Otto Nomicus

I don't believe you or anyone in this forum is really interested in science, just repeating what anyone can read in wikipedia. I certainly understand SR and I certainly don't need to learn about it from people on a forum, I read the actual articles. But I suppose you can explain Einstein's theory better than him. Anyway, I'm obviously wasting my time here, so go back to rehashing wikipedia pages to people who haven't heard it exists, because I won't be posting here anymore, ever, none of you are worth the effort.

Do you have a credible refutation of my posts or is your position that SR is right so I have to make everything work to conform to it or I didn't do it right? It has never actually been proved that the one-way speed of light is constant in all frames as viewed from any other. How could it when nobody has ever measured the one-way speed of light in the history of mankind? Two-way speed of light can be the same while each of its one-way components are completely different, ie; 1 and 3, 2 and 2, 1.5 and 2.5, 1.25 and 2.75, I could go on all day. I could make a two-way diagram but it would be a lot more complicated and wouldn't help your case any, it would actually make it worse. Why don't you make a two-way diagram, Swansont, if you're so fond of the notion, or can't you make a diagram, just gripe about the ones made by others not being up to your standards?

Interesting that Genady considers disproving Special Relativity as trolling. I suppose if he can't actually refute my posts intelligently then throwing insults around is the next best thing. Same for Lorentz Jr. You fail to realize that the moving frame is not chasing its own light, it's exactly like it's not moving at all, you just think it has to chase its own light. It also doesn't matter if I do a two way version, I've done that on other forums actually. I'll see If I can find a diagram.

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?

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.

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.

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.

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.

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.

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.

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.

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

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.

I already fixed my original first post description up in the diagram post. The text in that post had an error in the last paragraph but I corrected it in a later post. To avoid confusion please disregard that post and use the revised version which I will post right after this. Sadly, I can neither edit nor delete the first version. REVISED VERSION Here's a snapshot taken of the train car as it passes directly in front of a ground observer with an instantaneous camera. The red and green lines are continuous laser beams. The velocity is known to the ground observer to be 0.866 c because it was previously set up in the lab to move at that velocity in the lab frame. Length contraction has reduced the apparent length from 10 to 5 m, as the observer knows is predicted by SR, and the observer knows that time is slowed down to half the rate of his because the Lorentz factor for that velocity is 2 Through trigonometry, he knows that the diagonal line has gone from being 14.1421 m when the car was stationary to 11.1803399 m in the snapshot. Knowing that the speed of light in his world is 1 m/s, the ground observer knows that from his point of view a light beam traveling that distance would take 0.5 second but he knows that observers in the car must see it take 1 second of their time, because they see it travel 10 m instead of 5, the car being uncontracted in their frame. How does he make 0.5 second of his time look like 1 second in the car? He imagines a timer at the back wall of the car showing 0 and a timer at the front wall showing 0.75 second when the light beam is emitted from the back wall, so that by the time the beam reaches the front wall the timer will have progressed by 0.25 second in that frame, while 0.5 second has passed in his, and will then show 1 second. That would seem to reconcile things but he recalls that Einstein stated that clocks at the back of moving frames will be seen by a stationary observer to be ahead of ones at the front so he finds this perplexing, because he had to imagine the front clock being ahead of the back one instead. The ground observer then sees what would happen with the diagonal beam with that same timer desynchronization. He would see the 11.1803399 m beam take 1.11803399 second in his frame but since time passes at half the rate in the car frame he would see it take 0.559016995 second in that frame. So with the front timer being ahead of the back one by 0.75 second, it would show 1.309016995 for the travel time, but it was supposed to show 1.41421 second in the car frame so it was short by 0.105193005 second. That's the conundrum, in addition to the timers being desynchronized in reverse order from what Einstein stated.

But I don't have such a proof, because the reality of the diagram appears to indicate that in fact there can be physical paradoxes.