Force due to gravity vs force due to accelratio

In summary, a force due to gravity permeates the area around it with time dilation an object in free fall in a room on the Earth would be affected by time dilation because it is accelerating at 32 ft per s per s,while an object in free fall in a room that was accelerating at 32 ft per s per s would not because it would be still and not accelerating.
  • #1
Reuben Smith
9
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Lets start over and let me rephrase my question, a force due to gravity permeates the area around it with time dilation an object in free fall in a room on the Earth would be affected by time dilation because it is accelerating at 32 ft per s per s,while an object in free fall in a room that was accelerating at 32 ft per s per s would not because it would be still and not accelerating.My experiment would be to make two identical oscillating circuits one in the barrel of a machine gun that is rapidly firing ,one in a vacuum levitated by a magnetic field would either circuit show signs of time dilation?
 
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  • #2
Reuben Smith said:
Lets start over and let me rephrase my question, a force due to gravity permeates the area around it with time dilation an object in free fall in a room on the Earth would be affected by time dilation because it is accelerating at 32 ft per s per s,while an object in free fall in a room that was accelerating at 32 ft per s per s would not because it would be still and not accelerating.
I am not sure that I agree with this. Time dilation is essentially a coordinate effect. If you are using an accelerating coordinate system, e.g. Rindler coordinates, then an object in free fall in a room would indeed experience the same time dilation as an equivalent object in a gravitational field. If you are in a gravitational field then the equivalent coordinate system would be some sort of free-fall coordinates, e.g. the rain frame. In those coordinates the free falling object would not be subject to time dilation even in a gravitational field.
 
  • #3
Reuben Smith said:
Lets start over and let me rephrase my question, a force due to gravity permeates the area around it with time dilation an object in free fall in a room on the Earth would be affected by time dilation because it is accelerating at 32 ft per s per s,while an object in free fall in a room that was accelerating at 32 ft per s per s would not because it would be still and not accelerating.

Wrong, on several counts. It is true that a massive body influences the region around it. It is false that this is a 'time dilation field'. The rate of a time is always a function of the path of given 'clock' through spacetime. A clock sitting on the Earth is following a non-inertial path through space time (in its own frame of reference, one may describe its path as (0,0,0,t), but it is still a path through spacetime). A clock free falling toward the ground would advance at a faster rate than one sitting on the ground (because it is following an inertial path through spacetime), and the free falling clock would approximate the behavior of a clock in deep space.

Basically, your two clocks would behave the same, and must, within the approximate limits described in my earlier post. If this were not true, the whole framework of general relativity would be wrong.
 
  • #4
Let me see if Iv got this straight a clock on Earth would run slower than one in deep space away from all other bodies.
 
  • #5
DaleSpam said:
I am not sure that I agree with this. Time dilation is essentially a coordinate effect. If you are using an accelerating coordinate system, e.g. Rindler coordinates, then an object in free fall in a room would indeed experience the same time dilation as an equivalent object in a gravitational field. If you are in a gravitational field then the equivalent coordinate system would be some sort of free-fall coordinates, e.g. the rain frame. In those coordinates the free falling object would not be subject to time dilation even in a gravitational field.

That doesn't seem to agree with what was discussed in this thread:

https://www.physicsforums.com/showthread.php?t=433866

Where it was described that there would be gravitational time dilation inside a massive sphere even though there would be no acceleration because the gravitational potential would be non-zero inside the sphere.

I think that means there would be as you say: "free-fall coordinates." Am I missing a subtle (or even obvious) distinction?
 
  • #6
Reuben Smith said:
Let me see if Iv got this straight a clock on Earth would run slower than one in deep space away from all other bodies.

Yes.
 
  • #8
okay, now here is my question,three identical oscillating circuits one in the barrel of a machine gun monted on a rocket that compinsates for the recoil that is rapidly firing in deep space away from the Earth and all bodies ,one on Earth in a vacuum levitated by a magnetic field ,and one in deep space by itself ,would they oscillated at the same frequecy and if not what which ones would be slower in coparison ?
 
  • #9
inflector said:
That doesn't seem to agree with what was discussed in this thread:

https://www.physicsforums.com/showthread.php?t=433866

Where it was described that there would be gravitational time dilation inside a massive sphere even though there would be no acceleration because the gravitational potential would be non-zero inside the sphere.

I think that means there would be as you say: "free-fall coordinates." Am I missing a subtle (or even obvious) distinction?
I am not sure where you think the conflict is between the two threads. This thread and the previous thread that you linked to were both dealing with a misapplication of the equivalence principle. Here the equivalence principle was misapplied locally by comparing accelerating coordinates in curved spacetime (Schwarzschild coordinates) with inertial coordinates in flat spacetime. In the other thread the equivalence principle was misapplied by considering a large enough region where the spacetime curvature could not be neglected.
 
  • #10
DaleSpam said:
I am not sure where you think the conflict is between the two threads.

[SNIP]

In the other thread the equivalence principle was misapplied by considering a large enough region where the spacetime curvature could not be neglected.

Earlier, you stated:
DaleSpam said:
I am not sure that I agree with this. Time dilation is essentially a coordinate effect.

[snip]

If you are in a gravitational field then the equivalent coordinate system would be some sort of free-fall coordinates, e.g. the rain frame. In those coordinates the free falling object would not be subject to time dilation even in a gravitational field.

In the other thread, it was stated that:

1) Time dilation was due to the gravitational potential and not due to acceleration.

2) An object that was not accelerating could be experiencing time dilation because the acceleration is the derivative of the gravitational potential. You experience zero acceleration even when the gravitational potential is non-zero as long as the potential is the same as adjacent space so that the derivative is zero. So an object in an area of constant but non-zero potential would have time dilation.

3) An object in the center of a massive sphere would have non-zero but constant gravitational potential so there would be no acceleration but there would be time dilation.

This appears to conflict with your statement above that I have bolded because if one considers the free-fall coordinates of an object in the center of the massive sphere, it would be experiencing no acceleration but time dilation according to the other thread.

I certainly could have got this wrong (again) but it seemed to conflict to me so that is why I asked. I want to get this stuff right. Thanks for taking the time to look into this.
 
  • #11
An clock at rest in the center of a massive hollow sphere would experience gravitational time dilation relative to a clock at the surface. This is a large region of spacetime, not a small one, therefore the equivalence principle does not hold. As I said:
DaleSpam said:
The equivalence principle applies only to a small region of spacetime, sufficiently small to ignore the spacetime curvature. If we are performing experiments comparing two clocks within the hollow in the center there is no gravitational acceleration and no relative time dilation and the equivalence principle holds nicely. If we are performing experiments comparing a clock within the hollow and a clock far above the surface then there is gravitational time dilation but the clocks are too far apart for the equivalence principle to hold.
Is that clear now? Two local clocks at rest in the center of the hollow sphere experience no time dilation relative to each other and the equivalence principle applies. A clock at the center experiences time dilation relative to a clock at the surface and the equivalence principle does not apply.

Two free falling clocks which are at rest near enough to each other for the equivalence principle to hold would not experience any time dilation regardless of if they were in deep space, in the center of a hollow sphere, or even falling through the event horizon of a large black hole. There may be time dilation between two free falling clocks which are at rest wrt each other provided they are far enough apart that the equivalence principle does not hold.
 
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  • #12
Okay, it appears that I my understanding was correct but that I was reading too much into your statement. Thanks for the clarification.
 
  • #13
No problem. Sorry about any confusion I caused.
 
  • #14
I am not sure I agree with the analysis of the center of hollow gravitating sphere. Consider the following fundamental statement from:

http://relativity.livingreviews.org/Articles/lrr-2006-3/

"The outcome of any local non-gravitational experiment is independent of the velocity of the freely-falling reference frame in which it is performed."

"The outcome of any local non-gravitational experiment is independent of where and when in the universe it is performed."

This is clarified in the text to mean any frame over which tidal effects can be ignored. So, the question is, are there tidal effects inside the sphere or are these cancelled? Is there curvature inside the sphere? I have assumed that tidal effects are cancelled, and spacetime is flat inside. If they are cancelled, then it clearly follows that either an atom decays at the same rate in the sphere (assuming sphere and reference frame at its center are inertial) as far away from it, or the Einstein equivalence principle is violated.

A final comment, again assuming flatness inside the sphere: then the metric is purely locally minkowski. The proper time computed for an inertial clock in the center is pure minkowski. How can it come out different from a clock in deep space?

[EDIT] OK, I researched this a bit and see the error of my ways. One can say that the inertial frame inside the sphere acts like an inertial frame moving at constant speed away from a distant inertial frame at rest relative to the sphere. This is basically due to the red shift and time slowing that would happen climbing throught the space between the sphere and the distant observer.
 
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  • #15
The key in all cases is how you compare the clocks.

If you have a NIST_ standard clock, and some sort of oscillating machine gun platform or electronic circuit, both located at the center of a hollow planet, and you compare the two while they're side by side, they'll agree.

They'll also agree if you're in deep space (or anywhere else for that matter).

The first issue comes up when you compare the clock at the center of the hollow planet to the one in deep space. How do you compare it? Well, you probably use the obvious way - you send out a radio signal. And the propagation time for the radio signal is constant for a round trip.

Using this method, you find then that both the NIST clock and the machine gun contraption out in space seem to run faster than the one on the planet, when compared by radio signals.

So, case closed? Not quite.

Suppose now, we put our two clocks (and our oscillating contraption, which doesn't add much to the story since it always reads the same as a NIST standard clock, but is perhaps not quite as precise), in an accelerating rocketship.

Now, our rocketship is accelerating, but if you work out the round-trip time for a light signal, the round trip time is constant. But while you can see from the inertial frame analysis that the round-trip time is constant, you disagree about the rate at which the clocks are running.

What happens from the point of view of the inertial frame is that one clock accelerates and experiences a doppler shift - but there is no "gravitational field" in the inertial frame, and there is no "gravitational time dilation" in the inertial frame either.

So what's going on? It's just another example of how simultaneity is relative. One method of comparing the rate of the two clocks gives a different result than another method. This is due to the counter-intuitive fact that there isn't any absolute way to compare clocks at all, unless they are side by side. The notion of simultaneity in the acclerating frame is a different notion than the one in the non-accelerating frame - which shouldn't be too surprising, every inertial frame has a self-consistent but _different_ notion of simultaneity, and the accelerating's frame is sort of "stitched together" from different notions in different inertial frames.

BTW, the inertial frame analysis isn't any more or less "correct" than the non-inertial analysis in GR - because GR is designed to handle arbitrary coordinates. The idea that you do everything in inertial frames is fundamentally due to the Newtonian analysis.
 
  • #16
Let me pose the following question about these issues. My understanding of the moment is that a free fall clock near the Earth's surface would go faster than one on the ground. Locally, one is inertial, the other isn't, the inertial clock goes faster if they are 'close enough'.

However, I now suspect that the falling clock still run slower than a distant inertial clock because of the gravity well effect. Neither can really be said to inertial relative to the other because there are no inertial frames over this large distance. All comparisons between them are affected by the curvature over a large region.

Yet another way I can think about this is as follows. There is no 'equivalence' between two widely separated inertial observers not moving relative to each other, in a pure SR universe, with one of them surrounded by an accelerating rocket (soon to have floor hit them, but not yet); and two widely separated inertial observers in GR, one 'far from everything' the other free falling near a massive body. To compare such distant observers requires accounting for space between them, which is just not the same.
 
  • #17
DaleSpam said:
An clock at rest in the center of a massive hollow sphere would experience gravitational time dilation relative to a clock at the surface.
This is quoted ad infinitum at various places but never have I seen a formula demonstrating that.

It would be nice if we could have the formula so we have numerical proof that GR indeed predicts this.

Say we have a sphere of mass X and a cavity of size N (you could use the Schwarzschild R for simplicity). Then what is the time dilation wrt an observer at infinity?

PAllen said:
Let me pose the following question about these issues. My understanding of the moment is that a free fall clock near the Earth's surface would go faster than one on the ground. Locally, one is inertial, the other isn't, the inertial clock goes faster if they are 'close enough'.
Also a clock can free fall at various velocities.

One approach is the calculate the time dilation(r) for a stationary observer and then calculate the dilation(r) for free falling observers at various proper velocities wrt the stationary observer.

If we take rs=1 and radial motion for simplicity then for the stationary observer we have:

[tex]
{t}^{2}={\tau}^{2} \left( 1-{r}^{-1} \right) ^{-1}
[/tex]For free falling observers with an initial velocity v0 the proper velocity v(r) is:

[tex]
v(r) = \sqrt {1- \left( 1-{v_0}^{2} \right) \left( 1-{r}^{-1} \right) }
[/tex]

Applying the Lorentz factor would give us the time dilation wrt an observer at infinity.
 
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  • #18
DaleSpam said:
An clock at rest in the center of a massive hollow sphere would experience gravitational time dilation relative to a clock at the surface.

Actually, the theory on this that I found in reputable sources states that the clock inside the sphere would run slow compared to a distant observer with minimal motion relative to the sphere. Compared to one on ther surface, the whole derivation I saw does not apply. My guess would be that the rate on the surface would generally be the *same* as inside the sphere, but it might depend a little on the assumed thickness of the sphere.

Someone should double check this.
 
  • #19
PAllen said:
LThere is no 'equivalence' between two widely separated inertial observers not moving relative to each other, in a pure SR universe, with one of them surrounded by an accelerating rocket (soon to have floor hit them, but not yet); and two widely separated inertial observers in GR, one 'far from everything' the other free falling near a massive body. To compare such distant observers requires accounting for space between them, which is just not the same.
Yes, this is correct.

PAllen said:
Actually, the theory on this that I found in reputable sources states that the clock inside the sphere would run slow compared to a distant observer with minimal motion relative to the sphere. Compared to one on ther surface, the whole derivation I saw does not apply. My guess would be that the rate on the surface would generally be the *same* as inside the sphere, but it might depend a little on the assumed thickness of the sphere.

Someone should double check this.
Can you link to the derivation you are referring to? I am sure I can extend it to include a surface clock.
 
  • #20
DaleSpam said:
Can you link to the derivation you are referring to? I am sure I can extend it to include a surface clock.

I'll look for it, but that derivation obviously cannot be so generalized. It explained the whole effect as due to the integrated curvature or total redshift from the surface to the disant observer, and showed why the same would apply to a source inside the hollow sphere. In fact, this derivation strongly implies (but doesn't state) that there would be little or no difference from the internal clock compared to one on the surface of the hollow sphere.
 
  • #21
DaleSpam said:
Can you link to the derivation you are referring to? I am sure I can extend it to include a surface clock.

I can't seem to find the discussion I found before (which involved gluing the exterior schwarzchild solution to an interior solution). However, let me turn it around:

*All* discussions I have found on this make the claim about an observer inside the sphere versus a distant observer. *None* claim such an effect relative to an observer right on the surface. Can you find any source that claims this?
 
  • #22
I haven't searched for any derivation either way. It would probably be easier to derive it myself than look for someone else's derivation.
 
  • #23
Google turned up this paper which may be of interest, for an "exact solution for a static spherical shell of matter in vacuum":

http://www.cleibovitz.org/physics/Leibovitz (1969) An exact solution for a static spherical shell of matter in vacuum.pdf

I've only skimmed it and can't verify that it's correct at this point, but the solution derived in the paper does have the following interesting features:

(1) The metric exterior to the thin static spherical shell is the Schwarzschild metric, with M being the total mass of the shell (I think it's implicitly assumed that the outer radius of the shell is greater than 2M, so only the Schwarzschild exterior solution is used).

(2) The metric inside the static spherical shell is flat, but with the time coordinate scaled so that clocks inside the shell run slower than clocks at spatial infinity (outside the shell).

(BTW, the URL above is technically not well-formed so it may not open correctly in all browsers--it works OK in mine. The Google search that turned it up is:

http://www.google.com/search?q=metric+of+static+spherical+shell&ie=UTF-8&oe=UTF-8

It was the fourth hit down on the first page of search results when I ran the search just now.)
 
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  • #24
I could be misunderstanding this paper, but it seems to support my understanding:

1) significant effect on time rate between interior clock and clock at infinity (the only case they discuss)

2) if I interpret the solution right (?), minimal difference between the interior and surface, approaching zero as the shell's thickness decreases.

Note, this is also consistent with the idea that the effect is proportional to gravitational potential difference: minimal between inside of shell and outide, large from either of these to infinity. Note that I never thought that would be *no* difference between inside and surface, just that it would be small and related to shell thickness, which is exactly what I get out of this paper.
 
  • #25
Yes, that is correct, the result will depend on the thickness of the shell. I was considering a thick shell.
 
  • #26


PAllen said:
Note that I never thought that would be *no* difference between inside and surface, just that it would be small and related to shell thickness, which is exactly what I get out of this paper.

Intuitively, it makes sense, the potential has to be smooth and continuous. It doesn't make sense that you'd get a huge jump at the shell itself.

What I wonder, is if anyone has actually tested the prediction of constant time dilation with a dropping acceleration.

We don't have any large hollow spheres around, but it seems like a relatively easy test with today's clocks for a similar experiment. The Earth isn't a spherical shell but the idea should apply to a filled sphere as well. I haven't checked the math for a filled sphere (or seen it anywhere), but the gravitational potential should increase as you get to closer to the center of the Earth even though the acceleration would drop because the gravity from mass above would start offsetting the gravity from the mass below until you reached the center of the Earth where it would completely balance and you'd have no acceleration at all but plenty of potential.

So there should be a drop in the acceleration due to gravity as you go towards the center of the Earth without a corresponding drop in the time dilation. One could put a clock down a very deep mine shaft where the gravitational acceleration should be lower, while the potential goes up so there should be more time dilation, and compare it with the clock on the surface.

If Pound and Rebka could check the time dilation from a 22.5m tower in 1959, it ought to be much easier to check a 2 or 3km difference in a mine using today's clocks or http://en.wikipedia.org/wiki/Pound–Rebka_experiment" run down a vertical shaft of the mine if needed.

Has anyone seen any papers on an experiment like this?
 
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  • #27
Gravitational time dilation is a real enough phenomenon, one that we can easily measure on Earth. The latest experiments can detect a height difference as little as a foot.

http://www.physorg.com/news204470740.html

Philosophically, though, the thing to realize is that it's coordinate dependent. "Time dilation" can be thought of as the relation between coordinate time which is the abstract time used to label a map of space-time, and proper time, which is what clocks actually measure.

Pre-relativity, one tends to blur the two concepts together. Post relativity, it becomes important to separate them - it's part of the process of getting away from the "universal now" or "universal time" of pre-relativistic physics.

This is all terribly abstract, so a concrete example might help. To label coordinates on the earth, we use lattitude and longitude (among other possible systems). This is a purely spatial example of a curved surface, so our example will only have space, and distance, and not time.

The Earth's surface is curved, so one degree of longitude is not a constant distance. Why don't we make a coordinate system that has a constant relation between coordinates and distances everywhere? Because it's impossible - you can't do that on a curved surface.

So, we wind up with coordinates on our map that aren't evenly spaced. And we introduce something called a metric, which allows us to relate coordinate changes to physical distances.

What's going on in relativity isn't that much different. You have coordinate systems (usually there's the possibility of more than one, which makes things more confusing), and the relation between the spacing of the time coordinates at some point, and actual elapsed times there, is given by the local value of time dilation _in that coordinate system_, which gives us one of the components of the more general space-time metric.
 
  • #28
Despite the philosophy provided by pervect, to which I could add the obvious point (I'm sure he would agree) that in any scenario in which clocks separate and come back together, along different spacetime paths, their difference is not coordinate, but invariant. Thus, scenarios where a clock is sent into a mine, and then brought back, is measuring something invariant, not coordinate.

Inflector's experiment seems very interesting in that you should be able to easily measure that:

1) acceleration of gravity is lower in the mine than on the surface.
2) a clock will run slower there in an invariant way (will be behind one on the surface when brought back up).

This is a clear prediction that is different from the Pound-Rebka type experiments (where time dilation was proportional to acceleration of gravity). I did a quick search to see if anyone measured this and couldn't find anything. I found one reference to a 1960 paper for which no complete abstract was available which might have measured this. I would be very interested to see a measurement even though I 'know' what the result should be.
 
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  • #29
PAllen said:
Despite the philosophy provided by pervect, to which I could add the obvious point (I'm sure he would agree) that in any scenario in which clocks separate and come back together, along different spacetime paths, their difference is not coordinate, but invariant. Thus, scenarios where a clock is sent into a mine, and then brought back, is measuring something invariant, not coordinate.

That's true - if you specify two clocks that separate and meet again, their readings when they rejoin will be independent of coordinates.

Inflector's experiment seems very interesting in that you should be able to easily measure that:

1) acceleration of gravity is lower in the mine than on the surface.
2) a clock will run slower there in an invariant way (will be behind one on the surface when brought back up).

This is a clear prediction that is different from the Pound-Rebka type experiments (where time dilation was proportional to acceleration of gravity).

Time dilation isn't proportional to the acceleration of gravity in the Pound-Rebka experiment. It's proportional to "gravitational potential energy" (which is actually a Newtonian concept), i.e. mgh. However, g doesn't vary much, so what's being confirmed is that the rate of change of time dilation with height is g.

This is more generally true, the rate of change of time dilation , i.e. the metric coefficient g_00, with position is proportional to "gravity" in the sense that "gravity" is the acceleration an observer needs to remain stationary, i.e. what an observer at a constant coordinate (the coordinates used to measure the metric) will read on their accelerometer.
 

FAQ: Force due to gravity vs force due to accelratio

What is the difference between force due to gravity and force due to acceleration?

The force due to gravity is the force exerted by the gravitational pull of a massive object, such as the Earth, on another object. It is a constant force and is always directed towards the center of the massive object. On the other hand, force due to acceleration is the force required to accelerate an object. It is determined by the mass of the object and the rate of change in its velocity.

Which of these forces is responsible for keeping objects on the surface of the Earth?

The force due to gravity is responsible for keeping objects on the surface of the Earth. It is the force that pulls objects towards the center of the Earth, keeping them from floating away into space.

How are force due to gravity and force due to acceleration related?

Force due to gravity and force due to acceleration are related through Newton's Second Law of Motion, which states that force is equal to mass multiplied by acceleration. In other words, the force due to gravity acting on an object is equal to the object's mass multiplied by the acceleration due to gravity.

Is force due to gravity always the same?

Yes, force due to gravity is always the same for a particular object, regardless of its location. However, the force due to gravity may vary for different objects, as it is dependent on the mass of the object and the distance between them.

How does force due to gravity affect the motion of an object?

The force due to gravity is a crucial factor in determining the motion of an object. It is responsible for keeping objects in orbit, causing objects to fall towards the ground, and creating tidal effects. The magnitude and direction of the force due to gravity determine the acceleration of the object and, therefore, its motion.

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