# Curvature and Schwarzschild metric

• I

## Main Question or Discussion Point

Many text books use the space (spacetime, actually, but for now only space is good enough) around a spherically symmetrical Schwarzschild object to demonstrate curvature of space due to gravity.

Let’s consider two shells around such a Schwarzschild object (say a neutron star of 1 solar mass) whose coordinate radii are 10km and 11km, i.e., 1km apart. Since coordinate radius is being used, the gravitational bending effect is being avoided by measuring around object. Then they calculate the radial distance using the formula $$\Delta r_{shell} = \int_{r1}^{r2} \frac{dr}{\sqrt{1 - \frac{2k}{r}}}$$ where ##k=GM/c^2## and show this to be more than 1km, 1.18km in this example.

But my question is, isn’t the components of the metric tensor expected to take care of the geometry of the space? In particular, isn’t the component ##g_{11} = -\frac{1}{A(r)}## where ##A(r) = 1-\frac{2k}{r} ## supposed to account for the curvature of space due to gravity? If the space was Euclidean, in spherical coordinates, we would have used ## ds^2 = dr^2 + r^2 d\theta^2 + r^2 \sin^2\theta d\phi^2##, but in Schwarzschild space, we are using ## ds^2 = \frac{1}{A(r)}dr^2 + r^2 d\theta^2 + r^2 \sin^2\theta d\phi^2 ##. So, shouldn't the equation for ##\Delta r_{shell}##, which is derived by setting ##d\theta = d\phi = 0## given us the same value of 1km? Is the metric tensor for the Schwarzschild space incorrect as it does not match the observed/differently measured distance of 1km? What is the real separation distance between the 2 shells? 1km or 1.18km?

What am I missing?

Related Special and General Relativity News on Phys.org
PeterDonis
Mentor
2019 Award
isn’t the components of the metric tensor expected to take care of the geometry of the space?
Yes, by telling you how much physical length corresponds to a unit interval of coordinate length.

shouldn't the equation for ##\Delta r_{shell}##, which is derived by setting ##d\theta = d\phi = 0## given us the same value of 1km?
No, because the difference in the radial coordinate ##r_2 - r_1## is a coordinate length, not a physical length. So if ##r_2 - r_1## is 1 km, the physical length, which you are calling ##\Delta r_{shell}##, will be larger than 1 km, because ##g_{rr} > 1##.

vibhuav
PeterDonis
Mentor
2019 Award
What is the real separation distance between the 2 shells? 1km or 1.18km?
What does "real" mean?

Orodruin
Orodruin
Staff Emeritus
Homework Helper
Gold Member
You cannot separate time and space so easily. If you choose to foliate the Schwarzschild spacetime according to Schwarzschild time coordinate surfaces, those will be spacelike surfaces with the metric given by ##ds^2 = dr^2/A(r) + r^2 d^2\Omega##. This metric on the spacelike surface is exactly what defines distances. The ##r## coordinate is just a coordinate and a priori tells you nothing about the distance between to fixed ##r## shells. You need the metric for this. In Schwarzschild coordinates, the ##r## coordinate is chosen so that each shell has area ##4\pi r^2##, not so that the spatial distance between shells is given by ##\Delta r##.

vibhuav
I tried to follow-up based on the replies I got, but I am not completely convinced.

With the example in the OP, with ##k=1.477## (neutron star of ##1 M_\bigodot##) and ##r_1 = 10km, r_2=11km##, if I plug these coordinate radii into radial distance formula that I obtain by solving ## \int \frac{dr}{\sqrt{1-2k/r}} ##, I get 13.595 instead of 10km and 14.775 instead of 11km, so that the distance between the shells is 1.18km instead of 1km.

Redefining the ##r## as reduced circumference (or coordinate radius) instead of the radial distance, appears to me as a mathematical trick. “Since the distance equation near a Schwarzschild object (in particular, when ## d\theta = d\phi = 0##) does not match the observed distance, let’s redefine the ##r## as ##r = circumference /{2\pi}##.” Of course, I understand that using ##C/{2\pi}## avoids the bending effects in the ##\theta## and ##\phi## directions, but why does the original distance formula near the Schwarzschild object require a correction/redefinition? I am not able to relate this redefinition to physics. Could you please elaborate?

I could relate the time dilation near gravitating objects, which is similar to the length anomaly, to physics. A clock closer to the surface of the gravitating object reads less than a clock far away. Here, we are talking about two different clocks, at two different locations and so we can expect two different readings for time. But in case of the shells problem, it is the same length (distance between the two shells), but one method (## circumference/{2\pi}##) gives a different result than the other (direct radial distance method), and to fix the problem we simply redefine the meaning of ##r## in the formula for the distance around the Schwarzschild object?

PeterDonis
Mentor
2019 Award
Redefining the ##r## as reduced circumference (or coordinate radius) instead of the radial distance, appears to me as a mathematical trick.
It's not "redefining" ##r##, it's making a coordinate choice. You can choose any coordinates you like. The problem here is that you are thinking of ##r## as an actual physical radius when it isn't; it's just a coordinate, a number that labels 2-spheres.

“Since the distance equation near a Schwarzschild object (in particular, when ##d\theta = d\phi = 0##) does not match the observed distance, let’s redefine the ##r## as ##r = circumference /{2\pi}##.”
Who are you quoting here? Nobody responding in this thread has said this.

Also, you appear not to realize what the correct "distance equation" is. The correct "distance equation" is the equation for ##\Delta r_{shell}## that you wrote down in your OP. It is not ##r_2 - r_1##.

why does the original distance formula near the Schwarzschild object require a correction/redefinition?
There is no "original distance formula". See above.

one method (##circumference/{2\pi}##) gives a different result than the other (direct radial distance method),
##circumference / {2 \pi}## is not a "distance equation" for radial distances, because the two shells you are trying to get the radial distance between do not have the same circumference.

I don't know what you are referring to by the "radial distance method", but if by that you mean ##r_2 - r_1##, see above.

I wasn't quoting anybody, but my own understanding. Thanks anyway, i'll chew on your reply.

By distance equation/distance formula, I was referring (hopefully correctly) to the spacetime interval which, around a Schwarzschild object, would be ##c^2d\tau^2 = A(r) c^2 dt^2 -\frac{dr^2}{A(r)} - r^2 d\theta^2 - r^2 \sin^2\theta d\phi^2##, from which ##\Delta r_{shell}## was derived

PeterDonis
Mentor
2019 Award
By distance equation/distance formula, I was referring (hopefully correctly) to the spacetime interval
Ok, then your statement that "the distance equation does not match the observed distance" is incorrect. ##\Delta r_{shell}## is the observed distance.

vibhuav
pervect
Staff Emeritus
By distance equation/distance formula, I was referring (hopefully correctly) to the spacetime interval which, around a Schwarzschild object, would be ##c^2d\tau^2 = A(r) c^2 dt^2 -\frac{dr^2}{A(r)} - r^2 d\theta^2 - r^2 \sin^2\theta d\phi^2##, from which ##\Delta r_{shell}## was derived
##c^2 d\tau^2## gives you a positive number for the square magnitude of a timelike interval, and a negative number for the squared magnitude of a space-like interval with the sign convention you choose.

By taking the absolute value, we can find tine interval for any sign convetion, and for the distance s we can write ##s = \sqrt{\left| c^2 d\tau^2 \right|} = \sqrt{ \left| A(r) c^2 dt^2 -\frac{dr^2}{A(r)} - r^2 d\theta^2 - r^2 \sin^2\theta d\phi^2\right| } ##

Then it is a simple matter of substituting ##dt = d\theta = d\phi =0## into the above expression (which is essentially the one you just wrote) to find the same answer as the textbook answer, ##ds = dr/\sqrt{\left| A(r) \right|}##.

I think I understand now.

First of all, let me correct a mistake that crept in. I now understand that ##\Delta r_{shell}## given in the OP is, in fact, the observed/measured distance between the 2 shells. For two shells with coordinate radii 10 and 11 km, k = 1.477, ##\Delta r_{shell}## turns out to be 1.18 km.

The coordinate radius around a Schwarzschild objects are like tic marks on a graph paper - one tic mark every cm, and each cm represents, maybe, 100 km. Similarly, around a Schwarzschild object, we make tic marks, say every 1 km (which is defined as ##circumference/{2\pi}##).

We can think of making shells far away from the Schwarzschild object where there is no effect of gravitation. Concentric shells of 1, 2, 3... km radius are made far away, and transported and placed around the gravitating object symmetrically. Alternatively, these shells could have been marked off by taking a tape measure around the Schwarzschild object, and dividing by ##2\pi##, to be the coordinate distance.

Now consider a shell which is at the 10 km tic-mark (coordinate radius) around a neutron star with k = 1.477. A measurement of its radius by ##C/2\pi## method gives 10, whereas a tape measure from shell surface to the center, gives 13.595 km. Is that a discrepancy? No, because going around the Schwarzschild object avoids the gravitational effects of bending spacetime, whereas in the ##r## direction, the gravitational bending of spacetime (and hence causing curvature) is in effect.

PeterDonis
Mentor
2019 Award
I now understand that ##\Delta r_{shell}## given in the OP is, in fact, the observed/measured distance between the 2 shells.
Ok, good.

The coordinate radius around a Schwarzschild objects are like tic marks on a graph paper - one tic mark every cm, and each cm represents, maybe, 100 km. Similarly, around a Schwarzschild object, we make tic marks, say every 1 km (which is defined as ##circumference/{2\pi}##).
Yes. Or, to put it another way, we take a spacelike surface of constant time and treat it as a family of concentric 2-spheres; and we label each 2-sphere with a number ##r## equal to the circumference of the 2-sphere divided by ##2 \pi##.

We can think of making shells far away from the Schwarzschild object where there is no effect of gravitation. Concentric shells of 1, 2, 3... km radius are made far away, and transported and placed around the gravitating object symmetrically.
Careful. If you do this, you cannot be sure that the geometry of the shells will stay unchanged, because you're moving them, and the spacetime geometry where they end up is different from the spacetime geometry where they started out. So there's no way to be sure that, for example, if you make marks on a given shell separated by 1 meter, as measured by meter sticks far away from the gravitating body, that the physical distance between those marks will still be 1 meter after the shell is placed around the gravitating body.

Alternatively, these shells could have been marked off by taking a tape measure around the Schwarzschild object, and dividing by #2\pi##, to be the coordinate distance.
Yes, this is a better way of looking at things, because you aren't depending on the shell's geometry staying the same when it's moved.

going around the Schwarzschild object avoids the gravitational effects of bending spacetime, whereas in the ##r## direction, the gravitational bending of spacetime (and hence causing curvature) is in effect.
Again, you have to be careful how you state this. It is not the case that there is no spacetime curvature in the tangential direction in Schwarzschild spacetime; the Riemann tensor components in that direction are not zero.

What is the case is that, since Schwarzschild spacetime is spherically symmetric, you can choose to use the physical areas of 2-spheres as your standard for marking off radial tick marks, and things will look nice and simple when you write down the metric in those coordinates. But that does not mean such a choice is required; for example, in isotropic coordinates, the radial coordinate is not defined this way. And in isotropic coordinates, the spatial metric coefficients at a given event are the same in every direction (that's why the coordinates are called "isotropic"); the difference between the radial and tangential directions is not there. So this difference itself is coordinate-dependent, whereas spacetime curvature is not. That's why you have to be careful; you can't just "read off" curvature effects from the metric in a particular set of coordinates. You have to actually calculate the Riemann tensor.

vibhuav