Curvature of Time: How Do We Experience It?

[pervect]

Another way of visualizing the curved space-time, specifically the curved space-time due to a black hole (which will apply to any spherically symmetric massive body) is given in:

http://arxiv.org/PS_cache/gr-qc/pdf/9806/9806123v3.pdf

This is rather nice if one is familiar with special relativity, because it is an embedding diagram of space-time, i.e. gravity is no more and no less than drawing SR space-time diagrams on this curved surface.

Of course to fit this diagram into three dimensions, all but the r and t coordinates have been suppressed.

Unfortunately, I find the diagrams a bit hard to follow personally, though I've been playing around with them a little recently.

 

[robphy]

You might find Don's slides (and link to his course notes) useful:

Don Marolf - "Teaching Black Holes"
http://www.aapt-doorway.org/TGRUTalks/Marolf/Marolf1of5.htm
as part of last year's AAPT Topical Conference on Teaching General Relativity to Undergraduates (http://www.aapt-doorway.org/TGRU/ ),
which has a lot of interesting slides and posters.

 

[pervect]

OK, thanks.

I think I'm getting it - attached is my color coded version of the Schwarzschild geometry based on Marolf's paper.

http://arxiv.org/PS_cache/gr-qc/pdf/9806/9806123v3.pdf

It is an embedding of the r-t radial plane of a Schwarzschild black hole in a 3 dimensional Mikowski geometry, as described by the paper. The only advantage of this diagram over the ones in the link is that it's color coded. Drawing it was very useful to me in understanding the paper, however.

It has four regions, as it is the fully extended Schwarzschild spacetime, which is a non-traversable wormhole connecting two different asymptotically flat space-times.

The two asymptotically flat spacetimes are colored green and blue, which represent the exterior region of the black hole outside the event horizon. I think of the green region as "our" space-time (for no particularly good reason).

Note that these are the same four regions that are shown on a penrose diagram of a black hole. For readers unfamiliar with Penrose diagrams see for instance:

http://en.wikipedia.org/wiki/Image:PENROSE2.PNG

the diagram for the Schwarzschild geometry is the one labelled "static wormhole".

There are also two interior regions, colored red and pink. The pink region represents the interior of a white hole, the red region is the interior of a black hole. As T increases, any object in the pink region must eventually leave it and enter the blue or green regions.

The singularity itself is located at R=0, which corresponds to Y = -infinity (which also implies T = +infinity or T=-infinity, as per the Penrose diagram). The attachment is only drawn for Y greater than -5, however.

The event horizon, at r=1, is a lightlike (null) surface where differing colors intersect.

The coordinate labeled T is the time coordinate of the Minkowski geometry, X and Y are space coordinates. T increasing determines the direction of increasing time for any (timelike) worldline.

Lines of constant r are planes of constant Y on the diagram. Y>0 corresponds to r>2M, i.e. one of the two exterior regions. Y<0 corresponds to r<2M, i.e. one of the two interior regions. Y=0 is the event horizon.

The equations used to construct it are interesting, and unfortunately are a bit obfuscated in the paper (in my opinion).

We wish to create a map from (r,t) to (X,Y,T), where (r,t) are the Schwarzschild coordinates, and (X,Y,T) are the coordinates of our embedding. The Schwarzschild radius is assumed to be unity (i.e the mass of the black hole is 1/2 in geometric units).

The coordinate Y can be expressed as an integral, and depends only on r

$$\int_1^r \left( 1+1/R+1/R^2+1/R^3 \right) dR$$

By construction, when r=1, Y=0, i.e. the event horizon is located at Y=0.

The X and T coordinates are functions of both r and t. For the exterior region, the formula is:

$$X = \pm 2 \sqrt{1-1/r} \cosh t/2$$
$$T = 2 \sqrt{1-1/r} \sinh t/2$$

The plus and minus sign gives two separate regions, the green and blue, representing the two different asymptotically flat space-times in the exterior region.

In the interior region, the formula is slightly different
$$T = \pm 2 \sqrt{1-1/r} \cosh t/2$$
$$X = 2 \sqrt{1-1/r} \sinh t/2$$

Again, the plus and minus signs represent different regions.

I won't go through the algebra in detail, but one can confirm that

-dT^2 + dX^2 + dY^2

yields the Schwarzschild metric when re-expresed in terms of dr and dt, i.e. one substitutes

dT = (dT/dt)*dt + (dT/dr)*dr
dX = (dX/dt)*dt + (dX/dr)*dr
dY = (dY/dr)*dr

and gets the Schwarzschild metric

(-1+1/r) dt^2 + 1/(1-1/r) dr^2

One may note that r=1 from the above transformation equations corresponds to a point, T=0 and X=0, but the event horizon is actually a line. This happens because Schwarzschild coordinates are ill-behaved. The event horizon really has the topology of a null surface and because we are modeling only 1+1 dimensions said "null surface" is a pair of lines on the diagram.

 

[JDługosz]

We can all see what curvature of space looks like, just by throwing a ball and watching it follow the natural geodesic.

But what does curvature of time look like?

I disagree with that assessment. As I recall, it is the curvature along the time dimension that exactly reproduces Newton's gravity. Adding in the curvature of space is necessary to account for the precession of Mercury. Gravity Probe B is measuring the curvature of space around the earth; specifically that going all the way around is a little shorter than you would expect based on the mathematical value of π.

If you measure the distances between some points in a vertical plane around the trajectory of a thrown ball, you will get results that Euclid would agree with, even if the most accurate tools are used. But if you put clocks at those points, you will find the higher ones tick faster. The natural motion of the ball serves to minimize its "proper time". Or is it maximize? I don't remember. But the point is to show how it involves time, not curved space.

 

[A.T.]

I disagree with that assessment. As I recall, it is the curvature along the time dimension that exactly reproduces Newton's gravity.

It is obvious that curvature of space only, would only affect objects which are already moving through space. But gravitation also affects objects which are initially at rest, so you have to consider the curvature of space-time.

 

[pervect]

It is obvious that curvature of space only, would only affect objects which are already moving through space. But gravitation also affects objects which are initially at rest, so you have to consider the curvature of space-time.

How does this differ from what JDlugsloz is saying, or are agreeing with him?

 

[A.T.]

How does this differ from what JDlugsloz is saying, or are agreeing with him?

I agree with him. Just wanted to put it in simpler words, why gravity cannot be curvature of space only.