Time dilation: Gravity force vs Gravitational potential

Discussion in 'Physics & Math' started by Ultron, Apr 26, 2016.

  1. Ultron Registered Senior Member

    I have question about one weird aspect of General relativity.

    Lets imagine, that we have two twins with atomic clocks. One twin stays on Earth and second twin will be transported in space module in free space without gravity. And the space module will start to rotate and the second twin will live for 1 year in rotating space modul in form of cylinder, where the centrifugal force will be equal to 1 g.

    After one year they will meet again and their atomic clocks will have the same time, because based on the general relativity they had the same time dilation on Earth surface with 1g and rotating space module with 1 g.

    But lets move to the second example. The first twin will use fictional elevator to get to the center of Earth and will stay there in fictional module floating around without gravity. The second twin will get to the space module and will have the rotation set to achieve 3 g.

    After one year they will meet again and their atomic clocks will have the same time, because based on the general relativity they had the same time dilation in Earth center with 0 g and rotating space module with centrifugal force 3 g.

    Sounds weird? Yea, it sounds definitely weird for me, but what I know, this is the current interpretation of equotations of GR meaning that time dilation on Earth is not related to force of gravity, but to gravitational potential. Meaning that the time dilation in center of Earth is bigger than time dilation on surface. This is pure theory, because while time dilation was confirmed for surface of Earth and above it, it was never confirmed for situations below surface and unfortunately it is practically impossible to check it experimentally.

    What is your view on it? Im I right with the example?
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  3. Q-reeus Valued Senior Member

    No, it doesn't work that way. The peripheral rotational speed of module-bound twin is the appropriate determinant - not centripetal acceleration. One can indirectly work acceleration into the equation - by equating a pseudo 'gravitational potential drop' in the twin's moving from rotation axis to periphery, and equate such 'potential drop' to an equivalent drop in potential in a real gravitational field i.e. owing to moving radially in the g-field of a massive body.
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  5. Janus58 Valued Senior Member

    as already stated, the 1g force felt in the module will not guarantee the same time dilation as on the surface of the Earth. The only way to assure this is to have the radius of the module be such that the tanjential velocity of the clock is equal to the escape velocity from the surface of the Earth.( Your module would have to be 12,757.5 km in radius to acheive this) This is known as the "Clock Postulate"; the fact that a clock's tick rate is independent of any acceleration it experiences(centripetal or otherwise), and has been verified by puting radioactice samples in centrifuges and spinning them uo to thousands of gs.
    Again the g force felt by the clock is not what will determine its tick rate.
    You don't have to put a clock at the center of the Earth to check the validity of the equations that predict what would happen to a clock put there. There are plenty of experiments we can do that can be used to check the accuracy of the equations. The above mentioned centrifuge example is just one. The correction to GPS clocks is another, as are the Pound-Rebka experient and the Mossbaur effect. All of these give results that are consistant with gravitational time dilation being dependent on gravitational potential and also consistent with the fact that as a clock moves below the surface of the Earth towards the center it would tick slower.
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  7. Confused2 Registered Senior Member

    Looking at 'Gravitational Potential' or at least the potential difference between (say) A and B ... this is the work done moving a unit mass from A to B. Assume (correctly) I'm having a mental blank here... what is the work done moving a unit mass from the centre of the Earth to the surface (assuming the Earth isn't rotating)? Also what is the work done moving a unit mass from the centre of a rotating space station to the circumference where the acceleration at the circumference is 1g? Is this 'work done' sufficient so predict clock rates without actually having to know the velocity?
  8. danshawen Valued Senior Member

    Well done, Q-reeus!
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  9. Schneibster Registered Member

    This is incorrect. The clock that was in space will have lost time relative to the clock on Earth. The clock in space experienced no gravity; the clock on Earth experienced gravity. I note carefully that you did not say the second clock was in orbit around the Earth, but far from any gravity field; in orbit would have a different answer, though you would still be incorrect. Note that GPS satellites require relativistic correction both for their velocities relative to the GPS receiver on the ground, and for their higher position in the gravity field of Earth.

    The rotation of the module containing the clock in space, and its experience of the 1g centrifugal fictitious force mimicking the effects of a linear acceleration is unimportant to any time dilation. I will explain why, and why it's called a fictitious force, below in my answer to your second example.

    This is correct, but not for the reason you think.

    Let's carefully examine the effects of centrifugal force. What's really happening here is that the frame experiencing the force isn't undergoing a constant acceleration. It only appears so to the observer experiencing the force, and it would be clear to them if they could see outside that it was not a constant acceleration. From an outside frame, not experiencing the centrifugal force, the frame experiencing the force is undergoing a continuously varying acceleration, not a constant one. In fact, the centrifugal force frame experiences one acceleration at one side of the circle, and exactly the opposite acceleration at the opposite side. These two accelerations cancel out; and you can say the same thing for every point on the circle. Thus, despite experiencing centrifugal force, the net acceleration experienced by the frame experiencing the centrifugal force is zero! This is why centrifugal force is called a fictitious force. The observer on the space module would see one time dilation (due to their velocity, not the fictitious centrifugal force) at one side of the rotation, and the opposite time dilation on the other side. The two, over a whole number of rotations, cancel out leaving no net time dilation, just as there is no net acceleration.

    Now, why didn't the clock at the center of the Earth, deep in a gravity field, run slower? The answer is simple: at the center of the Earth, there is no net gravity field. The effects of one half of the Earth are offset by the effects of the other half. Time is not affected by position in a gravity field; it is affected by the curvature of space due to the net gravity field at the location of the clock.

    Thus, the total acceleration of the clock on the space module was zero, even though it was experiencing a 1g fictitious force, and the total gravity at the location of the clock at the center of the Earth was zero, even though it was deep in a gravity well, and (other than the effects of getting to and from the space module, and to and from the center of the Earth) the clocks come out the same.

    This was a very good question, though your first example was incorrect. I think you might have misunderstood the clock postulate. And the difference between real acceleration and apparent acceleration, particularly the difference between fictitious forces like centrifugal force and true net acceleration, needs to be better explained in popular science books.
    Last edited: Apr 27, 2016
  10. Schneibster Registered Member

    OK so far...

    I'm not sure about your reasoning here, but I had trouble following what you were thinking, so I can't say if this is right or wrong.

    You definitely have misunderstood the clock postulate, and the effects of circular motion. See my explanation of why centrifugal force is a fictitious force, and how the accelerations on opposite sides of the circle cancel. In fact both acceleration and gravity do cause time dilation; but that is due in acceleration's case to the changing velocity without cancellation as in the circular acceleration case. The Physics FAQ is unclear on this point, and since they brought up the centrifuge without explaining that the acceleration on one side is canceled by opposite acceleration on the other side, I consider this one of the less-well-written pages on what is otherwise a pretty good FAQ.

    After messing up by mentioning centrifuges without explaining why centrifugal force is a fictitious force, the author then compounds the error later in the page with a bad explanation of why this does not violate the equivalence principle.

    This subject gets pretty interesting when you consider that other parts of that same page make a true assertion, that for the infinitesimal moment when an accelerating body just matches the velocity of an inertial one, the rate of the passage of time for them is equal; the implications of this can really open GRT and SRT up to you if you can wrap your head around them.

    Here I think you've confused frames, as well as confusing fictitious forces with real ones. In the frame motionless at the center of the Earth, there is no gravity; in the frame of the rotating space module, there is no net acceleration. That's why the two clocks come out the same. If the first clock were on the Earth's surface, and the second clock were in a rocket accelerating at 1g (as opposed to in a space module experiencing a centrifugal fictitious force), then they would keep equal time also. And that time would be different from the proper time of an inertial observer, slower in both cases.

    I disagree that gravitational potential determines the time dilation; it is the degree of curvature of spacetime that determines the time dilation.
  11. Q-reeus Valued Senior Member

    To continue with your last part in #1; It's correct theory. As pointed out earlier in #3, verified via observation and experiments where redshift in frequency directly corresponds to gravitational potential - not 'field strength' (1st spatial derivative of metric) or 'curvature' (2nd spatial derivative of metric). Thus a clock inside the hollow interior of a spherical shell of matter experiences a completely flat i.e. uncurved spacetime region. But since the g_tt metric component is there depressed relative to far outside the shell, the clock will run slow relative to a clock far outside, by factor √g_tt. Gravitational redshift. Ignore ill-informed assertions elsewhere here claiming e.g. in #6:
    Above quoted is pure misleading bunkum. Other statements there about e.g. 'cancellation' of centripetal accelerations are likewise wrong-headed. Similarly for #7 - especially the last line there is totally in error. If you want a worked example for your specific centre-of-earth query:
  12. Q-reeus Valued Senior Member

    Here's a quick and lazy derivation valid in the slow velocity regime (which will always be the case for peripheral motion in a spinning space capsule).
    Centripetal acceleration at any radius r from rotation axis is just a = -(rω)²/r = -v²/r = -rω² - (1) (negative sign since a points inward)
    As angular speed ω is constant, (1) makes it clear there is a linear rise in a from axis where r = 0 to periphery at r = R. Hence the average a = -1/2v²/R - (2)
    Work done on a unit mass m = 1, slowly lowered from axis to periphery is then from (2), W = average force x distance = ma.R = -1/2v² - (3).
    Which is precisely the magnitude but opposite the sign of the KE gain in mass m moving from axis to periphery.
    The SR formula for time dilation relevant to above circular motion case is just 1/γ = √(1-v²/c²) ≈ √(1-2KE/c²) - (4) (as clock rate)

    Now compare with the usual GR expression for gravitational redshift i.e relative time dilation given at https://en.wikipedia.org/wiki/Gravi...rtant_features_of_gravitational_time_dilation
    t_0/t_f = √(1-2GM/(rc²)) - (5).
    But √(2GM/r) is the escape velocity v_e - that velocity gained by a test mass m falling from far out to radius r of a spherically symmetric mass M. Hence substituting, (5) becomes
    t_0/t_f = √(1-v_e²/c²) - (6)
    Equating v_e with the peripheral v in (2), (3), (4), it's seen there is a numerical correspondence.
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  13. Ultron Registered Senior Member

    I admit that Im not that good in math. What is the implication of this calculation? Does it mean that time dilation in rotating cylinder with 1 g is similar to 1 g on Earth?
  14. Schneibster Registered Member

    As a direct answer to your question, no, the rotating cylinder with 1g is not similar to 1g on Earth. The rocket accelerating at 1g is similar to 1g on Earth.

    The takeaways from the discussion above are:

    1. Centrifugal force, which is what the rotating space module's occupants (and clock) feel is a fictitious force. It is not what is meant in GRT by "net acceleration." You should avoid using it in your test cases because it adds confusion.

    2. Being at the center of the Earth is not "being in a gravity field." At the exact center of the Earth there is no net gravity field (except from the Sun and from the cosmological constant). In fact the gravity field strengthens from zero as you approach the Earth's surface from open space, and then weakens as you descend within the Earth, reaching zero at the center. I can explain the reasons for this if you like; just ask.
  15. Ultron Registered Senior Member

    1. I was reacting to calculation done by Q-reeus, which seems to conclude that the time dilation in rotating cylinder with 1 g calculated in Special relativity is nominally similar to time dilation on Earth with 1 g calculated in General relativity. But Im not sure, if I understand the calculation, maybe it is just my wrong interpretation.

    2. Yea, it is obvious for me, that there is no net gravity in center of Earth, I dont understand why do you think I dont know that.
  16. Schneibster Registered Member

    I cannot comment on any material posted by that individual.

    Your post appeared to me to imply that you were wondering why if someone is sitting in the middle of a gravity field they wouldn't experience time dilation. If you get it, that's fine then.
  17. Q-reeus Valued Senior Member

    No - as per example in #3, there would be equal time dilation, relative to a distant observer in deep space, only in the unrealistic case of a cylinder having that huge radius. Just plug in various specific parameter values for derived expressions in #9 to confirm the general case. Which is that always, it's the peripheral velocity that when used in transverse Doppler expression (that 1/γ = √(1-v²/c²) of (4) in #9), gives the correct time dilation - as reduced clock-rate.
    Centripetal acceleration a (artificial 'g') is only indirectly linked. Two cylinders, A & B, having equal peripheral speeds but with radii R in the ratio 1:2, will yield centripetal a's in the inverse ratio 2:1. The two products 1/2maR yielding energy change in moving from axis to periphery, are equal. As seen from working through #9. Smaller the cylinder, higher the artificial g for an equivalent energy change thus time dilation.
    This hopefully also settles your question 1. in #12. My #8 is correct and the article re link given last part there neatly further refutes nonsense being spouted by you know who. Make no mistake - gravitational redshift of distant stars is purely a function of depressed surface gravitational potential of such star (where final emission to free space occurs). Not surface gravity or especially not spacetime curvature i.e. tidal forces there. It so happens such redshift will be generally higher for a white dwarf than a normal star, and higher again for a neutron star. But the relative changes in redshift are nowhere near in relation to their respective surface gravity - far less in fact. And far less again relative to curvature i.e. tidal gravity there. And were it possible, traveling to the center where g is zero, would be the point of maximum time dilation relative to far outside or even to the surface.

    Just do a search using e.g. 'stellar gravitational redshift'. An example: http://www.einstein-online.info/spotlights/redshift_white_dwarfs
    Last edited: Apr 28, 2016
  18. Schmelzer Valued Senior Member


    Time dilation of a clock at rest in some system of coordinates is defined by $g_{00}$. Which is, for weak fields (like the Earth's) approximately defined by the Newtonian potential by $g_{00} \sim 1+ 2\Phi$, see for example formula (18.15c) of MTW.

    How the Newtonian potential looks inside the Earth one can see, for example, https://en.wikipedia.org/wiki/Gravitational_potential

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    Correspondingly, $g_{00}$ has its lowest value, thus, we have the highest time dilation, at the center. Even if there is no gravitational force at the center at all. The curvature is, instead, some expression depending on second order derivatives of the metric. I'm too lazy to compute this for this metric, but a simple modification makes clear that the time dilation has nothing to do with curvature.

    We simply have to make a large enough hole in the center of the Earth. Then the potential looks like fig. 2 of http://teacher.pas.rochester.edu/PHY235/LectureNotes/Chapter05/Chapter05.htm and is simply constant inside the hole. Thus, the whole metric will be approximately constant too, and curvature zero. But the time dilation will be nonetheless much greater than outside.
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  19. QuarkHead Remedial Math Student Valued Senior Member

    How do you separate the two? As Schmelzer says
    Which is true, and not hard to calculate.

    The consequence is that, if you assume the difference in the metric field at different spacetime points is equivalent to gravitational potential - as you must if you assume that gravitational time dilation refers to the fact that clocks located at different spacetime points cannot agree on tick rate - then in a region of spacetime where the metric field is constant (since the metric gives information about time as well as space), there is no gravitational time dilation, the curvature tensor vanishes and spacetime here is flat. Conversely, if the curvature tensor field in this region is non-zero, then the metric field here cannot be constant and clock rates at different spacetime points in this region cannot be the same.

    I do not see the point of difference between you two. Am I missing something?
  20. Confused2 Registered Senior Member

    If we have a hole at the centre of the Earth and the centre of the Moon (ignoring all rotations) with clocks in both. Both clocks in flat spacetime but not at the same potential.
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  21. Q-reeus Valued Senior Member

    • Passive-aggressive trolling is still trolling; not naming parties just make you more of a problem, not less.
    Exactly. It's hard to see how Quarkhead, last line in #16, could NOT see the obvious difference, given the very basic and clear errors of logic in #6, #7 [edit - not to forget #11, #13 too], pinned down clearly not only in #15 but my own earlier posts making the same and further critical observations. Yet there is no concession of error(s) from that poser - as expected.
    Last edited: Apr 29, 2016
  22. Schmelzer Valued Senior Member

    Assume there are two regions - far away outside and in the hole at the center - and in each of the regions the metric is approximately flat. The potential is, correspondingly, constant inside each of the two regions, but different between the two regions. The curvature is zero inside the regions, but not outside. There will not be a local difference between clocks at rest relative to each other inside the same region. But there will be a large difference between the clocks in different regions. The clock at the center of the Earth will be slower than the clock far away. So, to say it is not time-dilated is wrong. It is not time-dilated only relative to other clocks at the same hole in the center of the Earth.
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  23. Farsight Valued Senior Member

    Good stuff Schmelzer.

    Yes. Imagine you place optical clocks at a variety of locations throughout a circular equatorial slice through the Earth and the surrounding space. Wait a while, then collect you clock readings, then plot them such that higher clock readings appear higher in a 3D cylinder. Your plot ends up looking like this:

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    You are plotting the speed of light, which you typically refer to as gravitational potential. The force of gravity at some location depends on the gradient in the speed of light, which relates to the slope of the plot at that location. The curvature you can see in your plot relates to the tidal force, and to the Riemann curvature tensor. This is the "defining feature" of a gravitational field because without this curvature, your plot is flat and level, the speed of light is constant, and therefore light doesn't curve and your pencil doesn't fall down.

    It isn't. The difference in the metric field at different spacetime points gives you the gravitational gradient.

    No problem with that.


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