Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Some confusing things

  1. Feb 4, 2004 #1
    There is a couple of questions that have arisen after i read about the general theory of relativity.

    1. Should it be possible in principle for a photon to circle a star?

    2. Do binary stars rotating around a center of mass radiate gravitation waves?

    3. How are gravitational waves emitted?

    4. The twin trip example of the special theory of relativity seems to contradict the general theory of relativity. According to the second stated, a clock near a massive star ticks slower than a clock on a orbiting planet. But the twin trip states that a guy in a spaceship ages more slowly when leaving earth, hence stating that the clock ticks slower than the correspondin one on earth. Isn't gravity stonger on earth?
     
    Last edited: Feb 4, 2004
  2. jcsd
  3. Feb 4, 2004 #2

    NateTG

    User Avatar
    Science Advisor
    Homework Helper

    1. Yes, it should be possible for photons to be in orbits.

    4. First off, the twin paradox has to do with general relativity, not special relativity. One of the fundemental notions of General Relativity is that accelleration due to gravity is the same as acceleration due to other forces. One result is that time passes more slowly in more accelerated reference frames, so if the twin that goes into space does space travel with an acceleration of less than 1g then the twin that was left behind would be younger at the reunification.
     
  4. Feb 4, 2004 #3

    chroot

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Not a star, but a black hole. Light can orbit a black hole at a distance of

    [tex]\frac{3 G M}{c^2}[/tex]
    Yes. Any oscillating mass will radiate gravitational waves.
    Gravitational waves are just "ripples" in the curvature of space that propagate away from the source of the ripples. A spherical mass sitting still produces a static, symmetric curvature around it. If you suddenly move it a bit to the left, the curvature gets deformed by the motion (since gravitational waves do not propagate infinitely fast). If you move the mass back to the right, the deformation is in the opposite direction. If you wave the mass back and forth in steady motion, the deformations continue to propagate away from the mass in a regular pattern. This is same way that electromagnetic radiation is created, by waving a charge.
    In that context, the gravitational time dilation due to the earth is considered negligible. You are correct, though, that it is should be included in a complete calculation.

    - Warren
     
  5. Feb 4, 2004 #4
    Re: Re: Some confusing things



    Why do you say this? The event horizon is at

    [tex]\frac{2 G M }{c^2}[/tex]

    If the mass is all inside the photon sphere I don't know if that means that you are always talking about a black hole.

    dhris
     
  6. Feb 4, 2004 #5
    how can a photon sphere have a mass? Isnt light massless?
     
  7. Feb 4, 2004 #6

    chroot

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Re: Re: Re: Some confusing things

    Correct. Light does not orbit at the event horizon, however -- it orbits at a radius of 1.5 times the Schwarzschild radius, which is what I posted.

    - Warren
     
  8. Feb 4, 2004 #7
    Re: Re: Re: Re: Some confusing things

    I know. But you also said that it only happens for a black hole. My point is that since the light is not orbiting at the event horizon, why would it matter if the mass below the photon sphere has collapsed below the Schwarzschild radius?

    dhris
     
  9. Feb 4, 2004 #8

    chroot

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

  10. Feb 4, 2004 #9

    wolram

    User Avatar
    Gold Member

    may i with respect to other posters point out that gravitational
    radiation has not been found to date, and is hypothetical
    new detector might find it in a few years, but what if they
    do not?
     
  11. Feb 4, 2004 #10
  12. Feb 4, 2004 #11

    chroot

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The only mass that would affect the light is the interior mass. You are correct that, theoretically, the object would not have to be a black hole -- it would simply have to have all its mass within 3GM/c^2. This would be impossible to do in practice, however, without creating a black hole, as no normal form of matter is capable of being stable when compressed that much. Even a neutron star is a far cry from having all its mass within 3GM/c^2.

    - Warren
     
  13. Feb 4, 2004 #12
    That's my point. What did that google link have to do with it?
     
  14. Feb 4, 2004 #13

    chroot

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    It was just a calculation to show you the Schwarzshild radius of an object with the mass of the Earth -- it's centimeters. If you compressed the mass of the Earth into a volume centimeters in radius, you'd have a black hole.

    - Warren
     
  15. Feb 4, 2004 #14

    NateTG

    User Avatar
    Science Advisor
    Homework Helper

    Right, but if you compresed the Earth into a sphere 1.4 thimes that size, it need not collapse, but could still have a photon orbit.
     
  16. Feb 4, 2004 #15

    Janus

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    It does not matter how much G-force the astronaut undergoes when it comes to whether or not he will have aged less or more than his brother, as long as he attains a high enough relative velocity and maintains it long enough. If he attains .866c by accelerating at .5g he will still age less than his brother, even though his brother experienced 1g during this period.
     
  17. Feb 4, 2004 #16

    chroot

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Well, I see your point. If you put the mass of the earth within a sphere of radius 1.33 cm, it will certainly collapse to a black hole rather quickly. But, for a few fleeting moments, when the mass is within 3GM/c^2 but not within 2GM/c^2, you'd have a photon orbit without a black hole.

    Of course, that state would last a very very short time! If the mass of the earth were falling from 3GM/c^2 to 2GM/c^2 on a free-fall timescale, from rest, it would take only 30 milliseconds to become a black hole.

    In steady state, of course, you can't have a photon orbit around anything except a black hole, because anything that dense will become a black hole very quickly.

    - Warren
     
  18. Feb 5, 2004 #17
    For a naive calculation suppose you assume that the largest density that can exist is that of a neutron star where rho ~ 3x1017kg/m3. Set R = 2.5GM/c2. Then R ~ 3 million kilometers. It seems reasonable to assume that crushing the Earth to within the photonsphere will force it to collapse to a black hole. Although this was a rough calculation. It left out the contribution of gravitational forces due to pressure.
     
  19. Feb 5, 2004 #18

    chroot

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Arcon,

    I don't know what you're talking about. A typical neutron star is about the mass of the earth and is about 7 km in radius. The photonsphere for an object with the mass of the earth is 1.33 centimeters. I have no idea where you got 3 million kilometers from.

    - Warren
     
  20. Feb 5, 2004 #19
    Neutron stars do not have a mass ever close to that of the Earth. A neutron star is what is left after a high mass star goes supernova and has a mass of about the same as our sun.

    Note that I didn't say that the calculation was for a neutron star. I said assume that the maximum density was that of a neutron star and then set R = 2.5GM/c2. That means assume that you have a spherical object whose mass density is that of the center of the nucleus of an atom and assume that the radius allowed for a photonsphere. What is that radius? Again - assume a naive calculation by leaving out pressure - just to give a rouch idea of the size of such an object.

    The density of the nucleus = density of a neutron star = 3x1017kg/m3.

    [tex]M = \rho V = \rho (\frac {4}{3}\pi R^{3})[/tex]

    Substitute that into R = 2.5GM/c2 and solve for R. You'll get about 2.2x109 meters.

    The whole point was to illustrate that such an object is unlikely to exist.
     
  21. Feb 5, 2004 #20

    chroot

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    If you're asking:

    What would be the radius of an object, with density 3 * 10^17 kg/m^3, whose photonsphere is right at its edge?

    The answer is 18,901 meters.

    I still have no idea where you're getting 2.2 * 10^9 meters, and I have no idea why you're using a factor 2.5 instead of 3.

    - Warren
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?