Einstein's clock

[arfa brane]

Nope, you have changed the experiment.

You can look at the 6 steps I provided to AN.

Please point out how exactly I changed "the experiment". All I did was add a mirror so the 'location' of the emitted light pulse (your addition) could be known. It only assumes that t is known, c is constant, so the mirror should be at a distance ct/2.

But it isn't needed, you only need c, actually. You don't need to add an expanding sphere of light to Einstein's ansatz. You just assign coordinates, like L = (t,x,y,z) and L' = (t',x',y',z') to the stationary and moving clocks (respectively), and calculate the Lorentz shift between them.

 

[chinglu]

Please point out how exactly I changed "the experiment". All I did was add a mirror so the 'location' of the emitted light pulse (your addition) could be known. It only assumes that t is known, c is constant, so the mirror should be at a distance ct/2.

But it isn't needed, you only need c, actually. You don't need to add an expanding sphere of light to Einstein's ansatz. You just assign coordinates, like L = (t,x,y,z) and L' = (t',x',y',z') to the stationary and moving clocks (respectively), and calculate the Lorentz shift between them.

Where did i have a mirror in my experiment?

Deal with the experiment presented or run and hide.

 

[arfa brane]

Where did i have a mirror in my experiment?

FFS. You don't mention a mirror.

Why does adding a mirror change your experiment? What does it change?
You aren't very good at thinking, but you seem to be quite good at being stupid.
Let's face it why don't we. You're the one hiding, here on a science forum at the end of the social universe. Don't you have anything better to do than bring up totally pointless "thought" experiments.
What you thought, and what Einstein thought, are obviously antipodal.

You don't get it, you won't get it, your shitty attitude is getting in the way instead. Learn to accept your mistakes and bloodywell learn something. You don't have to be grateful or anything. You don't even have to shrug nonchalantly.

 

[brucep]

If this statement is true, then you can prove my conclusions are false.

But, since you have been completely unable to do that, good luck!!

You just won't accept your experiment is stupid. It's stupid because you don't even know what relativity predicts to start with. Relativity predicts when the observer who orbits the sun and returns to the observer who remained at rest on r_shell earth orbit they'll both measure the local coordinate speed of light to be invariant. It's a LOCAL measurement. Now that they're together again the tick ratio dt_shell observer 1 / dt_shell observer 2 = 1. They're at rest with respect to each other. So they use an 'identical clock' to measure the remote coordinate speed of the pulse of light they emitted a year ago including any local measurements of the coordinate speed of light the observers decide to do.. Both measure the remote coordinate speed of light exactly the same. The separation between observer 1 & 2 is nil when they make both measurements so both observer tick rates are synched are SYNCHED. The trip around the earth makes no difference when they come back together. The ratio dt_shell observer 1 / dt_shell observer 2 = 1. Observer 1 & 2 tick rates are SYNCHED. Synched... SYNCHED. Did I mention both observers 1 & 2 clocks are synched? IE SYNCHED.

 

[chinglu]

FFS. You don't mention a mirror.

Why does adding a mirror change your experiment? What does it change?
You aren't very good at thinking, but you seem to be quite good at being stupid.
Let's face it why don't we. You're the one hiding, here on a science forum at the end of the social universe. Don't you have anything better to do than bring up totally pointless "thought" experiments.
What you thought, and what Einstein thought, are obviously antipodal.

You don't get it, you won't get it, your shitty attitude is getting in the way instead. Learn to accept your mistakes and bloodywell learn something. You don't have to be grateful or anything. You don't even have to shrug nonchalantly.

You can call me all the names you want.

I do not use a mirror as that would be difficult in an experiment that is circular.

Anyway, do you have anything as to the specific 6 point steps I provided to AN?

 

[chinglu]

You just won't accept your experiment is stupid. It's stupid because you don't even know what relativity predicts to start with. Relativity predicts when the observer who orbits the sun and returns to the observer who remained at rest on r_shell earth orbit they'll both measure the local coordinate speed of light to be invariant. It's a LOCAL measurement. Now that they're together again the tick ratio dt_shell observer 1 / dt_shell observer 2 = 1. They're at rest with respect to each other. So they use an 'identical clock' to measure the remote coordinate speed of the pulse of light they emitted a year ago including any local measurements of the coordinate speed of light the observers decide to do.. Both measure the remote coordinate speed of light exactly the same. The separation between observer 1 & 2 is nil when they make both measurements so both observer tick rates are synched are SYNCHED. The trip around the earth makes no difference when they come back together. The ratio dt_shell observer 1 / dt_shell observer 2 = 1. Observer 1 & 2 tick rates are SYNCHED. Synched... SYNCHED. Did I mention both observers 1 & 2 clocks are synched? IE SYNCHED.

There is just one problem with your post.

Einstein claimed the moving clock would show a time of t/γ where t is the time on the clock of the stationary clock.

You refuted his assertion.

Now, go back through the experiment assuming Einstein's statement is correct and let me know the results.

 

[AlexG]

There is just one problem with your post.

Einstein claimed the moving clock would show a time of t/γ where t is the time on the clock of the stationary clock.

You refuted his assertion.

Now, go back through the experiment assuming Einstein's statement is correct and let me know the results.

How do you miss the part where Bruce explains they're at rest with respect to one another?

 

[chinglu]

How do you miss the part where Bruce explains they're at rest with respect to one another?

They are not.

How did you miss the point this is my experiment?

 

[AlexG]

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.

 

[brucep]

There is just one problem with your post.

Einstein claimed the moving clock would show a time of t/γ where t is the time on the clock of the stationary clock.

You refuted his assertion.

Now, go back through the experiment assuming Einstein's statement is correct and let me know the results.

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.

 

[Prof.Layman]

You can assume anything you want, but the case must show c'=c for any way you calculate c'.

Now, if you agree "c would, and not d/(t/y)", then you confess the 2nd observer did not measure c as the speed of the light pulse that was emitted when the 2 clocks were at first common.

So, you simply agreed with my position.

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.

 

[brucep]

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.

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.

 

[brucep]

They are not.

How did you miss the point this is my experiment?

They are to. Your experiment is the idiot wind that 'blows forth' from your anus. It doesn't matter whether one or both are moving along the shell. Any measurement of the local coordinate speed of light is invariant regardless of the motion of the receiver or emitter. There can be a delta between the remote coordinate speed of light measurement when the relative velocity between obs 1 and obs 2 is > 0. But that is expected [however miniscule for your experiment in the weak field] since the remote coordinate speed of light can be any velocity below 1. The remote coordinate speed of light is

dr/dt = 1 - 2M/r

For your dubious weak field experiment both observers moving along r_shell earth orbit get the same answer because their tick ratio is still as close to 1 as it gets. The reality IS any delta would be predicted by the theory of relativity. You can't understand or don't want to understand. Either way you should pick another subject.

Since this thread isn't about a 'alternative theory' and really nothing more than a stubborn case of intellectual dishonesty does the forum ever consign something like this to the dust bin?

 

[AlphaNumeric]

Well, since my math is very specific

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.

 

[RJBeery]

I think everyone should chill out with the name calling and Chinglu you should probably take the attitude down a notch as well.

The mind experiment doesn't have to be that complicated: 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? The question is, how is this explained?

The constancy of c is only valid for non-accelerating frames. Observer B is clearly accelerating and should expect to get a different answer.

Is there another explanation?

 

[arfa brane]

I do not use a mirror as that would be difficult in an experiment that is circular.

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?

 

[Neddy Bate]

I think everyone should chill out with the name calling and Chinglu you should probably take the attitude down a notch as well.

The mind experiment doesn't have to be that complicated: 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? The question is, how is this explained?

The constancy of c is only valid for non-accelerating frames. Observer B is clearly accelerating and should expect to get a different answer.

Is there another explanation?

I like that explanation, but I'd like to challenge it with this:

Change the path of observer B from a circle to a polygon. While observer B is moving at constant speed along each side of the polygon, he is at rest in an inertial frame, and therefore he finds the speed of light to be c. The amount of time he spends turning each corner of the polygon is theoretically zero, (assuming instantaneous acceleration), so observer B finds the speed of light to be c for the entire journey around the polygon.

Now if you increase the number of sides of the polygon to approach infinity, the polygon approaches the same shape as the circle. So, why is the constancy of c not valid for an accelerating frame?

 

[RJBeery]

I like that explanation, but I'd like to challenge it with this:

Change the path of observer B from a circle to a polygon. While observer B is moving at constant speed along each side of the polygon, he is at rest in an inertial frame, and therefore he finds the speed of light to be c. The amount of time he spends turning each corner of the polygon is theoretically zero, (assuming instantaneous acceleration), so observer B finds the speed of light to be c for the entire journey around the polygon.

Now if you increase the number of sides of the polygon to approach infinity, the polygon approaches the same shape as the circle. So, why is the constancy of c not valid for an accelerating frame?

It's because infinite acceleration is required at each vertex to be instantaneous, which breaks the math. Why not just say observer B bounces between two magic trampolines at nearly the speed of light? There is simply no way to change his frame while dodging the acceleration issue.

 

[brucep]

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.

Since I said it's derived from the metric of flat spacetime geometry [with the huge contribution by Maxwell] I'll write it down. Really simple.

In geometric units and setting theta = 0

dTau^2 = dt^2_shell - dr^2_shell - r^2(dphi)^2

Setting dphi = 0

dTau^2 = dt^2_shell - dr^2_shell = dt^2_shell - ds^2_shell

Setting dTau_light = 0

0 = dt^2_shell - ds^2_shell

ds^2_shell = dt^2_shell

ds^2_shell / dt^2_shell = (+or-) 1

The +or- indicates direction along the pulse of lights path.

 

[Neddy Bate]

It's because infinite acceleration is required at each vertex to be instantaneous, which breaks the math. Why not just say observer B bounces between two magic trampolines at nearly the speed of light? There is simply no way to change his frame while dodging the acceleration issue.

In that case, let's make the acceleration smaller and see what happens. Imagine the polygon is enormously large, and the speed of observer B is very small. Most of the time, observer B will be traveling along a straight line at a constant speed, so he finds the speed of light to be c. Every once in awhile he rounds a corner of the polygon and feels a slight gravity-like pull, but it goes away very quickly.

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?