Question on Penrose diagram for Schwazschild metric

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Hi,
I am just learning about Penrose diagrams and got confused:
Why is the singularity r=0 in the Schwarzschild spacetime depicted by a horizontal line in the Penrose diagram? I thought a surface like r=0 would be timeike(as in the Reisner-Nordstrom case) rather than spacelike but obviously I must be missing something here.
 

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  • #2
George Jones
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Inside the event horizon, what type of a coordinate is r?
 
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George Jones
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Inside the event horizon, what type of a coordinate is r?
What is the sign of

[tex]g \left(\frac{\partial}{\partial r} , \frac{\partial}{\partial r} \right)[/tex]

in:

Schwarzschild with [itex]r[/itex] outside the event horizon;

Schwarzschild with [itex]r[/itex] inside the event horizon;

Reissner-Nordstrom with [itex]r[/itex] outside the event horizon;

Reissner-Nordstrom with [itex]r[/itex] between the event horizon and the Cauchy horizon;

Reissner-Nordstrom with [itex]r[/itex] inside the Cauchy horizon.
 
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HI, Thanks for the immediate reply and thanks for pointing that out to me!
You are absolutely right inside the horizon r becomes the timelike coordinate so that the surface of the singularity r=0 is a timelike surface just that the timelike dirction is the horizontal one rather than the vertical one as r and t have interchanged roles, right?
 
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Now I have another question.
If my previous post is correct, than is it the case that a horizontal path inside the Schwarzschild horizon is a timelike path? If this is the case I am a bit confused as I think remembering the lecturer say that any allowed(timelike, null) path in any Penrose diagram makes an angle less or equal to 45° with the vertical. But maybe I got that wrong.
 
  • #6
George Jones
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In Minkowski sapcetime, what is the (hyper)surface t = 3 like?
 
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In minkowski space ds^2=-dt^2+dr^2+r^2(d(theta)^2+sin^2(theta)d(phi)^2)
so for the (hyper)surface t=const we get
ds^2=dr^2+r^2(d(theta)^2+sin^2(theta)d(phi)^2)
which is the Euclidean space R^3. However unfortunately I do not see your point. This is spacelike in any direction and if I am not mistaken it is represented in the Penrose diagram by a line joining i^0 (spacelike infinity) with the vertical line r=0, i.e. it is a spacelike curve in the Penrose diagram and it makes an angle bigger than 45° with the vertical.
@George: By the way I really like your way of asking questions instead of answering the question directly, as I think that I definitely learn much more that way. So if you could continue like this I would really appreciate it!
 
  • #8
George Jones
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I was trying to make an analogy.

In Minkowski spacetime, t is a timelike coordinate, and the surface t = const is a ...

Inside the eveny horizon, r is a timelike coordinate, and the surface r = const is a ...

Now for another (seemingly trivial) question: what does r = conts mean?
 
  • #9
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Hi, thanks for the reply.
I see what you are saying: In Minkowski t is timelike coordinate so t=const is spacelike surface. Then in Schwarzschild r is timelike coordinate inside the horizon so r=const is spacelike surface there, I see that now, thanks.
Actually I do not find your r=const question trivial at all:
In Minkowski r=R(const) implies
ds^2=-dt^2+R^2(d(theta)^2+sin(theta)^2 *d(phi)^2)
and I am not sure how to deduce the form of the hypersurface from there...
the last bit is a sphere of radius R and independence of t says that this does not change in time.However I have no idea how to think of such a 3-dimensional surface... Help would be appreciated
 
  • #10
George Jones
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In the Penrose diagram for Shwarzschild, two degrees of freedom (theta and phi) are suppressed, so a curve r = const is produced by fixing r and letting t vary. Thus, the nature of r = const is greatly influenced by the nature of t.

Now, a definition. A hypersurface is called timelike (spacelike) if the normal 4-vector to the surface is everywhere spacelike (timelike). This does not necessarily mean that, for example, if r is a timelike coordinate, the r = const is a spacelike hypersurface.

For example consider a two-dimensional Minkowski spacetime that is the span of the orthonomal basis {e_0 , e_1}, with e_0 (e_1) being timelike (spacelike). The set {e_0 , e_0 + 1/2 e_1} also spans this space. Use this latter set to coordinatize the space, that is, write any position vector as X^1 e_0 + X^2 (e_0 + 1/2 e_1). Both X^1 and X^2 are timelike coordinates.

Consider the surface (here, a line) X^2 = const. This means fixing X^2 and letting X^1 vary. This produces a timelike line parallel to the timelike X^1 axis, so X^2 = const is a timelike surface (since anything normal to something timelike is itself spacelike) even though X^2 is a timelike coordinate.

Now consider a three-dimensional Minkowski spacetime without (the standard) x^3, and define r^2 = (x^1)^2 + (x^2)^2 in the usual way. The surface r = const is the surface of a cylinder that goes up the t axis. On this cylinder, both spacelike and timelike motions are possible, but, by the above definition, the surface is timellike since its normal is spacelike. On a Penrose diagram, the polar angle theta = tan^-1(x^2/x^1) would be suppressed, so the cylindrical surface r = const would be the timelike line obtained by letting t vary. Howver, each point on the line, say at t = a, actually represents the circle obtainded by the intersection of the plane t = a with the the cylinder r = const. in the original spacetime.

Finally, back to Schwarzschild.

Inside the event horizon, t is a spacelike coordinate, so the Penrose diagram curve produced by fixing r and varying t is spacelike. Each point on the curve represents the two-dimensional surface of a sphere.
 
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