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Einstein Light Clock Conundrum


Otto Nomicus
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I drew a diagram to illustrate Einstein's moving light clock thought experiment but it doesn't seem to work out right with the time dilation factor for the velocity involved. The light clock is 0.5 m in height so the two-way light beam trip would be 1 m. The clock completes its up/down cycle in the time it takes for it to move to the right 0.866025403 m, which means the clock's velocity is 0.866025403 c. The time dilation factor for that velocity is 2. How do you get the zig-path down to 1 m, like it is inside the clock when it's not moving, by dividing it by 2? Even if I contract space and say that the clock's path is only 0.43301270189221932338186158537647 c, the time dilation factor for that velocity is 1.1094 while the length of the zig-zag would be 1.089724736 m. While it's close, it still doesn't work out exactly. Do I even need to contract the clock's path to half? I don't think I would have to but you tell me.

 

Light-Clock-4.png

I realized that I wouldn't need to contract the bottom line by a factor of 2 if the velocity wasn't even 0.866 c to start with, so that part is moot.

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In the clock's reference frame, the proper time for the light to cover distance l is l/c.

In the external reference frame where clock is moving with velocity v, the light covers the distance sqrt(l^2+t^2*v^2) in the time t:

t^2*c^2 = sqrt(l^2+t^2*v^2)

From this equation,

t=l/c*gamma, where gamma=1/sqrt(1-v^2/c^2).

Thus, t is the proper time times gamma.

No length contraction here.

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1 hour ago, Genady said:

In the clock's reference frame, the proper time for the light to cover distance l is l/c.

In the external reference frame where clock is moving with velocity v, the light covers the distance sqrt(l^2+t^2*v^2) in the time t:

t^2*c^2 = sqrt(l^2+t^2*v^2)

From this equation,

t=l/c*gamma, where gamma=1/sqrt(1-v^2/c^2).

Thus, t is the proper time times gamma.

No length contraction here.

Okay, thanks for that, but how would you get the equivalent of the zig-zag length being reduced to 1 m using relativity? I don't mean show me a bunch of confusing equations, I mean just logically how does a time dilation factor of 2 reduce 1.322875656 m to 1 m?

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1 minute ago, Otto Nomicus said:

Okay, thanks for that, but how would you get the equivalent of the zig-zag length being reduced to 1 m using relativity? I don't mean show me a bunch of confusing equations, I mean just logically how does a time dilation factor of 2 reduce 1.322875656 m to 1 m?

It does not. There is time dilation at work in this experiment, not a length contraction.

PS. If there is something confusing in the equations, ask for a clarification.

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42 minutes ago, Genady said:

It does not. There is time dilation at work in this experiment, not a length contraction.

PS. If there is something confusing in the equations, ask for a clarification.

I'm not disputing that there's no length contraction. The time dilation factor for 0.866 c is 2. How does slowing time down to half speed resolve the path of the beam being longer than 1 m and therefore seeming to need to exceed normal c by about 32%? 32% os not 100% so how could a time dilation factor of 2 be appropriate?

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11 minutes ago, Genady said:

How do you get v=.866c? Is it a given?

It's based on light in the clock, which travels 1 m vertically in total round trip. If the clock itself moves horizontally 0.866 m during the time it takes for the light to make that round trip then obviously the clock apparatus must be moving horizontally at 0.866 c. You don't even need to know the actual value of c, just how far the apparatus moves in relation to how far the light beam moves, which is 0.866 to 1, thus velocity 0.866 c.

Edited by Otto Nomicus
spelling correction
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1 minute ago, Otto Nomicus said:

If the clock itself moves horizontally 0.866 m during the time it takes for the light to make that round trip then obviously the clock apparatus must be moving horizontally at 0.866 c

This is incorrect. In the time the clock moved .866 m the light covered not 1 m but sqrt(1+.8662)=1.323 m, i.e., 1 m vertically and .866 m horizontally.

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42 minutes ago, Genady said:

This is incorrect. In the time the clock moved .866 m the light covered not 1 m but sqrt(1+.8662)=1.323 m, i.e., 1 m vertically and .866 m horizontally.

Exactly. The conundrum is how to make the 1.323 m zig-zag path appear to take the same amount of time as the 1 m vertical path took when the clock was stationary. You obviously can't do it using the time dilation factor for 0.866 c, which is 2, so how do you do it?

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The clock moves .866 m in the time the light moves 1.323 m. Thus the clock speed is v=.866/1.323=.655c.

The time dilation for this speed, gamma=1.323.

The time of the full trip of light is (1 m)/c in the clock's reference frame, and (1.323 m)/c in the external observer reference frame.

I don't see any conundrum. This is what time dilation is.

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6 hours ago, Otto Nomicus said:

The clock completes its up/down cycle in the time it takes for it to move to the right 0.866025403 m

You’re mixing frames.

The up/down cycle is 1 sec in the clock frame (t), but you can’t use 1 sec in the lab frame (t’) to say it moves 0.866 m, since t ≠ t’

It travels a distance L = vt’ 

The light travels ct’

For a half cycle, solve with the Pythagorean theorem.  vt’^2 + d^2 = ct’^2

 

t’ = d/(sqrt(c^2-v^2)

d=0.5m

For v =0.866c, you get 1 sec for the half-cycle and thus 2 sec for the complete trip. Exactly as expected

 

 

You can see the derivation here.

https://www.ck12.org/book/ck-12-physics---intermediate/section/22.2/

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21 hours ago, Genady said:

The clock moves .866 m in the time the light moves 1.323 m. Thus the clock speed is v=.866/1.323=.655c.

The time dilation for this speed, gamma=1.323.

The time of the full trip of light is (1 m)/c in the clock's reference frame, and (1.323 m)/c in the external observer reference frame.

I don't see any conundrum. This is what time dilation is.

You did get it work out, and without complicated equations involving squares, well done. The only reservation I have is that the velocity of the clock along its horizontal path had to be distorted for it to work. I understand the logic behind it, that the light beam was perceived as completing a longer path than 1 m during the cycle and that was used as the basis for gauging the clock's velocity, assuming that the beam had traveled that path at normal light speed so it took a longer time, thus the clock must have traveled a shorter distance relative to the cycle time.

What if the clock were traveling along a track set up in a lab of sufficient size and we could gauge its velocity along that track? We would find that it moved 0.866 m in the time the light took to complete its cycle, because the light beam really only moved the vertical height of the clock, not a slanted path. A laser beam doesn't propagate on a slant when directed vertically, because somebody would have noticed if it did. Why, then, would it be assumed that the beam had traveled a longer path simply because the clock moved horizontally at the same time? I don't think that's logically plausible. By that logic, if you dropped an object from a certain height and simultaneously threw another matching object horizontally then the one one dropped straight down would hit the ground before the horizontally thrown one, which is not actually the case. Do you consider the thrown object to have traveled a curved path to ground at a greater velocity than the dropped object traveled straight down, or do you just consider it to have traveled horizontally at a certain velocity at the same time as traveling vertically at the speed of gravity? It was two different velocity vectors, not one.

The clock moving horizontally is a similar situation, the light waves moved straight down at 1 c while also moving horizontally at 0.866 c, or at least appearing to move horizontally, but it's velocity was not actually a combination of the two, which would mean it exceeded normal light speed and was therefore seen as requiring correction to bring it back down to normal, using velocity contraction and time dilation. There was never a need for correction because the photons had never traveled any farther than 0.5 m up and down vertically, totally a round trip of 1 m, at a velocity of 299,792,458 m/s.

Now there's the question of whether the light waves/photons were actually inheriting the horizontal motion of the emitter after being emitted vertically. That would make it like a ball tossed vertically by a person on a moving train. What causes the ball to keep moving with the train after being tossed? The obvious answer is inertia. Does light possess inertia? How could it when inertia is a property of mass and light has no mass? So were the vertically propagating light waves really ever moving horizontally with the clock or was it an illusion? One thing is certain, the light was never propagating on a slanted path.

It appears that the moving clock should have been considered as if it weren't moving at all, in which case, photons would not be required to possess inertia to explain the situation. The obvious conclusion is that, when observing a moving frame with a vertical light beam in it, you should consider the situation to be that you see the light clock exactly as you would if neither of you were moving, no time dilation involved whatsoever. If you see the situation that way then how could you see the situation differently if there were two beams in the moving frame, one vertical and one horizontal? You couldn't use time dilation for the horizontal beam and not have it also effect the vertical beam, which never required time correction at all. So how is time dilation a real thing? Maybe muons just decay more slowly when accelerated, how would you know that wasn't the real explanation for muons making it to the earth's surface? Their mass would supposedly be increased, that theory may be valid, and maybe their rate of decay is slowed in direct proportion to that mass increase. That would remove the most popular proof cited for time dilation and length contraction being valid.

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I am sorry, but I don't understand the above description. It still looks to me that you are mixing different reference frames. Could you explain it referring to one frame at a time? There are two frames of interest in this situation that I see: 

Frame A is the one where the clock/mirror is not moving, and the light goes vertically. This is what an observer which is attached to the mirror observes.

Frame B is the one where the clock/mirror is moving to the right, and the light goes diagonally. This is what an observer moving to the left relative to the mirror observes.

What is your objection?

 

Edited by Genady
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1 hour ago, Otto Nomicus said:

You did get it work out, and without complicated equations involving squares, well done. The only reservation I have is that the velocity of the clock along its horizontal path had to be distorted for it to work. I understand the logic behind it, that the light beam was perceived as completing a longer path than 1 m during the cycle and that was used as the basis for gauging the clock's velocity, assuming that the beam had traveled that path at normal light speed so it took a longer time, thus the clock must have traveled a shorter distance relative to the cycle time.

What if the clock were traveling along a track set up in a lab of sufficient size and we could gauge its velocity along that track? We would find that it moved 0.866 m in the time the light took to complete its cycle, because the light beam really only moved the vertical height of the clock, not a slanted path. A laser beam doesn't propagate on a slant when directed vertically, because somebody would have noticed if it did. Why, then, would it be assumed that the beam had traveled a longer path simply because the clock moved horizontally at the same time? I don't think that's logically plausible. By that logic, if you dropped an object from a certain height and simultaneously threw another matching object horizontally then the one one dropped straight down would hit the ground before the horizontally thrown one, which is not actually the case. Do you consider the thrown object to have traveled a curved path to ground at a greater velocity than the dropped object traveled straight down, or do you just consider it to have traveled horizontally at a certain velocity at the same time as traveling vertically at the speed of gravity? It was two different velocity vectors, not one.

The clock moving horizontally is a similar situation, the light waves moved straight down at 1 c while also moving horizontally at 0.866 c, or at least appearing to move horizontally, but it's velocity was not actually a combination of the two, which would mean it exceeded normal light speed and was therefore seen as requiring correction to bring it back down to normal, using velocity contraction and time dilation. There was never a need for correction because the photons had never traveled any farther than 0.5 m up and down vertically, totally a round trip of 1 m, at a velocity of 299,792,458 m/s.

Now there's the question of whether the light waves/photons were actually inheriting the horizontal motion of the emitter after being emitted vertically. That would make it like a ball tossed vertically by a person on a moving train. What causes the ball to keep moving with the train after being tossed? The obvious answer is inertia. Does light possess inertia? How could it when inertia is a property of mass and light has no mass? So were the vertically propagating light waves really ever moving horizontally with the clock or was it an illusion? One thing is certain, the light was never propagating on a slanted path.

It appears that the moving clock should have been considered as if it weren't moving at all, in which case, photons would not be required to possess inertia to explain the situation. The obvious conclusion is that, when observing a moving frame with a vertical light beam in it, you should consider the situation to be that you see the light clock exactly as you would if neither of you were moving, no time dilation involved whatsoever. If you see the situation that way then how could you see the situation differently if there were two beams in the moving frame, one vertical and one horizontal? You couldn't use time dilation for the horizontal beam and not have it also effect the vertical beam, which never required time correction at all. So how is time dilation a real thing? Maybe muons just decay more slowly when accelerated, how would you know that wasn't the real explanation for muons making it to the earth's surface? Their mass would supposedly be increased, that theory may be valid, and maybe their rate of decay is slowed in direct proportion to that mass increase. That would remove the most popular proof cited for time dilation and length contraction being valid.

To your point of a laser beam propagating at an angle, It does.  This is a well understood concept called the aberration of light.  If I put a laser on a moving cart, aimed straight up, and the cart is moving relative to me, I would measure the laser as propagating at an angle other than straight up.  Of course since the speed of light is as fast as it is. the cart would have to be moving at a pretty good clip for me to notice it without very accurate measuring equipment.

A lot of the rest of your post revolves around motion, and appears to treat it as an absolute.  It is not.  This is something Galileo understood.

The "Moving clock" can considered to be at rest, and the "observer" as moving.  It doesn't matter which of the two you consider as moving, the observer will observe the same thing.  Relative to himself, the light travels at angle.  The total distance traveled as measured by them, is longer than that measured by the clock.  A postulate of SR is that light travels at c relative to frame of reference from which it is being measured.

Here's an animation comparing 2 light clocks, one moving relative to the frame, the other not. The white dots are the light pulses bouncing between two mirrors. The circles expand at c to represent the speed of light.

time_dil.gif.fa0d4b0060980654f46d01695c08b8e8.gif

Now as far as the moving clock is concerned, the it's pulse just goes up and down between the  mirrors at c, so it measures 1 "tick" to last the same length of time as the non-moving clock does in this animation.  So for example, both clocks would measure going from 0 to 1 as taking 1 us.

For a horizontal pulse, you need to take length contraction into account, as the
stationary clock would measure the moving clock as being length contracted:

length_con2.gif.2b298f0c8b70c353a3d75faeea405139.gif

You will also note that as far as the stationary clock goes, the horizontal pulse going in one direction take longer for the moving clock. This is an example of the Relativity of Simultaneity.

Also, if we were to switch our viewing frame, so that we saw the clock moving to the right as "stationary" and the other other clock moving to the left. Then it would be the clock moving to the right that would be seen as ticking slower (keeping in mind that we change nothing but which clock we are "following")

As far as muons go, their "mass" does not increase.  Their kinetic energy is high, and thus their momentum. But what is really happening is that the increase in both rise at a different rate than that predicted by Newtonian physics.  So, if you were to apply Newtonian formulas to them, it would seem as if their mass increased, but the Newtonian formulas don't apply properly here.

Besides that, the muon example, while one of the earlier tests of Relativity, was not the only and definitely not the last.  Countless of observations have been made, all giving results affirming Relativity.  You would have to come up with multiple explanations for them all. Explanations which conspired to produce the results of Relativity.

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2 hours ago, Otto Nomicus said:

The only reservation I have is that the velocity of the clock along its horizontal path had to be distorted for it to work.

No. We can all agree on the speed. Your error was in the duration

2 hours ago, Otto Nomicus said:

A laser beam doesn't propagate on a slant when directed vertically, because somebody would have noticed if it did. Why, then, would it be assumed that the beam had traveled a longer path simply because the clock moved horizontally at the same time? I don't think that's logically plausible. By that logic, if you dropped an object from a certain height and simultaneously threw another matching object horizontally then the one one dropped straight down would hit the ground before the horizontally thrown one, which is not actually the case. Do you consider the thrown object to have traveled a curved path to ground at a greater velocity than the dropped object traveled straight down, or do you just consider it to have traveled horizontally at a certain velocity at the same time as traveling vertically at the speed of gravity? It was two different velocity vectors, not one.

A better analogy would be tossing/dropping a ball while on a train. To someone on the train it goes straight down. To someone on the ground, it does not.

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17 minutes ago, Janus said:

To your point of a laser beam propagating at an angle, It does.  This is a well understood concept called the aberration of light.  If I put a laser on a moving cart, aimed straight up, and the cart is moving relative to me, I would measure the laser as propagating at an angle other than straight up.  Of course since the speed of light is as fast as it is. the cart would have to be moving at a pretty good clip for me to notice it without very accurate measuring equipment.

A lot of the rest of your post revolves around motion, and appears to treat it as an absolute.  It is not.  This is something Galileo understood.

The "Moving clock" can considered to be at rest, and the "observer" as moving.  It doesn't matter which of the two you consider as moving, the observer will observe the same thing.  Relative to himself, the light travels at angle.  The total distance traveled as measured by them, is longer than that measured by the clock.  A postulate of SR is that light travels at c relative to frame of reference from which it is being measured.

Here's an animation comparing 2 light clocks, one moving relative to the frame, the other not. The white dots are the light pulses bouncing between two mirrors. The circles expand at c to represent the speed of light.

time_dil.gif.fa0d4b0060980654f46d01695c08b8e8.gif

Now as far as the moving clock is concerned, the it's pulse just goes up and down between the  mirrors at c, so it measures 1 "tick" to last the same length of time as the non-moving clock does in this animation.  So for example, both clocks would measure going from 0 to 1 as taking 1 us.

For a horizontal pulse, you need to take length contraction into account, as the
stationary clock would measure the moving clock as being length contracted:

length_con2.gif.2b298f0c8b70c353a3d75faeea405139.gif

You will also note that as far as the stationary clock goes, the horizontal pulse going in one direction take longer for the moving clock. This is an example of the Relativity of Simultaneity.

Also, if we were to switch our viewing frame, so that we saw the clock moving to the right as "stationary" and the other other clock moving to the left. Then it would be the clock moving to the right that would be seen as ticking slower (keeping in mind that we change nothing but which clock we are "following")

As far as muons go, their "mass" does not increase.  Their kinetic energy is high, and thus their momentum. But what is really happening is that the increase in both rise at a different rate than that predicted by Newtonian physics.  So, if you were to apply Newtonian formulas to them, it would seem as if their mass increased, but the Newtonian formulas don't apply properly here.

Besides that, the muon example, while one of the earlier tests of Relativity, was not the only and definitely not the last.  Countless of observations have been made, all giving results affirming Relativity.  You would have to come up with multiple explanations for them all. Explanations which conspired to produce the results of Relativity.

You're actually showing one ray going straight down and a second ray going over at an angle, because it was emitted at an angle, it's two different light rays. You show it as circular waves but it could also be seen as rays coming out from the point in all directions, diverging from each other as they move farther from the source. A laser beam is not analogous to that at all, which is why it's a beam instead of looking like a light bulb. When you blow smoke in the area of a laser beam you see a thin pencil of light going directly straight out from the emitter. You're not showing a laser, you're showing essentially a mini-star. Now do an illustration with a laser beam emitted straight down and see what happens.

33 minutes ago, swansont said:

No. We can all agree on the speed. Your error was in the duration

A better analogy would be tossing/dropping a ball while on a train. To someone on the train it goes straight down. To someone on the ground, it does not.

 

Would the ball be moving faster than the speed of gravity in the vertical axis because the train is moving horizontally at the same time? 

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3 hours ago, Otto Nomicus said:

Would the ball be moving faster than the speed of gravity in the vertical axis because the train is moving horizontally at the same time? 

The speed along the vertical axis is the same in both frames. The analogy is meant to show the difference in path that you are denying.

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6 hours ago, Otto Nomicus said:

Would the ball be moving faster than the speed of gravity in the vertical axis because the train is moving horizontally at the same time? 

No, the acceleration due to gravity is the same whether you are moving or not.  In fact if you fired a perfectly horizontal gun and at the instant you fired the gun you dropped a bullet from the same height as the barrel, both bullets would hit the ground at the same time.  This scenario would actually probably only work perfectly in a vacuum.

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7 hours ago, Bufofrog said:

No, the acceleration due to gravity is the same whether you are moving or not.  In fact if you fired a perfectly horizontal gun and at the instant you fired the gun you dropped a bullet from the same height as the barrel, both bullets would hit the ground at the same time.  This scenario would actually probably only work perfectly in a vacuum.

Precisely, and that's why the cycle time of the moving clock does not require correcting using time dilation, it didn't take any longer just because a longer slanted path was imagined by Einstein. Thanks for agreeing that I am right, that's always nice.

10 hours ago, swansont said:

The speed along the vertical axis is the same in both frames. The analogy is meant to show the difference in path that you are denying.

That was my entire point though, that the clock's cycle time stays the same regardless of its horizontal motion, thanks for agreeing.

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It seems you did not understand Swansont's point:

18 hours ago, swansont said:

A better analogy would be tossing/dropping a ball while on a train. To someone on the train it goes straight down. To someone on the ground, it does not.

Let's try again with your laser and a train. On the very fast moving train, a light clock is clicking with a vertical 'bouncing' light ray from the laser. In the frame of the train the light clock of course stands still. That means the light beam always goes exactly vertically, because from the frame of the train the light clock is in rest, i.e. it is standing still.

Now you, on the ground frame, you see the train passing by at high speed. This means you see a zig-zag line. The laser has not to point in another direction.

E.g., imagine that the mirror is very small. In the frame of the train there is no problem: adjust the laser so that it points exactly at the small mirror i.e.the the laser is pointed exactly vertically downwards. As the train is standing still in its own frame, this can easily be done. Now you, from the ground frame, do you think you would see that the laser beam would miss the mirror? Of course not, that would be inconsistent. All observers agree on what is occurring physically, they just do not agree on when and where physical events happen, but they see the same physical events. That means for you,that you still see the laser is hitting the small mirror, just as for an observer on the train. But for you the train is moving, so you will see a the beam in a zig-zag line. And because this zig-zag line is longer than the vertical distance between laser and mirror, and c is still c, you see the light clock ticking slower. As Genady showed, you only need x = vt and Pythagoras to derive the correct formula for time dilation. And this time dilation is experimentally tested to the bone. 

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5 hours ago, Otto Nomicus said:

That was my entire point though, that the clock's cycle time stays the same regardless of its horizontal motion, thanks for agreeing.

I’m not agreeing. Are you agreeing that the path lengths are different?

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1 hour ago, Eise said:

It seems you did not understand Swansont's point:

Let's try again with your laser and a train. On the very fast moving train, a light clock is clicking with a vertical 'bouncing' light ray from the laser. In the frame of the train the light clock of course stands still. That means the light beam always goes exactly vertically, because from the frame of the train the light clock is in rest, i.e. it is standing still.

Now you, on the ground frame, you see the train passing by at high speed. This means you see a zig-zag line. The laser has not to point in another direction.

E.g., imagine that the mirror is very small. In the frame of the train there is no problem: adjust the laser so that it points exactly at the small mirror i.e.the the laser is pointed exactly vertically downwards. As the train is standing still in its own frame, this can easily be done. Now you, from the ground frame, do you think you would see that the laser beam would miss the mirror? Of course not, that would be inconsistent. All observers agree on what is occurring physically, they just do not agree on when and where physical events happen, but they see the same physical events. That means for you,that you still see the laser is hitting the small mirror, just as for an observer on the train. But for you the train is moving, so you will see a the beam in a zig-zag line. And because this zig-zag line is longer than the vertical distance between laser and mirror, and c is still c, you see the light clock ticking slower. As Genady showed, you only need x = vt and Pythagoras to derive the correct formula for time dilation. And this time dilation is experimentally tested to the bone. 

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? 

5 minutes ago, Bufofrog said:

Sorry, I don't agree, so I guess that's sad for you.

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.

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