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Schwarzschild Radius, Constantcy of Light speed, and Time Dialation

  1. Jul 9, 2010 #1
    First off, I'd like to point out that I am by no means an expert in this area, and I am only doing some casual research as a personal interest topic, and have some further unanswered questions that I'm unable to find reasonable answers for. These questions are all inter-related, so I'll post them together.

    The following assumptions immediately below may be flawed, or simply incorrect, so please correct the shortcomings in my understanding if you know better:

    1. From what I can interpret from the Schwarzchild radius, the event horizon of a black hole is effectively the proximity to said black hole where the exterior cumulative acceleration from to the force of gravity will generate a net change in velocity equivalent equal to the speed of light

    IE: the Schwarzchild radius is such that any particle traveling from free fall towards the black hole in question reaches the speed of light at this point, with respect to the frame of reference of a distant observer.

    IE: the total depth of the gravity well outside the Schwarzchild radius imparts a velocity of c to any distant object.

    2. The speed of light is a constant. Photons must always travel locally at the speed of light, regardless of any other forces on them.

    IE: a photon generated slightly above the surface of an event horizon of a Black Hole traveling away from the Black Hole will travel outwards at the speed of light, and if unobstructed, eventually escape the Black Holes gravitational influence, and continue traveling at speed c. (other effects such as frequency shift and energy loss could make it possible to characterize such a photon)

    3. Gravitational Time Dilation causes time to slow down, with respect to a distant observer, proportional to the speed of light.

    IE: as an observer approaches the speed of light, time slows down in comparison to a distant observers reference frame.

    ____________

    Given the above assumptions, here are the questions:

    1. I don't understand the mechanism by which this photon can not continuously (albeit slowly, and with much red-shift) continue to egress the Black-hole. If a photons velocity is constant at c, how is it possible that a black hole can effect its velocity, short of Gravitational Time Dilation, or Frame dragging? It is understandable to me that any photon not attempting to exit the black-hole perfectly normal to its surface could be tangentially bent back to the surface of the black hole, but short of allowing the speed of a photon to be locally effected, or invoking frame-dragging effects, shouldn't such a photon be able to escape from even the inside of the Schwarzchild radius?

    2. I don't understand how a singularity can occupy an infinitesimally small space. Would not the very same mechanism which causes the acceleration of objects towards the black-hole, and the frame-dragging effect, and the inhibition of any egress of mass or energy also cause extreme time dilation with the net effect that objects in the gravity well of the black-hole reach a speed of exactly 100% of c, thereby causing no passage of time locally? If so, would this mean that the singularity encapsulates the entire volume of the minimal state of the gravity well, and that the volume was finite (and potentially equally in size to the absolute horizon)?

    Thanks for your time,

    OverLOAD
     
  2. jcsd
  3. Jul 9, 2010 #2
    Great questions.
    Your assumptions:
    2 and 3 are about right.
    1) The event horizon is the location at which NO resistive force (e.g. rocket engine etc) can resist the pull of gravity. Remember that event horizons pop-op in general relativity, so thats the way to think about them. At the event horizon, space-time becomes curved to such an extent that there is NO SUCH THING as a path leading outward, only paths leading inward (towards the singularity).
    The schwarzschild radius is the point at which a stable ORBIT would have to be at the speed of light (and that only applies to a non-spinning [non-kerr] black-hole).

    Questions:
    1) Remember that acceleration is a change in velocity, not speed--so photons can be accelerated without changing their speed (thats the only way to accelerate them). Also, remember that gravity isn't JUST some force with some magnitude; inside the event-horizon, the bending of space-time REQUIRES any object (photon or otherwise) to travel inward.
    2) We don't know whats going on in/near the 'singularity'--thats why its called a 'singularity', the math that we generally use just stop working. According to some theories (e.g. string theory) the 'singularity' would indeed occupy a finite (incredibly small--planck scale) volume.
    The key to General relativity is the "Equivalence Principle," which means that all of the time-dilation effects/etc are AS VIEWED FROM AN OUTSIDE OBSERVER. An observer traveling into a black-hole sees nothing interesting (for the most part at least) as they pass the event horizon, they perceive time as proceeding normally, continually falling towards the center (until they're destroyed).
     
  4. Jul 9, 2010 #3

    Ich

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    "Frame of reference" is difficult.
    It reaches c wrt the local stationary frame of reference at the horizon. Another way of saying this: there is no local statioary reference frame at the EH, as it would have to travel at c wrt everything.
    The failure (impossibility to travel at c) is in the alleged stationary reference frame, not in the infalling things.
    Yes, wrt such stationary frames ("shell frames").
    As I said, there is no stationary frame where the photon moves at c at the EH. In Schwarzschild coordinates (the global frame, where time depends on position and the speed of light is variable), the photon has a velocity of 0 there.
    The local speed of light is c, but the respective frame cannot be stationary. This line of reasoning doesn't work.
    Again, travelling at c wrt a local stationary frame. There is no such frame inside the horizon. All frames are infalling, all relative speeds are smaller than c.
     
  5. Jul 9, 2010 #4
    OverLOAD is right, the "speed" of light is constant no matter what. Consider a photon leaving the black hole at some angle not normal to the surface of the event horizon. The 'velocity' of the photon has not only a component of speed, but also one for direction (hence velocity is a vector). Gravity 'accelerates' the photon by changing its 'direction' but not it's speed. The trade off in accelerating a photon is that you change it's energy; this correlates to changes in frequency and wavelength. A photon that does travel perfectly normal from the event horizon looses 'all' energy to the point that we can consider it as no longer existing (read Freedman's "Universe" textbook). You can 'picture' this in your head like a ripple on a large pond; eventually the crests of the wave become so small that it looks to you as though it has 'disappeared' into the larger body of water [again, this is an 'analogy' to visualize: proceed with caution!].

    Also food for thought is that the photon always travels the path of 'quickest time' between two points; on a piece of paper or 'flat' 3D-space the shortest distance between two points is a straight line; in 4D-spacetime the shortest 'path' between two points is called a geodesic which can be thought of as a 'straight line in spacetime' but it can be 'curved' when you only consider the 'space' between the two points. The photon path is 'bent' in space because it is traveling the 'quickest time' between two points in space-time. I'm not sure if this second paragraph helps, but I wish you my best with your casual research.
     
  6. Jul 9, 2010 #5
    A photon would be able to remain stationary at hte event horizon if its velocity vector was pointing radially outward, but I wouldn't call that an orbit.

    I think you were talking about the photosphere radius, the smallest radius where (unstable) circular orbits exist, which is 3/2 of the Schwartzschild radius. At this radius, photons may orbit the black hole.
     
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