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What is the minimum mass required to distort space and time ?

  1. Dec 20, 2009 #1

    SBC

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    Hi

    What is the minimum mass required to distort space and time ?

    According to Einstein's General theory of relativity .
    Mass distorts Space and Time.
    Well Quantum Physics does not satisfy with General theory of relativity.

    So, it asks a Question does G exist in quantum size world ?
    well ,

    My question is that .....

    "What is the minimum mass required to distort space and time ?"

    I mean Does asteroid can distort space and Time ?

    Does big comet can distort space and time ?

    Does astronaut can distort Space and time ?

    I mean What is the minimum MASS limit to distort Space and time ?

    I call it "Einstein space time distortion limit !"
     
  2. jcsd
  3. Dec 20, 2009 #2
    According to GR, all mass distorts spacetime. It is only a question of how much it distorts. There is no "limit".
     
  4. Dec 20, 2009 #3

    SBC

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    a particle would distort too ??
     
  5. Dec 20, 2009 #4
    Yes, but something like an electron makes such little distortion that it is negligable.

    In spherical coordinates, the radial component of the spacetime metric for a point mass is [tex]\frac{1}{1-\frac{2GM}{c^2r}}[/tex] where [tex]G=6.673(10)\cdot 10^{-11} m^{3} kg^{-1} s^{-2}[/tex] is the gravitational constant, [tex]c=2.99792458\cdot 10^{8}m\, s^{-1}[/tex] is the speed of light in vacuum, [tex]M[/tex] is the mass of the particle and [tex]r[/tex] is the distance from the particle. You can see that unless the mass is huge og you are close to the schwartzschild radius of the paricle, there really is no significant curvature.
     
  6. Dec 20, 2009 #5
    If I recall correctly, the r in the Schwarzschild equation is the radius of the particle. Or am misremembering things? Otherwise, solid explanation.
     
  7. Dec 20, 2009 #6
    The r in the Scwartzschild equation is definately not the radius of the particle. The equation is only valid outside the mass, and if the radius is smaller then the Scwartzschild radius, the equation only holds outside the Scwartzschild radius. There are other coordinates which describe the interior of the black hole, and all of them (as far as I know) break down at the center (r=0).
     
  8. Dec 20, 2009 #7
    Ack, you're most definitely correct. Sorry, carry on :)
     
  9. Dec 20, 2009 #8

    Nabeshin

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    Another way of thinking about it is that Einstein's bending of spacetime geometry is simply a different way of thinking about what we have classically thought of as the gravitational force. If you were to ask what is the smallest particle which interacts gravitationally, the answer would surely be "no matter how small the mass, all massive particles interact gravitationally."
     
  10. Dec 20, 2009 #9
    But there is a difference between "interacting with gravity (passive)" and "producing a gravitational field (active)". Laboratory experiments have demonstrated that leptons have passive gravitational mass. But there is no conclusive experimental evidence to show that leptons produce a gravitational field (bend space time). This is an assumption based on theory.
     
  11. Dec 20, 2009 #10
    If we are talking about single leptons, I think a theory of quantum gravity is neccesary. Quantum effects would certainly be a factor in measurements of the gravitational attraction between single leptons.
     
  12. Dec 20, 2009 #11

    Nabeshin

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    Interesting.

    My first thought is, probably quite naturally, to write it off as the fact that a lepton produces a terribly weak gravitational field which is extremely difficult to detect.

    If indeed a lepton did not produce a gravitational field, I have a feeling that would throw a wrench in something in GR but I can't quite put my finger on it. Equivalence principle, perhaps? Not sure. At any rate, thanks for the clarification!
     
  13. Dec 25, 2009 #12

    Vanadium 50

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    This thread has drifted into speculation. The answer was given in Post #2.
     
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