GR Effects on Satellites in Free Fall: Resolving the Confusion

[Buckethead]

Two very simple questions, but ironically I've listened to a number of podcasts and articles that leave me confused as they don't always sound consistent so...

1. Is a satellite that orbits a planet (i.e. a satellite in free fall) feel the GR effects of the gravity of the planet? In other words, disregarding SR for the moment, is a clock on a satellite ticking at the same rate as a clock in the middle of space. I'm guessing they will because neither feels gravity.

2. If you are standing on the North Pole and a satellite is orbiting around the equator, and disregarding any gravitational effects and only addressing SR effects, does a clock on the North Pole tick at the same rate as a clock on the satellite. I'm guessing it will since the distance between the satellite and observer never changes.

Thanks, hope to bury these questions once and for all.

 

[PAllen]

Two very simple questions, but ironically I've listened to a number of podcasts and articles that leave me confused as they don't always sound consistent so...

1. Is a satellite that orbits a planet (i.e. a satellite in free fall) feel the GR effects of the gravity of the planet? In other words, disregarding SR for the moment, is a clock on a satellite ticking at the same rate as a clock in the middle of space. I'm guessing they will because neither feels gravity.

2. If you are standing on the North Pole and a satellite is orbiting around the equator, and disregarding any gravitational effects and only addressing SR effects, does a clock on the North Pole tick at the same rate as a clock on the satellite. I'm guessing it will since the distance between the satellite and observer never changes.

Thanks, hope to bury these questions once and for all.

Both your suppositions are incorrect.

1) It is correct that a satellite in orbit is in free fall, and is following an inertial trajectory in GR. However, it is not correct in GR that all inertial trajectories mutually observe the same time rate. Further, if two events are connected by two different inertial trajectories, they will end up aging differently, producing a pure inertial version of twin differential aging. For example, suppose a clock is passing an orbiting satellite moving radially away from the planet at a speed such that it will fall back and cross the satellite's orbit again after one orbit. Both trajectories, separating and meeting again, are purely inertial. Yet, the radial free fall trajectory will be the one that shows greater elapsed time. You might be thinking of separating this into SR vs. GR effects, but GR is a unified theory including SR. However, to make some attempt to do this separation for your example, you don't get the result you expect because the orbiting satellite is at a different gravitational potential than a far away clock. Where GR can be approximated by a potential, time dilation can be separated into a relative speed component (call it SR) and component due to potential difference.

2) This is really confused. If you subtract gravity, the satellite around the equator is not in free fall, and is, instead, analogous to circular trajectory with constant centripetal acceleration. This would be analogous to particles in an accelerator ring, which would clearly be time dilated relative to an observer stationary with respect to the ring's center.

 

[Buckethead]

Both your suppositions are incorrect.

1) ... For example, suppose a clock is passing an orbiting satellite moving radially away from the planet at a speed such that it will fall back and cross the satellite's orbit again after one orbit. Both trajectories, separating and meeting again, are purely inertial. Yet, the radial free fall trajectory will be the one that shows greater elapsed time.

How does one determine that the radial unit shows the greater elapsed time? (excellent example BTW) Is it because the radial unit will have gone a lesser distance between the two meetups?

You might be thinking of separating this into SR vs. GR effects, but GR is a unified theory including SR.

My mistake. I was trying to think of this in the same way that a diagonal vector can be broken down into an x and a y component. So I assumed that time dilation could be broken down the same way into a "SR axis" and a "GR axis" in a way. As you are perhaps suggesting in the below quote.

However, to make some attempt to do this separation for your example, you don't get the result you expect because the orbiting satellite is at a different gravitational potential than a far away clock. Where GR can be approximated by a potential, time dilation can be separated into a relative speed component (call it SR) and component due to potential difference.

OK so a stationary object and an orbiting object at the same elevation show the same GR effects (same time dilation degree) and in addition the fact that one is moving relative to the stationary one adds even more time dilation. Correct?

2) This is really confused. If you subtract gravity, the satellite around the equator is not in free fall, and is, instead, analogous to circular trajectory with constant centripetal acceleration. This would be analogous to particles in an accelerator ring, which would clearly be time dilated relative to an observer stationary with respect to the ring's center.

I was not really trying to subtract gravity, only separate it into a gravity component and a relative velocity component, so are you saying that this is not allowed as it is meaningless to think of an object in both a free fall and a circular trajectory at the same time unless a gravity component is preserved (or in the case of the accelerator a mag field)? I think I'm seeing a deeper meaning here so let me ask this question:

In the case of the accelerator, there is no gravity so there has to be a force (mag field) to keep the particle in line. If I were to sit at the center and spin around so that the particle seemed to stand still, would I have to say the particle is still moving when determining its time dilation because of the required force to keep the status quo and not because of any relative speed between me and the particle? And does this extend to the example of the orbiting satellite as well in that I am not allowed to say the satellite is stationary just because I spin around on the North Pole due to the gravity of the planet keeping the satellite from flying off.

What I'm getting at with these questions is that I almost feel like there is always some kind of relative speed involved but in the case of GR this relative speed translates into the opposing force (gravity or the mag field in the accelerator) rather than an actual relative speed and because of this it doesn't make sense to think of an orbiting satellite in free fall unless you also include the required force. Am I on the right track?

Looks like I won't be burying any questions today.

 

[PAllen]

How does one determine that the radial unit shows the greater elapsed time? (excellent example BTW) Is it because the radial unit will have gone a lesser distance between the two meetups?

Well the real answer is to integrate the proper time for the paths using the Schwarzschild metric. You can justify by other arguments, but none covers the general case in GR. For example, note that the radial trajectory spends more time at higher gravitational potential. Posit that since both motions are inertial, this dominates, so the radial inertial trajectory elapses more time due to the higher potential. You can analyze many near spherically symmetric situations this way, and also weak gravity cases that are less symmetric, but it is not really GR - it is an approximation to GR that works in many simple cases.

My mistake. I was trying to think of this in the same way that a diagonal vector can be broken down into an x and a y component. So I assumed that time dilation could be broken down the same way into a "SR axis" and a "GR axis" in a way. As you are perhaps suggesting in the below quote.

This is only possible in very specialized cases. To the extent you do this, you are not really grappling with GR.

OK so a stationary object and an orbiting object at the same elevation show the same GR effects (same time dilation degree) and in addition the fact that one is moving relative to the stationary one adds even more time dilation. Correct?

Yes, for a spherically symmetric, nearly static field (that thus can be treated with a potential).

I was not really trying to subtract gravity, only separate it into a gravity component and a relative velocity component, so are you saying that this is not allowed as it is meaningless to think of an object in both a free fall and a circular trajectory at the same time unless a gravity component is preserved (or in the case of the accelerator a mag field)? I think I'm seeing a deeper meaning here so let me ask this question:

Yes, I think it is meaningless to talk about free fall circular trajectory without gravity.

In the case of the accelerator, there is no gravity so there has to be a force (mag field) to keep the particle in line. If I were to sit at the center and spin around so that the particle seemed to stand still, would I have to say the particle is still moving when determining its time dilation because of the required force to keep the status quo and not because of any relative speed between me and the particle?

Rotating frames are complex in relativity. However, yes, a clock spinning in the center at the same angular speed as a particle moving at some signficant radius will see the circular moving particle time dilated. Further, the circular moving particle will see the spinning central clock going fast. This is not a case mutual time dilation. This is all purely SR - no gravity inolved.

And does this extend to the example of the orbiting satellite as well in that I am not allowed to say the satellite is stationary just because I spin around on the North Pole due to the gravity of the planet keeping the satellite from flying off.

Actually, for a polar observer and an equatorial orbiting satellite you would have to balance two effects - the satellite's higher potential would make it go fast per the polar observer, but its speed would make it go slower. The altitude of the satellite would determine which wins. If in a low orbit, it would be dilated relative to the polar clock, if in a high orbit would run faster.

What I'm getting at with these questions is that I almost feel like there is always some kind of relative speed involved but in the case of GR this relative speed translates into the opposing force (gravity or the mag field in the accelerator) rather than an actual relative speed and because of this it doesn't make sense to think of an orbiting satellite in free fall unless you also include the required force. Am I on the right track?

Looks like I won't be burying any questions today.

For simple, symmetric, weak gravity situations you can pursue this direction. But the key to doing it right is to forget force and focus on potential. For a distant observer, a clock on a massive shell of matter is dilated; a clock inside the shell, experience no force, is even more dilated.

 

[Buckethead]

Rotating frames are complex in relativity. However, yes, a clock spinning in the center at the same angular speed as a particle moving at some signficant radius will see the circular moving particle time dilated. Further, the circular moving particle will see the spinning central clock going fast. This is not a case mutual time dilation. This is all purely SR - no gravity inolved.

Why is this pure SR? I would have thought it would be pure GR for the reason that 1) There is no relative speed between the particle and the axis (the distances between the two remain constant), 2) There is a constant accelerating force on the particle meaning that its clock will move slower than the axis clock due to the acceleration, 3) There is no mutual time dilation as there is in two inertial ships passing each other (just as there is no mutual time dilation between a clock in and out of a gravity field), and 4) Because there is a preferred frame of reference as in when the axis clock is spinning at the same rate as the particle, the axis knows it by its centriputal force.

Actually, for a polar observer and an equatorial orbiting satellite you would have to balance two effects - the satellite's higher potential would make it go fast per the polar observer, but its speed would make it go slower. The altitude of the satellite would determine which wins. If in a low orbit, it would be dilated relative to the polar clock, if in a high orbit would run faster.

I understand the time dilation due to gravity, but the time dilation due to the orbit still confuses me in that to the observer, whether or not the satellite is actually moving depends only on whether the observer is spinning around in his north pole chair. Since distance never changes between the observer and the satellite and since (if the observer is spinning with the satellite) there isn't even a relative movement of any kind between the two, it seems that gravity is the only thing that has any say in the matter, not SR. Now I understand that there is an underlying asymmetry in that the observer obviously knows he is spinning due to the forces that he feels and also because he knows that the satellite would fall to Earth if it were not orbiting (although he could equally just assume that the Earth had no mass I guess).

Part of the difficulty I'm having with some of this is the nagging question of how an object knows when it is spinning or to put it another way, how does it know it's suppose to feel centripetal forces when it rotates relative to the surrounding space when it can't possibly know if it is spinning or if the space around it is spinning and it is stationary. But perhaps that is a question for another thread or a question that can't be answered.

I'd like to just for a moment go back to the question of why there is a gravity potential for an orbiting satellite when according to Einstein's philosophy there is no gravity in the sense of a force but rather just a mass and along with that mass comes a curvature in spacetime and it is this curvature alone that determines the actions on orbiting objects. So if there is no force and only curvature, then doesn't this imply that a orbiting object and a linearly traveling inertial object should be at the same gravitational potential or is it that the degree of curvature in space is what dictates a "gravitational" potential and not really the massive object (other than of course the massive object is causing the curvature)?

 

[PAllen]

Why is this pure SR? I would have thought it would be pure GR for the reason that 1) There is no relative speed between the particle and the axis (the distances between the two remain constant), 2) There is a constant accelerating force on the particle meaning that its clock will move slower than the axis clock due to the acceleration, 3) There is no mutual time dilation as there is in two inertial ships passing each other (just as there is no mutual time dilation between a clock in and out of a gravity field), and 4) Because there is a preferred frame of reference as in when the axis clock is spinning at the same rate as the particle, the axis knows it by its centriputal force.

If gravity can be ignored, it is pure SR. The central clock and the clock revolving about it could be nearly massless in empty space. There is no significant gravity, so it is purely SR. SR distinguishes inertial frames from accelerating or rotating frames. Acceleration and rotation are absolute, not relative - they can be detected locally, inside a black box. Acceleration and rotation were analyzed using SR before GR existed.

The easy way to analyze this with pure SR is to pick any inertial frame for the analysis.

I understand the time dilation due to gravity, but the time dilation due to the orbit still confuses me in that to the observer, whether or not the satellite is actually moving depends only on whether the observer is spinning around in his north pole chair. Since distance never changes between the observer and the satellite and since (if the observer is spinning with the satellite) there isn't even a relative movement of any kind between the two, it seems that gravity is the only thing that has any say in the matter, not SR. Now I understand that there is an underlying asymmetry in that the observer obviously knows he is spinning due to the forces that he feels and also because he knows that the satellite would fall to Earth if it were not orbiting (although he could equally just assume that the Earth had no mass I guess).

An object always 'knows' if it is spinning. Close your eyes and ask someone to spin you on a chair without warning. You have no trouble knowing when you are spinning. This is an absolute, not relative, effect in both SR and GR.

However, the spin of a north pole clock is wholly insignificant - one rotation per day. Even if it were more rapid spin, for a sufficiently small, ideal, clock it makes no difference in time dilation between such a clock and a non-spinning inertial clock. What is important is that the clock knows it is spinning, and can measure this even inside a black box. Thus, it knows that something stationary at some distance is moving relative to an inertial frame. For emphasis: there is no principle of relativity in SR except between inertial frames.

Anyway, I prefer to integrate metrics for such problems. You wanted to try to factor effects. If you want to do that, you have to start with the notion of static world lines in the SC geometry. These define potential as a function of position. These world lines are not inertial - they experience proper acceleration. Then, any object moving relative to static world lines experiences relative motion time dilation compared to an adjacent static observer. For some other static observer, there is then a potential difference effect and a motion effect. The north pole observer, whether you worry about spin or not, is effectively a static observer at a fixed potential.

I'd like to just for a moment go back to the question of why there is a gravity potential for an orbiting satellite when according to Einstein's philosophy there is no gravity in the sense of a force but rather just a mass and along with that mass comes a curvature in spacetime and it is this curvature alone that determines the actions on orbiting objects. So if there is no force and only curvature, then doesn't this imply that a orbiting object and a linearly traveling inertial object should be at the same gravitational potential or is it that the degree of curvature in space is what dictates a "gravitational" potential and not really the massive object (other than of course the massive object is causing the curvature)?

Gravitational potential is the energy to get a static object from its current position to 'infinity'. The fact that potential difference for static observers corresponds to relative time dilation between them is derived special case of GR. You don't have to ever use this concept, and it isn't useful in the general case. The general method is to have the metric represented in some coordinates, and compute the proper time along world lines by integrating. This is how you derive the special features of static or stationary metrics, or of the weak field approximation.

 

[WannabeNewton]

Part of the difficulty I'm having with some of this is the nagging question of how an object knows when it is spinning or to put it another way, how does it know it's suppose to feel centripetal forces when it rotates relative to the surrounding space when it can't possibly know if it is spinning or if the space around it is spinning and it is stationary. But perhaps that is a question for another thread or a question that can't be answered.

It can certainly be answered: rotation is absolute not relative. I don't need to pay any attention to the distant stars to tell if I'm rotating or not. There are a myriad of ways I can locally measure rotation. For example I can carry a pronged sphere with beads inserted through the prongs and see if the beads are flung outwards by inertial forces, or see if local gyroscopes rotate relative to me since gyroscopes define what it means to be locally non-rotating.

 

[Buckethead]

If gravity can be ignored, it is pure SR. The central clock and the clock revolving about it could be nearly massless in empty space. There is no significant gravity, so it is purely SR. SR distinguishes inertial frames from accelerating or rotating frames. Acceleration and rotation are absolute, not relative - they can be detected locally, inside a black box. Acceleration and rotation were analyzed using SR before GR existed.

So if you are in an inertial frame in space (a space station) and you shoot off a rocket with a constant acceleration you can use SR alone to calculate the time dilation on the rocket during its acceleration? And the answer would be the same as if you had a stationary rocket sitting on a planet that had the same acceleration and used GR to determine its time dilation?

An object always 'knows' if it is spinning. Close your eyes and ask someone to spin you on a chair without warning. You have no trouble knowing when you are spinning. This is an absolute, not relative, effect in both SR and GR.

Yes, I recognize that it is a simple matter to know when something is rotating. What I was really asking is, if rotation or acceleration is absolute, what is it absolute to? Einstein said no to the stars (Mach's principle) and everyone says no to absolute space, so what's left? You can't be absolute relative to nothing at all. (although if it was it would be no stranger than quantum mechanics I suppose)

Anyway, I prefer to integrate metrics for such problems. You wanted to try to factor effects. If you want to do that, you have to start with the notion of static world lines in the SC geometry. These define potential as a function of position. These world lines are not inertial - they experience proper acceleration. Then, any object moving relative to static world lines experiences relative motion time dilation compared to an adjacent static observer. For some other static observer, there is then a potential difference effect and a motion effect. The north pole observer, whether you worry about spin or not, is effectively a static observer at a fixed potential.

Gravitational potential is the energy to get a static object from its current position to 'infinity'. The fact that potential difference for static observers corresponds to relative time dilation between them is derived special case of GR. You don't have to ever use this concept, and it isn't useful in the general case. The general method is to have the metric represented in some coordinates, and compute the proper time along world lines by integrating. This is how you derive the special features of static or stationary metrics, or of the weak field approximation.

I'm sorry to say this is a bit over my head but I think I get the general idea.

 

[WannabeNewton]

Yes, I recognize that it is a simple matter to know when something is rotating. What I was really asking is, if rotation or acceleration is absolute, what is it absolute to? Einstein said no to the stars (Mach's principle) and everyone says no to absolute space, so what's left? You can't be absolute relative to nothing at all. (although if it was it would be no stranger than quantum mechanics I suppose)

As noted, rotation of frames is with respect to locally non-rotating frames (i.e. Fermi transported frames). This is analogous to how acceleration of frames is with respect to locally inertial frames (i.e. freely falling frames).

So, in other words, the local standard of freely falling of a Lorentz frame $\{e_{\alpha}\}$, where $e_0 = u$ is the 4-velocity, is $\nabla_{e_0}e_0 = 0$ and the local standard of non-rotation is $\nabla_{e_0}e_i = g(e_i,a)e_0$ where $a = \nabla_{e_0}e_0$.

Therefore if we have an arbitrary Lorentz frame, we can say it is locally rotating if it rotates relative to a coincident Fermi transported frame and accelerating if it does so relative to a coincident freely falling frame. Rotation and acceleration as defined in this way are absolute in the sense that the above equations are covariant.

 

[Buckethead]

Therefore if we have an arbitrary Lorentz frame, we can say it is locally rotating if it rotates relative to a coincident Fermi transported frame and accelerating if it does so relative to a coincident freely falling frame. Rotation and acceleration as defined in this way are absolute in the sense that the above equations are covariant.

I'm sorry to say I don't understand physics math, but I suspect the answer lies in my understanding of what you mean by a Fermi Transported frame (I don't know what that is either). For the case of 2 simple coincident frames that are rotating relative to each other it is of course impossible to know which is rotating and which is free falling in determining which would feel a rotation and which would not. Apparently a Fermi Transported frame is a special case of a frame that is distinguishable from a simple rotating (or non rotating ) frame using some other piece of information other than relative rotations?

 

[WannabeNewton]

A Fermi transported frame is the same thing as a locally non-rotating frame. When we say a given frame is rotating, we mean that it is rotating relative to a coincident non-rotating frame. Physically, a locally non-rotating frame is one in which the spatial axes are gyroscopes.

 

[Buckethead]

A Fermi transported frame is the same thing as a locally non-rotating frame. When we say a given frame is rotating, we mean that it is rotating relative to a coincident non-rotating frame. Physically, a locally non-rotating frame is one in which the spatial axes are gyroscopes.

So a Fermi t. frame has no special properties other than its definition as a non rotating frame. This in my mind makes it no different than any other coincident frame so my original question still stands. How does one know when 2 frames are rotating relative to one another which one should be endowed with acceleration and which one should not. The obvious answer of course is that the one that feels acceleration is the one that should be endowed with acceleration but as you can see that's a circular answer.

 

[WannabeNewton]

Again, local non-rotation can be measured absolutely and rotation can be measured relative to locally non-rotating frames. The same goes for free fall vs. acceleration. The equations I wrote in post #9 do exactly this.

 

[Dale]

The obvious answer of course is that the one that feels acceleration is the one that should be endowed with acceleration but as you can see that's a circular answer.

No that isn't a circular answer, it is an experimental answer. You have a device for measuring proper acceleration, an accelerometer (or more generally an inertial guidance unit). You can use that to experimentally test whether or not a given coordinate system is inertial.

Coordinate systems have no inherent physical meaning. They are only labels for events in spacetime. The physical meaning is contained in the metric. To calculate the proper time along some worldline you simply integrate the metric. Because the metric contains the physics, that procedure works regardless of whether you are using an inertial frame in flat spacetime, some non-inertial frame in flat spacetime, or an arbitrary frame in curved spacetime.

The components of the metric are different in a Fermi frame than in other frames. That is why they are physically identifiable and meaningful. Due to the different metrics, there is no confusion about symmetry between rotating and non-rotating frames.

 

[PAllen]

So if you are in an inertial frame in space (a space station) and you shoot off a rocket with a constant acceleration you can use SR alone to calculate the time dilation on the rocket during its acceleration? And the answer would be the same as if you had a stationary rocket sitting on a planet that had the same acceleration and used GR to determine its time dilation?

Yes, you can use pure SR to get all possible observations for an accelerating rocket far from a gravitating body. Then, by the principle of equivalence you can conclude that a static rocket in gravitational field should have the same observations to the extent tidal gravity can be ignored. The principle of equivalence is thus always a local and approximate statement (tidal gravity is never exactly zero). It allows one to derive approximate predictions of GR using SR scenarios, and can be used to motivate equations of GR. It never requires that GR be used for a situation without significant gravity.

 

[Buckethead]

OK so a stationary object and an orbiting object at the same elevation show the same GR effects (same time dilation degree) and in addition the fact that one is moving relative to the stationary one adds even more time dilation. Correct?

Yes, for a spherically symmetric, nearly static field (that thus can be treated with a potential).

I'm sorry but after a very long walk in the woods today thinking about this I have to address it again as I find it contradictory or maybe I just don't understand your answer. Let's take the example of a man jumping off a tall building and half way down passes a woman watering her window flower pots. At the moment the man passes the woman, their wrist watches are ticking at different rates even though they are at the same elevation. The woman is under acceleration so her clock will be ticking more slowly (she is the particle in the accelerator) and the man's watch is going faster because he is not being subject to acceleration. So the answer to my above question has to be no. I still recognize though that this doesn't mean the satellite clock is ticking at the same rate as a distant free floating object.) I guess I'm just still confused why an object in free fall can have a different clock rate than one in orbit when from inside the objects in question there is no experiment that can be done to differentiate between a free floating object in space and a free falling object near a planet (according to Einstein). If there is a gravity potential difference then this can be used to create an experiment to differentiate between the actual state of these two objects simply by doing a communication and comparing clock rates.

 

[Buckethead]

No that isn't a circular answer, it is an experimental answer. You have a device for measuring proper acceleration, an accelerometer (or more generally an inertial guidance unit). You can use that to experimentally test whether or not a given coordinate system is inertial.

I recognize that an accelerometer can be used to measure acceleration, but my question relates not to what has acceleration, but why one has the acceleration and not the other, but I'll explain that more below.

Coordinate systems have no inherent physical meaning. They are only labels for events in spacetime. The physical meaning is contained in the metric. To calculate the proper time along some worldline you simply integrate the metric. Because the metric contains the physics, that procedure works regardless of whether you are using an inertial frame in flat spacetime, some non-inertial frame in flat spacetime, or an arbitrary frame in curved spacetime.

The components of the metric are different in a Fermi frame than in other frames. That is why they are physically identifiable and meaningful. Due to the different metrics, there is no confusion about symmetry between rotating and non-rotating frames.

My brain lacks the ability to understand metrics at this point but the fact that a Fermi frame has a uniquely identifiable metric leads me to believe this is a dominant feature of a frame that can elevate it to the status of a preferred frame of reference. I don't exactly have a problem with that, but I thought preferred frames of reference (with regard to rotating bodies) went out with Newton. I really am fond of the idea of everything being relative even with regard to rotation and acceleration and although this may seem like a pipe dream, I can actually give an example where it is workable at least in a thought experiment. I had this thought several months ago and I almost let it go as being nonsense, but I love that it abolishes preferred frames of references.

Relying heavily on the equivalency principle imagine abolishing the notion of inertia and instead replace all inertial references with the equivalent gravitational potential. There are two scenarios here, the first is a ship dead in space turning on its engines and accelerating and the second is a rotating Earth experiencing rotational forces. Replacing inertia with gravity it could be imagined that when the ship attempts to accelerate, the act of accelerating causes a curvature in space directly behind the ship acting as a gravitational field to resist the ships attemps. This is the inertia that is felt and why the ship requires fuel to proceed. When the ship cuts its engines and coasts, the space curvature dissolves. Remember this is just a thought experiment so cut me some slack.

Now with regard to the rotating Earth (or a rotating skater for that matter). Instead of the rotational forces being due to centripetal force, and because we are replacing inertia with gravity, we can think instead that what is occurring is similar to what is happening in the ship namely when the Earth accelerates in its rotation, at the points of intersection between the rotating Earth and the non rotating space around the Earth a gravitational field is created due to the acceleration. This field will take the shape of a ring around the equator and will resist the acceleration. In other words it will be a gravitational force pulling the equator outward.

The beauty of this abstract thought is that in both the case of the accelerating ship and the case of the rotating Earth, there need not be any preferred frame of reference. It doesn't matter if the ship accelerates or the space behind it is accelerating instead as the gravitational field being generated will simply pull the space and the ship together and the effect (on the astronauts for example) will be the same even with this perfect symmetry. With regard to the Earth it doesn't matter if you think of the Earth as rotating or the Earth being stationary and the spaced around it rotating as in either case the gravitational ring will ensure that the equator will feel an outward acceleration resulting in a bulge there. Centripetal force and absolute rotation are easily dismissed. Again, not proposing anything here, just kind of a mind game to try an eliiminate a preferred frame of reference when addressing any type of acceleration.

 

[PAllen]

Relying heavily on the equivalency principle imagine abolishing the notion of inertia and instead replace all inertial references with the equivalent gravitational potential.

This is absurd on its face. There is no equivalence between inertial frames and gravitational potential. What the equivalence principle says about inertial frames is the local physics is the same for all of them. Comparing clocks between different inertial frames (either separated, or adjacent in relative motion) is not local physics, and the equivalence principle does not apply.

 

[Buckethead]

This is absurd on its face. There is no equivalence between inertial frames and gravitational potential. What the equivalence principle says about inertial frames is the local physics is the same for all of them. Comparing clocks between different inertial frames (either separated, or adjacent in relative motion) is not local physics, and the equivalence principle does not apply.

Comparing an accelerating rocket with a rocket that has its engines on but is not moving because it is being held by a gravitational field is very much the equivalency principle.

 

[TumblingDice]

@Buckethead
You keep adding more variables to a situation that, if anything, could benefit from simplifying to better understand.

One problem appears to be that you feel, since gravity offsets the orbital acceleration, there should be no time dilation due to gravity. Let's look at this from an inertial reference frame - you standing outside. Imagine you can see a geostationary satellite above you. Same as if it was on a very, VERY tall mountain, right? Clock on the sat will tick faster than yours because of the component of the Earth's gravitational field, which is weaker.

Even though the sat appears to be stationary relative to our reference frame, we also know it has to travel at high velocity to maintain its geostationary position, and that is going to add another component, slowing down the tick of the sat clock. All of this has is fact, and a part of the clock adjustments on GPS satellite clocks. They tick 45 microseconds faster due to lesser gravitation, and 7 microseconds slower due to relative motion, per day, so are adjusted to tick 38 microseconds slower to keep synchronized w/Earth time. I think I got that right... Here's a link to check:
http://www.astronomy.ohio-state.edu/~pogge/Ast162/Unit5/gps.html

Also, as was mentioned previously, there is 'absolute' acceleration, and rotational acceleration is one type of this, and all can be measured with an accelerometer, even at times when you may feel weightless. The equivalence principle is only in effect if/when you cannot tell the difference with measurements. An accelerometer - even a simple gyroscope - can tell the difference in situations you've posed.

 

[PAllen]

Comparing an accelerating rocket with a rocket that has its engines on but is not moving because it is being held by a gravitational field is very much the equivalency principle.

Yes, but in this case, all it says is that physics inside each rocket is the same. In no way does it say anything about how two such rockets clock rates would compare.

There are two common formulations of the equivalence principle (that can be shown to be equivalent given certain other assumptions):

1) The local physics of two labs undergoing the same measured uniform proper acceleration is the same. Thus, top to bottom time dilation for a lab accelerated by a rocket in empty space is the same as for a lab sitting on a planet, with the same measured proper acceleration. Note that while this says something about altitude effect within each lab, it says nothing about comparing two such labs. That is completely outside the scope of the equivalence principle.

2) All labs in free fall (no internally measured acceleration) have the same local physics, that given by SR. Note, again, it says nothing at all about comparing clock rates between such labs, only the physics within each lab.

It helps to understand what a principle says before you try to (mis)apply it.

 

[PAllen]

I'm sorry but after a very long walk in the woods today thinking about this I have to address it again as I find it contradictory or maybe I just don't understand your answer. Let's take the example of a man jumping off a tall building and half way down passes a woman watering her window flower pots. At the moment the man passes the woman, their wrist watches are ticking at different rates even though they are at the same elevation. The woman is under acceleration so her clock will be ticking more slowly (she is the particle in the accelerator) and the man's watch is going faster because he is not being subject to acceleration. So the answer to my above question has to be no. I still recognize though that this doesn't mean the satellite clock is ticking at the same rate as a distant free floating object.) I guess I'm just still confused why an object in free fall can have a different clock rate than one in orbit when from inside the objects in question there is no experiment that can be done to differentiate between a free floating object in space and a free falling object near a planet (according to Einstein). If there is a gravity potential difference then this can be used to create an experiment to differentiate between the actual state of these two objects simply by doing a communication and comparing clock rates.

The facts of the matter are as follows:

Clock 1 is suspended at the 100th floor of a 200 floor building. Clock2 is dropped from the top of the building. An observer attached to clock 1 will see clock 2 going slow when it passes. An observer attached to clock 2 will see clock 1 going slow as they pass it. Due to collocation at moment of passing, potential effects are irrelevant, and you have simply reciprocal relative motion time dilation.

A more interesting question is what an observer at the top of the building sees. The key here is that their relation to clock 1 is static. Any comparison with clock 1 true now, will be true an hour or a day from now because of the static relationship between them. The observer at the top of the building sees clock 1 running slow; and observer at clock 1 sees a roof clock running fast. The falling clock, as of when it passes clock 1, is seen to be running slower than clock 1 both by the clock 1 observer and the rooftop observer.

Now, you may ask what a clock 2 observer sees about the rooftop clock? There are two competing effects, and the result would depend on the details: they would see the roof top clock ticking fast compared to clock 1, but since the roof top clock is moving away from them, they would see relative motion time slowing. How these competing effects balance depends on the details.

A similar competing situation is for a polar observer watching a clock in equatorial orbit. The higher the elevation of the satellite, the faster its clock ticks based on gravitational potential difference. Further, the higher its elevation, the slower its apparent speed to a polar observer, so there is less motion time rate slow down. As a result, for a near surface orbit, the polar observer sees little elevation effect but a large relative motion effect, so the clock is slow. At some altitude, the orbiting clock would be considered to run at the same rate as the polar clock by the polar observer. At higher orbits, the satellite clock would be seen to run faster.

 

[Buckethead]

Yes, but in this case, all it says is that physics inside each rocket is the same. In no way does it say anything about how two such rockets clock rates would compare.

There are two common formulations of the equivalence principle (that can be shown to be equivalent given certain other assumptions):

1) The local physics of two labs undergoing the same measured uniform proper acceleration is the same. Thus, top to bottom time dilation for a lab accelerated by a rocket in empty space is the same as for a lab sitting on a planet, with the same measured proper acceleration. Note that while this says something about altitude effect within each lab, it says nothing about comparing two such labs. That is completely outside the scope of the equivalence principle.

2) All labs in free fall (no internally measured acceleration) have the same local physics, that given by SR. Note, again, it says nothing at all about comparing clock rates between such labs, only the physics within each lab.

It helps to understand what a principle says before you try to (mis)apply it.

Thanks for the clarification and yes I agree with your last point. My amusing and possibly absurd thought experiment was not addressing delta clock rates between inertial systems, I was addressing the possibility of viewing accelerating systems as being completely relative and not absolute. It was really a tangent on the discussion and I was obliquely working toward something that I'll ask now.

First just to be clear. Was my correction above correct in that an orbiting satellite with have a faster clock rate than a stationary object at the top of a tower at the same elevation?

OK but because of the gravity potential near the Earth an orbiting sat will tick more slowly than an inertial clock far out in empty space. True?

If the orbiting satalite were stopped and held in space by retro rockets and a second Earth nearby and on the opposite side of the sat. and situated so that the stationary sat was half way between the 2 Earths, then would the gravity potential drop to zero? Or would it double? In other words, would the sat (which even though stationary would feel no gravity due to the opposing gravites) now have a clock rate the same as a stationary object out in empty space or would the difference be even greater?

edit: I wrote this before I saw your reply above.

 

[PAllen]

First just to be clear. Was my correction above correct in that an orbiting satellite with have a faster clock rate than a stationary object at the top of a tower at the same elevation?

No, it will be slower. In particular, each time it passes the tower satellite, its clock will be further and further behind.

OK but because of the gravity potential near the Earth an orbiting sat will tick more slowly than an inertial clock far out in empty space. True?

It will tick slower due to two different factors which work in the same direction: slower due to lower potential, slower more due to motion relative to a static observer at the same elevation.

If the orbiting satalite were stopped and held in space by retro rockets and a second Earth nearby and on the opposite side of the sat. and situated so that the stationary sat was half way between the 2 Earths, then would the gravity potential drop to zero? Or would it double? In other words, would the sat (which even though stationary would feel no gravity due to the opposing gravites) now have a clock rate the same as a stationary object out in empty space or would the difference be even greater?

edit: I wrote this before I saw your reply above.

Let's not deal with the complications of your final scenario when you are having so much trouble with simpler cases. When having a problem of understanding, simplify first; when simple case is solidly understood, you can try adding more complexity.

 

[Buckethead]

No, it will be slower. In particular, each time it passes the tower satellite, its clock will be further and further behind.

I get the gravity part of this now: Gravity affects both units the same way due to the identical elevations so is not considered in this experiment. But the inertial frame part of this still has me baffled. I can't count the acceleration of the object on the tower since the orbiting satellite feels the same acceleration by virtue of its identical elevation in a gravitational field (even though its weightless??!). So what's left is two ships passing each other out in space. But why will the sat be the slower one? There is no acceleration involved like the astronaut in the twin paradox effect so this can't cause the time difference. Is it the continual change in the direction of the satallite? but isn't this considered to be inertial but just in a curved space? Am I ever going to understand this??!

 

[pervect]

Comparing an accelerating rocket with a rocket that has its engines on but is not moving because it is being held by a gravitational field is very much the equivalency principle.

Note though that correctly analyzing a rocket held by an actual gravitational field requires general relativity. Analyzing a rocket in empty space can be done with just SR.

Analyzing a rocket held by an actual gravitational field without already knowing GR is going to lead to confusion.

 

[WannabeNewton]

Am I ever going to understand this??!

Not until you actually learn the math. Could you understand a French novel without knowing French in the first place?

 

[PeterDonis]

I can't count the acceleration of the object on the tower since the orbiting satellite feels the same acceleration by virtue of its identical elevation in a gravitational field (even though its weightless??!).

You are using the word "acceleration" incorrectly here. The orbiting satellite does *not* feel the same acceleration as the object on the tower: you obviously realize that since you realize that the orbiting satellite is weightless but the object on the tower is not. If you attached accelerometers to the two objects, the orbiting satellite would read zero acceleration but the object on the tower would read nonzero acceleration. So the two objects do *not* feel the same acceleration. Your intuitive sense that they both feel the same "acceleration due to gravity" is a leftover from Newtonian physics that is best discarded.

However, it should be noted that a correct understanding of "acceleration" here is *not* the resolution to the question of why the orbiting satellite's clock runs slower. See below.

So what's left is two ships passing each other out in space. But why will the sat be the slower one?

Because it is moving while the object on the tower is static. Here the words "moving" and "static" can be given an invariant meaning by referring them to the central gravitating body, the Earth--more precisely, to the Earth's center of mass. (Also we're ignoring the Earth's rotation here; including it would complicate matters.) The object on the tower sits still while the orbiting object moves around in a big circle. The fact that it moves around in a big circle, so that the two objects are co-located once per orbit, is why the difference in clock rates due to motion can be given an invariant meaning: the orbiting satellite's clock runs slower in an invariant sense because it ticks off less time between each successive pair of co-location events.

It is true that spacetime here is not flat, so the situation is not exactly the same as the analogous situation in flat spacetime, where one object is at rest and the other goes around in a big circle. In the flat spacetime case, the object at rest would feel zero acceleration and the object moving around in the circle would feel nonzero acceleration; in this case, as I noted above, it's the other way around. But that just illustrates, again, that differences in "acceleration" are not what explains why the orbiting satellite's clock runs slower. (See below for a further comment on that.)

What the spacetime not being flat *does* do is add a second effect, the effect of altitude (changing gravitational potential), in addition to the effect of motion, that can cause differences in clock rates. In this particular example, you've eliminated that by specifying that both objects are at the same altitude. But in general, for objects that are in the gravitational field of a single isolated massive body like the Earth, you have to include both effects.

There is no acceleration involved like the astronaut in the twin paradox effect so this can't cause the time difference.

Actually, as I noted above, it's worse than that: the apparent effect of acceleration is *backwards* compared to the standard twin paradox: the object that feels acceleration (the one on the tower) has a clock that runs *faster*. This is one way of seeing why using "acceleration" to explain differences in clock rates doesn't work in general; in some special cases it happens to give the right answer, but you can't use it as a general rule.

 

[PAllen]

To help accept the fundamental differences introduced by spacetime curvature, go back the the pure inertial twin 'paradox' I described early in this thread:

You have clock A in circular orbit. You have clock B in pure radial free fall trajectory passing A at some point. A and B synch their times as B passes. The outward radial speed of B is chosen so that B eventually falls back, such that A has completed one orbit when B falls back to this position.

Both paths are inertial all the the time. However, clock B will elapse more time between the meetings.

Such a situation is impossible without gravity (equivalently, in flat space time). You cannot have two inertial paths between the same events without gravity. Geometrically, you must have intrinsic curvature to allow multiple geodesics (free fall paths) between the same points (events). For timelike geodesics, one of these will be a path of maximum time between the events; the other, even thought it is geodesic, will only have a local maximizing property rather than a global one.

The local property still true for the orbit is that for two events sufficiently near each other in proper time along the orbit, the orbit is the path that maximizes proper time between those events. However, for the the orbit, this property breaks down for events far enough apart along the orbit.

 

[Dale]

I recognize that an accelerometer can be used to measure acceleration, but my question relates not to what has acceleration, but why one has the acceleration and not the other,

Well, there certainly is no mystery to that. One has proper acceleration and the other does not because that is what you specified when you described the scenario. You certainly could have specified a scenario where they both have acceleration (e.g. each attached to opposite ends of the same rope) or where neither has acceleration (e.g. orbiting each other with gravity). If you describe a different scenario then we can analyze that scenario, but once you describe the scenario the reason why that scenario is simply because that is what you chose to describe.

My brain lacks the ability to understand metrics at this point

I suspect that you will be able to understand if you make the effort, but if you really lack this capability then you simply will not be able to understand general relativity. There is no way to learn GR without metrics.

I don't exactly have a problem with that, but I thought preferred frames of reference (with regard to rotating bodies) went out with Newton.

No intertial frame is preferred over any other, but that doesn't imply that there is no distinction between inertial and non-inertial frames. Because there is no physical distinction between inertial frames and because the metric contains the physics, the metric for any inertial frame is the same.

Rotating frames are non-inertial, they are physically distinct from inertial frames, and the form of the metric is different in non-inertial frames. I don't know why you thought that "went out with Newton".

Relying heavily on the equivalency principle imagine abolishing the notion of inertia and instead replace all inertial references with the equivalent gravitational potential. There are two scenarios here, the first is a ship dead in space turning on its engines and accelerating and the second is a rotating Earth experiencing rotational forces. Replacing inertia with gravity it could be imagined that when the ship attempts to accelerate, the act of accelerating causes a curvature in space directly behind the ship acting as a gravitational field to resist the ships attemps. This is the inertia that is felt and why the ship requires fuel to proceed. When the ship cuts its engines and coasts, the space curvature dissolves. Remember this is just a thought experiment so cut me some slack.

So, curvature of spacetime is a coordinate independent quantity which describes the presence of tidal gravity.

The curvature that you are describing here is not a curvature of spacetime, but rather a curvature of your coordinates. This curvature of coordinates gives rise to fictitious forces like the centrifugal and Coriolis forces in rotating reference frames. This same quantity is the quantity which contains the g field of Newtonian gravity, so you could think of it this way. But, again, it is a curvature of coordinates, not a curvature of spacetime.

The beauty of this abstract thought is that in both the case of the accelerating ship and the case of the rotating Earth, there need not be any preferred frame of reference.

This thought doesn't change the simple physical fact that the metric has a different form in these different coordinates.

Again, not proposing anything here, just kind of a mind game to try an eliiminate a preferred frame of reference when addressing any type of acceleration.

If the criterion is that having a different metric indicates a preferred frame, then you have not eliminated it.

 

[PAllen]

To emphasize the importance of the metric, I will show just how clearly and simply it explains the orbital versus static temporal relationship for this case.

First, the relevant metric for this case is the Schwarzschild metric, which you can find here:

http://en.wikipedia.org/wiki/Schwarzschild_metric

However, if we restrict ourselves to constant r, and one plane through the center of a planet, this reduces to:

d$\tau$^2 = g00 dt^2 - r^2 d$\theta$^2

where g00 may be treated as a positive constant.

Note, that as Peter explained, the notion of constant positions (r,$\theta$) in these coordinates has invariant meaning (provided by lack of motion relative to the massive body) - even though the particular coordinate values and units are purely conventional.

You can see that for a static clock ($\theta$ not changing), the rate of proper time to coordinated time is √g00. For any circular motion, geodesic (orbit) or forced, you have the following ratio of proper time to coordinate time.

√(g00 - (r θ')^2) , where θ' is the angular speed.

No matter what the functional form of θ'(t), integrating this over t such that a circle is traversed, will produce a smaller result than √g00 times the coordinate time of the circular path. Thus, the circular moving clock will elapse less time, no matter what the nature of its motion relative to static clocks.

 

[WannabeNewton]

Just to add on to PAllen, we can easily calculate the ratios of proper times of the static observer and circularly orbiting observer between two events coinciding on both their worldlines. Imagine for example that two observers $O$ and $O'$ are in a ship together that is in circular orbit at some allowed radius $R > 3M$ around a static spherically symmetric star (the allowed radii can be determined by using conserved quantities and the definition of stable/unstable circular orbits as the minima/maxima of the effective potential). At some event $p$, observer $O$ steps out with a rocket and hovers in place while observer $O'$remains in the circular orbit. The two meet again at event $q$ after $O'$ makes a complete orbit starting from $p$, at which point they compare their clock readings (which were synchronized at $p$).

The 4-velocity of $O$ between $p$ and $q$ is just $u = (1 - 2M/R)^{-1/2}\partial_t = \gamma \partial_{t}$. The 4-velocity $\tilde{u}$ of $O'$ between said events takes a bit more work. The angular frequency of the orbit as measured by an observer at infinity will be $\omega := \frac{\mathrm{d} \phi}{\mathrm{d} t} = \sqrt{M/R^3}$ so $\tilde{u} = \tilde{\gamma} (\partial_t + \omega \partial_{\phi})$ where $\tilde{\gamma}$ is yet to be determined. Using the fact that $g(\tilde{u},\tilde{u}) = -1$ we have that $\tilde{\gamma} = (1 - 2M/R - R^2\omega^2)^{-1/2}$. This is basically a combination of the gravitational time dilation factor and the kinematical time dilation factor when boosting from the local Lorentz frame of a static observer at $R$ to the local Lorentz frame of a circularly orbiting observer at $R$.

The proper time $\Delta \tau_{pq}$ as measured by $O$ between $p$ and $q$ is then $\Delta \tau _{pq} = \int _{p}^{q}\gamma^{-1}dt = (1 - 2M/R)^{1/2}\Delta t_{pq}$ whereas the proper time $\Delta \tau'_{pq}$ as measured by $O'$ between said events is $\Delta \tau' _{pq} = \int _{p}^{q}\tilde{\gamma}^{-1}dt = (1 - 3M/R )^{1/2}\Delta t_{pq}$ where $\Delta t_{pq} = \frac{2\pi}{\omega}$ is just the period of the circular orbit as measured by an observer at infinity. Thus $\frac{\Delta \tau _{pq}}{\Delta \tau' _{pq}} = (\frac{1 - 2M/R }{1 - 3M/R })^{1/2}$; you can see that $\Delta \tau _{pq} > \Delta \tau' _{pq}$.

You may object to this because $O'$ is freely falling whereas $O$ is accelerating so it would seem that $O'$s clock should read more proper time between $p$ and $q$ than that recorded by $O$s clock since, as noted, $O'$ is freely falling. The thing is that between two events in space-time, a free fall worldline only maximizes proper time amongst worldlines in a sufficiently small neighborhood of this free fall worldline in the function space of all possible worldlines between the two events. In our case the circular orbit corresponds to a free fall worldline that is not close in the function space to the worldline of the static observer.

 

[Dale]

You may object to this because $O'$ is freely falling whereas $O$ is accelerating so it would seem that $O'$s clock should read more proper time between $p$ and $q$ than that recorded by $O$s clock since, as noted, $O'$ is freely falling. The thing is that between two events in space-time, a free fall worldline only maximizes proper time amongst worldlines in a sufficiently small neighborhood of this free fall worldline in the function space of all possible worldlines between the two events. In our case the circular orbit corresponds to a free fall worldline that is not close in the function space to the worldline of the static observer.

Which is exactly why there is no general substitute for integrating the metric along the worldline, and one of the many reasons why you simply cannot understand GR without understanding the metric.