Can Spherical Symmetry Be Achieved Without Varying Line Element?

[GR191511]

"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($d\theta^2+\sin^{2}\theta d\phi^2)$"
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!

 

[Ibix]

That's 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.

 

[GR191511]

That's 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.

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

 

[Dale]

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

A sphere is 2 dimensional, so $dx^2+dy^2+dz^2$ has too many dimensions.

 

[Nugatory]

What about $dx^2+dy^2+dz^2$?Does the line element vary when x and y and z are varied?

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]

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.

But the $d\theta^2+\sin^{2}\theta d\phi^2$ vary when $\theta$ are varied (because of $sin^{2}\theta$)

 

[Dale]

The key concept here is the Killing vector field: https://en.m.wikipedia.org/wiki/Killing_vector_field

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.

 

[PeroK]

But the $d\theta^2+\sin^{2}\theta d\phi^2$ vary when $\theta$ are varied (because of $sin^{2}\theta$)

Why is the $\sin^2 \theta$ term there?

 

[vanhees71]

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

That's of course also symmetric under rotations. When you parametrize $(x,y,z)$ with the usual spherical coordinates, you get
$$d\vec{r}^2=\mathrm{d} r^2 + r^2 (\mathrm{d} \vartheta^2 + \sin^2 \vartheta \mathrm{d} \varphi^2).$$
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.

 

[DrGreg]

Edit: vanhees71 beat me to it with a reply that says more or less the same thing as this post, but I won't delete this as it uses slightly different language so might still be helpful.

What about $dx^2+dy^2+dz^2$?Does the line element vary when x and y and z are varied?

If you perform the standard Cartesian-to-spherical-polar coordinate conversion$$
\begin{align*}
x &= r \sin\theta \cos\phi \\
y &= r \sin\theta \sin\phi \\
z &= r \cos\theta
\end{align*}
$$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 $r$.

 

[PeterDonis]

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.

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 are two linearly independent ways of "rotating about the poles", not one. Those two are the other two KVFs.

 

[PeterDonis]

"Spherical symmetry requires that the line element does not vary when$\theta$ and$\phi$ are varied

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]

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

Why"all the the other terms independent of $\phi$ and $\theta$"indicates spherical symmetry?

 

[Ibix]

Why"all the the other terms independent of $\phi$ and $\theta$"indicates spherical symmetry?

It doesn't. All the other terms being independent of $\phi$ and $\theta$ and the $\theta\theta$, $\theta\phi$ and $\phi\phi$ components being as quoted does.

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 not relate to revised longitude and latitude coordinates in the same way they relate to the normal ones. Only on a perfect sphere are you free to choose your origin of angular coordinates without changing your metric (and picking the origin seems to be what is meant - somewhat inaccurately - by varying $\theta$ and $\phi$ in your OP).

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.

 

[vanhees71]

Well, one should also note that the introduction of spherical coordinates make only part of spherical symmetry directly manifest, i.e., the rotations around the polar axis, with the azimuthal angle $\varphi$ as a holonomous coordinate, i.e., by introducing a preferred direction for the polar axis you apparently break the full rotation symmetry. This manifests itself that the metric components depend on the polar angle, $\vartheta$.