Gravity Effects on Rod Length - Reference Sources

[SinghRP]

General relativity finds that gravity dilates the wavelength of light and the time period of an atomic clock. What does gravity do to the length of a (material) rod? If you know the answer, please cite me a reference or two.

 

[Naty1]

In GR gravitation potential dilates time; In special relativity velocity dilates length and time.

See Einstein online...of wikipedia articlesd on relativity. All because it is the speed of light that is fixed, not time nor space.

"What does gravity do to the length of a (material) rod?"

Within a black hole, it crushes it to nothingness.

 

[SinghRP]

Special relativity is not in discussion here.
ln gereral relativity, here is how the frequencies of a vibrating particle are related:
(frequency at infinity) = (1 + k.m/r) (frequenct at r), where k is modified gravitational constant.
You can deduce dilatation of light wavelength and clock time period from the above.
Is that eq. true for a rod's length?

 

[Q-reeus]

Special relativity is not in discussion here.
ln gereral relativity, here is how the frequencies of a vibrating particle are related:
(frequency at infinity) = (1 + k.m/r) (frequenct at r), where k is modified gravitational constant.
You can deduce dilatation of light wavelength and clock time period from the above.
Is that eq. true for a rod's length?

You are no doubt feeling some frustration without any satisfactory response forthcoming. Has been my battle too - but have come to realize that in GR spatial distortion is a bit of a slippery eel thing. 'Go down there' to check and you shrink/bend/whatever same way as the rod, 'stay out here' and a host of issues re measurement and what it really means crop up. There is a short answer that may or may not satisfy you right here in PF: https://www.physicsforums.com/showthread.php?t=145285
A more lengthy and technical discussion, : https://www.physicsforums.com/showthread.php?t=404153
People seem far more comfortable in general with net results - just aply EFE's and out pops orbital dynamics etc, rather than thinking in terms of a 'clean' partitioning into time vs distance effects.

 

[SinghRP]

Thank you, Q-reesus for a fitting answer.
I will check the sites you suggested.
Meanwhile, if you are interested in knowing how I derived that in GRT, a rod is longer closer to mass, please let me know. I can't post it here unless I have the Forum's permission. I have been asking it because in SRT, a moving rod appears contracted
in the direction of motion to an observer.
Thanks again.

 

[PeterDonis]

SinghRP, the following thread, where Q-reeus and I have been discussing the effects of spacetime curvature around a gravitating body on radial vs. tangential distance measurements, might also be worth a look:

https://www.physicsforums.com/showthread.php?t=541317

It's also worth noting that there are some general issues with the way your question is posed, which may be of interest:

(1) What is the rod's state of motion? Usually, unless otherwise specified, it is assumed that the rod is "hovering"--that is, that it remains at a constant radial coordinate r. But this requires the rod to accelerate, in order to remain at a constant height against the pull of gravity. So if you want to see any effects of spacetime curvature on the rod's length, you have to first factor out the effects of the acceleration.

(2) Also, the usual argument for "length contraction" of a rod placed radially in a gravitational field is based on the metric coefficient g_rr in Schwarzschild coordinates being greater than 1, and getting larger as the radial coordinate r gets smaller. But this is not true in other coordinates; for example, in Painleve coordinates, g_rr is 1. This means, for example, that a rod freely falling towards the gravitating body, instead of "hovering" at a constant radial coordinate, will *not* see any "length contraction" even if the usual argument is correct.

(3) How is the "length" of the rod to be measured? The "obvious" way to do it is to make marks on the rod while the rod is very far away from all gravitating bodies, so spacetime in its vicinity can be assumed to be flat, and then measuring the distance between the marks once the rod is lowered into the gravity well and placed radially. But this begs the question, how do we measure the distance between the marks? We could use another rod, but the same question would arise for it. Or we could use light, say a laser rangefinder at one end of the rod and a mirror at the other, but gravity affects light too. There is simply no "distance measure" we can use that is unaffected by gravity, so whatever effect gravity has on the rod, it will have on the distance measure too.

(4) There is another way to judge the rod's length that avoids the above problem, by measuring stresses in the rod (or some other frame-invariant observable). For example, I make marks on the rod in some region of spacetime that is practically flat, after verifying that there are no measurable stresses in the rod. Then I place the rod alongside a second identical rod with identical marks on it, also unstressed. Then I place the first rod under known stresses, and measure its marks against the marks on the second, unstressed rod. This allows me to calibrate the rod's stress-strain ratio--i.e., how the distance between its marks changes with stress.

Now, having done all this, I lower the rod into the gravity well and place it radially. But GR predicts that, when I do this, I will measure *no* stress on the rod! (Strictly speaking, I will measure some stress due to the acceleration the rod has to sustain to "hover" at a constant radial coordinate, but we already talked about that above; once we correct for that, there will be no measurable stress in the rod. There may also be effects of tidal gravity, which I don't think we need to go into detail about here; they would need to be factored out the same way acceleration is.) When people in those various threads that were linked to above talk about there being no locally measurable effects of the difference in metric coefficients (radial vs. tangential, in Schwarzschild coordinates), this is the sort of thing they're talking about: whenever you look at a frame-invariant, physical observable, locally, you find that there is *no* difference radial vs. tangential.

It's also worth nothing that the case of "time dilation" in a gravity well is somewhat different, because there is a frame-invariant physical observable we can use to measure it--the gravitational redshift/blueshift between observers "hovering" at different heights in the field.

 

[Naty1]

Peter..some interesting comments above...

There is simply no "distance measure" we can use that is unaffected by gravity, so whatever effect gravity has on the rod, it will have on the distance measure too.

there is a frame-invariant physical observable we can use to measure it--the gravitational redshift/blueshift between observers "hovering" at different heights in the field

So can we place identical light sources on the ends of a radially placed rod in the vicinity of a mass and then observe the redshift (say) from a distant observational point...to measue distance??

Singh...

Special relativity is not in discussion here.

I don't understand what that means...You know that special relativity is a special case of GR I am sure...But questions of distance (length) get less precise in curved spacetime.

Other "tricks" to gain some insight would be Einstein's equivalance principle and consideration of uniform gravitational fields (no tidal effects).

Are you familiar with " Born Rigidity" ??

Good read...from prior post above:

A more lengthy and technical discussion, : https://www.physicsforums.com/showthread.php?t=404153

 

[PeterDonis]

So can we place identical light sources on the ends of a radially placed rod in the vicinity of a mass and then observe the redshift (say) from a distant observational point...to measue distance??

How will you translate the redshift measurements into distance measurements?

 

[ZealScience]

Basically, the effect of gravity on space-time is that it change the metric which characterize the property of space-time. In Newtonian approximation the time component of metric contribute to potential the most, which is affecting the wavelength.

But I think that it is based on approximation of relatively flat space, would there be effects on space component as well?

 

[Q-reeus]

Now, having done all this, I lower the rod into the gravity well and place it radially. But GR predicts that, when I do this, I will measure *no* stress on the rod! (Strictly speaking, I will measure some stress due to the acceleration the rod has to sustain to "hover" at a constant radial coordinate, but we already talked about that above; once we correct for that, there will be no measurable stress in the rod. There may also be effects of tidal gravity, which I don't think we need to go into detail about here; they would need to be factored out the same way acceleration is.) When people in those various threads that were linked to above talk about there being no locally measurable effects of the difference in metric coefficients (radial vs. tangential, in Schwarzschild coordinates), this is the sort of thing they're talking about: whenever you look at a frame-invariant, physical observable, locally, you find that there is *no* difference radial vs. tangential.

Neglecting possible higher order derivative effects, what is written above is of course a logical necessity simply because, as I put it in #4 "'Go down there' to check and you shrink/bend/whatever same way as the rod," If I'm not mistaken SinghRP understands that bit, but like myself is trying to get a handle on relating, in some physically meaningful way, effect of gravity on 'length down there' as it relates to us 'out here'. Bottom line - if curvature of spacetime has any real meaning for space as well as time (the uncontroversial redshift part), there must be evident effects observable to us 'out here'. And not just bending of light. For instance, there must be some sense in which say a neutron star will appear, after all mechanical stress/strain, gravitational lensing etc, is factored in, to have a different diameter based on 'warped length scale' - a geometric effect independent of any coordinate system used. So, the statement 'neutron star xyz has a diameter of 20km' refers to what - local measure or our measure? It was argued elsewhere that while something will be there, too much inherent 'freedom of measurement choice etc' for definite, unambiguous predictions. The slippery eel.

 

[PeterDonis]

Neglecting possible higher order derivative effects, what is written above is of course a logical necessity simply because, as I put it in #4 "'Go down there' to check and you shrink/bend/whatever same way as the rod,"

Remember that in the passage you quoted I was measuring distance by measuring stress. Stresses are invariant physical observables. So the "shrinking/bending" of the measuring device is irrelevant. I was making an even stronger statement: in terms of stresses, there is *no* "shrinking/bending" of measuring devices "down there" relative to "up here". The only effects of the K factor "down there" are *how many measuring devices* will fit between concentric spheres, relative to the Euclidean prediction. I don't know how I can make it any clearer.

If I'm not mistaken SinghRP understands that bit, but like myself is trying to get a handle on relating, in some physically meaningful way, effect of gravity on 'length down there' as it relates to us 'out here'.

It has no effect on *length*, per se. It has an effect on "how much length" is present, radially, in between concentric spheres, relative to the Euclidean prediction. That *is* a physically meaningful way of describing the difference between "down there" and "up here". The K factor, as I defined it, is a physical observable. I told you how to physically measure it.

Bottom line - if curvature of spacetime has any real meaning for space as well as time (the uncontroversial redshift part), there must be evident effects observable to us 'out here'.

There are; the K factor as I defined it is just as observable "out here" as "down there". The non-Euclideanness of packing of little objects can be observed from anywhere.

And not just bending of light. For instance, there must be some sense in which say a neutron star will appear, after all mechanical stress/strain, gravitational lensing etc, is factored in, to have a different diameter based on 'warped length scale' - a geometric effect independent of any coordinate system used. So, the statement 'neutron star xyz has a diameter of 20km' refers to what - local measure or our measure?

You would have to quote a specific instance of such a statement to answer this, and the answer would depend on the specific instance. As "diameters" are usually quoted, I suspect they are actually in "radial coordinate units", meaning they are really the square roots of areas divided by 4 pi, or they are circumferences divided by 2 pi (well, for "radius", not "diameter", but it's easy to relate the two). If actual "physical diameter" is meant, then yes, you would have to evaluate the K factor throughout the interior of the star to see how much actual, physical distance there was inside a sphere with the area of the star's surface, given the K factor.

 

[SinghRP]

My response to Q-reesus and PpeterDonis at Post 835.

Thank you for the dialogue. Let’s keep the issue simple.

The issue with length change in gravitational field has been on my mind ever since I read Adler, Bazin, and Schiffer’s “An Introduction to General Relativity” and George Gamow’s “Gravity” – that was a long, long time ago! I searched several books, journals, and websites. I did not find satisfaction.

The clearest statement I came to was what Gamow wrote in simple English: “A typist working on the first floor of the Empire State Building will age more slowly than her twin sister working on the top floor. The difference will be very small, however; it can be calculated that in ten years the girl on the first floor will be a few millionth of a second younger that her twin on the top floor. In the difference in gravity between the surface of the Earth and the surface of the Sun, the effect is considerably larger. A clock placed on the surface of the Sun would slow down by one ten-thousandth of a percent in respect to the terrestrial clock. Of course, nobody can pace a clock on the surface of the Sun and watch it go slow, but the expected slowdown was confirmed by observing the frequencies of spectral lines emitted by atoms in the solar atmosphere.” These statements are the clearest even for a physicist like me. In addition, it isolates gravity from the clutter of issues.

I knew that light produced on the sun is redshifted compared to the light produced at infinity. [Redshift: the wavelength is elongated. This is not Doppler or Wolf shift.]

So, I asked myself: “Does the girl on the first floor get taller or shorter relative to her twin on the top floor?” I will force myself now not to worry about the non-uniformity of gravitational field over the girls nor about whether they are accelerating or in relative motion to each other. So, I stared with very basic physics and mathematics and considered to ignore all perturbations and clutters.

I took Einstein’s equation about the frequencies (ν) of a vibrating particle from his book “ The Meaning of Relativity”: ν∞= (1 + κm/r) νr , where m is mass under whose gravity the vibrating particle is, r is the distance between the two, and κ = G/c2, where G is the gravitational constant.

Then I imagined a very thin wire, whose “particles” (charge e, mass δm << m) are separated by x. Under the electrostatic forces of its neighboring particles, each particle oscillates with a frequency. (Heisenberg’s uncertainty principle forces the oscillations.) The frequency of this oscillation can be derived elementarily to be: f = z/xq. Here z and q are parameters of the rod. For instance, for a one-dimensional rod dimensional rod, z = (1/π) (Q e2 /δm)1/2 and q = 3/2. Here Q is Coulomb’s constant.
Substituting the particle’s frequency into the Einstein’s equation, we have: xr = (1 + κ m/r)1/q x∞. This tells me that: the rod is longer closer to the mass; and the rod is flattened and disintegrated near a black hole (m/r → ∞). If the gravitational field is not uniform over the rod, the rod is deformed.

Now I answer my question: “The girl on the first floor gets taller relative to her twin on the top floor.”

So this is my “short” contribution to the dialogue. I don’t see any violation of basic physics in above. I hope I found a satisfactory answer to my original question. [By the way, I already moved to the first floor.]

What do you think? Please keep the discussions going on this topic.

 

[Q-reeus]

SinghRP: If you can get a hold of a copy of 'Gravitation and Relativity' - M.G.Bowler. Chapter 6 'The distortion of reference frames' goes into all that in what some might consider an unorthodox way. He seems to implicitly work in isotropic coordinates, and references *all* locally gravitationally effected quantities to their coordinate values. He obtains the opposite finding - shortening. I'm sure someone here will blast the approach taken there, but the results do give a consistent outcome. Typsetting is attrocious btw.

 

[PeterDonis]

I took Einstein’s equation about the frequencies (ν) of a vibrating particle from his book “ The Meaning of Relativity”: ν∞= (1 + κm/r) νr , where m is mass under whose gravity the vibrating particle is, r is the distance between the two, and κ = G/c2, where G is the gravitational constant.

I think this formula, as you give it, is approximate; it assumes that km << r. But that's only a technical point, and doesn't affect the argument; I just mention it in passing.

The more important point is that this equation relates *observed* frequencies at one point, to *emitted* frequences at the other. For example, if someone shines a laser emitting light of a known frequency from one point, the light will appear to have the frequency given by the equation at the other point. We say informally that light descending into a gravity well from far away is "blueshifted" (it's observed to have a higher frequency than the known frequency of emission) when it's observed at radius r, and light ascending out of a gravity well from radius r to someplace far away is "redshifted" (it's observed to have a lower frequency than the known frequency of emission). So this equation doesn't say anything about the "real" frequency at the point of emission; it only talks about how the signal carrying information about the frequency is changed as it passes through the intervening space.

Then I imagined a very thin wire, whose “particles” (charge e, mass δm << m) are separated by x.

How is x measured? Remember all the difficulties I raised in my last post about that.

Under the electrostatic forces of its neighboring particles, each particle oscillates with a frequency. (Heisenberg’s uncertainty principle forces the oscillations.) The frequency of this oscillation can be derived elementarily to be: f = z/xq. Here z and q are parameters of the rod. For instance, for a one-dimensional rod dimensional rod, z = (1/π) (Q e2 /δm)1/2 and q = 3/2. Here Q is Coulomb’s constant.

How are you deriving this from Heisenberg's uncertainty principle? First of all, I don't see Planck's constant anywhere. Second, oscillations due to the uncertainty principle are not the same as oscillations due to Coulomb force.

But even supposing, for the sake of argument, that you had a valid equation relating some observed frequency to the distance x between neighboring particles, just plugging that equation into Einstein's equation above about frequencies doesn't mean what you seem to think it means. See below.

Substituting the particle’s frequency into the Einstein’s equation, we have: xr = (1 + κ m/r)1/q x∞. This tells me that: the rod is longer closer to the mass; and the rod is flattened and disintegrated near a black hole (m/r → ∞). If the gravitational field is not uniform over the rod, the rod is deformed.

No, it's not telling you that. It's only telling you that, if the oscillations in the rod generate some sort of signal (for example, oscillations due to electrical forces inside the rod might generate a radio signal--the rod might be a radio antenna), the frequency of that signal if it is observed far away may be different than the frequency the signal had when it was emitted from the source. In order to derive information from the signal about the inter-particle distances x at the source, you would first have to correct for what happened to the signal during its travel, which means you would have to *undo* the effect predicted by Einstein's equation on the signal.

The general principle here is that, if you want to use information about frequencies to determine information about distances, you have to use local frequencies--frequencies measured at the same location as the distances you are interested in.

If the gravitational field is not uniform over the rod, the rod is deformed.

It's also worth nothing that nothing about what you have said above involves any non-uniformity of the field *over the rod*. It only involves non-uniformity of the field between the rod and the point where signals coming from the rod are being observed. Non-uniformity of the field over the rod itself is tidal gravity, and I understood that we were supposed to be leaving that out.

 

[SinghRP]

To PeterDonis:
You have raised good points. I will answer them later. As an old man I take time.
Thanks so much for keeping the dialogue alive.

 

[Q-reeus]

SinghRP: Just to make sure there was no misunderstanding, when I wrote in #13: "Typsetting is attrocious btw." - that was referring to Bowler's book, not your entry! And when I wrote "...references *all* locally gravitationally effected quantities to their coordinate values.", that meant considering the effect of gravitational potential on things like permeability, permittivity, mass, charge, velocity, speed of light, mommentum, even Planck's constant. Not as straight forward as it first seems, but an exhaustive cross checking procedure used there ties it all down in the end.

 

[SinghRP]

To PeterDonis at Post 856.
Sorry, I am late! I was busy grading papers. My answers follow after yours.

I think … in passing.

The complete formula is: vx = ((1+km/r)/(1+km/(r+x))vr. In the limit x → ∞, it’s: v∞ = (1+km/r)vr.

The more important … the intervening space.

You have a point. I will quote Einstein as he interprets the aforementioned limiting formula: “The rate of a clock is accordingly slower the greater is the mass of the ponderable matter in its neighbourhood. We therefore conclude that spectral lines produced on the sun’s surface will be displaced towards the red, compared to the corresponding lines produced on the earth.” Lindsay and Margenau in Foundations of Physics said the same. Personally I think that: Gravity affects atoms and associated emission and absorption of light (A particle vibrates at lower frequency closer to the mass); Gravity affects the passage of light through its field (Light is redshifted as it is rising and blueshifted as it is falling in the field.)

How is x measured? Remember … that.

There is no need to measure r at great separations. It is avoided by experiments such as Pound and Rebka’s. Physics is replete with overemphasis on measurements. Moreover, coordinate systems and observers are not a part of the laws of nature – E. A. Milne.

How are you deriving … Coulomb force.

HUP ensures that a particle shall move (Δx must not be zero.) Then, in a crystal, the repulsive forces of neighboring atoms take over the oscillations of the particular atom. (A finger has to prod the pendulum to move. Not at the macroscopic but at the microscopic level, HUP is a very much more effective finger.)

But even supposing, … See below.

I believe you and I already addressed it previously.

No, it’s not telling … on the signal. The general … interested in.

As discussed before, gravity does affect vibrators and what they emit or absorb. I am sure we can agree on those for the time being. All I wanted to have at this point is a relationship between the frequency of vibration of an atom and separation distance between atoms in a crystal. (Separations are infinitesimal and local.) So, that substitution is justified. An experiment similar to the Pound-Rebka’s may be conducted, where changes in Bragg’s reflection/diffraction patterns from a crystal (such as NaCl) may show whether spacings between the atoms are changed by gravity. Thermal and other non-gravitational effects must be sorted out. I am very much confident – just being a human here – that atoms in a crystal have larger separations in a stronger gravitational field or closer to a mass. On the other hand, I could be wrong.

It’s also worth … leaving that out.

The scenario in my statement is quite simple. The separations between the atoms in a crystal are affected differently in a non-uniform gravitational field, so the crystal will be deformed.

 

[SinghRP]

Sorry, I am late! I was busy grading papers. My answers follow after yours.

I think … in passing.

The complete formula is: vx = ((1+km/r)/(1+km/(r+x))vr. In the limit x → ∞, it’s: v∞ = (1+km/r)vr.

The more important … the intervening space.

You have a point. I will quote Einstein as he interprets the aforementioned limiting formula: “The rate of a clock is accordingly slower the greater is the mass of the ponderable matter in its neighbourhood. We therefore conclude that spectral lines produced on the sun’s surface will be displaced towards the red, compared to the corresponding lines produced on the earth.” Lindsay and Margenau in Foundations of Physics said the same. Personally I think that: Gravity affects atoms and associated emission and absorption of light (A particle vibrates at lower frequency closer to the mass); Gravity affects the passage of light through its field (Light is redshifted as it is rising and blueshifted as it is falling in the field.)

How is x measured? Remember … that.

There is no need to measure r at great separations. It is avoided by experiments such as Pound and Rebka’s. Physics is replete with overemphasis on measurements. Moreover, coordinate systems and observers are not a part of the laws of nature – E. A. Milne.

How are you deriving … Coulomb force.

HUP ensures that a particle shall move (Δx must not be zero.) Then, in a crystal, the repulsive forces of neighboring atoms take over the oscillations of the particular atom. (A finger has to prod the pendulum to move. Not at the macroscopic but at the microscopic level, HUP is a very much more effective finger.)

But even supposing, … See below.

I believe you and I already addressed it previously.

No, it’s not telling … on the signal. The general … interested in.

As discussed before, gravity does affect vibrators and what they emit or absorb. I am sure we can agree on those for the time being. All I wanted to have at this point is a relationship between the frequency of vibration of an atom and separation distance between atoms in a crystal. (Separations are infinitesimal and local.) So, that substitution is justified. An experiment similar to the Pound-Rebka’s may be conducted, where changes in Bragg’s reflection/diffraction patterns from a crystal (such as NaCl) may show whether spacings between the atoms are changed by gravity. Thermal and other non-gravitational effects must be sorted out. I am very much confident – just being a human here – that atoms in a crystal have larger separations in a stronger gravitational field or closer to a mass. On the other hand, I could be wrong.

It’s also worth … leaving that out.

The scenario in my statement is quite simple. The separations between the atoms in a crystal are affected differently in a non-uniform gravitational field, so the crystal will be deformed.

 

[PeterDonis]

The scenario in my statement is quite simple. The separations between the atoms in a crystal are affected differently in a non-uniform gravitational field, so the crystal will be deformed.

The word "non-uniform" is crucial here; for the spacing between atoms in the crystal to be affected, the field needs to be non-uniform *across the crystal*.

This is *not* required for the observed frequencies of light emitted by an object to be seen as redshifted from far away; it's only necessary that the field be non-uniform between the point of emission and the point of reception. The field can be uniform across the object that is emitting the light.

 

[SinghRP]

Some statements were made previously regarding special relativity. And I said that SR was not in discussion. As you know, SR and GR are two separate topics.

Gravity a real force – the fourth fundamental force, assuming there are no others. In GR, gravitation is due to the curvature which matter creates in the field of space-time geometry. The field of space-time geometry is the gravitational field. At the microscopic level, gravitons would be the quanta of gravitational field. No gravitons and gravitational waves have been detected so far. I never knew how matter (creates and) warps space-time!

The strong, the weak, and electromagnetic fundamental interactions are mediated by the color, the weak, and electromagnetic fields associated with the color, the weak, and electrical charges of matter or antimatter. There is no such analogy associated with gravitational interaction. I never knew why!

SR is an apparent, not real, effect:
1. If we observe a rod moving past us with a uniform velocity v, it will look contracted in the direction of its motion by a factor (1 – v2/c2)1/2.
2. If we observe a clock moving past us with a uniform velocity v, it will appear to be losing time in the direction of motion, its rate slowed by a factor (1 – v2/c2)-1/2.
If observer A is moving uniformly with v relative to observer B, then the latter is moving with –v relative to the former. The twin-paradox is not a paradox at all. If an observer is moving with the mesons, s/he sees no changes in their decay rate.

 

[A.T.]

SR is an apparent, not real, effect:
1. If we observe a rod moving past us with a uniform velocity v, it will look contracted in the direction of its motion by a factor (1 – v2/c2)1/2.

No. It will not always look contracted. It might even look elongated if it moves towards you
http://www.spacetimetravel.org/bewegung/bewegung3.html

However it will always be measured contracted. And that is what SR is about, not some apparent visual effects, as you seem to believe.

 

[pervect]

I think the best answer is that it is not really a good idea to think of gravity as "slowing down time" in any physical sense.

There are two types of time. One sort of time is coordinate time, in which you assign every event in space-time a time coordinate.

Another sort of time is proper time - this is the sort of time that a clock keeps.

It's a desirable goal of coordinate systems to have the coordinate time match proper time as much as possible But it turns out to be impossible to make this happen everywhere, the reason for this is ultimately space-time curvature.

So, on the Earth, we make coordinate time match proper time at sea level, for instance, when we define the coordinate time standard TAI time. At any other altitudes the two don't match, and the ratio is the time dilation factor at that altitude.

When proper time differs from coordinate time, the result is typically referred to as "time dilation". It is rather common to mis-interpret time dilation in terms of some sort of underlying "absolute time", to think of the time that clocks measure as being slowed down, some how, from this "absolute time".

Of course this idea is well known not to work even in special relativity, the twin paradox, for instance, demonstrate that trying to think of time dilation in terms of an underlying absolute time will lead to paradox. However, people seem to be determined to do it anyway, why I don't know.

Fortunately, there doesn't seem to be any similar confusion about distance, for the most part. Textbooks even go out of their way to explain that the Schwarzschild r coordinate is "just a coordinate".

In any event, you'll find a lot of places that give you the metric coefficients of a gravitating body. So the physics is well documented, the argument is mostly about semantics and philosophy.

So, ultimately the important thing is that given a metric, the metric defines both a way to assign coordinates to events based on physical data (such as the proper time reading of clocks and the reception of radar signals), and a way to go in reverse, given coordinates of events to find the associated physical readings of observers according to their local clocks and/or rulers.

ANd I'm not quite sure what to do about people who don't know how to use the metric to perform this sort of computation, except to encourage them to try to learn - and perhaps to focus their efforts on actually learning how to use the metric as a space-time map, how to go from measurements to coordinates and vica-versa, rather than getting lost somewhere along the way.

 

[PeterDonis]

Gravity a real force – the fourth fundamental force, assuming there are no others. In GR, gravitation is due to the curvature which matter creates in the field of space-time geometry. The field of space-time geometry is the gravitational field. At the microscopic level, gravitons would be the quanta of gravitational field.

This is our best current belief, yes, based on the belief that there is a consistent theory of quantum gravity (we just haven't found it yet). But if it's true, then this...

The strong, the weak, and electromagnetic fundamental interactions are mediated by the color, the weak, and electromagnetic fields associated with the color, the weak, and electrical charges of matter or antimatter. There is no such analogy associated with gravitational interaction.

...is *not* correct. The graviton mediates the gravitational interaction in the same way that the gluons, weak bosons, and photon mediate the strong, weak, and electromagnetic interactions.

 

[Q-reeus]

As discussed before, gravity does affect vibrators and what they emit or absorb. I am sure we can agree on those for the time being. All I wanted to have at this point is a relationship between the frequency of vibration of an atom and separation distance between atoms in a crystal. (Separations are infinitesimal and local.) So, that substitution is justified. An experiment similar to the Pound-Rebka’s may be conducted, where changes in Bragg’s reflection/diffraction patterns from a crystal (such as NaCl) may show whether spacings between the atoms are changed by gravity. Thermal and other non-gravitational effects must be sorted out. I am very much confident – just being a human here – that atoms in a crystal have larger separations in a stronger gravitational field or closer to a mass. On the other hand, I could be wrong.

It’s also worth … leaving that out.

The scenario in my statement is quite simple. The separations between the atoms in a crystal are affected differently in a non-uniform gravitational field, so the crystal will be deformed.

SinghRP: Your #1 entry suggested, by the use of analogy with differential aging wrt height in a gravitational potential, length also a relational measure (inferred in that one could not notice anything by travelliong from one location to another). It was on that basis I referred you to Bowler's book. Not clear now what your conception of length change relates to - but seems increasingly you mean a locally measured length change. It is further not clear whether that locally measured length change is owing to graivitational potential, or tidal forces. If the former, you will have to explain how it could be locally detected - after all any 'ruler' is subject to change just as the crystal of NaCl or whatever one is using as 'detector'. Do you imply then that local measurement is possible owing to material dependence - 'ruler' made of steel changes differently to crystal made of NaCl? If on the other hand you just mean the effect of tidal forces - that has no direct linkage to redshift. A compact gravitating mass will generate far greater tidal forces on your suspended rod than for a less dense body - given radial separations achieving the same relative redshift factor. One is a function of potential the other is a function of second order derivatives of the same. Please then clarify your conception of exactly what length change means, and what generates it.

 

[Q-reeus]

I think the best answer is that it is not really a good idea to think of gravity as "slowing down time" in any physical sense.

pervect: Several times now I have given the example of sending down a clock to the surface of a planet, leaving it a set coordinate time, and retrieving. There will be a noticed difference in elapsed time. Sure the sending down and retrieving parts create some difficulty if a one-shot procedure is employed - just separating SR from 'pure' GR effects is one. All that's required to eliminate such difficulties is to repeat the procedure a second time, leaving the clock a different period of elapsed coordinate time. Provided sending and retrieving are carried out just as before, we can easily eliminate everything but the true difference in clock rates. How can it be said there is no precise definition of 'gravity slowing down time' - and of course it goes without saying that has to be on a relational - 'here' vs 'there' basis.

btw I got your hint later on in your entry:
"Fortunately, there doesn't seem to be any similar confusion about distance, for the most part. Textbooks even go out of their way to explain that the Schwarzschild r coordinate is "just a coordinate"." As sometimes happens, in departing the scene, a fresh perpective eventuates. i will in due course be taking up the issue again in that other thread.

 

[yoron]

'Gravity' may 'slow down' the time relative your clock, observing it from outside the gravitational well. But what happens if you follow it down? Will it slow down then? Do you expect yourself to find yourself getting a greater life span relative your local clock, if staying put inside a gravity well?

All time dilations are definitions relative other 'frames of reference'. But you won't notice them as such, until you do a 'twin experiment'. Although you might argue that the twin coming back now has a longer lifespan, relative what he would have had if he had stayed at home, he won't find it to be so himself. Instead it will to him be as if it is you that aged 'faster than normal' while he was gone, and that his own measure of time never changed, locally.

If you expect a 'time dilation' to be a phenomena based solely on where you are, ignoring other 'frames of reference', then it seems to me that you also expect SR to be wrong? Because, according to SR you're constantly 'time dilated' relative different stars and other objects 'relative motion'.

And as some moves faster relative Earth, and others move slower you then either can define yourself 'simultaneously younger' as well as 'older', depending on what star you measure yourself against, also depending on how you define that relative motion relative Earth. Or you define it such as SR is wrong, and a 'time dilation' is an 'absolute effect', depending on 'gravity', with 'frames of reference' becoming another type of definition. But if you do you also put into doubt Einsteins equivalence principle for GR, equalizing a uniform constant acceleration with a constant 'gravity'. And that acceleration is what the 'traveling' twin also need, to become 'time dilated' relative Earth when being back.

 

[Q-reeus]

'Gravity' may 'slow down' the time relative your clock, observing it from outside the gravitational well. But what happens if you follow it down? Will it slow down then? Do you expect yourself to find yourself getting a greater life span relative your local clock, if staying put inside a gravity well?
All time dilations are definitions relative other 'frames of reference'.

I will assume yoron this is all addressed to me - I will have to assume that because you didn't quote anyone, and mine was the previous entry. And what exactly is this saying any different from my own qualifier in #25:
"How can it be said there is no precise definition of 'gravity slowing down time' - and of course it goes without saying that has to be on a relational - 'here' vs 'there' basis."? There is imo a rampant disease afflicting many PF contributors - failing to carefully follow what the other has actually said, and what follows is knocking down a straw man. My comments addressed in #25 began with a particular quote - which to me could easily give a wrong impression about what can and cannot be inferred about 'gravitational time dilation'.

But you won't notice them as such, until you do a 'twin experiment'. Although you might argue that the twin coming back now has a longer lifespan, relative what he would have had if he had stayed at home, he won't find it to be so himself. Instead it will to him be as if it is you that aged 'faster than normal' while he was gone, and that his own measure of time never changed, locally.

If you expect a 'time dilation' to be a phenomena based solely on where you are, ignoring other 'frames of reference', then it seems to me that you also expect SR to be wrong? Because, according to SR you're constantly 'time dilated' relative different stars and other objects 'relative motion'.

Same remarks as above.

And as some moves faster relative Earth, and others move slower you then either can define yourself 'simultaneously younger' as well as 'older', depending on what star you measure yourself against, also depending on how you define that relative motion relative Earth. Or you define it such as SR is wrong, and a 'time dilation' is an 'absolute effect', depending on 'gravity', with 'frames of reference' becoming another type of definition. But if you do you also put into doubt Einsteins equivalence principle for GR, equalizing a uniform constant acceleration with a constant 'gravity'. And that acceleration is what the 'traveling' twin also need, to become 'time dilated' relative Earth when being back.

On that last matter. You can therefore explain why a watch whizzed around in a small circle (large centripetal acceleration), ticks identically to when whizzed around in a large circle (low centripetal acceleration) - rim speed being identical?

 

[yoron]

On what last matter do you mean? That you need a acceleration for the twin to get a 'speed' different from Earths? As for the clocks :) That was a tricky one. Are you saying that if we had two clocks spinning around, in loops of different diameters, they must show a same time presuming their 'rim spin' to be of the same speed?

Are you suggesting something similar to two rockets accelerating, but being 'at rest' relative each other? But then we have the fact that the outer clock would have needed a greater acceleration to get into that apparent 'sync' with the inner clock spinning around that you assume.

And as 'speed' always is measured by a clock relative a ruler they can't be at a same speed relative that center, if they are in different orbits relative some 'center'. Also, if you imagine both clocks 'at rest' relative each other the outer clock must have needed a greater acceleration to get to that speed.

And then you assume their clock rate to be the same relative what? A 'inertial observer'?
Or do you mean that, if we exchange the clocks for rocket ships on strings, :) A and B, that A can look at B and find B:s clock to 'tick identically' to A? And that the same will go for B looking at A?
=

But yes, if what you meant was that 'gravity' do slow 'clocks' down, relative the observer, we have ample proof of that in NIST latest experiments here on earth, proofing 'time dilations' existing at so short distances as, around half a meter? Or was it shorter? But the main point here, to me, is that it always will be 'relative the observer', meaning that you can choose to place yourself at any of those clocks and still find your life measure to be of the exact same length relative your own clock, or that clock you now 'superimposed' yourself on, well not 'superimposed', but still, trying to :) What I'm meaning is that 'times arrow' only have one 'rate' to serve you locally, no matter where you go, or how fast you are relative something else.

 

[Q-reeus]

Are you saying that if we had two clocks spinning around, in loops of different diameters, they must show a same time presuming their 'rim spin' to be of the same speed?

I would prefer the word orbiting rather than spinning, but yes that's what I meant. Implying that centripetal acceleration in this setting has no effect on time dilation measured once the 'twins' are brought back together - just relative speed on a time averaged basis. Of course from the point of view of anyone of the orbiting two that won't hold at any particular instant, because relative motion is continually changing hence computed instantaneous relative clock-rate, but it will overall. At all instants they both equally age less relative to a 'stationary' observer watching on, say at the common center of curvature of those two orbits. It could be argued this only works for centripetal acceleration where v.a = 0, and Einstein did use an argument tying in a linear acceleration/deceleration twins arrangement to 'gravitational time dilation' via equivalence principle - but that approach has it's critics: e.g. http://www.ias.ac.in/currsci/dec252005/2009.pdf

Are you suggesting something similar to two rockets accelerating, but being 'at rest' relative each other? But then we have the fact that the outer clock would have needed a greater acceleration to get into that apparent 'sync' with the inner clock spinning around that you assume.

Sure, the 'getting into sync' part requires a linear acceleration component during spin-up and spin-down phase, but that can be an arbitrarily small part of the orbiting 'twins' total time in relative motion. And just as for the clock example in #25, we can precisely deal with that complication by performing the run twice, with two different durations for constant orbital speed.

But yes, if what you meant was that 'gravity' do slow 'clocks' down, relative the observer, we have ample proof of that in NIST latest experiments here on earth, proofing 'time dilations' existing at so short distances as, around half a meter? Or was it shorter? But the main point here, to me, is that it always will be 'relative the observer', meaning that you can choose to place yourself at any of those clocks and still find your life measure to be of the exact same length relative your own clock, or that clock you now 'superimposed' yourself on, well not 'superimposed', but still, trying to :) What I'm meaning is that 'times arrow' only have one 'rate' to serve you locally, no matter where you go, or how fast you are relative something else.

Think we're basically on the same wavelength here. I would just put it that GPS works so well because the bods who run it have a perfectly good notion of how to account for both SR and GR contributions to time dilation. :zzz:

 

[yoron]

Yep, that's my notion too, it's not the acceleration per se, but the speed that will create it, although accelerations should play a part here too depending on length/duration of it. It would have been easier in some ways if one could place a 'time dilation' on the fact of that acceleration though. Even though one might argue that every 'speed' should include a acceleration, somewhere.

I would still expect a different 'time dilation' , but it depends on how wide away those 'clocks' are relative each other also. Imagine one clock rotating ten meters from that center, and the other clock being a light year away, both orbiting the center 'at rest' with each other, according to some 'still observer' at the center. If you mean that the speed relative that center will be the main cause creating this time dilation I think I will agree.

And yes, I've also wondered about the equivalence there, how one G on a planet will correspond to one G constantly uniformly accelerating, relative their 'time dilation' relative some arbitrarily set 'frame of reference' in uniform motion. That's one of the trickiest, and most interesting questions I know. It's about how far you can take this equivalence. When Einstein defined that acceleration as a constant inertia/gravity acting on the accelerating frame I definitely agree, but their 'proportionality' seems much harder to define in form of time dilations and Lorentz contractions.

 

[Q-reeus]

And yes, I've also wondered about the equivalence there, how one G on a planet will correspond to one G constantly uniformly accelerating, relative their 'time dilation' relative some arbitrarily set 'frame of reference' in uniform motion. That's one of the trickiest, and most interesting questions I know. It's about how far you can take this equivalence. When Einstein defined that acceleration as a constant inertia/gravity acting on the accelerating frame I definitely agree, but their 'proportionality' seems much harder to define in form of time dilations and Lorentz contractions.

Just to clarify my previous comments about linearly accelerated twin and relating that to gravitational time dilation via equivalence principle. From the point of view of an inertial observer, a clock on the end of a spring performing sinusoidal linear accelerations will on a time averaged basis age less than the inertial observer (who is stationary relative to the clocks center of motion) only on the basis of the time averaged speed of the clock. To add any acceleration term would yield an error. As many would no doubt point out, mathematically it gets down to computing and comparing the world lines of the 'twins', and that is a function of relative velocities over time, with acceleration coming in only incidentally as means of generating the changed velocities. Must go.

 

[jk22]

In GR gravitation potential dilates time; In special relativity velocity dilates length and time.

Do you have an indice explaining the transition from velocity (in special relativity) to potential (in general relativity) ?

indeed if velocities would affect space-time in general relativity, the metric would be time dependent which were not reasonable.

 

[pervect]

I think you're missing the point.

In general relativity, the metric coefficent for g_00 can be apprxoimated as (1-2U), where U is the Newtonian potential.

It's the metric coefficeint which causes time dilation and it's not velocity dependent.

In sepcial relativy, the metric coefficeints are always unity. So the time dilation in SR isn't coming from the metric coefficients.

 

[PeterDonis]

And yes, I've also wondered about the equivalence there, how one G on a planet will correspond to one G constantly uniformly accelerating, relative their 'time dilation' relative some arbitrarily set 'frame of reference' in uniform motion. That's one of the trickiest, and most interesting questions I know. It's about how far you can take this equivalence. When Einstein defined that acceleration as a constant inertia/gravity acting on the accelerating frame I definitely agree, but their 'proportionality' seems much harder to define in form of time dilations and Lorentz contractions.

The equivalence between acceleration and gravity is only "local". Over any significant distance or any significant length of time, the equivalence breaks down because gravity curves spacetime. See below.

As many would no doubt point out, mathematically it gets down to computing and comparing the world lines of the 'twins', and that is a function of relative velocities over time, with acceleration coming in only incidentally as means of generating the changed velocities.

In flat spacetime, yes. In curved spacetime, no. As pervect pointed out, the metric coefficient g_00 can also cause time dilation if it is different from 1. This time dilation can't be due to relative motion, since two objects at rest relative to each other at different heights will experience different rates of "time flow". But it can't be due to "acceleration" either, as you point out, because it's easy to construct examples of observers experiencing the same acceleration but with different rates of time flow.

The resolution of this apparent dilemma is that, in the presence of gravity, spacetime is curved; and in particular, around a static gravitating body, spacetime is curved in such a way that the worldlines of static observers at lower heights are shorter than the worldlines of static observers at higher heights. This difference in length appears, physically, as a difference in rate of time flow, because the worldlines in question are timelike.

(A more precise way of stating this is: if we pick as our simultaneity convention the simultaneity of observers very, very far away from the gravitating body, which is the simultaneity convention of the Schwarzschild time coordinate t, then all the static observers at various heights above the gravitating body will agree on which events are simultaneous, but they will disagree on how much time elapses between two given sets of simultaneous events; observers at lower heights will see less time elapse than observers at higher heights.)

 

[Passionflower]

This time dilation can't be due to relative motion, since two objects at rest relative to each other at different heights will experience different rates of "time flow". But it can't be due to "acceleration" either, as you point out, because it's easy to construct examples of observers experiencing the same acceleration but with different rates of time flow.

Actually that is an interesting proposal.
Can I challenge you demonstrate it with numbers and formulas? I think this would be of great didactic value.

Indeed two stationary test observers at different r values in a Schwarzschild solution show different time flow as you call it but can you demonstrate that is not due to the difference in proper acceleration?

Similarly with various test observers radially free falling at different relative velocities, their proper time derivative is all different when they at an instant all fly by at the same location, but can you demonstrate that is not due to the Lorentz factor?

For instance:

A stationary test observer at a given r coordinate value undergoes constant acceleration to resist the escape velocity at that location.

The escape velocity at a given r is (rs is the Schwarzschild radius):
\Large v_{{{\it escape}}}=\sqrt {{\frac {{\it rs}}{r}}}

While the time dilation is:
\Large d\tau_{{r}}=\sqrt {1-{\frac {{\it rs}}{r}}}

However if we simply take the escape velocity and apply the Lorentz factor we get the same time dilation.
Thus how would you want to prove any gravitational time dilation through this method?
And the same situation arises for the various free falling test observers.

 

[PeterDonis]

Actually that is an interesting proposal.
Can I challenge you demonstrate it with numbers and formulas? I think this would be of great didactic value.

Observers at varying radii on disks rotating at varying angular velocities. By varying the radius and angular velocities appropriately you can pretty much achieve any combination of proper acceleration and linear velocity relative to the center of the disk that you want, subject of course to restrictions on velocities not reaching the speed of light.

Indeed two stationary test observers at different r values in a Schwarzschild solution show different time flow as you call it but can you demonstrate that is not due to the difference in proper acceleration?

Sure, because the different time flow is due to g_00, while the acceleration is due to the radial rate of change of g_00. By varying the mass M of the central body and the radius r, you can achieve pretty much any combination of rate of time flow and acceleration you like.

Similarly with various test observers radially free falling at different relative velocities, their proper time derivative is all different when they at an instant all fly by at the same location, but can you demonstrate that is not due to the Lorentz factor?

If observers are in relative motion, the relative motion always contributes to their comparative rates of time flow, but in general it won't be the only contribution. If all the observers are also at exactly the same radial coordinate r above the same central mass M, obviously their relative velocities are the only respect in which they differ, so in that particular case, that's the only thing that *can* contribute to a difference in rate of time flow. That's not the type of case we've been talking about, or at least I don't think it is; but I agree that it's always good to clarify exactly what can affect what.

 

[Passionflower]

Peter I mean by demonstrating using a test situation and formulas to prove it. It is so because it is so, is just a tautology.

I just added some formulas to my prior posting to show you what I am getting at.

 

[PeterDonis]

However if we simply take the escape velocity and apply the Lorentz factor we get the same time dilation.
Thus how would you want to prove any gravitational time dilation through this method?
And the same situation arises for the various free falling test observers.

So far in this thread we've been talking about static observers only, or at least I think we have. Saying that a static observer's time dilation arises from "resisting escape velocity" is fuzzy reasoning; I don't understand how it validates your derivation of the time dilation factor, since a static observer is not moving at escape velocity in the coordinates in question.

You are correct that freely falling observers are different; in fact, since they *are* moving inward at escape velocity, you could argue that your time dilation formula should apply to them, *not* to static observers. Unfortunately, you left out an important piece of information: a freely falling observer is changing his r coordinate as well as his t coordinate, so to calculate his rate of time flow you have to integrate dtau along his worldline using both the dt^2 and the dr^2 terms in the line element; you can't just look at g_00 like you can for static observers. This is easier to do in Painleve coordinates, particularly if you want to carry the computation inside the horizon. You've posted such calculations in these forums before, so I know you know how to do them; such a calculation is also given on the Wikipedia page on Painleve coordinates:

http://en.wikipedia.org/wiki/Gullstrand–Painlevé_coordinates

 

[Passionflower]

So far in this thread we've been talking about static observers only, or at least I think we have. Saying that a static observer's time dilation arises from "resisting escape velocity" is fuzzy reasoning; I don't understand how it validates your derivation of the time dilation factor, since a static observer is not moving at escape velocity in the coordinates in question.

They would be moving wrt to a rain frame in GP coordinates. It is simply a matter of perspective and coordinates.

 

[Passionflower]

If observers are in relative motion, the relative motion always contributes to their comparative rates of time flow, but in general it won't be the only contribution.

Again I wait for you to show a case with formulas to prove what you claim.

How about this:
Let's take three observers:

rs=1

O1 = Stationary at R1
O2 = Free falling at escape velocity at R1
O2 = Free falling at 0.5* escape velocity at R1

If we consider their resp tau differentials at that location can we prove their differences are gravitational instead of velocity based?

Can you show it with math?

 

[PeterDonis]

Peter I mean by demonstrating using a test situation and formulas to prove it. It is so because it is so, is just a tautology.

I just added some formulas to my prior posting to show you what I am getting at.

I described how to set up the "test situations" you speak of, but sure, I'll post a few formulas as well to give more detail on how it goes.

(1) Observers at radius r on a disk rotating with angular velocity \omega, in flat spacetime, have a linear velocity v and proper acceleration a of:

v = \omega r

a = \frac{\omega^{2} r}{1 - \omega^{2} r^{2}} = \frac{\omega v}{1 - \omega^{2} r^{2}} = \omega \frac{v}{1 - v^{2}}

It should be obvious from the above that I can first pick whatever v I like (subject to the constraint 0 <= v < 1), and then adjust \omega appropriately so as to make a assume any value > 0 that I want, by adjusting r in concert with \omega so as to keep their product constant and equal to v. So I can choose v and a independently and achieve any combination of the two within the constraints, as I said.

(2) Observers "hovering" at a constant radial coordinate r above a gravitating body experience time dilation and proper acceleration of:

\frac{d\tau}{dt} = \gamma = \sqrt{1 - \frac{2 M}{r}} = \sqrt{1 - 2 U}

a = \frac{M}{r^{2} \sqrt{1 - \frac{2 M}{r}}} = \frac{U}{r \gamma} = \frac{1}{r} \frac{U}{\sqrt{1 - 2 U}}

So again, it should be obvious that I can pick any value for \gamma that I like (subject to the constraint that \gamma &gt; 1), which fixes the ratio M/r = U, and then use the second formula to adjust r appropriately, adjusting M in concert with r to keep U constant, to make a assume any value > 0 I want. So again I can choose \gamma and a independently and achieve any combination of the two within the constraints, as I said.

 

[PeterDonis]

They would be moving wrt to a rain frame in GP coordinates. It is simply a matter of perspective and coordinates.

No, static observers are not moving relative to GP coordinates. The GP radial coordinate r is defined the same as it is for Schwarzschild coordinates, so the orbits of static observers, which are lines of constant r, theta, phi in Schwarzschild coordinates, are also lines of constant r, theta, phi in Painleve coordinates. The only thing that changes between Schwarzschild and Painleve coordinates is the surfaces of simultaneity; in Schwarzschild they are orthogonal to the static observers' worldlines, but in Painleve they are orthogonal to the worldlines of observers freely falling inward "from infinity".

 

[PeterDonis]

How about this:
Let's take three observers:

rs=1

O1 = Stationary at R1
O2 = Free falling at escape velocity at R1
O2 = Free falling at 0.5* escape velocity at R1

If we consider their resp tau differentials at that location can we prove their differences are gravitational instead of velocity based?

Obviously not, since the case you posed has all three observers at the *same* radial coordinate, and I already said that in that special case, the relative velocity *is* the only contribution. I was merely saying that if we have observers at *different* radial coordinates, who also happen to be in relative motion, there will be a contribution from their relative velocity *and* a contribution from the difference in heights. The GPS satellites are an example; the adjustment to their clock frequencies to make GPS time run at the same rate as UTC includes an adjustment for height *and* an adjustment for relative velocity.

 

[Passionflower]

I was merely saying that if we have observers at *different* radial coordinates, who also happen to be in relative motion, there will be a contribution from their relative velocity *and* a contribution from the difference in heights.

It depends on the chosen coordinates if the difference in heights constitutes a velocity or not.

 

[Passionflower]

No, static observers are not moving relative to GP coordinates. The GP radial coordinate r is defined the same as it is for Schwarzschild coordinates, so the orbits of static observers, which are lines of constant r, theta, phi in Schwarzschild coordinates, are also lines of constant r, theta, phi in Painleve coordinates. The only thing that changes between Schwarzschild and Painleve coordinates is the surfaces of simultaneity; in Schwarzschild they are orthogonal to the static observers' worldlines, but in Painleve they are orthogonal to the worldlines of observers freely falling inward "from infinity".

Rain frame Peter, I am talking about a rain frame in GP coordinates.

I am fully aware of your point I merely wanted to demonstrate that from another perspective it is velocity based instead of gravitational based.

 

[PeterDonis]

It depends on the chosen coordinates if the difference in heights constitutes a velocity or not.

Huh? I'm not sure what you're getting at here. If you mean that the term "relative velocity" does not have a unique well-defined meaning in curved spacetime for observers that are spatially separated, yes, you're correct; one would have to settle on a specific meaning for the term. The one I had in mind was velocity relative to the Earth-Centered Inertial frame, since that's the relative velocity definition that was used, for example, to analyze the results of the Hafele-Keating experiment, but I should have made that explicit.

 

[Passionflower]

If you mean that the term "relative velocity" does not have a unique well-defined meaning in curved spacetime for observers that are spatially separated, yes, you're correct; one would have to settle on a specific meaning for the term.

I think there is no need to settle anything. I think that considering multiple viewspoints of things tend to enrich understanding.
That was all I wanted to show. :)

 

[PeterDonis]

I am talking about a rain frame in GP coordinates.

You mean rain frame as defined on the Wiki page?

http://en.wikipedia.org/wiki/Gullstrand–Painlevé_coordinates

As that line element is written, it mixes two different radial coordinates, so I'm not sure what it's supposed to mean physically. I don't believe such a frame is valid globally. I see the Wiki page references Taylor-Wheeler, I'll have to break open my copy and re-read their discussion of it.

Edit: Oops, read "Taylor-Wheeler" and immediately thought "Spacetime Physics" without reading the actual book title on the Wiki page. I don't have Exploring Black Holes but it's on my list to get.

 

[Passionflower]

You might also want to search for Lemaître observers.

Sulu: Captain! The stars. they are gone! We can't navigate!
Spock: Captain, I am locating an unidentified large object rapidly approaching us, with increasing speed. (Lemaitre)
Captain: Scottie, full speed reverse.
Scottie: Aye captain! Captain, we keep the engines at full speed and we are now just staying ahead of this 'thing'. Soon we will deplete our dilithium crystals. (Static)
Spock: Captain, I have a theory, it might work.
Scottie: We are just about to burn all of them out captain!
Spock: We should stop the engines and give some small lateral impulse power.
Captain: Sulu do it now!
Sulu: Captain it works, but now this thing is following us at a constant distance. (Hagihara)
Captain: Following us? Is it intelligent Spock?

It all depends on the point of view!

 

[jk22]

Similarly with various test observers radially free falling at different relative velocities, their proper time derivative is all different when they at an instant all fly by at the same location, but can you demonstrate that is not due to the Lorentz factor?

For instance:

A stationary test observer at a given r coordinate value undergoes constant acceleration to resist the escape velocity at that location.

The escape velocity at a given r is (rs is the Schwarzschild radius):
\Large v_{{{\it escape}}}=\sqrt {{\frac {{\it rs}}{r}}}

While the time dilation is:
\Large d\tau_{{r}}=\sqrt {1-{\frac {{\it rs}}{r}}}

However if we simply take the escape velocity and apply the Lorentz factor we get the same time dilation.
Thus how would you want to prove any gravitational time dilation through this method?
And the same situation arises for the various free falling test observers.

This is what I meant, using the equivalence principle to compute the speed of the local inertial frame.


This velocity of escape is calculated with a classical kinetic energy : U = 1/2mv^2

However shouldn't we use a relativistic expression : U = mc2*(1/Sqrt(1-v2/c2) - 1) ?

Then we find the velocity throuh equating with the potential energy : GMm/r2