Einstein's clock

[RJBeery]

Since the polygon is so large, observer B's clock will still lag behind observer A's clock after one trip around the polygon. Yet both observers found the speed of light to be c (practically) the whole time, right?

No. You're just taking the acceleration observer B would experience if he were to traverse around a similarly large circle and concentrating it to tighter arcs separated by stretches of inertial movement. You can't just dismiss the "slight gravity-like pull" of turning through these arcs (at relativistic speeds) and then claim that "practically the whole time" he was travelling with no acceleration. He certainly would experience very significant acceleration if he's moving fast enough to induce a time-dilation effect on his local clock compared to observer A, and this is going to be true regardless of the size of the circle (or polygon).

 

[Neddy Bate]

You're just taking the acceleration observer B would experience if he were to traverse around a similarly large circle and concentrating it to tighter arcs separated by stretches of inertial movement.

I agree with that.

You can't just dismiss the "slight gravity-like pull" of turning through these arcs (at relativistic speeds) and then claim that "practically the whole time" he was travelling with no acceleration.

Actually I had changed the speed to be slower, so maybe it's not very relativistic.

He certainly would experience very significant acceleration if he's moving fast enough to induce a time-dilation effect on his local clock compared to observer A, and this is going to be true regardless of the size of the circle (or polygon).

You seem to be saying that the time dilation effect will depend only on the speed, and not on the size of the polygon. If so, then I disagree. I can make the polygon larger without changing the speed, and I would expect observer B to have a greater time dilation effect.

Anyway, your explanation that the speed of light is not c in the accelerating frame still works, it's just that it only holds for the corners of the polygon, not the sides. Yet your explanation holds for the entire circle, so I suppose it is a good enough explanation.

 

[RJBeery]

Anyway, your explanation that the speed of light is not c in the accelerating frame still works, it's just that it only holds for the corners of the polygon, not the sides. Yet your explanation holds for the entire circle, so I suppose it is a good enough explanation.

Chinglu has become conspicuously quiet. Perhaps he agrees.

 

[chinglu]

You haven't provided any maths. I asked you to do the specific calculations. All you can do is arm wave. This is why you've failed to make your case. You haven't justified your position, you've just repeatedly asserted it. And as we've all seen previously with your other light sphere claims or claims about Cantor, you're quite happy to assert as irrefutable things which are easily refuted. If you could formalise your claims you'd see them to be false but you aren't able to do so.

Your ignorance is blinding you.

I have already been through this.

I gave you a 6 step procedure to the conclusion.

I'll ask you again, which step in the logic is false?

 

[chinglu]

This has all been explained to you, time and again, at great length, and you simply deny it.

I don't know why anyone bothers.

Nothing has been explained to me that is valid.

Now, I gave AN a 6 step set of operation that lead to the conclusion.

Which step is false?

If you can't provide an answer, clearly I am offering logic beyond your grasp.

 

[chinglu]

You're just an ineducable dunce. You should find another hobby. The time dilation doesn't matter goofball. I wrote the whole thing out in detail at the beginning of this stupid thread. All you're doing is putting !!! on either your stupendous intellectual dishonesty or ineducable stupidity. Probably both. Good for you you're stubborn ignorance wins out.

If you are claiming time dilation does not matter, then you are claiming Einstein's statement regarding time dilation and a clock moved in a circle is false.

And, I already refuted any contention you had that contradicted my conclusions.

Thus far, you are only flailing around.

Now, I provided a 6 step argument for AN of which no one can refute the ultimate conclusion that SR is false.

Can you refute the 6 steps I provided, yes or no.

 

[chinglu]

actually, i think i just realized that c = d/(t/y) it would have too in order to maintain the constant speed of light. Where y is gamma or sqrt(1-v^2/c^2). If you ever tried to determine c by the dialated distance and the dialated time you would get something like c=d'/t', but that statement couldn't be true because it has two gammas. They would both cancel and then get the same answer if there was no time dialation at all. So then there must be some missing link or rule that get's rid of one of the gamma's. Thanks, i will have to rethink my own interpretation of sr.

ok...

 

[chinglu]

In geometric units this is the local coordinate speed of light

dr/dt=1

Gamma has nothing to do with the derivation of the local coordinate speed of light. It's derived from the geometry of spacetime [the metric]. And gamma isn't what you said it is.

This is gamma

y = 1/(1-v^2)^1/2 = (1-v^2)^-1/2

Just think about this: What's the value of v^2 for 1 = dr/(dt/y). Figure it out.

First, I gave the gamma given by Einstein which was simply an approximation given by him.

Next, where you have failed is that let's assume each measures c with different times on their clocks. Then given the fact that each must measure the same units in the y direction, your conclusion implies one light sphere is located at 2 different places on a common y axis for the two clocks, which is a contradiction of nature.

So, your logic is a complete failure.

 

[chinglu]

Why? What's difficult about having a mirror at a fixed distance from the stationary clock?

Without a mirror to reflect the emitted spherical wave, how does either of the observers know where it is after t seconds, or how to measure c?
I introduced it so you have a precisely defined interval of distance: ct/2, along with the interval t.
But the clock that moves in a closed curve sees this distance as length contracted. It doesn't matter how the stationary clock defines distance, because the moving clock will see any distance in the frame defined by t as Lorentz-contracted.

But you obviously have no idea what rigour means. You have your dumbass interpretation which is clearly confused and doesn't agree with Einstein's actual gedanken. That's all you got.
Plus of course, your belief that it's wrong plus your inability to engage at any sort of meaningful level.

I'd say your looking at three strikes. What's that? You need more alcohol?

If you want to use mirrors, be my guest and open a new thread.

This thread is about the experiment I provided and the precise 6 step sequence of reasoning I provided to AN.

Now, which step is false?

 

[AlphaNumeric]

Now, I provided a 6 step argument for AN of which no one can refute the ultimate conclusion that SR is false.

So submit it to a journal and claim your Nobel Prize.

I gave you a 6 step procedure to the conclusion.

You didn't provide a single drop of actual SR calculation. The onus is on the person making the claim, the onus is on you. I have had enough of walking you through remedial maths and physics and you refusing to listen. If you cannot justify your claim we have nothing to discuss. Asserting is not justifying.

Unfortunately I believe you are simply too stupid to grasp relativity. You go around saying things like "The conclusion is therefore I have refuted relativity" or "The conclusion is therefore I have refuted Cantor" when you haven't done anything of the sort. Your willingness to lie, either through malice or ignorance, about our discussions about Cantor shows you have no intellectual integrity.

If you're so sure you're right why are you here? Why are you telling us when you could be telling journals? You've already shown you won't listen to us when we correct you, both in terms of maths and concepts, and you believe we're all insufficiently capable to grasp what relativity or Cantor supposedly imply. You don't want a discussion, you want a monologue.

You've given 6 points. 6 assertions. Demonstrate they are valid and lead to the conclusion you claim they do by doing the maths. You haven't done it yet so we have nothing to discuss other than your dishonesty.

 

[chinglu]

No. You're just taking the acceleration observer B would experience if he were to traverse around a similarly large circle and concentrating it to tighter arcs separated by stretches of inertial movement. You can't just dismiss the "slight gravity-like pull" of turning through these arcs (at relativistic speeds) and then claim that "practically the whole time" he was travelling with no acceleration. He certainly would experience very significant acceleration if he's moving fast enough to induce a time-dilation effect on his local clock compared to observer A, and this is going to be true regardless of the size of the circle (or polygon).

Make sure you include Einstein's statement in with your reasoning or indicate why it is false.

Otherwise, the angular acceleration is too small to consider. And otherwise, give its effect on the moving clock as t'.

I have already been through this.

Then, remove it from the calculations and you are left with SR time dilation.

 

[chinglu]

So submit it to a journal and claim your Nobel Prize.

You didn't provide a single drop of actual SR calculation. The onus is on the person making the claim, the onus is on you. I have had enough of walking you through remedial maths and physics and you refusing to listen. If you cannot justify your claim we have nothing to discuss. Asserting is not justifying.

Unfortunately I believe you are simply too stupid to grasp relativity. You go around saying things like "The conclusion is therefore I have refuted relativity" or "The conclusion is therefore I have refuted Cantor" when you haven't done anything of the sort. Your willingness to lie, either through malice or ignorance, about our discussions about Cantor shows you have no intellectual integrity.

If you're so sure you're right why are you here? Why are you telling us when you could be telling journals? You've already shown you won't listen to us when we correct you, both in terms of maths and concepts, and you believe we're all insufficiently capable to grasp what relativity or Cantor supposedly imply. You don't want a discussion, you want a monologue.

You've given 6 points. 6 assertions. Demonstrate they are valid and lead to the conclusion you claim they do by doing the maths. You haven't done it yet so we have nothing to discuss other than your dishonesty.

You are not telling the truth.

I provided the complete mathematics to solve the problem. Here it is for the 3rd time.

Also, I gave you a 6 step procedure to my conclusion. You have still been completely unable to refute any of the steps that lead to the fatal conclusion that SR is logically false. What is the problem?

It is quite clear, here is the math refute it or submit.

Further, in this thread, I am very specific, a light pulse is emitted when the clocks are common.

One clock moves in a circle and returns to the other clock.

Here is Einstein's statement.

If we assume that the result proved for a polygonal line is also valid for a continuously curved line, we arrive at this result: If one of two synchronous clocks at A is moved in a closed curve with constant velocity until it returns to A, the journey lasting t seconds, then by the clock which has remained at rest the travelled clock on its arrival at A will be second slow.

Next, since the y-axis is perpendicular to the line of travel, it is not length contracted. So, assume both frames the distance the pulse traveled is d.

Also, assume the time on the clock with the stationary observer is t.

By SR, c = d/t.

However, since the moving clock moves in a circle, then there exists some very very small time differential from the stationary clock, say t'.

Then, we must apply Einstein reasoning, the moving clock shows a time of t/γ.

So, the actual time on the moving clock is t' + t/γ.

According to Einstein, c is a constant between the frames and time dilation is a result of this assumption.

Now, since t' is absolute, we can remove t' from the calculations and all we have left is what Einstein claimed as the time on the moving clocks as t/γ.

But, that means, c' = d/(t/γ) for the moving clock.

We also have c = d/t for the stationary clock. But, under SR all observers must measure c as the speed of light.

Hence, c = d/t = c' = d/(t/γ). This means γ=1.

But, if γ=1, then v = 0, which is a contradiction.


There is the specific math and logic.

Your turn with your math.

 

[Prof.Layman]

I don't see the contradiction, I would think that the only time d/t=d/(t/y) is when y=1. That would in a sense be the only place the two lines intersect. At this point the v=0 and at all other points d/t =/= d/(t/y). This would be because the dialated frame of reference has become the undilated frame at this velocity of zero. I think you are treating two coordinate systems as if they are the same system, you have to figure out how to treat them as two different systems. After all, that is what the prime symbol represents. So then these two coordinate systems only produce the same answer at v=0. That makes perfect sense.

 

[arfa brane]

Next, since the y-axis is perpendicular to the line of travel, it is not length contracted.

That is only true for the stationary clock. It's been explained to you several times: the moving clock (or the clock that moved) sees any distance in the t frame as length-contracted.
so

assume both frames the distance the pulse traveled is d.

does not follow.

But you're too attached to "Einstein was wrong", aren't you? Noone can point out a mistake to someone with wilful ignorance as a constant companion.
What you have is a failure of logic and the stubbornness to maintain you don't. Congratulations; I bet you're a big hit at parties.

 

[chinglu]

I don't see the contradiction, I would think that the only time d/t=d/(t/y) is when y=1. That would in a sense be the only place the two lines intersect. At this point the v=0 and at all other points d/t =/= d/(t/y). This would be because the dialated frame of reference has become the undilated frame at this velocity of zero. I think you are treating two coordinate systems as if they are the same system, you have to figure out how to treat them as two different systems. After all, that is what the prime symbol represents. So then these two coordinate systems only produce the same answer at v=0. That makes perfect sense.

v=0 means the clock never left.

 

[chinglu]

That is only true for the stationary clock. It's been explained to you several times: the moving clock (or the clock that moved) sees any distance in the t frame as length-contracted.
so does not follow.

But you're too attached to "Einstein was wrong", aren't you? Noone can point out a mistake to someone with wilful ignorance as a constant companion.
What you have is a failure of logic and the stubbornness to maintain you don't. Congratulations; I bet you're a big hit at parties.

I have tried to teach you SR. Maybe Einstein can.

Thus, whereas the Y and Z dimensions of the sphere (and therefore of every rigid body of no matter what form) do not appear modified by the motion
http://www.fourmilab.ch/etexts/einstein/specrel/www/

 

[arfa brane]

I have tried to teach you SR. Maybe Einstein can.

Please, please teach me what the correspondence between motion (of say, a clock), a rigid sphere, and propagation of light is.

But if you're implying that a spherical wavefront of light is like a rigid sphere, then you got it wrong.
You twat.


In the clock-moving-in-a-closed-curve scenario, if the y axis of the stationary clock is perpendicular to the xz plane the moving clock moves in, the moving clock doesn't see this axis as perpendicular does it? In that case, the moving clock has to account for length contraction in the y direction, relative to the coordinates of the stationary clock. In fact distances in the x and z directions from the stationary clock are also length-contracted in the moving clock's frame, because the moving clock moves away from all three of the stationary clock's 'axes' in R[sup]3[/sup] as it traverses the xz plane relative to same. Doesn't it?

 

[RJBeery]

Place a mirror 1LY away and perpendicular from the ground, flash a light at it just as observers A and B start their synced clocks. Observer A sits patiently for 2 years while observer B quickly runs in circles. When the flash of light returns observer B will declare that the light traveled for less than 2 years implying that its velocity exceeded c. I don't think anyone denies that this would be the result of this experiment, correct?

Make sure you include Einstein's statement in with your reasoning or indicate why it is false.

Otherwise, the angular acceleration is too small to consider. And otherwise, give its effect on the moving clock as t'.

Don't change the subject, Chinglu. Einstein's specific quotes supporting this effect are irrelevant because his SR postulate on the constancy of light only applies to inertial frames. This is well known. Haven't you ever heard of the time dilation effects of a centrifuge? All of these experiments are basically variants of the centrifuge concept. If you're just looking for attention then I'm done feeding the troll but it's already been explained to you why you haven't dismantled Relativity.

And you cannot just dismiss angular acceleration with a wave of your hand like that! hahaha
This is from another forum

The answer to that quesiton is that you can either use a gravitational time dilation formula (if you use a rotating frame of reference), or you can use a velocity-dependent gamma formula (if you use a non-rotating frame of reference). The former approach requires GR, the later approach requires only SR. The result from either approach is the same - the clock that's not accelerating ticks faster than the clock that is accelerating, and the magnitude of the effect can be given by 1/sqrt(1-(w*r)^2/c^2).

The formulas that apply are either

t/tau = 1/sqrt(1-v^2/c^2) (the SR approach)
t/tau = 1/sqrt(1-2*U/c^2), a weak-field GR approach where U = -Phi is the negative of the Newtonian potential energy.

Either approach alone is correct - applying a "double whammy" would give the wrong answer (there is no approach where it makes sense to have both a velocity red-shift and a gravitational one).

 

[Neddy Bate]

But, that means, c' = d/(t/γ) for the moving clock.

We also have c = d/t for the stationary clock.

Yes, in both clocks' coordinate systems, the distance the light travels up the y axis is d, as long as the y axis is always held perpendicular to the direction of motion of the moving clock. Of course that means moving the y axis constantly. That works for a light sphere which was emitted at the center of the circle, for example.

And it is also true that the stationary clock will measure the time as t, while the moving clock will measure the time as t/γ.

But, under SR all observers must measure c as the speed of light.

Well, that statement is where you are getting yourself cornfused. It's just a little more complicated than that. All inertial observers measure the speed of light to be c, but an accelerating observer only measures the speed of light to be c locally. The way you have your thought experiment set up, the light is not local to the accelerated observer (the moving clock). You can see this clearly if you use a polygon instead of a circe, and let the traveling clock move at constant speed along each side of the polygon. The speed of light is clearly c for that clock, but its simultaneity changes every time it rounds a corner of the polygon.

Oh, and you cannot talk about a clock in orbit and also say that you want to disregard its acceleration due to gravity. If you get rid of the gravitational field, the clock will travel a straight line instead of a circle. So you would have to include a little rocket engine to force the clock to travel a circular path instead of a straight line, and once you do that you have included acceleration again.

 

[kx000]

Figure rhythm to the universe. Universal time, truly.