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General formula for space-time distortion?

  1. Sep 2, 2007 #1
    Ok let me begin by saying I'm far from an expert on this subject... but I'm doing some personal research and have a question. Please answer without being TOO technical xD (I can do some calculus and somewhat advanced math but not like insanely complicated math.)

    Anyways... does anyone know any general formulas for the distortion of time from a certain amount of mass density and the resulting gravity potential well? For instance I heard that in one year, two atomic clocks - with one clock being one mile above the other - results in 5 milliseconds in distortion. The gravitational potential in this case would be something like...

    (2000m*10kg*9.81m/s^2) = 196200J = 5ms distortion / year. I don't even know what an atomic clock weighs or even if what I did is kosher math but that's the general idea.

    So can anyone help me on how I would be able to do calculations on space-time distortions on a larger scale [ie, outer space]? Thanks
     
    Last edited: Sep 2, 2007
  2. jcsd
  3. Sep 2, 2007 #2
    I wouldn't call it "distortion of time"..., but it is true that clocks at high altitude go faster than clocks at lower altitude (i.e., lower gravitational potential). The rate of the clock at elevation h is [itex] 1 + gh/c^2 [/itex] times faster than at the sea level. In the case h=2000m, this factor is about 1 + [itex]2 \cdot 10^{-13} [/itex], which translates into roughly 5us (5 microsecond) per year.

    Eugene.
     
  4. Sep 3, 2007 #3

    pervect

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    See for instance http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/gratim.html

    To rephrase it slightly

    [tex]T_0 / T = \sqrt{1 - \frac{2GM}{Rc^2}} [/tex]

    [tex]T_0[/itex] is the time interval on the close clock, and T is the time interval on a reference clock "at infinity". R is a measure of the location of the "near clock".

    This basically follows from the Schwarzschild metric. See for instance

    http://en.wikipedia.org/w/index.php?title=Schwarzschild_metric&oldid=151657557

    For weak fields, you can take R to be the distance that the clock is away from the center of the massive body. For strong fields, it's better to think of R as the Schwarzschild coordinate, which is defined by the fact that a circle of constant radius R has a circumference of 2 pi R, and a sphere of radius R has a surface area of pi R^2, as was mentioned in another thread.

    Note that meopemuk gives you an approximate formula. It's OK for the Earth, but if you try to apply it to an extreme enough astronomical situation, it may fail.
     
  5. Sep 3, 2007 #4
    The formula that I gave is an approximation of the order [itex]c^{-2}[/itex]. It can be derived from Newtonian gravity and basic quantum mechanics without any involvement of general relativity.

    In QM we know that any non-stationary process (e.g., a clock) involves two or more stationary energy levels. In the simplest case, we can consider a two level system, where [itex]E_0 [/itex] and [itex]E_1 [/itex] are energies of the ground and excited levels respectively; [itex]m_0 = E_0/c^2 [/itex] and [itex]m^1 =E_1/c^2 [/itex] are their masses, and the characteristic frequency of the isolated clock is

    [tex] f = \frac{2 \pi}{\hbar}(E_1 - E_0) [/tex]

    Let us now consider the same clock in the Earth gravitational field. The total energies of this system in the ground and excited states, respectively, are

    [tex] E_0' = Mc^2 + m_0c^2 - \frac{GMm_0}{R} [/tex]
    [tex] E_1' = Mc^2 + m_1c^2 - \frac{GMm_1}{R} [/tex]

    Then the clock's frequency in the gravitational field is

    [tex] f' = \frac{2 \pi}{\hbar}(E_1' - E_0') = f(1 - \frac{GM}{Rc^2}) [/tex]

    For two clocks with the height difference of H near the Earth surface, the difference of their [itex] \frac{GM}{R} [/itex] factors can be approximated by [itex] gH [/itex], so the gravitational time dilation factor becomes [itex] 1 - \frac{gH}{c^2} [/itex].

    Eugene.
     
  6. Aug 6, 2010 #5
    There are numerous paradoxes in Special Relativity, you will need to study Minkowski's
    theory to be able to reconcile them all. Im not sure but I think that General Relativity
    followed on from Minkowski's theory. Until these theories were developed no one could
    make sense of SR, including I suspect 'the great man himself' who developed the theory.
    The problem was that he could not find a 'Simple Universal' solution for every 'Relitavistic
    System' occurring in space time problems. Fortunately Minkowski's theory is not too difficult
    to understand, though I have not read it through myself. I am trying to find my own solution to the paradoxes invoked by SR, but it would be wise to follow my advice and do as I say,
    and not as I do.
     
  7. Aug 6, 2010 #6

    bcrowell

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    This is a completely incorrect description of both the history and the physics.
     
  8. Aug 8, 2010 #7
    Thank you for your input bcrowell, but could you please be a little more specific, I am very interested and keen to learn. I think that you may find that Special Relativity is incomplete
    without Minkowskis "4 Vector Theory".
    The Relativistic Formula "y=1/((1-(v^2/c^2))^1/2" seems insatisfactory to me, in so far as it
    is not able to reconcile the paradoxes invoked by the basic theory.
    If this formula is applied to more complex sytems with 3 velocity components or more it
    reveals inconsistencies in the basic theory. It is much more likely that all four of the
    relativistic quantities, time dilation, lorentz contraction, momentum, and simultaniety work
    together in a very specific way.
     
    Last edited: Aug 8, 2010
  9. Aug 8, 2010 #8
    Hi How come the surface of the sphere is pi r2 and not 4 pi r2 ????
    Thanks
     
  10. Aug 8, 2010 #9
    can be approximated by [itex] gH [/itex], so the gravitational time dilation factor becomes [itex] 1 - \frac{gH}{c^2} [/itex].
    I am trying to understand Rindler so is the g in this notation the acceleration??
    Is this applicable to Rindler observers and coordinates??

    Thanks
     
  11. Aug 8, 2010 #10

    Mentz114

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    What paradoxes ?

    How can you know it's not too difficult to understand if you haven't read it through ?

    Don't believe you. Why haven't you done it then ?
     
  12. Aug 8, 2010 #11

    OK Mentz
    Seems that whatever I say you are just simply going to contradict, however I will qualify one of my statements a little more for you; when I said that "Minkowskis 4 Vector Theory was
    not too difficult to understand" by that I meant that it did not involve any advanced maths.
    I had prior instruction to that effect before examining the theory in any detail.
    I hope that clears up that specific point for you. You were correct to make that observation.

    However if you were not aware of paradoxes in SR then it would
    seem logical to me that you would never have heard of Minkowski anyhow.

    As for your last point, it is both unqualified and irrelevant.
     
  13. Aug 8, 2010 #12

    Mentz114

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    This is just hand-waving. What are you talking about ? Name one of these paradoxes.
     
  14. Aug 11, 2010 #13
    I dont think anyone ever listed them by name, the only parodox that was named was "The Twin Parodox" but i dont think that was a parodox, it was too simple and was only based
    on 2 inertial frames with just 1 of those frames changing to a different frame. SR is really
    only of any use in non changing inertial frames. Where non inertial i.e 'accelerated frames'
    are used the system breaks down completely when there are more than 3 reference frames.
    If you dont believe me try doing some thought experiments with the help of a few diagrams
    im sure you will see my point. You will get contradictory results. I have also noticed that the
    Lorentz Contraction cannot be used arbitrary in place of Time Dilation as is commonly thought, i can explain this to you if you wish.
     
  15. Aug 11, 2010 #14

    Mentz114

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    Nonsense. SR is internally consistent, there aren't any paradoxes or mathematical contradictions.

    Don't bother.
     
  16. Aug 13, 2010 #15
    Your problem, not mine. As one day you will find out. END OF DEBATE.
     
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