Curved space/ spacetime

yuiop

Hi Kev. I think you have these two parts slightly wrong. The Born-accelerated observers on your 'floors' do not agree that their respective clock rates are the same, because their clocks get more and more out of sync during the acceleration. Think what will happen in your Rindler diagram when the acceleration is stopped - don't you think the clocks will then need to be resynchronized?
I analysed this in some detail in this old thread (with yet more diagrams) https://www.physicsforums.com/showthread.php?t=216113 and concluded that two clocks would be out of sync after born rigid accleration when they reach same final constant velocity.

In this thread I am considering the slightly different situation where the two clocks maintain constant born rigid acceleration indefinitely. I would be interested in your opinion of my conclusion in that old thread ;)

Your purple lines of "instantaneous simultaneity" is when the clocks on all floors read the same time, but they realize that the clocks get out of sync. Light signals send forward and backward during the acceleration are red-shifted and blue-shifted respectively, just like in a gravitational field. It's only the tidal effects that differ, AFAIK.

-J
The light signals sent forward and backward should behave as any other light signal in a time space diagram moving at 45 degrees as I have shown on this diagram (the blue lines). If the signals are sent at 0 seconds by the upper and and lower observers and both arrive at 1 seconds as shown by their respective clock and then return at 2 seconds as shown by their respective clocks then presumably the upper and lower clocks must remain in sync (while constant proper born rigid acceleration is maintained) as far as I can tell.

pmb_phy

I thought that parallel was synonymous with "same 4-velocity". So since they both initially start out at rest their geodesics would start out parallel. Am I missing some subtle point here?
To be called "the same 4-velocity" one must be able to compare the 4-velocities when the particles are located at different points in spacetime. However since there is no way to compare 4-vectors which are located at different points in spacetime the 4-velocities cannot be said to be "the same." This is easily vizualied by using the two-dimensional surface of a sphere such as the Earth. Start with a tangent vector located at the equator which points eastward. First parallel transport this vector to the North Pole by moving along the a great circle which passes through the poles and through the initial position of the tangent vector. Then picture what would happen if you first parallel transported one quarter the way around the sphere and then parallel transport along the great circle which passes through the poles as well as through this new location. You will see that the end results result in tangent vectors which have a 45 degree angle between them.

Pete

Dale

Mentor
Yes, that makes sense. I guess to talk about parallel world lines converging or diverging you have to start with two lines close enough together that the spacetime between them at the beginning can be considered flat in order to avoid such problems. Then you have to move them far enough through the curved spacetime that you can see them converge or diverge since it won't be apparent right away.

Jorrie

Gold Member
I analysed this in some detail in this old thread (with yet more diagrams) https://www.physicsforums.com/showthread.php?t=216113 and concluded that two clocks would be out of sync after born rigid accleration when they reach same final constant velocity.
As far as I checked, I agree with your conclusions in that thread.
The light signals sent forward and backward should behave as any other light signal in a time space diagram moving at 45 degrees as I have shown on this diagram (the blue lines)
Yep, that's in the inertial reference frame. In an accelerating frame, light does not move isotropically like that, as I'm sure you know.
If the signals are sent at 0 seconds by the upper and and lower observers and both arrive at 1 seconds as shown by their respective clock and then return at 2 seconds as shown by their respective clocks then presumably the upper and lower clocks must remain in sync (while constant proper born rigid acceleration is maintained) as far as I can tell.
I don't think so. Your 'old thread' has shown that when the acceleration stops and both ends of the lab are moving at the same velocity, their clocks will be out of sync. Since the two ends are not experiencing the same proper acceleration, they will reach that 'same velocity' at different times, not only in the reference frame, but also in their own frames. To me that makes their clocks never synchronized during the acceleration (after the start, that is). But I may be wrong...

-J

pmb_phy

Flat Tilted Plane and a Uniform Gravitaional Field

my_wan = If I've got this right then I imagine that you're visualizing gravity using an embedding diagram. The one for the Earth is like a flat space with a bowling ball placed on it and then depresses into the surface and thus "curving" it. I would imagine that you're thinking of the diagram in terms of a little ball rolling on the surface. Where the surface is level there is no acceleration and where it dips down due to the bowling ball then it accelerates "down" the surface. The meaning of an embedding diagram is to represent the increases in distances as a function of the decrease in the radial coordinate r. If one were to use this vizulization for a uniform gravitational field then you should imagine a tilted plane. In the rest frame of the observer in free-fall all he measures is a flat level plane. Notice that any observer in free-fall can transform the gravitational field away by moving to a free-fall frame. This is to be vizualized by picturing the location a little ball is from the view where the surface normal is directed upward. In a small enough region the the surface looks flat and since the surface is not titled then there is no gravitational field present at that location from that observers point of view.

MeJennifer

If one were to use this vizulization for a uniform gravitational field then you should imagine a tilted plane. In the rest frame of the observer in free-fall all he measures is a flat level plane.
How could an observer possibly measure a flat plane? If the Riemann curvature tensor vanishes it is flat, and there is nothing to measure. So are you saying there is some kind of force still active that causes gravitation?

Again Peter, what you say about gravitation in flat spacetime does not make any sense to me, I gladly attribute it to my lack of understanding of the matter.

yuiop

Is the spacetime background experienced by an observer in a rocket experiencing constant proper acceleration flat or curved?

MeJennifer

Is the spacetime background experienced by an observer in a rocket experiencing constant proper acceleration flat or curved?
There is a difference between something being equivalent and identical.

An accelerating rocket is not identical to gravitation.

yuiop

...
Yep, that's in the inertial reference frame. In an accelerating frame, light does not move isotropically like that, as I'm sure you know.
I agree, the diagram in post #45 is the point of view of an inertial observer in flat spacetime and to him light follows straight lines in 3 space and in 4D spacetime, as shown in that diagram. Sure, in the reference frame of the observers in the accelerating lab light follows a curved path in 3 space. However, the events are the same from any observers point of view. If A sends a signals at one second intervals as measured by A's clock and they arrive at B at one second intervals as measured by B's clocks then those two clocks are syncronised in the reference frame of A and B. Since the readings are the proper times of those observers then any observer would have to agree that is what they measure.

...

I don't think so. Your 'old thread' has shown that when the acceleration stops and both ends of the lab are moving at the same velocity, their clocks will be out of sync. Since the two ends are not experiencing the same proper acceleration, they will reach that 'same velocity' at different times, not only in the reference frame, but also in their own frames. To me that makes their clocks never synchronized during the acceleration (after the start, that is). But I may be wrong...

-J
I disagree. To the inertial observer in the diagram the observers A and B have different velocities at any given moment. In the reference frame of A and B they consider themselves to have the same velocity at any given instant (which is why they consider their spatial seperation to be constant in their reference frame. The line of simultaenity is not only the the line that connects events that simultaneous in the reference frame of the accelerating observers, it is also the line that connects equal velocity as measured by the inertial observer (although of course he does not consider A and B to have equal velocity simultaneously). In the reference frame of the accelerating observers they can consider their velocity to be zero while experiencing different proper accelerations. This is analogous to two observers that are stationary with respect to each but at different altitudes in a tower on the Earth for example.

A subtle difference is that the observers in the Earth tower measure their clock rates to be different while the observers in the accelerating lab do not. This is a tidal effect of the curved spacetime of the Earth that is not present in the lab experiencing artificial constant proper acceleration.

yuiop

There is a difference between something being equivalent and identical.

An accelerating rocket is not identical to gravitation.

You have avoided the question. Is the spacetime in an accelerating rocket flat or curved?

I did not ask what it is equivalent to.

pmb_phy

How could an observer possibly measure a flat plane? If the Riemann curvature tensor vanishes it is flat, and there is nothing to measure. So are you saying there is some kind of force still active that causes gravitation?
This subject has been discussed in the relativity literature to quite an extend. You can read this in Exploring Black Holes by Taylor and Wheeler (It was yours truly who came up with the title in fact! That part is online in www.eftaylor.com. See the download section and look through the available chapters that are there for you to download.

The idea of locally flat is similar to people standing on Earth. Suppose the Earth was a perfect sphere with a perfectly smooth surface. Then people standing on the Earth who are examining only the region in the nearby space around him (surrounded by, say, a small loop ~ 1 mile in diameter) might conclude (if his instruments weren't sensitive enough) that he was standing on a plane. When applied to general relativity it becomes the equivalence principle. Indeed if one were to parallel transport a vector around such a loop then the the difference between the initial vector and the final vector would decrease with the size of the loop. For the person standing on a smooth sphere, as mentioned above, he wouldn't have enough accuracy in his instruments to detect the curvature. However, whether the curvature is detected or not will depend on the accuracy of the instruments he is using. Thus "local" means that your instruments are not accurate enough to detect the curvature. Ask yourself this; if you are on a lake in a boat and the water is so calm that the surface is as smooth as glass then how would you, with no instruments, detect the curvature? Here the only instruments you are using will be your own eyes.
Again Peter, what you say about gravitation in flat spacetime does not make any sense to me, I gladly attribute it to my lack of understanding of the matter.
Then answer this form me; Why do you associate gravity with spacetime curvature? As far as your lack of understanding I recommend that you read Einstein's relativity book in the subject. Its called Relativity: Special and General, by Albert Einstein. It covers almost everything we've discussed. However, while Einstein does explain the relative existance of a gravitational field he does not get into gravitational fields which have tidal gradients in it (i.e. spacetime curvature). Also ask yourself what Einstein meant when he stated in his 1916 review paper on GR that one can "produce" a gravitational field by merely changing coordinates.

I do have one last question for you - Do you have a solid understanding of the differences in the definition and meaning of tidal forces and the gravitational field, at least within Newtonian mechanics?

Thanks

Pete

pmb_phy

There is a difference between something being equivalent and identical.

An accelerating rocket is not identical to gravitation.
In this case it means identical. I.e. there is no possible way to determine whether one is at rest in a uniform gravitational field or in an accelerating frame of refernce. The metrics, which actually defines the field, are identical. Consider this; what experiment could you do within the rocket cabin (which is at rest with respect to the accelerating rocket) to determine you were accelerating as opposed to being at rest in a uniform gravitational field? Looking out side the cabin won't help you because there might just be a uniform field whose source you can't see. The firing of the engines won't help you since its just as possible that the engines are merely making sure that you don't fall but remain at rest in the field.

A good example of gravity in flat spacetime is a vacuum domain wall. This is an object which finds use in early cosmology. I think that there is a Scientific American article on them as well as cosmic strings. Such a wall generates repulsive gravitational field (i.e. antigravity). Objects in free-fall near the wall will be gravitationally accelerated away from the wall. The reason is due to there being tension in such a wall. The tension is in two dimensions which gives two terms for the pressure in the wall. There is one term for the mass density. Since each is a source of gravity, when they are added together the resulting active gravitational mass is a negative value, hence antigravity. Antigravity like this also explains the accelerating expansion of the universe as well as inflation theory. Oops! I see I went off on a wee bit of a tangent, huh?

Pete

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yuiop

In this case it means identical. I.e. there is no possible way to determine whether one is at rest in a uniform gravitational field or in an accelerating frame of refernce. The metrics, which actually defines the field, are identical. Consider this; what experiment could you do within the rocket cabin (which is at rest with respect to the accelerating rocket) to determine you were accelerating as opposed to being at rest in a uniform gravitational field? Looking out side the cabin won't help you because there might just be a uniform field whose source you can't see. The firing of the engines won't help you since its just as possible that the engines are merely making sure that you don't fall but remain at rest in the field.

A good example of gravity in flat spacetime is a vacuum domain wall. This is an object which finds use in early cosmology. I think that there is a Scientific American article on them as well as cosmic strings.

Pete
Hi pete,

Is there a requirement that the cabin is infinitesimal?

If not, then you could observe that acceleration at different "heights" within the cabin is proportional to 1/R while the acceleration due to any normal gravitational body is proportional to 1/(R^2). As I mentioned in an earlier post, the gravitational attenuation perpendicular to an infinitely long cylindrical body is also proportional to 1/R. To make the two analogies completely equivalent someone would have to show that gravitational time dilation with respect to "height" within the rocket is the same as for the cylindrical massive body. It is interesting to note that the observers in the accelerating rocket even see an apparent event horizon below which they can never receive light signals, as long as they maintain constant proper acceleration.

pmb_phy

Hi pete,

Is there a requirement that the cabin is infinitesimal?
Any cabin will have a finite size so I see no way to create an experiment that will measure the size of a cabin to be infinitesimal. So no, there is no such requirement. "Local" refers to the restriction of size and space of a region of spacetime. That means that not only must the volume of space be limited to a certain size but so too does the time it takes to run an experiment, i.e. this is a limitation in both space and time.
If not, then you could observer that acceleration at different "heights" within the cabin is proportional to 1/R while the acceleration due to any normal gravitational body is proportional to 1/(R^2).
Einstein never said that a curved spacetime is equivalent to a uniformly accelerating frame of reference. Einstein was speaking about a uniform gravitational field. Seems to me you have a particular field in mind, e.g. that outside a spherically symetric body. It is not limited to that. Consider what I said about a cavity inside a sphere with uniform mass density whose center is not the center of the Earth. Inside the cavity the strength of the field has the same value everywhere inside the cavity. Hence it does not vary according to 1/R. By the way, the gravitational field varies as 1/R2. It is the gravitational potential that varies as 1/R. This, of course, refers to the weak field limit. When using the strong field exact values it becomes hopelessly complex and not very interesting.
As I mentioned in an earlier post, the gravitational attenuation perpendicular to an infinitely long cylindrical body is also proportional to 1/R.
Ah! I see where that 1/R comes from now. Thanks for clarifying that for me.

As far as the rest, please see my comments above regarding the equivalence principle. Note that there are two different forms, one called the weak form while the other is called the strong form.

Pete

yuiop

....

Einstein never said that a curved spacetime is equivalent to a uniformly accelerating frame of reference. Einstein was speaking about a uniform gravitational field. Seems to me you have a particular field in mind, e.g. that outside a spherically symetric body. It is not limited to that.

Pete
The point I am getting at is that the observers inside the cabin should consider that the measurements they make within the cabin could be exactly "duplcated" by some gravitational body even if that gravitational body is only hypothetical. If there is no such hypothetical body that can duplicate the apparent field within the cabin, then they can clearly declare that they are artiificially accelerating (rocket power) and not in a gravitational field by measuring those observations that cannot be exactly duplicated by any (even hypothetical) gravitational body. That is why it is pleasing to me that an infinite massive cylinder can duplicate the 1/R attenuation of born rigid acceleration. With a bit a luck the cylindrical body can duplicate all other measurements inside the accelerating rocket. If they can can not be duplicated then all we have to do is to point to the differences and use them to answer the question "What possible measurements can you make inside the cabin that can tell you, you are not in a gravitational field?" Our hypothetical gravitational body must be free of tidal effects because they are not measured within the cabin, with the possible exception of red shift of light signals sent up and down the cabin.

Jorrie

Gold Member
In the reference frame of the accelerating observers they can consider their velocity to be zero while experiencing different proper accelerations. This is analogous to two observers that are stationary with respect to each but at different altitudes in a tower on the Earth for example.

A subtle difference is that the observers in the Earth tower measure their clock rates to be different while the observers in the accelerating lab do not. This is a tidal effect of the curved spacetime of the Earth that is not present in the lab experiencing artificial constant proper acceleration.
Hi Kev. Isn't proper acceleration that changes over the height of the Born-rigid accelerated lab equivalent to a tidal effect?

On clock rates, check Prof. http://www.csupomona.edu/~ajm/professional/talks/relacc.ppt" [Broken], where he comments on born rigid acceleration:
• Since the vertex distances are different, so are the accelerations
• The "front" object, B, has a smaller proper acceleration than the "rear" object, A
• A and B agree at all times on matters of simultaneity
• A and B agree at all times on their common velocity
• A and B agree that their proper separation is constant
• A and B agree that B’s clock runs faster in direct proportion to their respective vertex distances.
Perhaps not a refereed publication, but rigorously correct, AFAIK.

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pmb_phy

If they can can not be duplicated then all we have to do is to point to the differences and use them to answer the question "What possible measurements can you make inside the cabin that can tell you, you are not in a gravitational field?" Our hypothetical gravitational body must be free of tidal effects because they are not measured within the cabin, with the possible exception of red shift of light signals sent up and down the cabin.
Tidal effects are not required in a gravitational field. Redshifted light signals also do not imply the presence of tidal forces/spacetime curvature.

Why do you associate tidal forces with a gravitational field? I gave you a nice example of a body which generates a field with no tidal forces in it. Tidal forces and gravitational forces (or tidal gradients and gravitational acceleration) are defined quite differently. The former is defined through the tidal force tensor and the second defined through an acceleration 3-vector.

Pete

yuiop

Hi Kev. Isn't proper acceleration that changes over the height of the Born-rigid accelerated lab equivalent to a tidal effect?

On clock rates, check Prof. http://www.csupomona.edu/~ajm/professional/talks/relacc.ppt" [Broken], where he comments on born rigid acceleration:

Perhaps not a refereed publication, but rigorously correct, AFAIK.
I have a problem with item 4 in the following list:

Since the vertex distances are different, so are the accelerations
The "front" object, B, has a smaller proper acceleration than the "rear" object, A.
1) A and B agree at all times on matters of simultaneity
2) A and B agree at all times on their common velocity
3) A and B agree that their proper separation is constant
4) A and B agree that B’s clock runs faster in direct proportion to their respective vertex distances.

This quote from Einstein's "ON THE ELECTRODYNAMICS
OF MOVING BODIES" https://www.fourmilab.ch/etexts/einstein/specrel/www/#SECTION11 seems to disagree.

"Let a ray of light start at the A time'' Ta from A towards B, let it at the B time'' Tb be reflected at B in the direction of A, and arrive again at A at the A time'' Ta'.

In accordance with definition the two clocks synchronize if

Tb - Ta = Ta' - Tb

We assume that this definition of synchronism is free from contradictions, and possible for any number of points..."

The above relation holds in the accelerating rocket frame. A signal leaving A at Ta = 1 second arrives at B at time Tb = 2 seconds and returns to A at time Ta' = 3 seconds, then Tb - Ta = Ta' - Tb = 2-1 = 3-2 = 1 showing the clocks are synchronised according to Einstein's definition. We can further check that a signal leaving A at Ta = 3 seconds arrives at B at time Tb = 4 seconds and returns to A at time Ta' = 5 seconds, then Tb - Ta = Ta' - Tb = 4-3 = 5-4 =1.

Condition 1 in the list states that A and B agree at all times on matters of simultaneity. This suggests that they agree their clocks advance simultaneously.

We can also add a third inertial observer C and note the following:

1) A and B agree at all times on matters of simultaneity.
1) If A and B agree that two events are simultaneous then C does not agree those events are simultaneous.
2) A and B agree at all times on their common velocity.
2) C does not agree that their velocities are equal at any given instant.
3) A and B agree that their separation is constant.
3) C does not agree that the separation of A and B is constant.
4) A and B agree that B’s clock runs faster in direct proportion to their respective vertex distances.
4) C agreee that B’s clock runs faster in direct proportion to their respective vertex distances.

Odd that C disagrees with all measurements that A and B make except for item (4). It seams the measurements of A and B should be swapped with those of C for item 4 to make a consistent list of observations.

On page 13 of the presentation you linked to, the author does agree that flashing clocks at positions A and B will be flashing simultaneously. That contradicts item 4 of the list.

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Jorrie

Gold Member
Odd that C disagrees with all measurements that A and B make except for item (4). It seams the measurements of A and B should be swapped with those of C for item 4 to make a consistent list of observations.

On page 13 of the presentation you linked to, the author does agree that flashing clocks at positions A and B will be flashing simultaneously. That contradicts item 4 of the list.
Hi kev, I don't quite follow your logic with the extended list that you supplied and how it contradicts Mallinckrodt. Can you explain a little more?

Slide 13 of Mallinckrodt seems to hold that "flashing synchronously is not the same as flashing at a definite time interval because the clocks run at different rates". So how does that contradict the different clock rates?

I must confess that Rindler frames are quite slippery on some of these concepts and it is always possible that even the professor did slip up! I'll give it another thought in the next few days. Maybe the moderators can help us out?

-J

Jorrie

Gold Member
Slide 13 of Mallinckrodt seems to hold that "flashing synchronously is not the same as flashing at a definite time interval because the clocks run at different rates". So how does that contradict the different clock rates?
Hi again kev. I understood what Mallinckrodt wrote as: the rod flashes at the same clock times all along the rod, but for rear- and front end observers the intervals between the flashes are different. It should be like that if the proper length remains the same for those observers, yet they experience different proper accelerations. The front clock records less acceleration because it runs faster than the rear clock. How else?

-J

Antenna Guy

We can also add a third inertial observer C and note the following:

1) A and B agree at all times on matters of simultaneity.
1) If A and B agree that two events are simultaneous then C does not agree those events are simultaneous.
bzzzt - not necessarily.

Let A and B be displaced equidistant from an origin O along x.
Let C (and D) be the same distance from O along y.
Let two events that appear as simultaneous to A and B occur on the z-axis.
C (and D) will not only agree on simultaneity, they will also agree with A and B regarding the time that the simultaneous events occured (relative to O).

Regards,

Bill

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yuiop

bzzzt - not necessarily.

Let A and B be displaced equidistant from an origin O along x.
Let C (and D) be the same distance from O along y.
Let two events that appear as simultaneous to A and B occur on the z-axis.
C (and D) will not only agree on simultaneity, they will also agree with A and B regarding the time that the simultaneous events occurred (relative to O).

Regards,

Bill
In the particular situation we were discussing, A and B are non inertial observers onboard an accelerating rocket and are spatially separated along the axis parallel to the accelerating motion of the rocket. Both A and B feel and measure proper acceleration. C is an inertial observer not on board the accelerating rocket. If you wish to introduce a fourth inertial observer D that is spatially separated from C, then C and D should be on a line parallel to the line joining A and B. The situation we were discussing is the Rindler spacetime drawn on a Minkowski diagram with one space dimension and one time dimension. Under those conditions, my original assertion:

1) A and B agree at all times on matters of simultaneity.
1) If A and B agree that two events are simultaneous then C does not agree those events are simultaneous.
remains true and can be extended to :

1) A and B agree at all times on matters of simultaneity.
1) If A and B agree that two events are simultaneous then C and D do not agree those events are simultaneous.

Your counter argument is based on all the observers (A,B, C and D) being inertial observers which is not the situation that was under discussion.

yuiop

Hi again kev. I understood what Mallinckrodt wrote as: the rod flashes at the same clock times all along the rod, but for rear- and front end observers the intervals between the flashes are different. It should be like that if the proper length remains the same for those observers, yet they experience different proper accelerations. The front clock records less acceleration because it runs faster than the rear clock. How else?

-J

Hi Jorrie,

I am at a slight disadvantage because for some unusual reason my browser does not display any of the diagrams or equations in the Powerpoint presentation you linked to :(

The exact quote from Mallinckrodt was "Note that within the frame of the rod, flashing synchronously is not the same as flashing at a definite time interval because the clocks run at different rates."

His statement is a little vague without being able to see the diagram. If I add comments in brackets his statement reads as "Note that within the frame of the rod, flashing synchronously is not the same as flashing at a definite time interval (in which frame, accelerated or inertial?) because the clocks run at different rates.(in which frame, accelerated or inertial?"

The first part is clear. The flashes occur simultaneously within the frame of the rod.

The second part is not clear. The clocks run at different rates is an observation made by an inertial observer not within the frame of the rod. As far as I am concerned, the clocks run at the same rate as measured by accelerating observers within the frame of the accelerating rod.

For example if observer A is onboard the rocket (near the tail) and sends signals at intervals of one second as measured by his local clock, then an observer B also onboard the accelerating rocket but near the nose will detect the signals as arriving at intervals of once per second using his local clock. The same happens if B sends signals to A in the opposite direction. By Einstein's definition the clocks are synchronised as far as observers A and B are concerned.

Now what is a little puzzling is that the signals will appear to be redshifted to A and blue shifted to B. The prof seems to be wrongly assuming that a redshifted signal automatically implies the clocks are running at different rates. This is not always true. What is true is that velocity of the emmitter at the time the signal was emmited was slower than the velocity of the receiver at the time the signal was received. as viewed by the inertial observer outside the rocket. Remember the observers inside the rocket can consider themselves to be stationary within a gravitational field so they do not consider themselves to have a velocity.

As mentioned before, observers A and B can prove to themselves that their clocks run at the same rate and remain synchronized. Knowing that, presumably they must conclude that the wavelength of the light wave is length contracted as it falls from the nose to the tail in the apparent "gravitational field" they are experiencing.

Jorrie

Gold Member
His statement is a little vague without being able to see the diagram. If I add comments in brackets his statement reads as "Note that within the frame of the rod, flashing synchronously is not the same as flashing at a definite time interval (in which frame, accelerated or inertial?) because the clocks run at different rates.(in which frame, accelerated or inertial?"
Hi kev, I attach .JPGs of the ST diagrams he referred to; hope they're viewable. I think he meant the "instantaneous comoving inertial frame", in which the two accelerating observers are at rest for a moment (x-axis along the line of "instantaneous simultaneity" through the origin).

Remember the observers inside the rocket can consider themselves to be stationary within a gravitational field so they do not consider themselves to have a velocity.
But, in their 'apparent uniform gravitational field', their clocks would run at different rates, not so. I think we have a problem with what 'simultaneity' means along a lengthwise accelerated rocket.

Prof. Mallinckrodt also said in his conclusion:
• Observers in a rigidly accelerated frame agree on almost everything except what time it is. Clock rates are proportional to vertex distances.
• When a rigid body returns to the velocity at which all of its clocks were synchronized, the clocks regain synchronization.
• I believe this material is accessible, surprising, uncontroversial, but nevertheless not well known.
I still have a feeling that the prof is right...

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yuiop

Hi kev, I attach .JPGs of the ST diagrams he referred to; hope they're viewable. I think he meant the "instantaneous comoving inertial frame", in which the two accelerating observers are at rest for a moment (x-axis along the line of "instantaneous simultaneity" through the origin).
The observer in the "instantaneous comoving inertial frame" does not make the same measurements/ observations that the accelerating observers make. The first diagram you posted shows that the rocket stops accelerating and as you know, I have already concluded in another thread that when the rocket stops accelerating the clock will no longer be syncronised in the rocket frame. While the rocket continues to accelerate indefinitely, the accelerating observers onboard the rocket will consider the clocks to remain synchronized.

But, in their 'apparent uniform gravitational field', their clocks would run at different rates, not so.
I agree that in a curved gravitational filed such as on the Earth that a clock higher up a tower would appear to run faster than a clock lower down. It would be impossible to get two such clocks to remain sychronised without artificially processing the rate of one of the clocks to slow it down or speed it up. In the rocket the same is not true. The 'apparent uniform gravitational field' they observe on the rigidly accelerating rocket is not the same as the curved gravitational field of the Earth. Another way of expressing the spacetime onboard the rocket is saying it is equivalent to a parallel uniform gravitational field to separate it from a curved gravitational field, which is less confusing than calling it a flat spacetime that seems to suggest to some people that no acceleration or gravity is present.

I think we have a problem with what 'simultaneity' means along a lengthwise accelerated rocket.
...
Well, I am using Einstein's definition of simultaneity where clocks are syncronized by sending timing light signals, as I posted earlier. I am also interpretating the "line of simultaneity" within the rocket frame to mean the line along which observers in the accelerated frame consider events to be simultaneous. If signals are sent simultaneously (as measured by observers in the rocket) then they arrive simultaneously (as measured by the observers onboard the rocket), because the signals arrive on the same on the same line of simultaneity.

I have a feeling this question might have to become a thread of its own :P

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