Acceleration or gravitational time dilation the same?

[rcgldr]

First case, gravity. A clock near sea level on the earth, an identical clock in open space, the clock in open space runs "faster" than the clock on the earth.

Second case, acceleration. A rotating space station applies 1g of centrepital acceleration on a clock, and an identical clock is at located at the center of the space station, experiencing no acceleration (or rotation).

Will there be the same difference in clock speeds in the space station case as the Earth / open space case?

 

[Dale]

No. In fact, the time dilation for the second case depends only on the tangential velocity which is related to the acceleration by a = v^2/r

 

[MeJennifer]

In fact, the time dilation for the second case depends only on the tangential velocity which is related to the acceleration by a = v^2/r

Yes, but that does not show it is not equivalent. From a rotating frame perspective it is gravitational time dilation.

 

[Dale]

Yes, but that does not show it is not equivalent. From a rotating frame perspective it is gravitational time dilation.

I guess I was not clear. The time dilation between the Earth at sea level and space has one fixed value. The time dilation on a space station with 1g centripetal acceleration can have any value greater than 1 depending on r. There would only be one value of r for which the time dilation would be the same.

 

[rcgldr]

I was only interested in the acceleration effect, not velocity, so change the second case.

New version of second case. Two rockets, with identical clocks. One is not accelerating, the other is accelerating at 1g.

Is the speed difference between the two clocks on the rockets the same as the speed difference between a clock at sea level on earth, and an identical clock out in open space?

 

[Dale]

I was only interested in the acceleration effect, not velocity, so change the second case.

New version of second case. Two rockets, with identical clocks. One is not accelerating, the other is accelerating at 1g.

There is no acceleration effect in SR according to an inertial observer. This has been tested up to 10^18 g. See the http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html#Clock_Hypothesis".

Is the speed difference between the two clocks on the rockets the same as the speed difference between a clock at sea level on earth, and an identical clock out in open space?

Again, the time dilation factor in case 1 is a constant fixed value whereas the time dilation factor in case 2 (original and modified) can assume any value greater than 1 depending on the instantaneous velocity of the accelerating clock.

 

[garth007]

An interesting thread...
The speed of the two clocks at sea level and Earth orbit is 'constant'.
The speed of the non-accelerating rocket is 'constant'.
The speed of the (1g) accelerating rocket is 'increasing'.
Shouldn't the speed difference of the clock in the accelerating rocket automatically be increasing?...vis-a-vis Einsteins twins paradox?

 

[garth007]

Searchin' for Eotvos

I spotted this thread while looking for info on the "Re-analysis of the Eotvos experiment" by Ephraim Fischbach and the subsequent paper/experiment published by C.W. Stubbs and Eric Adelberger called "Limits on composition-dependent interactions on a laboratory source; Is there a fifth force coupled to isospin?" I had a question about the second experiment's experimental relevance...and if anyone 'in-the-know' about it could help.

Or should I just start a new thread?

Gratzi.

 

[rcgldr]

I'm not interested in the velocity differences, only if acceleration in GR results in the same time dilation as gravity, so for the rocket case I'm only interested in the time dilation effect at the exact moment when the accelerating rocket has the exact same velocity as the non-accelerating rocket.

Using GPS satellites as an example, most of the the time dilation is due to the difference in gravity, and not due to the speed of the satellites versus speed at the surface of the earth.

A calculation using General Relativity predicts that the clocks in each GPS satellite should get ahead of ground-based clocks by 45 microseconds per day.

Special relativity predicts that the on-board atomic clocks on the satellites should fall behind clocks on the ground by about 7 microseconds per day because of the slower ticking rate due to the time dilation effect of their relative motion.

The combination of these two relativitic effects means that the clocks on-board each satellite should tick faster than identical clocks on the ground by about 38 microseconds per day (45-7=38)

http://www.astronomy.ohio-state.edu/~pogge/Ast162/Unit5/gps.html

As another analogy, I could use the Earth and gps satellites for one case, and a very large rotating space station with clocks position so that their relative acceleration and velocites correspond to the gravitational pull and relative velocities of the clocks at sea level and in GPS orbit. Now given this, is the realtive rate of clocks for both cases the same?

The bottom line question is: Does acceleration have an identical GR time dilation affect as gravity?

 

[pmb_phy]

No. In fact, the time dilation for the second case depends only on the tangential velocity which is related to the acceleration by a = v^2/r

The time dilation due to a difference in speed is exactly the value which is found if one uses gravitational acceleration. They are not different phenomena, they are the same phenomena interpreted differently. Note that there is a 1-to-1 relation between speed and acceleration in this problem, hence the two equivalent ways to iinterpret the phenomena.

This is not an SR problem either. This is a GR problem in the sense defined by Einstein. An observer iat rest in the rotating frame at a distance r from the center of rotation will determine that he is at rest in a gravitational field. As such there will be a gravitational time dilation effect. The reason this is a GR problem is because the topic regards observations in a non-inertial frame of reference.

Pete

 

[garth007]

What a great question!
The relative gravitational mass effects on time dilation versus the velocity effects on time dilation...

I should think there would be no difference if the proper gravitational mass exactly balanced the inertial velocity at a given point in time.

 

[garth007]

Ah...I was slow on the draw. Great answer Pete!

 

[jimgraber]

Short answer: Yes, they are exactly the same.
That is what the principle of equivalence means.

A much more detailed and easy-to-understand explanation is at this URL.

http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html

Read especially the section "Inside the Rocket".

If you want more, google
"acceleration special relativity".

Hope this helps.

Jim Graber

 

[Dale]

I'm not interested in the velocity differences, only if acceleration in GR results in the same time dilation as gravity,
...

The bottom line question is: Does acceleration have an identical GR time dilation affect as gravity?

Even in GR time dilation is not directly related to acceleration/gravity. It is related to the gravitational potential. So if two different planets have the same g on their surface, but different radii (different densities) they will have different time dilation factors.

 

[pmb_phy]

Even in GR time dilation is not directly related to acceleration/gravity. It is related to the gravitational potential. So if two different planets have the same g on their surface, but different radii (different densities) they will have different time dilation factors.

If you take a close look at the gravitational potential you will see that there is a 1-to-1 relationship between them, e.g. \Phi = gz where g = a = acceleration. For this reason it is readily seen that acceleration generates the gravitational potential.

Pete

 

[Dale]

If you take a close look at the gravitational potential you will see that there is a 1-to-1 relationship between them, e.g. \Phi = gz where g = a = acceleration.

Uhhh, that's not a 1-to-1 relationship. For one value g there are an infinite number of possible values of \Phi depending on z.

The point is that to get a specific value of time dilation in either SR or GR it is insufficient simply to define g. You also need to specify some other aspect. This should be apparent by considering both the clock hypothesis and the equivalence principle. Since by the clock hypothesis time dilation does not depend directly on acceleration in SR then by the equivalence principle it cannot depend directly on g in GR.

So going back to the OP. The gravitational scenario specified on Earth at sea level, which specifies both g and z. Similarly, to get equal time dilation in the centrifugal case you need to specify both g and r.

 

[Hans de Vries]

The ceiling in a house on Earth ages faster as the floor, while the ceiling in an accelerating
rocket also ages faster as the floor. They do so by the same amount if both g and h(eight)
are the same. (The potential difference is g*h)

The effects cancel in a space station rotating around the earth, (the acceleration
is downwards) The ceiling and the floor of the space station age at the same rate.
There is zero-g inside the space station.

The cancellation is due to the equivalence principle.


So, the answer to the OP's question in the title is Yes (the equivalence principle upholds)
while the answer to the question in the post is No (because g is the same but h isn't)


Regards, Hans

 

[kahoomann]

Uhhh, that's not a 1-to-1 relationship. For one value g there are an infinite number of possible values of \Phi depending on z.

The point is that to get a specific value of time dilation in either SR or GR it is insufficient simply to define g. You also need to specify some other aspect. This should be apparent by considering both the clock hypothesis and the equivalence principle. Since by the clock hypothesis time dilation does not depend directly on acceleration in SR then by the equivalence principle it cannot depend directly on g in GR.

So going back to the OP. The gravitational scenario specified on Earth at sea level, which specifies both g and z. Similarly, to get equal time dilation in the centrifugal case you need to specify both g and r.

What r do we need to specify? r is just the distance where the clock is located from the center of the rotating space station

 

[yuiop]

a factor of 2 problem

I set up some equations in a spreadsheet and found a factor of 2 problem.

Starting with a planet the gravitational time dilation factor is:

\gamma_g=\left(1-\frac{2GM}{Rc^2}\right)^{(-0.5)}

The kinetic time dilation factor is given by:

\gamma_v=\left(1-\frac{v^2}{c^2}\right)^{(-0.5)}

From the above the equivalent velocity to produce the same time dilation as the gravitational case is given by

v=\left(c^2-\frac{c^2}{\gamma_g^2}\right)^{(0.5)}

By setting the radius of the space station equal to that of the planet the centripetal acceleration given by v^2/R works out to be twice the surface acceleration of the planet given by GM/R^2. In other words the two situations are only equivalent if the planet and space station have the same radius and it assumed the surface acceleration of the planet is given by 2GM/R^2 rather than just GM/R^2, or if GM/R^2 is assumed then the spacestation would have to have twice the radius of the planet to be equivalent.

Why the factor of 2?

Is it because escape velocity as calculated from (1/2)mv^2 = GMm/R is equivalent to sqrt(2GM/R)?

 

[pmb_phy]

Uhhh, that's not a 1-to-1 relationship. For one value g there are an infinite number of possible values of \Phi depending on z.

If you're thinking about the arbitrary additive constant that one normally assciates with \Phi then you should note that in GR that constant is chosen such that at z = 0 \Phi = 0. Otherwise clocks at the observers location would not tick at the same rate as the clocks at z = 0 and that would be a contradicton.

The point is that to get a specific value of time dilation in either SR or GR it is insufficient simply to define g. You also need to specify some other aspect. This should be apparent by considering both the clock hypothesis and the equivalence principle. Since by the clock hypothesis time dilation does not depend directly on acceleration in SR then by the equivalence principle it cannot depend directly on g in GR.

The clock hypothesis refers to the rate at which accelerating clocks tick as measured by observers in an inertial frame. That rate does not depend on acceleration. The equivalence principle is not referring to that.

Pete

 

[pmb_phy]

I set up some equations in a spreadsheet and found a factor of 2 problem.

Starting with a planet the gravitational time dilation factor is:

\gamma_g=\left(1-\frac{2GM}{Rc^2}\right)^{(-0.5)}

It should be pointed out that the (weak) equivalence principle refers to the equivalence between a uniform gravitational field and a uniformly accelerating frame of reference. The gravitational field of the Earth has a curved spacetime. The accelerating frame does not since it is accelerating with respect to an inertial frame in flat spacetime. The strong equivalnce principle refers to the change from partial derivative to the covariant derivative when going from equations in Lorentz coordinates in flat spacetime (in an inertial frame) to coordinates in an arbitrary frame and in general a curved spacetime. There is no general equivalence between an accelerating frame and a gravitational field.

Pete

 

[yuiop]

If you're thinking about the arbitrary additive constant that one normally assciates with \Phi then you should note that in GR that constant is chosen such that at z = 0 \Phi = 0. Otherwise clocks at the observers location would not tick at the same rate as the clocks at z = 0 and that would be a contradicton.

The clock hypothesis refers to the rate at which accelerating clocks tick as measured by observers in an inertial frame. That rate does not depend on acceleration. The equivalence principle is not referring to that.

Pete

Your last statement that clock rates do not depend on acceleration means the answer to the OP is that clocks on the space station experiencing 1g will not necessarily tick at the same rate as clock on the Earth experiencing 1g, which is what Dalespam has been saying all along.

They only tick at the same rate, when the space station has a specific radius.

(What that specific radius is, depends on the answer to the factor of 2 problem in my previous post).

One other factor that complicates things a bit is that the local proper acceleration experienced on the surface of a planet is given by \gamma_gGM/R^2 rather than the Newtonian GM/R^2. I assume the local proper acceleration on the perimeter of the spacestation is relativistically adjusted in a similar way?

 

[yuiop]

It should be pointed out that the (weak) equivalence principle refers to the equivalence between a uniform gravitational field and a uniformly accelerating frame of reference. The gravitational field of the Earth has a curved spacetime. The accelerating frame does not since it is accelerating with respect to an inertial frame in flat spacetime. The strong equivalnce principle refers to the change from partial derivative to the covariant derivative when going from equations in Lorentz coordinates in flat spacetime (in an inertial frame) to coordinates in an arbitrary frame and in general a curved spacetime. There is no general equivalence between an accelerating frame and a gravitational field.

Pete

I (sort of) understand that, but I was just curious why the difference between a uniform gravitational field and the curved spacetime of a spherical planet turns out to be exactly a factor of 2?

A similar question was asked in this thread here https://www.physicsforums.com/showthread.php?t=240405 but no one has answered it yet.

 

[pmb_phy]

Your last statement that clock rates do not depend on acceleration means the answer to the OP is that clocks on the space station experiencing 1g will not necessarily tick at the same rate as clock on the Earth experiencing 1g, which is what Dalespam has been saying all along.

That is not what I wrote in my post. The difference between the acceleration of the clock and the acceleration of the frame of reference should be kept distict. If the actual rate at which a clock ticked was a function of acceleration then clocks at the origin would not run at the proper rate. The clock hypothesis rules this out. Clocks at the observers location run at the correct rate, i.e. at the rate at which time passes at the origin. If it were otherwise then there would not be an equivalence principle. The fact that the acceleration of a clock does not affect the rate at which it ticks is crucial in this regard.

A clock in free-fall, with speed zero, at the origin of the observers origin of coordinates will run at the same rate as a clock at rest at the observers origin of coordinates. That is the essence of the clock hypothesis in this context.

Pete

 

[pmb_phy]

I (sort of) understand that, but I was just curious why the difference between a uniform gravitational field and the curved spacetime of a spherical planet turns out to be exactly a factor of 2?

I read your concern and have not addressed it yet. I'll have to sit down and work this out with pen and paper. Right now I'm busy with something else, i.e. watching Deep Impact on TV.

I'll get to your concern later. Thanks for your patience.

Pete

 

[yuiop]

I read your concern and have not addressed it yet. I'll have to sit down and work this out with pen and paper. Right now I'm busy with something else, i.e. watching Deep Impact on TV.

I'll get to your concern later. Thanks for your patience.

Pete

Watch away! ..cool film

 

[Dale]

So, the answer to the OP's question in the title is Yes (the equivalence principle upholds)
while the answer to the question in the post is No (because g is the same but h isn't)

Exactly! Thank you.

Pete, let me use math instead of english, not for a derivation, just for clarity:

The expression for gravitational time dilation in the Schwarzschild metric is
\frac{t_g}{t_f} = \sqrt{1 - \frac{2GM}{Rc^2}} = \sqrt{1 - \frac{2gR}{c^2}}
Where t_g is the proper time between events A and B for a slow-ticking observer at rest within the gravitational field, t_f is the proper time between events A and B for a fast-ticking observer located at rest far away (outside the field), M is the mass of the object and R is the radial Schwarzschild coordinate of the slow-ticking observer, and g is the gravitational acceleration at the slow-ticking observer.

The expression for time dilation in a spinning space station in flat spacetime is
\frac{t_s}{t_f} = \sqrt{1 - \frac{v^2}{c^2}} = \sqrt{1 - \frac{gr}{c^2}}
Where t_s is the proper time between events A and B for a slow-ticking observer at rest on the space station floor, and r is the radius of the space station (from the center to the floor).

In both cases the time dilation depends on something besides just g, and they are only equal if r=2R.

 

[Dale]

I set up some equations in a spreadsheet and found a factor of 2 problem.

I get the same factor of 2, but I don't think there is any problem. I don't know where it comes from in the Swartzschild solution but they are talking about different distances, so I don't think there is any conundrum.

 

[pmb_phy]

Watch away! ..cool film

Back to work!

You had commented that

By setting the radius of the space station equal to that of the planet the centripetal acceleration given by v^2/R works out to be twice the surface acceleration of the planet given by GM/R^2. In other words the two situations are only equivalent if the planet and space station have the same radius and it assumed the surface acceleration of the planet is given by 2GM/R^2 rather than just GM/R^2, or if GM/R^2 is assumed then the spacestation would have to have twice the radius of the planet to be equivalent.

From what I see you are requiring the acceleration in the space station to be GM/R^2 and are finding a problem when you don't get this result. The reason that you don't get this result is because the equivalence principle does not require the acceleration to be GM/R^2. Notice that R has a different meaning in each case. For the Earth a decreasing value of R leads to a smaller value of g while in the rotating frame a decreasing value of R (angular velocity being constant) leads to smaller values of g.

To employ the equivalence principle require only that the acceleration due to gravity equal the local acceleration in the frame of reference. In the rotating frame there is a time dilation effect but there is also a frame dragging effect. If one measures the distance above the origin by z then an increasing z leads to a decreasing r. Trying making this coordinate transformation and see what you get.

Pete

 

[rcgldr]

My main question was if the equivalence principle of gravity and acceleration applies to time dilation as well.

The main difference between gravity and acceleration is that gravity changes with distance from the source, while acceleration remains constant. So in an enclosed environment, putting two clocks at the top and bottom of a chamber, if the clocks don't run at the same speed, it's gravity (from a point source), and if they're the same, it's acceleration.