everybody with you.

 

The outline scheme of though in analysing the relativity of the light clock runs as follow.

1. The light clock is a self contained unit the output of which is not tracking the light path but a 1 second 'tick'.
2. This tick measures 1 second when compared against the tick used by the stationary observer's ie the two clocks stationary are side by side.
3. When the clock (and it is the whole clock not part of it) is in motion relative to the stationary observer, the observer receives signals of the number of ticks on the moving clock.
4. When he times those ticks using his own clock he finds that they are longer than those of his own clock.
5. This difference is calculated by relativity using distance information derived from the moving clock's motion. 
6. This is why others have said the primary observation is time dilation here.

Really? I've never heard of this "relativity" thing you speak of, thanks for telling me about it.

Incidentally you have been stridently claiming that the angle of incidence within the clock does not vary.

 I don't recall ever using the term "angle of incidence"

This is not observationally true.
The correct analysis, backed up by experiment,  is not normally taught in elementary Special Relativity and leads to a formula for transforming between the angles of incidence observed by a stationary observer, θ, and by a moving observer θ', which are different, just as with the time ticks.

The invariant in this case is the number of ticks.

cosθ=cosθ′+vc1+vccosθ′
This formula is also associated with observed frequency changes to the light, known as the radia and the transverse doppler effects.

I thought it was a simple matter of finding out the velocity of the train by how far along the track we see it move in a certain amount of time, and seeing that it's 0.866 m per light cycle cycle in our frame, then using a time dilation calculator to get the factor 2, then dividing the length of the zig-zag path we saw traced out, 1.3228 m, by that number. Oh dear, it doesn't seem to work, it made the speed of light come out slower than normal. Gee, how did light get slowed down like that?

 

Edited January 28, 2023 by Otto Nomicus

[swansont]

6 hours ago, Otto Nomicus said:

I didn't dispute the obvious or say what you apparently think I said somewhere at some time, being that I'm not 5 years old. What I disputed was the idea that the clock moving horizontally meant that photons had to travel a longer path in the same amount of time

You haven’t clearly confirmed that you agree that they travel a longer distance.

Quote

In Einstein's day, lasers hadn't been invented yet so apparently he could actually get people to believe that light emitted vertically could travel on a slant. What's YOUR excuse for believing it long after lasers were invented? Do you actually think that's logical, that a laser beam could change its direction of wave propagation from vertical to 40 degrees from vertical just because the emitter is moving sideways? You think that makes perfect sense huh? Interesting. 

It’s a photon. It doesn’t have to be from a laser. But yes, I believe a laser will do this, because the lasers I had in my lab stayed aligned despite the rotation of the earth. If you observed the light path from the appropriate vantage point, it would be “slanted” (the target moves, and the laser still hits the target)

The direction you observe depends on your frame of reference. The light hits the mirror, so how can the light not be going along the slanted path, when observed from the lab frame? Just like the ball in the moving cart - it can’t drop back into the cart unless it’s traveling along a parabolic path. We know the impulse it gets is straight up.

[Bufofrog]

3 hours ago, Otto Nomicus said:

Did the light beam propagate on the slanted line just because you could draw a slanted line between the points where it left the top of the clock and where it hit the bottom of the clock? That was you drawing a slanted line, not photons propagating in the direction of the slanted line. The light clock cycle takes the same amount of time no matter who is looking at it from where, which is my point. 

So according to you the observer moving with the light clock would see that the light clock was operating properly, but the observer not moving with the clock would see that the light clock does not work because the light would miss the mirror?

[Otto Nomicus]

2 hours ago, swansont said:

You haven’t clearly confirmed that you agree that they travel a longer distance.

It’s a photon. It doesn’t have to be from a laser. But yes, I believe a laser will do this, because the lasers I had in my lab stayed aligned despite the rotation of the earth. If you observed the light path from the appropriate vantage point, it would be “slanted” (the target moves, and the laser still hits the target)

The direction you observe depends on your frame of reference. The light hits the mirror, so how can the light not be going along the slanted path, when observed from the lab frame? Just like the ball in the moving cart - it can’t drop back into the cart unless it’s traveling along a parabolic path. We know the impulse it gets is straight up.

Did the fact that the ball in the video was moving horizontally at a certain velocity when the launcher was pushed change its vertical velocity? Obviously not. Did the fact that the ball was moving vertically at a certain velocity change its horizontal velocity? Obviously not. Was either the vertical velocity of the light beam or the horizontal velocity of the clock higher than 299,792,458 m/s? Obviously not. Did anything need to violate the universal speed limit of 299,792,458 m/s? Obviously not. The only two vectors involved were vertical and horizontal, there never was a slanted vector, you just imagined one.

 

6 hours ago, Eise said:

So here lies your problem. If the photons propagated also vertical in the ground frame, they would stick to the tree for you. 

There's no tree sticking involved. Now please answer this question, did the wire's length contract in our view when we were riding along with the positively charged particle that was traveling near the wire at the same speed as the electrons were moving, yes or no?

[swansont]

1 hour ago, Otto Nomicus said:

Did the fact that the ball in the video was moving horizontally at a certain velocity when the launcher was pushed change its vertical velocity? Obviously not. Did the fact that the ball was moving vertically at a certain velocity change its horizontal velocity? Obviously not. Was either the vertical velocity of the light beam or the horizontal velocity of the clock higher than 299,792,458 m/s? Obviously not. Did anything need to violate the universal speed limit of 299,792,458 m/s? Obviously not. The only two vectors involved were vertical and horizontal, there never was a slanted vector, you just imagined one.

Did you know that you can add vector components to get a resultant? That this wll be at some slanted angle with respect to the axes? Apparently not.

The ball can have a higher speed owing to having a vertical velocity and a horizontal one, but light can’t. Light’s speed is fixed, so its behavior will be somewhat different.

 

[Otto Nomicus]

8 hours ago, swansont said:

Did you know that you can add vector components to get a resultant? That this wll be at some slanted angle with respect to the axes? Apparently not.

The ball can have a higher speed owing to having a vertical velocity and a horizontal one, but light can’t. Light’s speed is fixed, so its behavior will be somewhat different.

 

What if we take the viewpoint of the ball in the video? Was the ball launcher traveling on a longer parabolic path? What if we take the viewpoint of the ball on the train? Was the train traveling on a longer parabolic path? What if we take the viewpoint of the photon? Was the clock traveling on a longer zig-zag path? Now did the ball launcher, the train or the clock travel different and longer paths REALLY or did we just imagine that they did? Did we feel a need to adjust time so that the imagined longer paths of the ball launcher, the train or the clock would take the same amount of time as the real paths?

I know we weren't moving with the launcher, the train or the clock in the thought experiments but the same logic would apply. We could consider the clock to be taking the zig-zag path, could we not? The most ridiculous scenario is the one involving a speeding locomotive taking a parabolic path, did that really happen? In our imagination it happened, but in reality, pretty sure not. Isn't that the whole idea of "relativity", that we can take either view and everything works the same reciprocally? That's the principle of reciprocity between inertial frames and it mustn't be ignored.

Edited January 29, 2023 by Otto Nomicus

[Eise]

17 hours ago, Otto Nomicus said:

There's no tree sticking involved.

Then how do you explain that for the train frame the photons go vertical, but for you from the ground frame, according to you, the photons arrive back at the laser when the train is at the railway crossing? If you think a slant is needed, i.e. the observer/experimenter on the train must point the laser a bit forward, from the train frame it would miss the small mirror. Your viewpoint is logically inconsistent. There is no way the light clock would work for different inertial frames at the same time, including the rest frame of the light clock (the train).

And I am shocked you are so sure about your viewpoint, where you obviously have no idea what vectors are, and how one adds them. But obviously you are not here to learn anything.

17 hours ago, Otto Nomicus said:

Now please answer this question, did the wire's length contract in our view when we were riding along with the positively charged particle that was traveling near the wire at the same speed as the electrons were moving, yes or no?

I won't, because I am an 'amateur physicist'. But I am pretty sure the example of the wire and the length contraction is very simplified. I think it does nothing more than give a some intuition how special relativity, electricity and magnetism are related. E.g. from the example of the wire, I do not see how the magnetic field would be circular around the wire.

Maybe some of the real physicists here can give a short comment on that?

I have some notion of where the limits of my knowledge are. You obviously don't. You are saying that a theory that is already accepted and used for a century, and is basic to nearly all of fundamental physics, is wrong. Here is a list of empirical tests, that all confirm special relativity is correct. 

About the wire example, ChatGPT says the following:

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The example is meant to provide an intuitive understanding of how special relativity affects the behavior of electric charges, leading to the concept of a magnetic field. However, it is not a physically correct explanation as it ignores other factors such as the motion of electrons in a wire and electromagnetic forces. A more complete and accurate description of the relationship between special relativity, electric fields, and magnetic fields would require a deeper understanding of the theory of electromagnetism.

While the length contraction explanation may give a rough idea of how special relativity affects the behavior of electric charges, it is not a complete or accurate explanation of the relationship between electric and magnetic fields, and cannot be used to predict the direction of movement of an electrical charge in a magnetic field.

@swansont: is ChatGPT right?

Edited January 29, 2023 by Eise