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Gravitational time dilatation

  1. Oct 1, 2012 #1
    Here http://curious.astro.cornell.edu/question.php?number=278 [Broken] I see the following:

    Do I understand right that mentioning slower time passage at the bottom of the tower here is slightly inappropriate? Red-shifting of the light here is totally accounted for by photons loosing some energy crawling up against gravity (or space near the bottom of the tower being stretched by gravity).

    Gravitational time dilatation would be visible if you blinked light up the tower exactly each second and the observer at the top would register blinks slightly more spaced out. Am I right?
     
    Last edited by a moderator: May 6, 2017
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  3. Oct 1, 2012 #2

    Bill_K

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    No. Redshift in the wavelength of light and time dilatation measured by the space between blinks both occur. They are two aspects of the same phenomenon.
     
  4. Oct 1, 2012 #3
    Both occur, I agree, but they are two separate different aspects of the same phenomenon (that space-time is curved by mass).

    Slowing down of time has nothing to do with red-shift, I believe, since photons are not emitted by oscillator that is slowed down but are just portions of energy that atoms wants to get rid of and they are emitted instantly.
     
  5. Oct 1, 2012 #4

    tom.stoer

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    The gravitational redshift can be described via a "clock" defining a frequency f carried with the particle, so there is no need to refer to energy E = hf. It should be possible to detect interference phenomena which test the frequency directly.

    Energy E=hf is a derived quantity in this classical = non-quantum (!) context.
     
  6. Oct 1, 2012 #5

    PAllen

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    Classically, light is emitted by dipole oscillation. And QED must match this classical behavior. Thus, frequency of light must behave the same as a sequence of signals.
     
  7. Oct 1, 2012 #6

    Bill_K

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    Yes they are. To an observer living down in the gravitational well, of course, the atom looks and acts perfectly normal. The Rydberg is the same, and Planck's constant is the same. But g00 is not the same. You don't notice the difference until you compare the elapse of time on Earth and out in space. And you get the same difference no matter how you do it - measuring the frequency of light that has come up from the surface, or counting the beeps of a time clock - it's the same effect. Suppose each beep was sent from Earth to coincide with a trillion cycles of the light. That same relationship will hold in space.

    The idea that the light "loses energy" as it rises makes it sound incorrectly as if there's something physical happening to the light, but the space-time geometry affects everything - both light frequency and time intervals.
     
  8. Oct 1, 2012 #7

    tom.stoer

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    Einstein always tried to keep quantum mechanics and relativity seperate research areas; afaik there's not one single statement which relates both theories. Therefore E=hf is not required in any reasoning in the context of relativity.
     
  9. Oct 1, 2012 #8

    Bill_K

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    E is the 4th component of the energy-momentum 4-vector Pμ. And f is the 4th component of the frequency 4-vector kμ. They are both parallel-transported along null geodesics. In a spacetime with a Killing vector ξμ, the constants ξμPμ and ξμkμ evaluated at different points along a light ray express energy loss and redshift respectively. One may define h to be the ratio between these two constants.

    Well if you insist, one might imagine a nonquantum world in which the value of h was different on different geodesics. :uhh:
     
  10. Oct 1, 2012 #9
    What about a free faller from infinity nearing an event horizon? There is no stress per se , would this exempt him from the extreme dilation occurring within a momentarily passing static observer's clock?
    Specifically, what would be the relative periodicities of the freefalling clock and one at infinity??
     
  11. Oct 1, 2012 #10

    pervect

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    You are probably imagining the static observer being a material observer, one who can carry a clock and have a frame of reference. In formal language, you're imagining that the static observer is a time like observer.

    This viewpoint is acceptable anywhere outside the event horizon, though the "static" observer has to accelerate heavily to "maintain station", his proper acceleration to hold station approaches infinity as he approaches the event horizon.

    This viewpoint of the static observer is not sensible at the event horizon, because the event horizon is the surface traced out by a beam of light. Formally, we say that the vent horizon is a null surface. If you've red the FAQ's, you know that there is no such thing as a "frame of reference" for an observer moving at the speed of light, and that no material observer can move that fast. So there is no "static observer" at the event horizon.

    As far as what the free-faller from infinity observers. The energy-at-infinity of such an observer is zero if his velocity at infinity is zero. Two quantities might be of interest - the red shift from a signal emitted at infinity, and the red shfit from a co-located static observer.

    The observed red shift from infinity turns out to be a finite factor of 2:1 (from memory, I'd have to dig up the detailed thread where this was derived - but following the calculation would require enough familiarity with GR to do actual computations with it so I won't bother digging it up if it's a moot issue or not of interest).

    The observed red-shfit from a co-located static observer approaches infinity as one approaches the event horizon.

    Thinking in terms of "time dilation", however, is going to be totally fruitless if you do not also know that, because of special relativity, simultaneity is relative. This goes as far back as the twin paradox in special relativity. A can think B's clock is running slow, while B thinks that A's clock is running slow. Time dilation depends on the observer, there is no observer-independent significance to the notion of time dilation!

    The Schwarzschild coordinates, represent the viewpoints of static observers. There are other coordinates (such as Painleve coordinates) that are better suited to express the viewpoint of infalling observers. As an added bonus, they don't suffer from some of the pathologies that the Schwarzchild coordinates have at the event horizon.
     
  12. Oct 1, 2012 #11

    PAllen

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    There is also the following interesting comparison, showing asymmetries. Consider three clocks:

    C1: at stationary 'near infinity' in SC spacetime
    C2: stationary near horizon in SC spacetime
    C3: free fall from infinity near horizon, adjacent to C2

    C1: sees C2 extremely slow and C3 even slower
    C2: sees C1 extremely fast and C3 extremely slow.
    C3: C1 at a reasonably normal speed, C2 extremely slow.

    Note the key asymmetry: C1 sees C3 slowing and 'stopping', while C3 sees C1 moving normally, at a normal time rate. C3 will see a specific, finite, time on C1 when C3 reaches the singularity.

    This is in response to Austin0:

    "Specifically, what would be the relative periodicities of the freefalling clock and one at infinity??:"

    The question is ill formed. The answer depends on which one you ask. Time dilation is not a field, it is a coordinate feature. Observed time rate = doppler, is a function of source world line, target world line, and intervening geometry, and is not generally symmetric in GR.
     
  13. Oct 1, 2012 #12
    1---- When you talk about 0 velocity at infinity are you referring to the reduction in the coordinate speed of light relative to an observer at infinity ,dropping to zero at the horizon??
    Are you saying that approaching the horizon the freefallers velocity would also drop to approaching zero relative to infinity? I had wondered about this.

    2----As far as time dilation regarding static locations in gravity this seems to be a real evaluation. It is a frame independent observable comparable to acceleration , which we consider real by this same criteria of frame agreement. Additionally dilation in gravity can be observed simply by relocating identical clocks with no coordinate system at all , yes??

    in another thread it seemed that it was assumed that the geometry did effect the falling observer. That at any point the dilation factor for his clock was a combination of velocity dilation AND a dilation factor equal to the factor affecting any momentarily co-located static clock. WRT infinity.
    Now i am somewhat confused about this issue.
    You may be right about Schwarzchild coordinates but I want to try and get a handle on their workings and implications. It appears that everybody accepts that they accurately describe the physics right up to the horizon,
     
  14. Oct 1, 2012 #13

    PAllen

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    Any coordinates may be used to accurately describe physics - as long as you don't interpret coordinate quantities as physics. To compute physics, you need to compute metric invariants, which will come out the same in all coordinates.
     
  15. Oct 1, 2012 #14

    PAllen

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    "2 Time dilation depends on the observer, there is no observer-independent significance to the notion of time dilation!"

    This statement from Pervect is key: it is what I was trying get across with the specific clock comparisons I proposed.
     
  16. Oct 2, 2012 #15

    tom.stoer

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    All what you are saying is correct, but you don't need the relation E=hf in order to derive the gravitational redshift; that's what I am saying.
     
  17. Oct 2, 2012 #16
    That site is a little inaccurate as it's not the field strength but potential that matters. However, the mentioning of slower time passage at the bottom of the tower is perfectly appropriate. As a matter of fact, Einstein based his redshift prediction on that assumption. And photons loosing energy doesn't accomplish spacing out of blinks of light.

    This was discussed on this forum incl. reference last year:
    https://www.physicsforums.com/showthread.php?t=476946
    and also very recently here:
    https://www.physicsforums.com/showthread.php?t=634870
     
    Last edited by a moderator: May 6, 2017
  18. Oct 2, 2012 #17

    pervect

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    No, I was talking about the energy of the infalling body. Consider the Newtonian case, where total energy E is the gravitational potential energy -Gm/r plus the kinetic energy .5 m v^2. The total energy E = -Gm/R + .5 m v^2 is a constant of motion as the body falls in.

    If the velocity at r=infinity is zero, the total energy is zero. IF the body had a nonzero velocity at infinity, it would have a positive energy. (The velocity would have to be inwards directed for the orbit to apporach the central massive body). If the body had zero velocity at some finite r, less than infinity, E would be negative.

    The above equations were Newtonian, but there is also in the GR in the special case of a static central gravitating body a conserved orbital energy., the formulae are slightly different fromt he Newtonian counterparts, however. There is some detailed discussion online at https://www.fourmilab.ch/gravitation/orbits/, but I think it would be worth a different thread if you want to inquire as to more details about this.

    So in summary, my statement wasn't about what you said at all, and rather a statement about the energy of the infalling body to specify the orbit more exactly.

    No.

    I generally prefer to specify velocities relative to another co-located object, because that's the only coordinate independent way to do it. When you can specify things in a coordinate independent manner, it both provides more physical insight, and you also don't have to go through the precision of specifying exactly what coordinate system your remark is true in, making the exposition clearer.

    Because the concept of releative velocity in GR is coordinate dependent (which is unlike the situation in SR), I would specify not the "freefaler velocity relative to infinty", as you described, but something that makes the choice of coordinates clear, for instance "dr/dt in the Schwarzschild coordinate system for an infalling body approaches zero as r approaches the Schwarzschild radius".

    It is unfortunately possible and common to ascribe too much physical significance to statements such as the above. However, they seem to be related to your question, so I'll give you the answers, and hope you won't ignore the cautions about their coordinate dependence.

    Time dilation regarding static observers is perfectly unambiguious as long as your observers are static. It's when you start talking about observers that aren't static (such as infalling observers), and continue to apply the concepts of time dilation that you used for static observers to the non-static ones, that you run the very real risk of confusion.

    For instance, if one regards time dilation as "real" in special relativity, the next step is usually a question about the twin paradox. The same issues apply in GR, except that they are much more complicated than they are in SR. Thus it is useful , perhaps even essential, to understand the SR issues first. Unfortunately, sometimes people may ask questions about GR when they don't know SR well, we have no control over when they choose to ask the question or what background they have when they ask their quesiton. Most of the time we have to guess what their background is, as they don't want to reveal it ...

    If you take the remarks as correct for one coordinate system, but only applicable only to that one particular coordinate system, I hope at least some of the confusion will dissappear.

    For instance a different choice of coordinate systems might have a totally different value of time dilation, given that time dilation is the ratio of proper time to coordinate time.

    Schwarzschild coordinates are a poor approach to understand the horizon, because they become singular there. They are a good way to understand things outside the horizon, though, for instance the equations of motion are most easily derived in Schwazschild coordinates. So if you don't have any horizon crossing bodies, Schwarzschild coordinates are just fine and the easiest to work with (which doesn't mean imply that it's not rather involved). If you are concerned about horizon crossing bodies, then Schwarzschild coordinates are not the best approach.
     
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