Gravitational Time Dilation: A Thought Experiment

[lalbatros]

Hi,

I have now the feeling that I was wrong in my answer at the beginning of the thread.
I said that synchronisation was not possible.
I have doubt about the meaning of that now.
I would like to know what is the usual meaning for synchronisation.

My doubt came from reading Landau.
By a change of variables, if I understood correctly, it is always possible to bring the component of the metric so that: gi0 = 0 and to also g00 = 1.
Looks like synchronisation in some sense, isn't it?

Michel

 

[RandallB]

Er, you do realize that GR predicts the clocks run at different rates, and that experimental evidence confirms GR, right? Such evidence has been provided in this very thread!

What are you talking about? You’re the one that called for a uniform gravitational field! Which to be uniform in a vertical direction means it would provide a constant acceleration through the vertical fall. Gravity on Earth doesn’t do that right. You do know you weight less at the top of a mountain than you do at the bottom of a valley right. To build a uniform vertical gravitational field you need to collect a large mass of some depth or thickness that extends out flat in all directions to about infinity. Then you will weigh the same no matter how high you are and accelerate at one uniform rate instead of an accelerating acceleration as you get when you pile up your mass in a ball like the earth. And the high clock runs the same as the lower. (No mountains or valleys we need to keep the surface flat.)

As to as to figuring out when things are truly simultaneous (other than co-located events) and being a “straw man”, I’ll leave it as my evidently taking Einstein’s points about simultaneity too seriously for current science and leave it at that.

 

[Hurkyl]

And the high clock runs the same as the lower. (No mountains or valleys we need to keep the surface flat.)

Er, no?

A coordinate chart where proper time is computed by

<br /> c^2 d\tau^2 = A^2 z^2 c^2 dt^2 - dx^2 - dy^2 - dz^2,<br />

is one with a constant, uniform gravitational field oriented along the z axis. (A is a constant depending on the strength of the field, and I assumed that z is strictly positive)

Let "up" denote the direction of the positive z-axis.

Clearly, in this chart, the higher clock a clock is, the faster it ticks¹. Of course, we'd rather have an experiment than talk about coordinates.

Because the metric is symmetric under time translation, it's easy to see that if we performed an experiment

(1) Start with two stationary¹ clocks located at the same place, and simultaneously reset them.
(2) When both clocks read 1 hour, lift clock A 10 meters¹.
(3) When clock B reads 2 hours, lift clock B 10 meters¹ in the same way that you lifted clock A.

then clock A will have a greater reading than clock B.

Surely this result is inconsistent with whatever you actually mean by the two clocks running at the same speed?


1: As measured in (t, x, y, z)-coordinates.

 

[pervect]

Er, no?

A coordinate chart where proper time is computed by

<br /> c^2 d\tau^2 = A^2 z^2 c^2 dt^2 - dx^2 - dy^2 - dz^2,<br />

is one with a constant, uniform gravitational field oriented along the z axis. (A is a constant depending on the strength of the field, and I assumed that z is strictly positive)

This is the standard definition of a uniform gravitational field in the literature. But I thought I should point out that this standard definition can be confusing, because the definition measured with an accelerometer as a function of z will not be constant. I.e. if one defines an orthonormal basis of one forms

(A z c) dt, dx, dy, dz

(These differentials are one forms because they associate a scalar, i.e Lorentz interval, with a vector).

the dual of these one-forms gives an ONB of vectors that constitutes a frame field

\left[ \frac{1}{A z c} \frac{\partial}{\partial t}, \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right] = \left[ \hat{t}, \hat{x}, \hat{y}, \hat{z} \right]

and we can compute in this frame-field (a non-coordinate but orthonormal basis) that

\Gamma^{\hat{z}}{}_{\hat{t}\hat{t}} = -1/z

This connection coefficient is what I'm talking about when I say "what an accelerometer will measure".

In a coordinate basis (which is not normal) for comparison we get

\Gamma^z{}_{tt} = A^2 z

 

[hover]

Hey everyone,
speaking of time dilation what equation can predict how slow a clock will run when it is accelerated??

thanks!

 

[pervect]

Hey everyone,
speaking of time dilation what equation can predict how slow a clock will run when it is accelerated??

thanks!

If you are at the same location as the clock, neither the clock's acceleration or your acceleration will matter.

I believe there is a sci.physics.faq entry on this, if I find it I'll edit this post to enter it later.

[add]
The physics FAQ on this topic is at http://math.ucr.edu/home/baez/physics/Relativity/SR/clock.html

I'll give a short excerpt and encourage hoover and other interested parties to read the original in its entirety.

The clock postulate can be stated in the following way. First, we take the rate that our frame's clocks count out their time, and compare that to the rate that a moving clock counts out its time. Before the clock postulate was ever thought of, all that was known was that when the moving clock has a constant velocity v (measured relative to the speed of light c), this ratio of rates is the Lorentz factor

gamma = 1/sqrt(1-v2)

The clock postulate generalises this to say that even when the moving clock accelerates, the ratio of the rate of our clocks compared to its rate is still the above quantity. That is, it only depends on v, and does not depend on any derivatives of v, such as acceleration. So this says that an accelerating clock will count out its time in such a way that at anyone moment, its timing has slowed by a factor (gamma) that only depends on its current speed; its acceleration has no effect at all.

 

[MeJennifer]

This is the standard definition of a uniform gravitational field in the literature. But I thought I should point out that this standard definition can be confusing, because the definition measured with an accelerometer as a function of z will not be constant.

Indeed very confusing and an excellent opportunity to clear all this up.

I think it is not only acceleration versus proper acceleration that is confusing.

For instance if we say a constant uniform gravitational field do we mean that the gravitational potential is everywhere the same or decreasing with increasing values of z? And depending on the answer then how would we call the gravitational field inside a hollow shell?
And how do we argue that something is curved in both those fields, and how do we reason time (if any) dilations for different positions in those fields?

 

[RandallB]

A coordinate chart where proper time is computed by

<br /> c^2 d\tau^2 = A^2 z^2 c^2 dt^2 - dx^2 - dy^2 - dz^2,<br />

is one with a constant, uniform gravitational field oriented along the z axis. (A is a constant depending on the strength of the field, and I assumed that z is strictly positive)

Let "up" denote the direction of the positive z-axis.

Clearly, in this chart, the higher clock a clock is, the faster it ...

Well sure your formula here is nice but your description of "constant" is wrong. The gravitational field here is only constant in the x & y directions (zero), not z.
The equivalence principal tells you that if you want to compare the constant acceleration of a spaceship in the z direction with a constant gravitational force in the z direction, that force needs to be constant; your factor for z in the first term needs to be z^0 not z^2.

 

[pervect]

Hurkyl's description of the metric of an accelerating observer is not wrong. I definitely didn't want to get into the middle of RandallB's argument with Hurkyl - as in the case of most arguments of a non-science advisor with a science advisor, the smart money is on the science advisor. But I did want to clear up some potential confusion - it appears, though, that I've added to it. I'll try one more, however.

To define a metric, one needs a coordinate system. While to some extent coordinate systems are arbitrary, the coordinate system associated with the metric Hurkyl presented is the "obvious" one.

One simple procedure to construct the coordinate system associated with Hurkyl's metric goes something like this:

1) Define the notion of "simultaneity" for an accelerated observer as the same notion of "simultaneity" for an instantaneously co-moving observer at the same location in space. Physically, this choice of simultaneity is equivalent to saying that the speed of light appears to be isotropic near the accelerating observer. "Near" means << c^2/g, for an observer accelerating at 1 g any distance under 1 light year is "short".

While all choices of simultaneity are to some extent "conventional", this is the obvious and standard choice.

2) Having defined the notion of simultaneity, a unique definition of the direction of distance is defined by insisting that space must be orthogonal to time (except for rotation as Hurkyl has already mentioned). One generally choses a non-rotating spatial basis. Non-rotating can be defined further mathematically, but it's too advanced a topic for this thread. This defines both the notion of simultaneity and the notion of constant position.

3) Having defined both constant time and constant position, the Lorentz interval allows one to measure the lengths of curves of constant time, defining space intervals (distances), and the lengths of curves of constant position, defining time intervals (times). Having defined the curves that define constant time and constant position, the process of assigning coordinates to a specific point is just the process of measuring the Lengths of specific curves. Perhaps a diagram is needed, but the process is no different here than it is on any piece of graph paper - on a standard piece of graph paper, one can define coordinates as the lengths of "horizontal" and "vertical" line segments.

Given an accelerating obserer, and going through the above procedure, one gets the above metric, which is an example of Rindler metric. The point I wanted to make is that different observers in this metric measure different acclerations on their local accelerometers depending on their positions. Also, co-located inertial observers would measure different accelerations for observers depending on their Rindler coordinates. This relates to the points I was making earlier - if you have a pair of spaceships, and they maintain a constant distance frpm each other, the two spaceships are not accelerating at the same rate.

 

[Hurkyl]

Well sure your formula here is nice but your description of "constant" is wrong. The gravitational field here is only constant in the x & y directions (zero), not z.
The equivalence principal tells you that if you want to compare the constant acceleration of a spaceship in the z direction with a constant gravitational force in the z direction, that force needs to be constant; your factor for z in the first term needs to be z^0 not z^2.

If it was z^0 then, of course, gravitational force would be everywhere zero. You're right, though: I didn't mean to say "constant, uniform", I merely meant to say "uniform".

Hurkyl's description of the metric of an accelerating observer is not wrong. I definitely didn't want to get into the middle of RandallB's argument with Hurkyl - as in the case of most arguments of a non-science advisor with a science advisor, the smart money is on the science advisor.

Feel free to jump in! I forgot how much I hate doing these computations, especially without a CAS to do them for me.

 

[MeJennifer]

If it was z^0 then, of course, gravitational force would be everywhere zero.

Actually, is that fully correct?

 

[Hurkyl]

Actually, is that fully correct?

Yah -- if we replace z^2 with z^0, then the geodesics are all straight¹ lines. In fact, the metric is exactly the metric of SR where we use different units for measuring time.


1: as measured by the coordinate chart

 

[MeJennifer]

Yah -- if we replace z^2 with z^0, then the geodesics are all straight¹ lines. In fact, the metric is exactly the metric of SR where we use different units for measuring time.

Outside this field the gravitational potential is different right?

 

[pervect]

Feel free to jump in! This thread has reminded me why I hate partial differential equations, especially without a CAS to do the calculations for me.

I love my Maple CAS - except for the problems it occasionally has with domains and symbolic integration. When you pass a real valued function and ask Maple to symbolically integrate it, there's no guarantee over what domain the result will be real-valued (at least from my experience). I don't know if Mathematica is any better in this regard. GRTensor + Maple is especially powerful for GR and differential geometry.

 

[Hurkyl]

Outside this field the gravitational potential is different right?

I don't understand the question. What do you mean by "outside this field"?

(Incidentally, gravitational potential is only an approximate notion in GR)

 

[RandallB]

Hurkyl's description of the metric of an accelerating observer is not wrong. I definitely didn't want to get into the middle of RandallB's argument with Hurkyl - as in the case of most arguments of a non-science advisor with a science advisor, the smart money is on the science advisor.

Well if the smart money is against me that must mean I'd get odds on a c-note, I could use an extra grand on getting published! Anyway for this part of the argument, the claim Hurkyl makes is the front "higher" clock in a rocket ship experiences a different rate of time than a clock in the back "lower" end as the ship is accelerating. His "proof" is your post #5.
Which is ridiculous, The ship clocks both have a common identical acceleration and will see time pass at the same rate. The clocks at different heights in the tower on Earth do not see the same force of gravity (equivalent to acceleration). His metric does not describe a "constant" gravitation or acceleration; it describes a changing gravity with respect to altitude like you find on Earth and your example in post #5. That is not a "constant" acceleration but an accelerating acceleration; which is another derivative important to NASA in working out the dynamics of a launch, I think they call it "jerk".

if you have a pair of spaceships, and they maintain a constant distance from each other, the two spaceships are not accelerating at the same rate.

That of course depends on the frame you make that measurement from. Such as the to clocks in the front of back of the ship maintaining a constant acceleration & rate of time change and a fixed distance between each other in their reference frames. But to maintain a constant distance in any other frame would require a the ship change real "intrinsic" length.

 

[Hurkyl]

Which is ridiculous, The ship clocks both have a common identical acceleration and will see time pass at the same rate.

Why would the front of the rocket have the same coordinate acceleration as the tail of the rocket?

(All measurements performed in an inertial frame where the rocket is initially at rest)

We know that, as the rocket gains speed, this frame observes that the rocket length contracts. Length contraction cannot happen if the front and tail of the rocket undergo identical accelerations.

 

[MeJennifer]

I don't understand the question. What do you mean by "outside this field"?

Assume a region in space with such a constant uniform gravitational field. You claim there is no gravitational force in that field. But clearly a light ray going out of that field would turn out to be redshifted.

 

[Hurkyl]

Assume a region in space with such a constant uniform gravitational field. You claim there is no gravitational force in that field. But clearly a light ray going out of that field would turn out to be redshifted.

You mean outside of that region of space-time -- whatever happens in a region of space-time with a different metric depends on that metric.

 

[MeJennifer]

You mean outside of that region of space-time -- whatever happens in a region of space-time with a different metric depends on that metric.

Yes that is quite obvious.
So, when a photon emited from a region with such a constant uniform gravitational field and absorbed in a region with no gravitational field it would be observed redshifted correct? So, how then how can you conclude there is no gravitational force in this field?

 

[Hurkyl]

Yes that is quite obvious.
So, when a photon emited from a region with such a constant uniform gravitational field and absorbed in a region with no gravitational field it would be observed redshifted correct? So, how then how can you conclude there is no gravitational force in this field?

"constant uniform gravitational field" -- you're still talking about the field that would correspond to the metric

A^2 c^2 (dt)^2 - (dx)^2 - (dy)^2 - (dz)^2?

I've already stated that, with this metric, the corresponding gravitational field would be zero. So, I'm not sure just what contrast you're trying to make. Why would anything be redshifted?


How can I conclude there is no gravitational force in these coordinates with this metric? Well, I can compute the coordinate acceleration of any geodesic. Since the geodesics are lines, their coordinate acceleration is zero. Thus, any particle experiences zero gravitational force.

 

[yogi]

Not sure I really want to get back into the issue of equivalence of the accelerating rocket-uniform G field (at least I don't want to plow old ground). So consider the situation with a clock mounted to the circumference of a rotating wheel - experiments indicate the rate of an ideal clock is not influenced by the acceleration - the measured decrease in the accumulated time corresponds to the velocity v = rw But it is also known clocks do not run slow because they are moving - all clocks run at their proper time in their own frame...so time dilation really establishes nothing that indicates clocks run slow in relatively moving inertial frames. What is verified is the invariance of the interval - the motion of the clock mounted to the circumference logs less time during each successive revolution because some of its spacetime journey during each revolution is spatial - whereas the clock at the center (axis of rotation) travels only in time.

So much for the preface everyone already knows. The question posed, in view of the above, is whether there is any residual time difference due to acceleration - specifically is the residual effect (time difference after the wheel is stopped and the clocks compared) totally explainable as a consequence of the invariance of the spacetime interval of SR?

 

[RandallB]

(All measurements performed in an inertial frame where the rocket is initially at rest)

We know that, as the rocket gains speed, this frame observes that the rocket length contracts. Length contraction cannot happen if the front and tail of the rocket undergo identical accelerations.

This assumes that "this frame" is a satisfactory Preferred Frame that can correctly judge the simultaneous time and position of both ends of the rocket and declare a real change in length and distance for the rocket not just an apparent length contraction from the view of “this frame”.

 

[Hurkyl]

So much for the preface everyone already knows. The question posed, in view of the above, is whether there is any residual time difference due to acceleration - specifically is the residual effect (time difference after the wheel is stopped and the clocks compared) totally explainable as a consequence of the invariance of the spacetime interval of SR?

That the wheel-mounted clock reads less time is a consequence of actually measuring the interval along the worldlines of the two clocks. In the particular frame you're using, you could say that this fact is a consequence of the form of the interval.

That the form of the interval is invariant under Lorentz transforms is irrelevant -- that only matters when you're comparing the form of physical laws in different inertial reference frames.

This assumes that "this frame" is a satisfactory Preferred Frame that can correctly judge the simultaneous time and position of both ends of the rocket and declare a real change in length and distance for the rocket not just an apparent length contraction from the view of “this frame”.

Who says it's "preferred"? It's just a frame. I don't understand your objection -- a frame in SR is quite literally a function that assigns spatial and temporal coordiantes to any event.

 

[RandallB]

Who says it's "preferred"? It's just a frame. I don't understand your objection -- a frame in SR is quite literally a function that assigns spatial and temporal coordiantes to any event.

You did when you chose to use that particular frame rather then finding a new frame where the two clocks are moving at the same speed simultaneously as seen from that new frame at whatever future instant time. Sure it is not easy to find but if you don’t work out details like that your not taking simultaneity into account.

 

[Hurkyl]

You did when you chose to use that particular frame rather then finding a new frame where the two clocks are moving at the same speed simultaneously as seen from that new frame at whatever future instant time. Sure it is not easy to find but if you don’t work out details like that your not taking simultaneity into account.

Huh? That doesn't make it a preferred frame. That simply makes it the frame in which I chose to analyze the problem. It's basic coordinate geometry: if you want to solve a problem, you can impose a choice of coordiantes, express everything in terms of those coordinates, and algebraically work out the solution.

I've attached a diagram of the problem. The two black curves are the worldlines of the tail and head of the rocket. They are both hyperbolas whose center is the origin of the red coordinate system. The red lines indicate the axes of one inertial reference frame. The green lines indicate the axes of another inertial reference frame. The grey line is a light-like line I used to help me draw the diagram freehand -- it is an asymptote to the two black curves.

The red lines indicates the coordinate axes I used to do all the analysis: left-right is the spatial axis, and up-down is the temporal axis. I chose to normalize the proper time along the worldlines to be zero at t=0[/color] in this chart. I intent to compute a duration, so the normalization doesn't affect anything. In this frame, the two rockets are at rest at t=0[/color]. I've indicated that the length of the rocket is L[/color] as measured in this frame at t=0[/color].

The green lines are the coordinate axes to another inertial reference frame: the one that is comoving with the bottom clock when it reads a time of a. It turns out that both clocks are at rest at t = 0[/color]. I've indicated a line segment of length L[/color] at t = 0[/color]. If the rocket were uniformly accelerating, then we would expect the head of the rocket to be at the other end of this line segment -- in fact, this is exactly the assumption I made when I first worked through this problem. (Though there are other ways to arrive at the solution to this problem)

The Lorentz metric c^2 d\tau^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2 tells us how to compute the actual proper time experienced by a clock if we know its coordinate trajectory in an inertial frame. It's clear from the diagram that, between the red and green horizontal lines, \Delta t_{head} &gt; \Delta t_{tail} and \Delta x_{head} &lt; \Delta x_{tail}. And thus we would expect \Delta \tau_{head} &gt; \Delta \tau_{tail}. If we actually integrate it to get the exact answer, our expectation is correct. In fact, it turns out there is a constant K so that (given the normalization I chose earlier) \Delta \tau_{head} = K \cdot \Delta \tau_{tail}, no matter what I happened to choose for a.

 

[pervect]

Assume a region in space with such a constant uniform gravitational field. You claim there is no gravitational force in that field. But clearly a light ray going out of that field would turn out to be redshifted.

You'll have to be a bit more specific to avoid misunderstanding. For instance, http://math.ucr.edu/home/baez/physics/Relativity/SR/spaceship_puzzle.html suggests a specific metric that may or may not be what you have in mind:

d\tau^2 = e^{2x}dt^2 - dx^2

This metric has the property that \Gamma^{\hat x}{}_{\hat{t}\hat{t}} = -1. Though note that \Gamma^{\hat x}{}_{\hat{t}\hat{x}} is also nonzero (and I don't think this is avoidable).

But this is not a spaceship moving through a vacuum - one can compute the Einstein tensor for this metric, and show that the stress-energy tensor is not zero. In fact, if my calculations are right, this metric requires "exotic matter" to create.

 

[MeJennifer]

You'll have to be a bit more specific to avoid misunderstanding. For instance, http://math.ucr.edu/home/baez/physics/Relativity/SR/spaceship_puzzle.html suggests a specific metric that may or may not be what you have in mind:

d\tau^2 = e^{2x}dt^2 - dx^2

This metric has the property that \Gamma^{\hat x}{}_{\hat{t}\hat{t}} = -1. Though note that \Gamma^{\hat x}{}_{\hat{t}\hat{x}} is also nonzero (and I don't think this is avoidable).

But this is not a spaceship moving through a vacuum - one can compute the Einstein tensor for this metric, and show that the stress-energy tensor is not zero. In fact, if my calculations are right, this metric requires "exotic matter" to create.

While this is a very interesting field I was not talking about this one.

Instead think for instance of the inside of a hollow shell.
Do you think there is a gravitational field in the middle?
And if no, how would you explain the time dilation there (relative to someone far away removed from this shell).

 

[Garth]

There is no time dilation between two separate observers who are both inside the hollow shell. They would conclude there is no gravitational field inside.

There is time dilation between an observer inside the shell and one outside it. They would conclude there is a gravitational field.

Garth

 

[MeJennifer]

There is time dilation between an observer inside the shell and one outside it. They would conclude there is a gravitational field.

And that was exactly the point that I brought up several postings ago.

In general relativity one cannot locally determine if one is in a gravitational field if this field is flat.