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Why is Schwarzschild singularity not real

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  1. Nov 3, 2015 #1
    It seems to me that the Schwarzschild singularity is generated by the metric function which is an invariant and so has the same value in any coordinate system, so why is it not equally valid?
     
  2. jcsd
  3. Nov 3, 2015 #2

    Nugatory

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    There are two singularities here, and it's not clear which one you're talking about it.

    The one at ##r=0## is real.
    The one at the event horizon where ##r=R_S## is not real, or more accurately it's a "coordinate singuarity" - it appears only if you use Schwarzchild coordinates, and then only because those coordinates are defined in such a way that they don't make sense there. It's analogous to the singularity we find at the north and south poles when we use latitude and longitude as our coordinates - there's nothing wrong with the metric at the poles, there's no strange physics going on there, you can walk around on the curved surface just as you would anywhere else... But you don't have a longitude there.

    We know that the singularity at the event horizon is a coordinate singularity because it goes away if you just use a different coordinate system (Kruskal, for example).
     
  4. Nov 4, 2015 #3
    I have read that before but my question is how do I square it with D (the metric 'distance') being an invariant? You cannot get rid of it if it is invariant.
     
  5. Nov 4, 2015 #4

    Nugatory

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    By "metric distance" you mean the space-time interval calculated from the metric tensor? If so, I'm sorry, but I'm not sure I understand your question. That distance is indeed an invariant, and with appropriate choice of coordinate system we can calculate its invariant value at the event horizon and anywhere except at the central singularity.
     
  6. Nov 4, 2015 #5
    "That distance is indeed an invariant, and with appropriate choice of coordinate system we can calculate its invariant value at the event horizon and anywhere except at the central singularity" - precisely. I calculate the distance in Schwarzschild coordinates and it comes to infinity and hence this is true in any coordinate system as you rightly point out.
     
  7. Nov 4, 2015 #6

    martinbn

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    Which distance did you calculate? Between which two point/events?
     
  8. Nov 4, 2015 #7
    Tthe space-time interval - I don't know the best term for this.
     
  9. Nov 4, 2015 #8

    martinbn

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    Yes, but between which two events?
     
  10. Nov 4, 2015 #9
    Since it is infinite it will be infinite between any point outside the black hole and a radial point on the event horizon.
     
  11. Nov 4, 2015 #10

    martinbn

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    Schwarzschild coordinates do not apply for events on the horizon.
     
  12. Nov 4, 2015 #11

    Nugatory

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    Try it for a point near but not exactly on the horizon, compare with the interval as calculated using some other coordinates - you'll find the interval is finite and invariant.

    The infinity you're finding with Schwarzschild coordinates is happening because those coordinates don't cover events on the horizon itself, so you can't use them if one endpoint is on the horizon. Although it's not obvious from the way the Schwarzschild metric is usually written, Schwarzchild coordinates cover two non-overlapping patches of spacetime, once outside the horizon and the other inside.
     
  13. Nov 4, 2015 #12

    PeterDonis

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    Please show your work. Or no, I have a better idea; I'll show you the correct calculation.

    As Nugatory says, Schwarzschild exterior coordinates don't actually cover the horizon, so the best way to really make this calculation easy is to use other coordinates that do, such as Painleve coordinates. But I'll use a different approach, which is to compute the radial distance between two points, one of which is very close to the horizon, in Schwarzschild coordinates, and then take the limit as the radial coordinate of the lower point goes to the radial coordinate of the horizon.

    Since the only coordinate that will vary is ##r## (a purely radial line in a surface of constant coordinate time), the line element is

    $$
    ds^2 = \frac{1}{1 - 2M / r} dr^2
    $$

    So the radial distance is the integral of ##ds = \sqrt{ds^2}##, or

    $$
    s = \int ds = \int_{\rho}^{R} \sqrt{\frac{r}{r - 2M}} dr
    $$

    where ##r = \rho## is the inner point (the one whose radial coordinate we'll let go to ##2M##) and ##r = R## is the outer point, the one we'll hold constant. Looking up the antiderivative in a table of integrals (for example, see here), we obtain

    $$
    s = \left[ \sqrt{r \left( r - 2M \right)} + 2M \ln | \sqrt{r} + \sqrt{r - 2M} | \right]_{\rho}^{R}
    $$

    This is easily evaluated (I have rearranged the log terms to make things look a bit neater, and removed the absolute value signs since everything inside them is nonnegative):

    $$
    s = \sqrt{R \left( R - 2M \right)} - \sqrt{\rho \left( \rho - 2M \right)} + 2M \ln \frac{\sqrt{R} + \sqrt{R - 2M}}{\sqrt{\rho} + \sqrt{\rho - 2M}}
    $$

    We can now take the limit as ##\rho \rightarrow 2M##, and we get an obviously finite answer:

    $$
    s = \sqrt{R \left( R - 2M \right)} + 2M \ln \left( \sqrt{\frac{R}{2M}} + \sqrt{\frac{R}{2M} - 1} \right)
    $$

    Note, also, that as ##R \rightarrow 2M##, ##s## goes to zero, as it should.
     
    Last edited: Nov 5, 2015
  14. Nov 5, 2015 #13

    Dale

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    Then you calculated it wrong. All spacetime intervals are finite at the horizon.
     
  15. Nov 6, 2015 #14
    Thanks so much for the helpful guidance. It has given me a great deal to think about. Particularly Peter Donis who went to so much trouble to show me I was wrong! This is not a mistake that I will make again.
     
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