- #1

- 76

- 6

I wonder why:

"the line element does not vary when##\theta## and##\phi## are varied"could##\Rightarrow##"##\theta##and ##\phi##only occur in the line element in the form(##d\theta^2+\sin^{2}\theta d\phi^2)##"?Thanks!

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

In summary, spherical symmetry requires that the line element does not vary when ##\theta## and##\phi## are varied, so that ##\theta## and ##\phi## only occur in the line element in the form (##d\theta^2+\sin^{2}\theta d\phi^2)##. This is the metric of a 2-sphere, which is invariant under a change of pole and meridian. Any other metric represents something less symmetric, so will have some preferred direction or something. The key concept is the Killing vector field, which describes the symmetries of a manifold. For a flat 2D Euclidean metric, there are three Killing vectors, while for a spherical 2D manifold

- #1

- 76

- 6

I wonder why:

"the line element does not vary when##\theta## and##\phi## are varied"could##\Rightarrow##"##\theta##and ##\phi##only occur in the line element in the form(##d\theta^2+\sin^{2}\theta d\phi^2)##"?Thanks!

Physics news on Phys.org

- #2

- 11,178

- 12,641

- #3

- 76

- 6

What about ##dx^2+dy^2+dz^2##?Does the line element vary when x and y and z are varied?How?Ibix said:

- #4

Mentor

- 34,820

- 12,877

A sphere is 2 dimensional, so ##dx^2+dy^2+dz^2## has too many dimensions.GR191511 said:What about ##dx^2+dy^2+dz^2##?Does the line element vary when x and y and z are varied?How?

- #5

Mentor

- 14,602

- 8,898

No, and you can see that because ##x##, ##y##, and ##z## do not appear in the metric so it is the same everywhere. Compare with, for example, ##x^{-2}dx^2+dy^2+dz^2## where the metric is different at points with different ##x## coordinates.GR191511 said:What about ##dx^2+dy^2+dz^2##?Does the line element vary when x and y and z are varied?

- #6

- 76

- 6

But the ##d\theta^2+\sin^{2}\theta d\phi^2## vary when ##\theta## are varied (because of ##sin^{2}\theta##)Nugatory said:No, and you can see that because ##x##, ##y##, and ##z## do not appear in the metric so it is the same everywhere. Compare with, for example, ##x^{-2}dx^2+dy^2+dz^2## where the metric is different at points with different ##x## coordinates.

- #7

Mentor

- 34,820

- 12,877

This vector field describes the symmetries of a manifold. The algebraic form of the metric in some coordinates may obscure the nature and existence of the Killing vector fields.

For a flat 2D Euclidean metric there are three Killing vectors. Two are translations in different directions and one is rotation. That specifies the symmetries of a flat plane, which is the maximally symmetric 2D manifold.

For a spherical 2D manifold there are two Killing vector fields. One is a rotation about the poles and the other is a rotation about the equator.

It is these Killing vectors that distinguish the manifolds, not the coordinates.

- #8

- 25,774

- 17,345

Why is the ##\sin^2 \theta## term there?GR191511 said:But the ##d\theta^2+\sin^{2}\theta d\phi^2## vary when ##\theta## are varied (because of ##sin^{2}\theta##)

- #9

- 24,155

- 14,658

That's of course also symmetric under rotations. When you parametrize ##(x,y,z)## with the usual spherical coordinates, you getGR191511 said:What about ##dx^2+dy^2+dz^2##?Does the line element vary when x and y and z are varied?How?

$$d\vec{r}^2=\mathrm{d} r^2 + r^2 (\mathrm{d} \vartheta^2 + \sin^2 \vartheta \mathrm{d} \varphi^2).$$

Rotations around the origin only affect the angles ##\vartheta## and ##\varphi## and keeps ##r## invariant, and thus the most general pseudo-metric invariant under rotations is

$$\mathrm{d} s^2=g_{00}(t,r) \mathrm{d} t^2 - g_{11}(t,r) \mathrm{d} r^2 -g_{22}(t,r) (\mathrm{d} \vartheta^2 + \sin^2 \vartheta \mathrm{d} \varphi^2).$$

You could also make a more general ansatz with off-diagonal elements in the ##(t,r)##-plane, but these you can "gauge away", i.e., the ansatz above is the most general ansatz for a spherically symmetric spacetime.

- #10

Science Advisor

Gold Member

- 2,456

- 1,856

If you perform the standard Cartesian-to-spherical-polar coordinate conversion$$GR191511 said:What about ##dx^2+dy^2+dz^2##?Does the line element vary when x and y and z are varied?

\begin{align*}

x &= r \sin\theta \cos\phi \\

y &= r \sin\theta \sin\phi \\

z &= r \cos\theta

\end{align*}

$$you find that$$

\text{d}x^2 + \text{d}y^2 + \text{d}z^2 = \text{d}r^2 + r^2 (\text{d}\theta^2 + \sin^2\theta \, \text{d}\phi^2 )

$$It is the presence of the term ##\text{d}\theta^2 + \sin^2\theta \, \text{d}\phi^2##, with all the the other terms independent of ##\phi## and ##\theta##, that indicates spherical symmetry, but the presence of the multiplier ##r^2## in front of that term, and the presence of ##\text{d}r^2## with a multiplier of 1, adds further constraints to the metric (i.e. the form of dependence on ##r##). You could have different dependence on ##r## but still have spherical symmetry, e.g. ## A(r) \, \text{d}r^2 + B(r) \, (\text{d}\theta^2 + \sin^2\theta \, \text{d}\phi^2 )## where ##A## and ##B## are positive-valued functions of ##r##.

- #11

Mentor

- 43,886

- 21,299

No, there are three KVFs, not two, corresponding to the three generators of the rotation group SO(3), which is the relevant symmetry group. One is rotation about the equator (by which I assume you mean rotation that leaves the equator invariant), yes, but there areDale said:For a spherical 2D manifold there are two Killing vector fields. One is a rotation about the poles and the other is a rotation about the equator.

- #12

Mentor

- 43,886

- 21,299

Is this supposed to be a quote from somewhere? If so, please give a reference. (I suspect it's not, because it's wrong.)GR191511 said:"Spherical symmetry requires that the line element does not vary when##\theta## and##\phi## are varied

- #13

- 76

- 6

Why"all the the other terms independent of ##\phi## and ##\theta##"indicates spherical symmetry?DrGreg said:It is the presence of the term ##\text{d}\theta^2 + \sin^2\theta \, \text{d}\phi^2##, with all the the other terms independent of ##\phi## and ##\theta##, that indicates spherical symmetry

- #14

- 11,178

- 12,641

It doesn't. All the other terms being independent of ##\phi## and ##\theta##GR191511 said:Why"all the the other terms independent of ##\phi## and ##\theta##"indicates spherical symmetry?

The point is that for spherical symmetry we require that the line element have the same form wherever we choose to place the pole and the ##\phi=0## line. Does that make sense? If we idealise the Earth as a sphere, we could place the equator anywhere and distance measurements wuld depend on latitude and longitude in the same way. However that is not the case for the Earth in reality - the distance around the equator is different to a transpolar circuit, so (sufficiently precise) distance measurements would

So if you want a spherically symmetric spacetime and you want to express it in polar coordinates then your ##\theta## and ##\phi## had better behave like they do on a sphere. Hence ##r^2d\theta^2+r^2\sin^2\theta d\phi^2##.

Furthermore you need the metric components associated with your other two coordinates to be independent of ##\theta## and ##\phi##. Otherwise you have something non-spherically summetric.

- #15

- 24,155

- 14,658

Share:

- Replies
- 10

- Views
- 800

- Replies
- 3

- Views
- 704

- Replies
- 16

- Views
- 1K

- Replies
- 4

- Views
- 255

- Replies
- 3

- Views
- 520

- Replies
- 9

- Views
- 587

- Replies
- 62

- Views
- 3K

- Replies
- 10

- Views
- 1K

- Replies
- 7

- Views
- 1K

- Replies
- 9

- Views
- 711