The Two Light Beam Simultaneity Conundrum

[Lorentz Jr]

19 minutes ago, Otto Nomicus said:

The problem with showing a moving version is that t would take different amounts of time for the straight and the slanted beams to reach the end of the car, it would be rather complicated.

It does take different amounts of time for the straight and the slanted beams to reach the end of the car, and calculating the (apparent) desynchronization of moving clocks is somewhat complicated.

19 minutes ago, Otto Nomicus said:

The snapshot version is equally valid without the complications.

The snapshot version is worthless if you ignore the train's motion.

19 minutes ago, Otto Nomicus said:

The beams are considered to be continuously emitted, not pulses.

That makes no difference at all. A continuous beam can just as easily be thought of as a series of "pulses" with no empty spaces between them.

19 minutes ago, Otto Nomicus said:

All you need to do is calculate how long they would take to complete their paths in both frames, I already contracted the length for you, you know the time dilation factor is 2, easy peasy.

No. The light beam has to catch up to the front of the train while the front is moving away from it.

19 minutes ago, Otto Nomicus said:

If somebody wants to make a moving version themselves and make things a lot more complicated, go for it, I'm just not going to do it myself because there's no good reason to.

I'm sorry, Otto, but that's simply not true.

Edited February 7, 2023 by Lorentz Jr

[Otto Nomicus]

1 hour ago, Lorentz Jr said:

It does take different amounts of time for the straight and the slanted beams to reach the end of the car, and calculating the (apparent) desynchronization of moving clocks is somewhat complicated.

The snapshot version is worthless if you ignore the train's motion.

That makes no difference at all. A continuous beam can just as easily be thought of as a series of "pulses" with no empty spaces between them.

No. The light beam has to catch up to the front of the train while the front is moving away from it.

I'm sorry, Otto, but that's simply not true.

Okay, making a moving version then. I can try but no promises.

[Lorentz Jr]

25 minutes ago, Otto Nomicus said:

I can try but no promises.

For the horizontal beam, a "sketch" of the train can be as simple as a horizontal bar that represents the car's 5m contracted length, and the light beam can be a dot that starts at the left end of the first bar. Then a series of sketches at .1sec intervals would each be 86.6cm to the right of the previous one, and the dots would be separated by 1m.

Even simpler, the entire diagram of the car could be an x-vs-t chart of a parallelogram with sides 5m apart, to represent the moving ends of the car, and sloping to the right at 86.6cm for every .1 second. Then the light beam would be a diagonal line (also x-vs-t) that starts at the left corner of the parallelogram and slopes to the right at 1m per .1 second. Whatever works for you. 🙂

Edited February 7, 2023 by Lorentz Jr

[Otto Nomicus]

46 minutes ago, Lorentz Jr said:

For the horizontal beam, a "sketch" of the train can be as simple as a horizontal bar that represents the car's 5m contracted length, and the light beam can be a dot that starts at the left end of the first bar. Then a series of sketches at .1sec intervals would each be 86.6cm to the right of the previous one, and the dots would be separated by 1m.

Even simpler, the entire diagram of the car could be an x-vs-t chart of a parallelogram with sides 5m apart, to represent the moving ends of the car, and sloping to the right at 86.6cm for every .1 second. Then the light beam would be a diagonal line (also x-vs-t) that starts at the left corner of the parallelogram and slopes to the right at 1m per .1 second. Whatever works for you. 🙂

It's just too difficult to do because the slanted beam would actually take on a curved shape because it's moving downward while the car is moving horizontally forward, it wouldn't be a straight slanted line. If you just made a straight slanted line from the starting upper corner to the ending lower corner, the middle part will have been pulled up from where it really was halfway through the trip. You would need some kind of special program to calculate and draw it, it would be something like a brachistochrone curve. It might just be like a part of a large circle whose center is up above the car somewhere. It would be quite difficult to figure the whole thing out really, I can't just crank it out like the other diagram. Maybe someone else here is good at that sort of thing. The 0.855 c is an odd velocity to work with too, maybe 0.5 c would be easier but the Lorentz factor wouldn't be a convenient 2 then, it would be 1.1547, which by the way is sqrt 3 divided by 1.5 while 0.866 is sqrt 3 divided by 2, which makes it easy to bring those precise long form numbers up on a calculator instead of the rounded off forms I just wrote.

[Mike-from-the-Bronx]

forum.txt

[studiot]

3 hours ago, Otto Nomicus said:

The problem with showing a moving version is that t would take different amounts of time for the straight and the slanted beams to reach the end of the car, it would be rather complicated. The snapshot version is equally valid without the complications. The beams are considered to be continuously emitted, not pulses. All you need to do is calculate how long they would take to complete their paths in both frames, I already contracted the length for you, you know the time dilation factor is 2, easy peasy. If somebody wants to make a moving version themselves and make things a lot more complicated, go for it, I'm just not going to do it myself because there's no good reason to.

If anyone has a problem with my instantaneous camera, it's a thought experiment, instant cameras are available in thought experiments. Regarding time taken for the light to reach the camera, Einstein never included the time taken for the light to reach the stationary observer's eyes. Again, it just complicates things. Just pretend I'm Einstein, nobody griped about him doing it that way. I suspect that most of his thought experiments would have quite different results if he had included that. My advice is to simply accept the obvious fact that his theory doesn't work when there is horizontal beam and a slanted beam at the same time, nobody's perfect, including Einstein.

 

 

 

The problem with showing a moving version is that t would take different amounts of time for the straight and the slanted beams to reach the end of the car,

 

OK you have outlined what you see as the problem viz that a horizontal beam takes a different time from a diagonal beam to reach the front of the train.

What you have yet to tell us is exactly what your reasoning is to arrive at the conclusion that this amounts to a conundrum. ?

So please explain with your working. Just pulling figures out of your hat is not enough.

Show your working.

 

As to your dismissive second paragraph.

I will remain polite and hust point out the for the last century Astronomers have been doing exactly this to correctly place the stars in the sky.

An absolutely vital exercise.

[Genady]

4 hours ago, Otto Nomicus said:

Regarding time taken for the light to reach the camera, Einstein never included the time taken for the light to reach the stationary observer's eyes.

This is a direct opposite to the truth. In the 1905 paper where Einstein has introduced SR (einstein_electrodynamics_of_moving_bodies.pdf (umd.edu)), he has started with this exact problem:

Quote

We might, of course, content ourselves with time values determined by an observer stationed together with the watch at the origin of the co-ordinates, and co-ordinating the corresponding positions of the hands with light signals, given out by every event to be timed, and reaching him through empty space. But this co-ordination has the disadvantage that it is not independent of the standpoint of the observer with the watch or clock, as we know from experience.

Then he has solved this problem by including the time taken for the light to travel from position to position in the definition of simultaneity:

Quote

If at the point A of space there is a clock, an observer at A can determine the time values of events in the immediate proximity of A by finding the positions of the hands which are simultaneous with these events. If there is at the point B of space another clock in all respects resembling the one at A, it is possible for an observer at B to determine the time values of events in the immediate neighborhood of B. But it is not possible without further assumption to compare, in respect of time, an event at A with an event at B. We have so far defined only an “A time” and a “B time.” We have not defined a common “time” for A and B, for the latter cannot be defined at all unless we establish by definition that the “time” required by light to travel from A to B equals the “time” it requires to travel from B to A. Let a ray of light start at the “A time” t_A from A towards B, let it at the “B time” t_B be reflected at B in the direction of A, and arrive again at A at the “A time” t'_A.

In accordance with definition the two clocks synchronize if

t_B − t_A = t'_A − t_B

 

[Lorentz Jr]

2 hours ago, Lorentz Jr said:

Even simpler, the entire diagram of the car could be an x-vs-t chart of a parallelogram with sides 5m apart, to represent the moving ends of the car, and sloping to the right at 86.6cm for every .1 second. Then the light beam would be a diagonal line (also x-vs-t) that starts at the left corner of the parallelogram and slopes to the right at 1m per .1 second.

Edited February 7, 2023 by Lorentz Jr

[Markus Hanke]

18 hours ago, Otto Nomicus said:

But I don't have such a proof

A physical paradox can arise only if there is an inconsistency in transformations between frames. Here’s what I mean by this. Say you perform a Lorentz transform to get you from frame A to A’ - by doing so, lengths and times in A’s choice of coordinates will change, since a Lorentz transform is nothing other than a combination of hyperbolic rotations and boosts. A linear transformation, in other words. But what happens if you perform a second Lorentz transformation, identical to the first one, only with a negative argument (-v instead of v)? The already transformed lengths and times associated with A’ will transform again - since this new transform is the same one as the original one, only in the opposite relative direction, one should recover the original frame A. If one doesn’t, the situation is not internally self-consistent.

So, consistency in this context means (L denotes Lorentz transform):

L(v)A -> A’

L(-v)A’=L(-v)L(v)A -> A

A applies L, and arrives at A’. Likewise, A’ applies L, and arrives at A. There is perfect symmetry between these frames. In a situation where such consistency holds, it is not possible to construct any physical paradoxes based on this linear transformation alone, because any pair of inertial frames will always agree on how they are related to one another.

So, in order to show that one cannot construct physical paradoxes in Minkowski spacetime based on the axioms of SR, it is sufficient to show that every Lorentz transformation has a unique and well defined inverse, such that \(L L^{-1}=I\), wherein I is the identity element. Suffice to say that this is indeed the case, and if you want I can present the formal proof here (or you can just Google it yourself).

Edited February 8, 2023 by Markus Hanke

[Otto Nomicus]

I guess the slanted beam couldn't really come out as curved but however it would come out, it would be ridiculously difficult to figure out or illustrate. For instance, do you use the real angle, 45 degrees, or the angle with the box contracted into a 10 x 5 rectangle? In fact, would you gauge the end of the trip to be when the back of the box reaches a certain distance from where it started or when the front end of it does? It's contracted to half width so either the back or the front would have traveled a different distance than if it was the original square box. This is how ridiculous things get with SR, you really don't know what the heck you're supposed to use for distances traveled and therefore velocities either. You don't what shape of the object to use for either of those, and it can't be the same for the contracted and uncontracted versions. If we can't depend on physical objects to retain their actual shape then we can't do much of anything with them with any certainty.

Suppose you start with a 10 m x 10 m box and a 20 m track. If in some reference frame it was moving at 0.866 c, when the front end gets to the end of the track, how many box lengths did it move, 1 or 3? How do you calculate its velocity, as moving 10 m in a certain amount of time or as moving 15 m in the same amount of time? There's a 50% difference in the two velocities, which one is the real velocity? Or does the object contract from both ends toward the middle? In that case, in both forms its middle would move 20 m, if it started with its middle at the start of the track and ended when its middle was at the end of the track, but both ends of it would have moved a different distance depending on whether it was square or rectangular. I guess we would have to go with the contracted version so it's front end would reach a point 2.5 m in front of the end of the track and then when it stopped there, its front end would thrust out another 2.5 m and the back end would thrust 2.5 m backward. How far did its front end really travel then, and how far did its back end travel, obviously 5 m less than the front end. The whole thing is rather ludicrous.

Edited February 8, 2023 by Otto Nomicus

[studiot]

I still don't see any answer to my question.

What exactly do you see and the conundrum itself, given that all agree the transit times of the horizontal and diagonal beams are different ?

 

[Otto Nomicus]

21 minutes ago, studiot said:

I still don't see any answer to my question.

What exactly do you see and the conundrum itself, given that all agree the transit times of the horizontal and diagonal beams are different ?

 

The fact that when you draw a diagram of the contracted box the timers require two different time offsets to work with each beams. It doesn't matter that one beam takes slightly longer than the other for its trip, the time offset would be the same, because the two clocks run at the same rate though their times are offset by a constant amount. I also maintain that the "snapshot" diagram I made is the only valid way to test Einstein's theories, because it's actually impossible to make a depiction of the box moving, because which shape would you use, square or rectangular, and which angle would you use for he beam, 45° or 26.5650512°? According to Einstein's theory, the snapshot is exactly what you would see at any instant in the box's transit so it is an entirely valid way to check his theories, time dilation, length contraction and relativity of simultaneity. It looks to me like they failed the test. I must conclude that the whole thing is pure fantasy.

[studiot]

15 minutes ago, Otto Nomicus said:

The fact that when you draw a diagram of the contracted box the timers require two different time offsets to work with each beams. It doesn't matter that one beam takes slightly longer than the other for its trip, the time offset would be the same, because the two clocks run at the same rate though their times are offset by a constant amount.

 

I see no mention of 'offsets' in Einstein's theory, can you reference it ?

However I would ask why you would not expect them to be different? They two beams follow different trajectories through any observers spacetime.

Just saying they are different (as everyone agrees) does not make it a conundrum, contradiction or paradox.

You need to establish a connection between the two beams which when applied leads to a contradiction.

It is your claim not mine, so it is up tp you to work it out.

I asked several times for your step by step reasoning, preferably with the maths spelled out. I have checked the very small amount of maths you have done and posted where I agreed.

[Lorentz Jr]

5 hours ago, Otto Nomicus said:

I guess the slanted beam couldn't really come out as curved

Good! I'm glad you figured that out yourself, Otto. Only gravity or an observer in a noninertial reference frame can make a light beam appear to be curved.

5 hours ago, Otto Nomicus said:

but however it would come out, it would be ridiculously difficult to figure out or illustrate.

Difficult? Yes, if you don't know the rules. Ridiculously? Not really. You just have to forget about common sense for a moment, focus on events, and use the Lorentz transformations.

5 hours ago, Otto Nomicus said:

For instance, do you use the real angle, 45 degrees, or the angle with the box contracted into a 10 x 5 rectangle?

You use 45 degrees inside the train (because that's something you know), and then you apply the Lorentz transformations to the end of the diagonal light beam's path.

Inside the train, [math]x' = L[/math].

The distance traveled by the diagonal beam is [math]s' = \sqrt{2}L[/math],

so [math]\displaystyle{ c = \frac{s'}{t'} = \frac{\sqrt{2}L}{t'}}[/math]

[math]\displaystyle{ t' =  \frac{\sqrt{2}L}{c}}[/math]

Now transform to the ground frame:

[math]\displaystyle{ x = \gamma (x' + vt') = \gamma \left(L + v \frac{\sqrt{2}L}{c}\right) = 2(10m)(1 + .866\sqrt{2}) = 44.5m  }[/math]

[math]\displaystyle{ t = \gamma \left(t' + \frac{vx'}{c^2}\right) = \gamma \left(\frac{\sqrt{2}L}{c} + \frac{vL}{c^2}\right) = \frac{2(10m)}{10m/s}\left(\sqrt{2} + .866\right) = 4.56s  }[/math]

 

5 hours ago, Otto Nomicus said:

In fact, would you gauge the end of the trip to be when the back of the box reaches a certain distance from where it started or when the front end of it does?

The end of the trip is when the light beam reaches the front of the car.

5 hours ago, Otto Nomicus said:

It's contracted to half width so either the back or the front would have traveled a different distance than if it was the original square box. This is how ridiculous things get with SR, you really don't know what the heck you're supposed to use for distances traveled and therefore velocities either.

You're not supposed to use anything except the Lorentz transformations and whatever specific information the problem gives you.

5 hours ago, Otto Nomicus said:

The whole thing is rather ludicrous.

The process can be a bit tricky and counterintuitive, but it's not horribly bad once you get used to it. 🙂

Edited February 8, 2023 by Lorentz Jr

[Otto Nomicus]

In the diagram below, the dimensions are in meters, the velocity is to the right at 0.866 c and the value for c is 1 m/s. 

For the travel time of the horizontal beam, we saw the box move 4.330127 m while the beam moved 5 m, so we saw that take 5 seconds. Note that although the beam appears in the diagram to be stretched out to 9.330127 m, we did not see 9.330127 seconds pass on our timer. If we had seen that much time pass while the box moved 4.330127 m then the box would only have been moving at a velocity of 0.4641 m/s, but it was really moving at 0.866 m/s, because we set the value of c in this thought experiment at 1 m/s and the velocity of the box at 0.866 of that value for c. We can't arbitrarily change the box's velocity at this point because if we don't have a predefined velocity then we can't do anything, the length contraction and time dilation factors would have to be different so the box wouldn't even appear to be the shape shown in the diagram. The Lorentz factor for 0.4641 c is 1.128946, so the length of the box would be 10 divied by that= 8.8578196 m. The same thing would then happen with the revised diagram so it would be an infinite series of redrawing the diagram with different velocities and therefore different Lorentz factors.

With the fact now amply established that we would see 5 seconds pass during the horizontal beam's trip, we need to get it to look like 2.5 seconds in the box frame, because Einstein says we need to dilate time in the box frame by a factor of 2, so we need the front timer to be 2.5 seconds behind the back timer. 

 During the travel time of the slanted beam, we saw the box move 9.68245832 m at a rate of 0.866 m/s, so we saw it take 11.1803398 seconds for the slanted beam's trip. Note that we did not see it take 17.764419 seconds, which is the length of a slanted line, in meters, made by joining the front end of the beam where it is at various points along its trip while the box moves to the right. The magenta slanted line shows how far the beam got at the time when the horizontal beam had reached the front end of the box, so that's how we got the 17.764419 long slanted line shown in blue. That is not the light beam itself, just the positions of its front end as time progressed. This is the same principle as with the horizontal beam not really being the stretched length it appears to be in the diagram.
 
 Inside observers saw the real slanted beam take 14.1421 seconds in their time, because that's the length of the beam in the uncontracted box, but we saw it take 11.1803398 seconds, and we need to time dilate it to half that, so we then have 5.5901699 seconds so we need to get 11.1803398 to look like 5.5901699. Subtracting 5.5901699 from 11.1803398 gives us a time of 5.5901699 which needs to be subtracted from the front timer. That's a different time offset (desynchronization) than we got for the horizontal beam, which was 2.5 seconds, so Special Relativity failed the test for validity. It also doesn't work if we use the stretched versions of both beams, so that wouldn't save the theory. You can work it out yourself if you want, I did but I don't want to have to write it all out.

The fact is, Einstein made the same error as I explained above when he did his moving light clock thought experiment and equated the light pulse positions at various points along the clock's horizontal travel to a slanted beam, when that was not the light beam at all, or the path of the light pulse, whichever way you want to put it. Einstein failed to notice the Lorentz factor "doom loop" that would result from that error, or he just purposely never mentioned it because he knew it would invalidate his whole theory, you would have to ask him which it was.
 
 

[Lorentz Jr]

59 minutes ago, Otto Nomicus said:

In the diagram below, the dimensions are in meters, the velocity is to the right at 0.866 c and the value for c is 1 m/s. 

The distance of the horizontal beam's trip only depends on L and v/c, both of which you've kept the same, so it still has to be 37.3 meters. And t = v/c, so the time is 37.3 seconds.

59 minutes ago, Otto Nomicus said:

With the fact now amply established that we would see 5 seconds pass during the horizontal beam's trip, ...

After five seconds, the front of the car is at 9.33 meters and the beam is at 5 meters. The beam still hasn't reached the front.

In other words, you've made exactly the same mistake you made in the c = 10m/s example.

And the same thing applies for the diagonal beam: The distance is the same as before and the time is ten times longer:

[math]\displaystyle{ x = \gamma L \left(1 + \sqrt{2} \frac{v}{c}\right) = 2(10m)(1 + .866\sqrt{2}) = 44.5m  }[/math]

[math]\displaystyle{ t= \gamma \frac{L}{c}\left(\sqrt{2} + \frac{v}{c}\right) = \frac{2(10m)}{1m/s}\left(\sqrt{2} + .866\right) = 45.6s  }[/math]

Edited February 9, 2023 by Lorentz Jr

[swansont]

8 hours ago, Otto Nomicus said:

For the travel time of the horizontal beam, we saw the box move 4.330127 m while the beam moved 5 m, so we saw that take 5 seconds

If “we” can see the box move, then “we” are in the lab frame. “we” would not see it take 5 sec for the light to hit the far wall, because the wall moves - the light travels more than 5 m. 

 

We can look at this like a light clock if we do a round trip

The light travels ct1 to the right, and the travels a distance d + vt1 (distance across the car + distance the wall travels), so these are equal

(c-v)t1 = d  

37.32m, or 37.32 sec

On the return trip the light travels ct2 to the left while the wall is moving vt2 to the right

(c+v)t2 = d

2.68 seconds

Total is 40, while in the train-car frame, the light travels 20m round-trip. The moving car clock runs slow by a factor of 2, as expected

[Lorentz Jr]

8 hours ago, Otto Nomicus said:

The fact is, Einstein made the same error as I explained above when he did his moving light clock thought experiment and equated the light pulse positions at various points along the clock's horizontal travel to a slanted beam, when that was not the light beam at all, or the path of the light pulse, whichever way you want to put it. Einstein failed to notice the Lorentz factor "doom loop" that would result from that error, or he just purposely never mentioned it because he knew it would invalidate his whole theory, you would have to ask him which it was.

The fact is, you have been either unable or unwilling to solve the problems in this thread correctly, and all of your claims and suppositions have turned out to be incorrect.

Your "analysis" of these problems is equivalent to saying that a car will collide with a slightly slower-moving car ahead of it, simply by reaching the point on the road where the leading car was when you started your analysis. Even in the old physics of Isaac Newton, that's not how moving objects work. The leading car has moved on, so the trailing car still needs to catch up to it.

Edited February 9, 2023 by Lorentz Jr

[Otto Nomicus]

No, my post is entirely correct, no idea what either of the above replies are talking about, but it sure isn't reality. Oh the horizontal beam took 5 seconds alright, it's obvious, you just don't realize where Einstein made his grievous error. I think I explained it clearly enough, I suggest reading as many times as required to comprehend it. I will illustrate the problem with this diagram of a square light clock  moving at 0.866 c and c being 1 m/s. There are two beams, the usual vertical one, magenta, but with an additional horizontal one, red. The square is really a 10 m x 20 m rectangle when stationary but is shown length contracted. Both beams are the same length in the contracted version so how did they get to take two different amounts of time to complete their trips across the square? Obviously, the green line is not in fact the vertical beam, the magenta line is, and the red line is not really the horizontal beam, it's a distorted version. How could you correct the red line and the green line using the same timer desynchronization? They're not the same length, nor is the red line twice the length of the green line, they appear as distorted versions of their true lengths, but distorted in different ways. This presents an unresolvable conundrum for Special Relativity and shows how Einstein's light clock thought experiment is flawed and invalid as the basis for his theory.

Edited February 10, 2023 by Otto Nomicus

[Lorentz Jr]

1 hour ago, Otto Nomicus said:

No, my post is entirely correct, no idea what either of the above replies are talking about, but it sure isn't reality. Oh the horizontal beam took 5 seconds alright, it's obvious, you just don't realize where Einstein made his grievous error.

Well, I can't stand this "conversation" anymore, so here's another pretty picture for anyone who's interested, just for fun:

Edited February 10, 2023 by Lorentz Jr

[Otto Nomicus]

7 hours ago, Lorentz Jr said:

Well, I can't stand this "conversation" anymore, so here's another pretty picture for anyone who's interested, just for fun:

Yeah that's real pretty, too bad you can't refute my last two diagrams credibly though. No idea what your diagram is supposed to depict, sure doesn't look like anything square or rectangular though. My diagrams are simple and logical, yours appear to be based on a severely distorted version of reality. By the way, I know that the vertical beam in my last diagram wouldn't be affected by the timer desynchronization but it still illustrates how Einstein made the mistake of thinking the actual light beams, or light pulse path, was stretched out just because the light clock was moving horizontally but how, if that were true, it would change the velocity of the light clock, same as I explained in the previous post.

Do you seriously think it's scientifically acceptable to have velocity change dramatically just to shoehorn a theory into seeming to work in other respects? Either we know the velocity of the clock or we don't, which is it? What's the velocity of the clock. 0.866 c or 0.4641 c? I posted another thread a while back about the "light clock conundrum" and the only way anybody could explain how the Lorentz factor could be valid for the light clock diagram in that one was to do just that, suggest that the velocity was different than what it obviously was and therefore a completely different Lorentz factor was appropriate. Does Special Relativity contract velocity? I must have missed that part of Einstein's article, please quote the part about velocity contraction, or admit that there is an unresolvable conundrum.

[Lorentz Jr]

39 minutes ago, Otto Nomicus said:

What's the velocity of the clock. 0.866 c or 0.4641 c?

Where do you get 0.4641 c from?

[Otto Nomicus]

50 minutes ago, Lorentz Jr said:

Where do you get 0.4641 c from?

In my square light clock diagram, the square moved 8.660254 m while the length of the stretched out horizontal light beam was 18,660254 m, dividing the former by the latter provides the velocity of the square relative to the speed of light, 0.46410161405091270461806146904538, but we know it was really 0.866 light speed, thus, a conundrum exists That's how you know that the stretched out appearance is not the actual light beam, just like the zig-zag line in Einstein's light clock thought experiment is not the actual light pulse, or beam if we use a modern form of the thought experiment with a laser beam. Obviously the light beam in my diagram only moved 10 m, not 8.660254, and therefore only 10 seconds had passed, if we set c at 1 m/s, so no time dilation is required, it's exactly how long you would expect it to take. It doesn't matter how it looks, that's how long it had to take unless velocity contraction is a thing in SR, which it's not.

Edited February 10, 2023 by Otto Nomicus

[Lorentz Jr]

1 hour ago, Otto Nomicus said:

Do you seriously think it's scientifically acceptable to have velocity change dramatically just to shoehorn a theory into seeming to work in other respects? Either we know the velocity of the clock or we don't, which is it? What's the velocity of the clock. 0.866 c or 0.4641 c?

1 hour ago, Lorentz Jr said:

Where do you get 0.4641 c from?

33 minutes ago, Otto Nomicus said:

In my square light clock diagram, the square moved 8.660254 m while the length of the stretched out horizontal light beam was 18,660254 m, dividing the former by the latter provides the velocity of the square relative to the speed of light, 0.46410161405091270461806146904538

Then why did you make all those comments about "Do you seriously think it's scientifically acceptable" and "Either we know the velocity of the clock or we don't, which is it?" 0.4641 c is from your calculation, not mine or anyone else's, and the light beam only travels ten meters in ten seconds, not 18.66 meters. This "stretched out light beam" idea is just another one of your little fantasies.

Edited February 10, 2023 by Lorentz Jr

[Otto Nomicus]

1 hour ago, Lorentz Jr said:

Then why did you make all those comments about "Do you seriously think it's scientifically acceptable" and "Either we know the velocity of the clock or we don't, which is it?" 0.4641 c is from your calculation, not mine or anyone else's, and the light beam only travels ten meters in ten seconds, not 18.66 meters. This "stretched out light beam" idea is just another one of your little fantasies.

I think you're probably the only person who didn't comprehend what I wrote, but I'll humor you. What do you say the velocity of the square and rectangle in the two diagrams were? If it's not 0.866 c or 0.4641 c then what is it? Are you going to make up some random velocity now?

Edited February 10, 2023 by Otto Nomicus