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My second shallow question - about black hole¬

  1. Oct 26, 2004 #1
    Dear all,

    I am writing because I found a contradiction between the universial gravitational law and black hole.

    We all know that the universial gravitational law is: [tex]F=G\frac{m1m2}{r^2}[/tex]

    But because the mass of a photon is zero, so the gravitational force acted on a photo should be zero as well, and that leads to the obvious result: light could escapa from black hole whatever the black hole mass is.

    I understand that my hypothesis should be 99.999999% wrong, because unless that means the Lucasian professor of Cambridge is an idiot¬
    ... Finally, by the way, is this related to the general relativity, because I guess for explaining this sort of large scale problem, we should account on it...

    Plz give me a hand¬ :bugeye: :uhh: :rofl:
    Last edited: Oct 26, 2004
  2. jcsd
  3. Oct 26, 2004 #2


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    Well 'ThomasJoe40' you are right and you are wrong! You are right to ask "is this related to the general relativity", but wrong to assume that gravity does not act on a photon.

    There is a 'hand waving' explanation and the proper one!
    The hand waving explanation is to point out that, according to Special Relativity, even though photons do not have rest mass, they do have energy and their energy has a mass equivalent [tex]E=m{c^2}[/tex]. So gravity acts on their energy and thus prevents them from escaping from a Black Hole.

    Actually black holes were thought about before special or general relativity was developed because it was known that light had a finite velocity. As the velocity of escape of an object depends on its mass and its radius, it was known that a sufficiently massive and compact object would have an escape velocity greater than the speed of light. Light would not be able to escape from such an object and hence the object would be 'black'.

    The proper explanation however is the GR one, which depends on its description of gravitation not as a Newtonian force but as the effect of freely falling objects travelling or 'falling' on straight lines, known as geodesics, along a curved hypersurface of four dimensional spacetime. Spacetime is curved by the presence of mass and energy. If the gravitating mass becomes massive and compact enough certain straight lines or geodesics of even an outward-bound photon are curved so much that they 'bend back on themselves' and the photon cannot leave the mass, it is a black hole.

    There is a further complication in that unless the mass of the black hole is very large, much larger than a solar mass, then the density required is so great, and the gravitational attraction so large that no known physical force can withstand it. The matter forming the gravitating mass then collapses down into zero volume. It that case the black hole consists of an event horizon at which escape velocity equals light velocity, and a singularity at the centre.
    - Garth
    Last edited: Oct 26, 2004
  4. Oct 26, 2004 #3


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    Its not a knock on Newton to say he didn't know everything: he did, after all, do his work several hundred years ago. Today, we know his theory is incomplete. In fact, the observation of gravity affecting light is one of the reaons we know it.
  5. Oct 26, 2004 #4


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    I would think that it would be clearer just to say that a large mass (black hole) bends the frame of spacetime and the massless photon is just following a normal straight line that is curved back in on itself. Gravity isn't "attracting" or "grabbing" the photon(s), gravity is just determining the shape of the local paths. (?)

    As far as photons having mass, see:
  6. Oct 27, 2004 #5


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    Well said, lab. The photon is just an innocent bystander. It takes the path of least resistance. Think of water flowing through a flexible tube. Bend the tube and the water follows the curvature.
  7. Oct 27, 2004 #6
    I don't agree with the idea that Newton would have thought that light has no mass and is therefore unaffected by gravity. Galileo demonstrated that all bodies fall with the same acceleration, whatever their mass. It seems perfectly reasonable to include zero mass within 'whatever their mass', so zero mass particles should be affected by a gravitational field just the same.

    This was known when Eddington' observed the bending of light during a solar eclipse. The important thing is that GR predicts twice the effect that Newtonian theory would, and the results agreed with GR. (Roughly, this is because GR is about the curvature of spacetime rather than just space)
  8. Oct 27, 2004 #7
    Thanks, thanks so much guys, all of you owns a greater deal of knowledges than I had... so first thing to say is my hypothesis is 1000% wrong, and the Lucasian Professor of Cambridge is still a genius...

    Chrono, I do not quite agree with what you said after viewed other people's opinions about GR...

    Well, [tex]G=ma[/tex], the mass of a photon is [tex]m=\frac{E}{c^2}[/tex] which is very small, so the gravitational pull on it is [tex]f=\frac{GMearthMphoton}{r^2}[/tex]. To calculate the gravitational acceleration of a photon is by [tex]g=\frac{f}{Mphoton}[/tex], substitute the universal gravitational formula in, [tex]\frac{GMearth}{r^2}[/tex], which is a constant - app. [tex]9.81\frac{m}{s^2}[/tex]

    Sorry, Chrono, you are right, brilliant¬¬ :rofl: :rofl: :rofl:

    But what I want to ask is, will this truly happen - every photon falls towards the Earth at this acceleration??? :grumpy: :uhh:
  9. Oct 27, 2004 #8


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    ThomasJoe40 - are you mixing up a Newtonian and a GR way of deriving the photon's acceleration?

    In GR four dimensional spacetime is considered to be like a 'sheet' in a higher dimension, this is a visual not a rigourous way of looking at it, and the presence of a mass bends that sheet. Imagine a bowling ball placed at the centre of a rubber sheet. There will be a dip in the middle. Photons and freely falling particles travel along staight lines along (actually in) that curved surface. Their paths 'bend' because space-time is 'bent'. The effect of this when observed from a supported (not free falling)observer is that all objects fall to the ground at the same rate. However in GR there is no gravitational force as there is in Newtonian gravity, instead the effect of a force is explained as the effect of the curvature of space-time.

    Your question,
    can be answered in three ways.
    1. According to the principles of GR photons will 'fall at the same rate' as all freely falling particles.
    2. If we go outside the confines of GR for a moment there may be forces other than gravitational, i.e. curvature, which acts on freely falling bodies and photons differently. In this case photons will not fall at this same rate, but this view would be very contentious.
    3. The answer to this question is to be determined by an experiment. This specific experiment, asking the question "Do photons 'fall at the same rate' as particles?" is yet to be done.

    Last edited: Oct 27, 2004
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