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Old 13th October 2020, 03:42 PM   #921
Myriad
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Originally Posted by Reality Check View Post
"...orthonormal world line to the acceleration?" gibberish?

It's correct in the intended context. The textbook is evaluating the accelerated observers' observations (including e.g. the Rindler horizon) by looking at "a given point" in its world line (that is, one unspecified point that will have the same mathematical properties as any other point) and then extending the analysis to the whole trajectory using the tools of calculus. At that point, the observations of a "comomving observer" are considered, which is an inertial observer that happens to have the same position and velocity as the accelerating observer at the given point/moment. Of course, that inertial observer is only comoving at that one precise point/moment.

At the given point, the two observers (just for that instant) are both stationary in the comoving observer's inertial reference frame. Therefore their world lines (just for that instant) are moving only in t, not in x (or y or z), and the t coordinate is perpendicular to x (and y and z). Since the acceleration is in x, it's orthogonal to the t axis, and thus to the world lines, at that instant. That fact is part of the basis for further analysis.

If one looks at a different point, one can posit a different co-moving inertial observer at that point, whose world line is also orthogonal to the acceleration at that point. One can, in the course of the analysis, even posit such a comoving observer at every point. But there is no single inertial observer whose world line is orthogonal to the acceleration at every point.

What SDG is having trouble with, in this particular case, is simply the methodology of calculus.
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Old 13th October 2020, 04:04 PM   #922
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Originally Posted by Myriad View Post
It's correct in the intended context. The textbook is evaluating the accelerated observers' observations (including e.g. the Rindler horizon) by looking at "a given point" in its world line (that is, one unspecified point that will have the same mathematical properties as any other point) and then extending the analysis to the whole trajectory using the tools of calculus. At that point, the observations of a "comomving observer" are considered, which is an inertial observer that happens to have the same position and velocity as the accelerating observer at the given point/moment. Of course, that inertial observer is only comoving at that one precise point/moment.

At the given point, the two observers (just for that instant) are both stationary in the comoving observer's inertial reference frame. Therefore their world lines (just for that instant) are moving only in t, not in x (or y or z), and the t coordinate is perpendicular to x (and y and z). Since the acceleration is in x, it's orthogonal to the t axis, and thus to the world lines, at that instant. That fact is part of the basis for further analysis.

If one looks at a different point, one can posit a different co-moving inertial observer at that point, whose world line is also orthogonal to the acceleration at that point. One can, in the course of the analysis, even posit such a comoving observer at every point. But there is no single inertial observer whose world line is orthogonal to the acceleration at every point.

What SDG is having trouble with, in this particular case, is simply the methodology of calculus.
Please, have a look at the diagram.
Is the back to front light round-trip going to be the same time interval as the front to back light round-trip?
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Old 13th October 2020, 04:30 PM   #923
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Originally Posted by SDG View Post
Please, have a look at the diagram.
Is the back to front light round-trip going to be the same time interval as the front to back light round-trip?

Please explain why I should care what the answer to that question is. Also, if you claim to know the correct answer, show the math.
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Old 13th October 2020, 04:55 PM   #924
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Originally Posted by Myriad View Post
Please explain why I should care what the answer to that question is. Also, if you claim to know the correct answer, show the math.
SDG has no idea what the diagram is about even when it is captioned in clear English.

The answer is unrelated to Fig. 12.13. Fig. 12.13 is the null geodesics associated with a single uniformly accelerated observer O. L0 is their worldline.

Consider a uniformly accelerating spacecraft with an observer Alice in the front and an observer Bob in the back. Obviously, there will be a "gravitational time dilation" between Alice and Bob because they are in the equivalent of a gravitational field.
Why do accelerating bodies have an event horizon? explains this better than I can.
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Old 14th October 2020, 08:00 AM   #925
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Originally Posted by Myriad View Post
Please explain why I should care what the answer to that question is. Also, if you claim to know the correct answer, show the math.
We don't have to care what the answer is.
Our lives will go on regardless...
If we want to find out the truth then we might start to care.
Before getting the correct answer we need to understand what is wrong here.
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Old 14th October 2020, 08:05 AM   #926
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Originally Posted by Reality Check View Post
SDG has no idea what the diagram is about even when it is captioned in clear English.

The answer is unrelated to Fig. 12.13. Fig. 12.13 is the null geodesics associated with a single uniformly accelerated observer O. L0 is their worldline.

Consider a uniformly accelerating spacecraft with an observer Alice in the front and an observer Bob in the back. Obviously, there will be a "gravitational time dilation" between Alice and Bob because they are in the equivalent of a gravitational field.
Why do accelerating bodies have an event horizon? explains this better than I can.



The worldline at x=0 has the same length as the worldline at x=1.7.
There is not time dilation.
There is no gravity here, the equivalence principle is wrong.
The time at the front of the spaceship does not flow faster.
The light clock (the length of the spaceship) is not going to measure any delta.
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Old 14th October 2020, 04:33 PM   #927
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Originally Posted by SDG View Post
https://i.imgur.com/rsQQtnu.png

The worldline at x=0 has the same length as the worldline at x=1.7.
There is not time dilation.
There is no gravity here, the equivalence principle is wrong.
The time at the front of the spaceship does not flow faster.
The light clock (the length of the spaceship) is not going to measure any delta.
Five sentences, all of them wrong as far as I.can see.

Are you saying that the world lines for the front and back of the ship would be parallel on this graph?

What do you think the "a" on the axes refers to?
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Old 14th October 2020, 04:54 PM   #928
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Exclamation More lies from SDG when he has a relativity textbook

Originally Posted by SDG View Post
The worldline at x=0 has the same length as the worldline at x=1.7....
Persistent ignorance about Fig. 12.13 and lies from SDG.
The caption states that the worldline of observer O is L0 which is a vertical line at x = 0. The other lines are not worldlines . The caption starts with "Null geodesics in the plane...". Obviously the green lines are those null geodesics. A null geodesic is the path of a massless particle such as a photon. Those green lines are the paths of light. They show that light emitted closer and closer to the Rindler horizon take longer and longer to get to the observer O.

A "There is not time dilation." lie. There is always time dilation when an observer moves relative to another. An accelerating observer will also have time dilation relative to an inertial observer.

A "There is no gravity here" lie when I did not write there was gravity.

A "the equivalence principle is wrong." lie when general relativity works and the equivalence principle passes its experimental tests !

A "The time at the front of the spaceship does not flow faster." lie when his relativity textbook probably has a section deriving that time will pass faster and I gave a source showing this from an undergraduate relativity textbook!

A bit of "The light clock (the length of the spaceship) is not going to measure any delta." gibberish which may be the length of the spacecraft does not change. No one says it does. Born rigidity is a fundamental part of special relativity scenarios.
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Old 14th October 2020, 06:54 PM   #929
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Originally Posted by SDG View Post
We don't have to care what the answer is.
Our lives will go on regardless...
If we want to find out the truth then we might start to care.
Before getting the correct answer we need to understand what is wrong here.

You have not succeeded in showing anything wrong. You've repeatedly claimed contradictions in Relativity, but the only contradictions you've shown are between what relativity predicts occurs, and your own opinion (unsupported by any mathematics, analysis, or coherent model) of what you think should occur.

Each of your attempts to state "what is wrong here" has contained errors of your own, which makes your claims unconvincing.

Now you're once again trying to apply the assumptions applicable to an inertial reference frame (such as, that clocks comoving with the frame would be observed to all tick at the same rate regardless of their locations, by any other inertial observer) to an accelerating reference frame (by definition not inertial). Many replies to your posts have warned you that this is an error, and even explained why it is an error, but you keep doing it anyway. You are not going to overturn basic undergraduate level physics that's been known and taught for well over a century with such poor work.
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Old 15th October 2020, 09:06 AM   #930
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Originally Posted by SDG View Post
The worldline at x=0 has the same length as the worldline at x=1.7.
The lines are both infinite. You have to pick a segment of the lines if you want a length. And if you want to calculate the proper time along those segments, you need to use the metric, because the proper time isn't generally equal to the coordinate time.

So we have three questions you need to answer:
1) what segments of the lines are you talking about?
2) what metric are you using to calculate the lengths?
3) what answer are you getting for these length calculations?

Absent these three things, your claim becomes meaningless.

Quote:
There is not time dilation.
Says who?

Again, what's the metric here? Do you even know what it is?

Do you even know what a metric means?

Quote:
There is no gravity here
True.

Quote:
the equivalence principle is wrong.
False and also irrelevant since there's no gravity.

As always, SDG, you can't even get your questions correct.
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Old 15th October 2020, 09:12 AM   #931
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Originally Posted by Robin View Post
Five sentences, all of them wrong as far as I.can see.

Are you saying that the world lines for the front and back of the ship would be parallel on this graph?

What do you think the "a" on the axes refers to?
Six months, eh?! Welcome back.
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Old 15th October 2020, 12:32 PM   #932
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Originally Posted by Robin View Post
Five sentences, all of them wrong as far as I.can see.

Are you saying that the world lines for the front and back of the ship would be parallel on this graph?

What do you think the "a" on the axes refers to?
The "a" is constant, how about a=1?
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Old 15th October 2020, 01:07 PM   #933
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Originally Posted by SDG View Post
The "a" is constant,
Yes, it is. But what is that constant? Not what's its value, but where does it come from? It has meaning.
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Old 15th October 2020, 01:21 PM   #934
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Originally Posted by Ziggurat View Post
The lines are both infinite. You have to pick a segment of the lines if you want a length. And if you want to calculate the proper time along those segments, you need to use the metric, because the proper time isn't generally equal to the coordinate time.

So we have three questions you need to answer:
1) what segments of the lines are you talking about?
2) what metric are you using to calculate the lengths?
3) what answer are you getting for these length calculations?

Absent these three things, your claim becomes meaningless.



Says who?

Again, what's the metric here? Do you even know what it is?

Do you even know what a metric means?



True.



False and also irrelevant since there's no gravity.

As always, SDG, you can't even get your questions correct.

1. For example the length of the spaceship.





2. Any observer on the spaceship has to agree on l0.
That's the point of the Rindler horizon, infinite radius, continuous frame jumping to have a comoving inertial observers.
The l0 is not changing for the accelerated observer, the comoving inertial observer is continuously 'frame jumping'.
The argument that there was a jump and time moved does not hold.
Any past or future comoving inertial observers are not orthonormal to the acceleration after any tiny delta time jump.




3. The light clock works based on a light crossing a known distance.
The distance between the back and the front of the spaceship is just one distance l0 that is the same for both observers (the back and the front).
The back observer sends a light beam to the front.
The front observer has a mirror and reflects the light beam back.
At 'the same time' the front observer sends his own front light beam to the back traveling next to the back light beam that was reflected.
There is not physical reason this beams would separate.
They will keep bouncing between the back and the front with the same period, time interval.


The equivalence principal 'states' (simplification for the readers) there is no difference between uniformly accelerated rocket at 1g and the rocket standing on the Earth at 1g.
There is no experiment that would prove this claim at the moment.
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Old 15th October 2020, 01:23 PM   #935
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Originally Posted by Ziggurat View Post
Yes, it is. But what is that constant? Not what's its value, but where does it come from? It has meaning.
It cannot be different for different x if the radius is infinite.
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Old 15th October 2020, 01:24 PM   #936
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Originally Posted by SDG View Post
It cannot be different for different x if the radius is infinite.
That doesn't answer my question.
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Old 15th October 2020, 01:33 PM   #937
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Originally Posted by Myriad View Post
You have not succeeded in showing anything wrong. You've repeatedly claimed contradictions in Relativity, but the only contradictions you've shown are between what relativity predicts occurs, and your own opinion (unsupported by any mathematics, analysis, or coherent model) of what you think should occur.

Each of your attempts to state "what is wrong here" has contained errors of your own, which makes your claims unconvincing.

Now you're once again trying to apply the assumptions applicable to an inertial reference frame (such as, that clocks comoving with the frame would be observed to all tick at the same rate regardless of their locations, by any other inertial observer) to an accelerating reference frame (by definition not inertial). Many replies to your posts have warned you that this is an error, and even explained why it is an error, but you keep doing it anyway. You are not going to overturn basic undergraduate level physics that's been known and taught for well over a century with such poor work.
Yet, why the light would travel differently if it is returning from the front to the back after a reflection compared to the light emitted at the front going back.
The same applies the other way.
Why the light would travel differently if it is returning from the back to the front after a reflection compared to the light emitted at the back going forward.

Does the l0 hold or not?
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Old 15th October 2020, 01:37 PM   #938
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Originally Posted by Ziggurat View Post
That doesn't answer my question.

It is a magnitude of an acceleration as observed by a comoving inertial observer.



Last edited by SDG; 15th October 2020 at 01:38 PM.
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Old 15th October 2020, 02:25 PM   #939
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Originally Posted by SDG View Post
It is a magnitude of an acceleration as observed by a comoving inertial observer.


https://i.imgur.com/XSsWYLb.png
Acceleration of what? Front or back?

The textbook you quote only has a single accelerated observer, but you keep referring to a second observer at the other end of the spaceship without showing the maths for it.

What is your justification for claiming that the world line of both ends of the spaceship will be parallel in this diagram?
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Old 15th October 2020, 02:37 PM   #940
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Exclamation More lies from SDG when he has a relativity textbook

Originally Posted by SDG View Post
The "a" is constant, how about a=1?
SDG writes a lie about the question which was What do you think the "a" on the axes refers to?, not what is its value.
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Old 15th October 2020, 02:45 PM   #941
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Originally Posted by SDG View Post
It is a magnitude of an acceleration as observed by a comoving inertial observer.
No. Different co-moving observers will experience different accelerations. "a" refers to only one of these.

Which one?
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Old 15th October 2020, 02:52 PM   #942
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Originally Posted by Robin View Post
Acceleration of what? Front or back?

The textbook you quote only has a single accelerated observer, but you keep referring to a second observer at the other end of the spaceship without showing the maths for it.

What is your justification for claiming that the world line of both ends of the spaceship will be parallel in this diagram?
Very good question.
The diagram shows the uniformly accelerated observer O worldline.
The worldline at the front (x=1.7) will be parallel because for every tiny time delta the horizontal spaceship length l0 will be 1.7.
The l0 is not changing.
The uniformly accelerated observer O has a comoving inertial observer for every tiny time delta.
The accelerated observer cannot 'stay' with the comoving inertial observers.
The orthogonality breaks after every tiny time delta by definition, therefore the 'frame jumping'.
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Old 15th October 2020, 02:58 PM   #943
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Originally Posted by Ziggurat View Post
No. Different co-moving observers will experience different accelerations. "a" refers to only one of these.

Which one?
For the one where t*=0, for the one where l0 is the shortest (t*=0), for the orthogonal one.
There is only one orthogonal comoving inertial observer at the time.
The acceleration is constant for that observer.
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Old 15th October 2020, 03:02 PM   #944
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Exclamation More lies from SDG when he has a relativity textbook

Originally Posted by SDG View Post
1. For example the length of the spaceship.
....
SDG lies about his relativity textbook , especially Fig. 12.13 which has no spaceship, only observer O as in the caption !
A "Any observer on the spaceship has to agree on l0." lie. There is no spaceship. L0 is the worldine of observer O.
A "That's the point of the Rindler horizon, ..." lie. The point of the Rindler horizon is that is where light cannot reach observer O.
A "...continuous frame jumping to have a comoving inertial observers" lie when his textbook explicitly states there is 1 inertial observer comoving to O at 1 point.
A "argument that there was a jump and time moved does not hold." when there is no such argument shown from his textbook. A uniformly (not jumping) accelerating observer O passes an inertial observer O'.

More irrelevance about an imaginary spaceship that is not in his textbook and SDG has not analysed using any math or physics.
A "There is not physical reason this beams would separate." fantasy when not one thinks two beams of light sent between the front and end of a spaceship will "separate".

A "There is no experiment that would prove this claim at the moment." lie about the equivalence principle which SDG knows has been tested. SDG's fantasy that we have to test a rocket accelerating at 1 g against a rocket sitting in 1 g does not make this extensive testing vanish.

Last edited by Reality Check; 15th October 2020 at 03:04 PM.
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Old 15th October 2020, 03:16 PM   #945
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Exclamation More lies from SDG when he has a relativity textbook

Originally Posted by SDG View Post
It is a magnitude of an acceleration as observed by a comoving inertial observer....]
SDG lies with an image from his relativity textbook that does not say what he writes. What the text says is it is convenient to introduce an inertial observer O*, an observer O and they have tangent worldlines at a point O(0). His textbook implies that we do not need observer O*! They are a convenient way to derive the physics being discussed. It may be that another textbook does not have them. The text refers to an earlier acceleration a.

SDG posted images of the start of the section that states O has that acceleration a. Observer O has been assigned that acceleration and it is easy for thm to measure it if that was needed.

Last edited by Reality Check; 15th October 2020 at 03:20 PM.
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Old 15th October 2020, 03:26 PM   #946
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Originally Posted by Elagabalus View Post
Six months, eh?! Welcome back.
One more little drink can't hurt can it? I can stop any time I want, honest!
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Old 15th October 2020, 04:29 PM   #947
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Originally Posted by SDG View Post
The worldline at the front (x=1.7) will be parallel because for every tiny time delta the horizontal spaceship length l0 will be 1.7.
Tiny delta on which clock?

If you are talking about the t referenced in the axis then this is not true.
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Old 15th October 2020, 05:39 PM   #948
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Originally Posted by SDG View Post
For the one where t*=0, for the one where l0 is the shortest (t*=0), for the orthogonal one.
There is only one orthogonal comoving inertial observer at the time.
The acceleration is constant for that observer.
Observers are not orthogonal. That doesn't even make sense. You are just piling words together that you've seen, even though you clearly don't know what they mean.

Your wrongness streak remains unbroken.
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"As long as it is admitted that the law may be diverted from its true purpose -- that it may violate property instead of protecting it -- then everyone will want to participate in making the law, either to protect himself against plunder or to use it for plunder. Political questions will always be prejudicial, dominant, and all-absorbing. There will be fighting at the door of the Legislative Palace, and the struggle within will be no less furious." - Bastiat, The Law

Last edited by Ziggurat; 15th October 2020 at 05:40 PM.
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Old 15th October 2020, 05:53 PM   #949
Reality Check
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Originally Posted by SDG View Post
..
The worldline at the front (x=1.7) will be parallel because for every tiny time delta the horizontal spaceship length l0 will be 1.7.
Reading this post again shows that SDG is still ignorant about what his textbook contains.

How do we get a new worldline for an new observer at x = 1.7 from a spacetime diagram with a Rindler horizon at x = -1 and an observer at x = 0? We do not do as SDG ignorantly does and just draw a line at x = 1.7 and leave the horizon at x = -1. We shift the entire diagram to x = 1.7 and superimpose the result. We now have 2 observers, say Alice at x = 0 and Bob at x = 1.7. Alice has a Rindler horizon at x = -1 and cannot receive signals from there or beyond. Bob has a Rindler horizon at x = 0.7 and cannot receive signals from there or beyond.

Alice and Bob cannot send signals to each other but that is what SDG demand they do!

This is fundamental to Rindler coordinates. Each observer has their own horizon. For an observer at the front of a spaceship and one at the back, the horizon for the front observer can be between them. It is a special relativity equivalent of observers either side of an event horizon.

If SDG understood what he read then that spaceship would be at x < 1. There would be no problem with light traveling between Alice and Bob. Their clocks will still be dilated by a "gravitational time dilation".
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Old 15th October 2020, 06:23 PM   #950
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To clarify, two observers separated in space can have the same Rindler Horizon but they will not have the same acceleration (correct me if I am wrong)
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Old 16th October 2020, 11:26 AM   #951
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Originally Posted by Ziggurat View Post
Observers are not orthogonal. That doesn't even make sense. You are just piling words together that you've seen, even though you clearly don't know what they mean.

Your wrongness streak remains unbroken.
Right, writing too fast ...
The 4-velocity of the uniformly accelerated observer and the comoving observer are orthogonal to the 4-acceleration only at t*=0.
So the uniformly accelerated observer O has 4-acceleration a(0) orthogonal to 4-velocity u(0).
A question - when the uniformly accelerated observer O is not at u(0), a(0)?
Is it OK to assume that t=0 'all the time'? The time does not move?
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Old 16th October 2020, 11:42 AM   #952
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Originally Posted by Robin View Post
To clarify, two observers separated in space can have the same Rindler Horizon but they will not have the same acceleration (correct me if I am wrong)
They have constant space like vector between them.
How is a beam light going to cross the same constant space like vector between them (the back observer and the front observer) at different times when crossing of this space like vector by a light beam is the definition of the time itself?
If there is no delta in time then there is no delta in 4-velocity and 4-acceleration.
How can a 'black hole' with infinite radius have a different acceleration at different radius.
There is no radius change.
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Old 16th October 2020, 02:18 PM   #953
Robin
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Originally Posted by SDG View Post
They have constant space like vector between them.
How is a beam light going to cross the same constant space like vector between them (the back observer and the front observer) at different times when crossing of this space like vector by a light beam is the definition of the time itself?
If there is no delta in time then there is no delta in 4-velocity and 4-acceleration.
Suppose you have the spaceship initially in an inertial frame, no acceleration and clocks synchronised.

The front observer sends a light beam backward at the same clock time as the back observer sends a light beam forwards.

So we put half silvered mirrors at the point the beams cross.

Now each observer sees two photons arrive at the same time at each return.

What happens if the spaceship begins to accelerate just as the observers send the photons?

They will still cross at the point the midway of the ship had been if no acceleration had happened, but the midway of the ship will no longer be there and the photon which bounces off the half silvered mirror will no longer arrive back simultaneously with the photon from the other end.

So, with these photons no longer arriving together and since the light journeys are the definition of time, how can the observers hold on to the opinion that time is passing at the same rate everywhere on their ship?
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Old 16th October 2020, 04:52 PM   #954
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Originally Posted by Robin View Post
Suppose you have the spaceship initially in an inertial frame, no acceleration and clocks synchronised.

The front observer sends a light beam backward at the same clock time as the back observer sends a light beam forwards.

So we put half silvered mirrors at the point the beams cross.

Now each observer sees two photons arrive at the same time at each return.

What happens if the spaceship begins to accelerate just as the observers send the photons?

They will still cross at the point the midway of the ship had been if no acceleration had happened, but the midway of the ship will no longer be there and the photon which bounces off the half silvered mirror will no longer arrive back simultaneously with the photon from the other end.

So, with these photons no longer arriving together and since the light journeys are the definition of time, how can the observers hold on to the opinion that time is passing at the same rate everywhere on their ship?

You described a change from inertial observers to accelerated observers.
The clocks become desynchronized, a new clock synchronization is required.



So the front observer will send the front light beam at M.
The MA2 is going to be the same for the back beam returning and the new front beam going from M to A2.
A2M2, then M2A3, then A3M3, ...
The light beams will be next to each other all the time.

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Old 16th October 2020, 05:48 PM   #955
Robin
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Originally Posted by SDG View Post
You described a change from inertial observers to accelerated observers.
It is not.the change from inertial to acceleration which causes this, it is the continuous acceleration.
Quote:
The clocks become desynchronized, a new clock synchronization is required.
That does no.good because as long as the acceleration continues the clocks are ticking at a different rate.

Again think about that half silvered mirror half way. If you attempt to set the clocks the same at the front and back the photon reflected from midway will not arrive with the photon from the end at the very next round trip.

Trying to say that the clocks on the accelerating space ship.are ticking at the same rate always leads to a contradiction.
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Old 17th October 2020, 12:44 PM   #956
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Originally Posted by Robin View Post
It is not.the change from inertial to acceleration which causes this, it is the continuous acceleration.


That does no.good because as long as the acceleration continues the clocks are ticking at a different rate.

Again think about that half silvered mirror half way. If you attempt to set the clocks the same at the front and back the photon reflected from midway will not arrive with the photon from the end at the very next round trip.

Trying to say that the clocks on the accelerating space ship.are ticking at the same rate always leads to a contradiction.
One more time.
The assumption from your post: "Suppose you have the spaceship initially in an inertial frame, no acceleration and clocks synchronised."

There is an observer at the back and at the front.
The a acceleration starts.
Is there any reason the acceleration is different at the back and at the front at the first delta time?
No, there is not.
Both observers have 1m light clocks with them, they start to measure time when the acceleration starts.
Why there would be a delta on these two clocks?
The observers will send their time to the other observer every 1000 seconds.
Is the delta going to grow over time or there is going to be only communication time delay after each exchange?

Another variation.
The back and the front observers are in closed rooms. No signal from the outside.
They have 1m light clocks and 2 additional 0.1m light clocks at the back and at the front of the 1m clocks.
Let us assume the time ticks faster at the front compared to the back of the 1m clocks.
Is there going to be a different delta for the back and the front observer on their respective 0.1m clocks?
Why?
If the first initial a acceleration is the same what makes the back observer preferred one?
Why not to place x=0 to the front observer?

If an observer is placed in the middle as you suggested then the space and time interval is being broken to parts.
There is going to be a sequence of comoving inertial observers.
Relatively moving inertial observers do not agree on the simultaneity.
The proper time and simultaneity requires very careful analysis.
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Old 17th October 2020, 01:36 PM   #957
Robin
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Originally Posted by SDG View Post
There is an observer at the back and at the front.
The a acceleration starts.
Is there any reason the acceleration is different at the back and at the front at the first delta time?
No, there is not.
If the acceleration at front and back is the same then proper distance between the front and back is no longer maintained, ie the space ship would stretch.

If you want the spaceship to be rigid then there has to be a different acceleration profile front and back.

In neither case would you expect to maintain light signals synchronised.

Quote:
If an observer is placed in the middle as you suggested then the space and time interval is being broken to parts.
There is going to be a sequence of comoving inertial observers.
Relatively moving inertial observers do not agree on the simultaneity.
The proper time and simultaneity requires very careful analysis.
So you are agreeing that the clock in the middle of the spaceship would not be in synch with the clock at either end, but you still maintain the clocks at either end will be in synch with each other?

What if there was a spaceship half as long travelling alongside at exactly the same acceleration with fronts aligned, would it's front and back clocks be in synch?
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Old 20th October 2020, 03:32 PM   #958
Reality Check
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Originally Posted by Robin View Post
To clarify, two observers separated in space can have the same Rindler Horizon but they will not have the same acceleration (correct me if I am wrong)
Looking at Fig. 12.13 that SDG has been posting from his textbook, what I get is.
It has labels "a c t" and "a x" where a is acceleration. The caption states the horizon is at ax = -1. So observers with different accelerations will have different horizons. Observers at the same x with the same a will have the same horizon. Move an observer from that x and their horizon also moves.
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Old 20th October 2020, 03:43 PM   #959
Robin
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Originally Posted by Reality Check View Post
Looking at Fig. 12.13 that SDG has been posting from his textbook, what I get is.
It has labels "a c t" and "a x" where a is acceleration. The caption states the horizon is at ax = -1. So observers with different accelerations will have different horizons. Observers at the same x with the same a will have the same horizon. Move an observer from that x and their horizon also moves.
Yes, my mistake, sorry.
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Old 21st October 2020, 08:59 AM   #960
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Originally Posted by Robin View Post
Yes, my mistake, sorry.

That's the reason I asked this. It is a fair question.

Quote:
If the first initial a acceleration is the same what makes the back observer preferred one?
Why not to place x=0 to the front observer?
... and the assumption from your post stands: "Suppose you have the spaceship initially in an inertial frame, no acceleration and clocks synchronised."
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