To get straight to the question, skip to the 'This is what spherical symmetry really means' paragraph.

I recently inquired about the shell theorem but this is a different matter (link), so I hope it is ok to make a new post. There is an elegant way of deriving the shell theorem by taking advantage of the spherical symmetry of the situation. But instead of just saying "by symmetry", I was pondering how exactly to describe that precisely. I have some thoughts but would like to know what more experienced people think.

So suppose $\rho: \mathbb{R}^3 \to \mathbb{R}$ is some spherically symmetric mass distribution; spherically symmetric means for any $T \in \operatorname{SO}(3)$ and $r \in \mathbb{R}^3$ one has $\rho(Tr) = \rho(r)$. The gravitational field at a point $r$ is $$ g(r) = -G \int \rho(s) \frac{r - s}{|r - s|^3} dV $$ so that at the point $Tr$ we have $$ g(Tr) = -G \int \rho(s) \frac{Tr - s}{|Tr - s|^3} dV. $$ Now we can apply the change of variables theorem. Since $\det T = 1$ it gives us $$ g(Tr) = -G \int \rho(Ts) \frac{Tr - Ts}{|Tr - Ts|^3} dV$$ Note $\rho(Ts) = \rho(s)$, and $|Tr - Ts| = |r - s|$ since $T$ is a rotation, so we can pull $T$ out of the integral to get $$g(Tr) = T \left( - G\int \rho(s) \frac{r-s}{|r-s|^3}dV \right) = Tg(r).$$

This is what spherical symmetry really means. Now let $\Sigma$ be some sphere centered around the origin, and let's say large enough that all of the mass $\rho$ lies inside of it. Then knowning that $gT = Tg$ does immediately give us that the magnitude of $g$ is the same everywhere in $\Sigma$, since rotations preserve magnitude. But it does not seem to immediately tell us, from a mathematical standpoint at least, that $g$ points inward everywhere.

I thought of a way out: in $\Sigma$ you can decompose the field $g = g_{\_} + g_{\perp}$ into its tangent and normal components (and the action of $\operatorname{SO}(3)$ respects this decomposition). Now $g_{\_}$ is a tangent vector field on the 2-sphere $\Sigma$. By the Hairy ball theorem it has a zero somewhere. But since rotations carry 0 to 0, it must be 0 everywhere! So $g$ only has a perpendicular component.

This is nice, but it only works because we are in an odd-dimensional space. So would the shell theorem not work for gravity in a plane? Is there a way to argue this not involving the hairy ball theorem?

  • 1
    $\begingroup$ I'm not sure I completely understand the question: are you asking how to show that the field points only in the radial direction for a spherically symmetric distribution? $\endgroup$
    – Philip
    Jul 9, 2020 at 17:17
  • $\begingroup$ Yes, that is what I am asking (I would have just typed Yes but there is a limit on how small comments can be...) $\endgroup$
    – Pedro
    Jul 9, 2020 at 17:34
  • 2
    $\begingroup$ I think you might be right that spherical symmetry by itself doesn't mean you would have a purely radial field in 2 spatial dimensions; a circulating field would still satisfy the condition you wrote down. But there field configuration is also invariant under reflections about the origin, doesn't this get rid of the non-radial component? $\endgroup$
    – Andrew
    Jul 9, 2020 at 21:12
  • 2
    $\begingroup$ Doing a quick check, seems to me you might need a reflection over an axis to get rid of all non-radial components. Full space inversion might not do it in 2D. $\endgroup$
    – secavara
    Jul 9, 2020 at 22:24
  • 1
    $\begingroup$ @secavara Yes, agreed, I was too quick, thanks! $\endgroup$
    – Andrew
    Jul 9, 2020 at 22:28

1 Answer 1


Theorem. Let $n > 2$ be a positive integer. Assume that in the space $\mathbb{R}^n$ (or possibly in $\mathbb{R}^n \setminus \{0\}$) we have a vector field $g(r)$ with the property that for any $T \in \text{SO}(n)$ $$T\,g(r) = g(Tr)$$ Then $g(r)$ is radial, i.e. there exists a scalar function $w : \mathbb{R}_{+} \to \mathbb{R}$ such that $$g(r) = w(|r|)\, r$$ Furthermore, if we define the function $$V(r) = W(|r|) \,\, \text{ where } \,\, W(\lambda) = \int_{\lambda_0}^{\lambda} \,s\, w(s) \, ds$$ then for any $r\in \mathbb{R}^n \setminus \{0\} $ $$\nabla \, V(r) = g(r)$$ i.e. all rotationally-invariant vector fields on $\mathbb{R}^n \setminus \{0\}$ are both radial and potential fields. $$ $$

Proof: Let $r \in \mathbb{R}$ be an arbitrary non-zero vector. Define the satbilizer of $r$ from the rotation group $\text{SO}(n)$ $$\text{Stab}(r) = \{\, T \in \,\text{SO}(n) \, : \, T\,r = r\}$$ Furthermore, denote by $L$ the orthogonal complement of the vector $r$, which is by definition $$L = \{\, v \in \mathbb{R}^n \, : \, v \cdot r = 0 \,\}$$ Then $\dim L = n-1$ and it is $\text{Stab}(r)$ invariant, i.e. for any $v \in \mathbb{R}^n$ such that $v \cdot r = 0$ and for any $T \in \text{Stab}(r)$ \begin{align*} T\,v \cdot r = T\,v \cdot T\, r = v \cdot r = 0 \end{align*} so $T\,v \in L$. Decompose the space $$\mathbb{R}^n = L \oplus \mathbb{R} \,r$$ Then $v \in \mathbb{R}^n$ decomposes uniquely as $v = v_L + \lambda \,r$, with $v_L \in L$. For any $T \in \text{Stab}(r)$ $$T\, v = T(v_L + \lambda \,r) = T\, v_L + \lambda \,T \, r = w_L + \lambda \, r$$ where $w_L = T\,v_L \in L$. Since $\text{Stab}(r)$ is a subgorup of $\text{SO}(n)$ and as such preserves the dot product on $\mathbb{R}^n$, when restricted to $L$ it preserves the dot product on the $n-1$ dimensional subspace $L$. Moreover, if you take any linear transformation $T_L$ on $L$ that preserves the dot product, it can be extended to $T \in \text{SO}(n)$ as $$T \,v = T\, (v_L + \lambda\, r) = T_L\, v_L + \lambda\, r$$ Thus, $\text{Stab}(r)$ is the full rotation group of the $n-1$ dimensional subspace $L$ and is therefore isomorphic to $\text{SO}(n-1)$. Form the properties of $\text{SO}(n-1)$ we know that if $v \in L$ is any non-zero vector, then its orbit under $\text{Stab}(r)$ is an $n-2$ dimensional sphere, so there is always a transformation $T \in \text{Stab}(r)$ such that $T \,v \neq v$.

Now, let us focus on the $\text{SO}(n)$-invariant vector field $g(r)$. Take any non-zero vector $r \in \mathbb{R}^n$. Split the space $$\mathbb{R}^n = L \oplus \mathbb{R} \,r$$ as explained above. Then $g(r)$ decomposes uniquely into $$g(r) = g_L + w\, r$$ where $g_L \in L$ and $w \in \mathbb{R}$. For any $T \in \text{Stab}(r)$ $$T\,g(r) = g(T\,r) = g(r)$$ which in decomposed form translates into $$T\,g(r) = T(g_L + w\, r) = T\,g_L + w\, T\,r = T\,g_L + w\, r = g_L + w\,r$$ and when we cancel out the term $w\,r$ from both sides of the last identity, we find out that for any $T \in \text{Stab}(r)$ $$T\,g_L = g_L$$ But this is possible if and only if $g_L = 0$, because as I already mentioned before, for a non-zero vector $g_L$ from $L$, there is always a transformation $T \in \text{Stab}(r)$ such that $T \,g_L \neq g_L$ (because the orbit is a proper $n-2$ dimensional sphere). Hence, for any non-zero $r \in \mathbb{R}^n$ there exists a real number $u(r) \in \mathbb{R}$, that varies with respect to $r$, such that $$g(r) = u(r)\, r$$ i.e. the vector field is radial.

Now, by invariance, for any $T \in \text{SO}(n)$ and for any non-zero $r \in \mathbb{R}^n$, $$T (u(r)\, r) = u(r)\, T\,r = T\,g(r) = g(T\,r) = u(T\,r) \, T\,r$$ which, because of the identity $u(r)\, T\,r = = u(T\,r) \, T\,r$, is possible if and only if $u(r) = u(T\,r)$.

Fix one unit vector $r_0 \in \mathbb{R}^n$ ($|r_0| = 1$). Take any other non-zero vector $r \in \mathbb{R}^n$. Both vectors $\frac{r}{|r|}$ and $r_0$ lie on the unit sphere and since $\text{SO}(n)$ acts transitively on the unit sphere (in fact with stabilizers always isomorphic to $\text{SO}(n-1)\,$) there exists an orthogonal transformation $T \in \text{SO}(n)$ such that $$T \, \frac{r}{|r|} = r_0$$ Now do the following: \begin{align} u(r) =& u(T\,r) \\ u(r)\, =& \, u\left(\,T \left(|r| \, \frac{r}{|r|}\right)\,\right) \\ u(r) \, =&\, u\left(|r| \, T \, \frac{r}{|r|}\right) \\ u(r) \, =&\, u\big(\,|r| \, r_0\,\big) \end{align} It follows from the last identity that there exists a single-variable function $$w : \mathbb{R}_{+} \to \mathbb{R}$$ $$w(\lambda) = u(\lambda \, r_0)$$ such that $$u(r) = w(|r|)$$ All of this allows us to conclude that for $n > 2$, any $\text{SO}(n)$-invariant vector field $g(r)$ on $\mathbb{R}^n$ is radial and can be expressed as $$g(r) = w(|r|)\, r $$ for some suitable function $w : \mathbb{R}_{+} \to \mathbb{R}$. If we set $$W(\lambda) = \int_{\lambda_0}^{\lambda} \, s\, w(s)\,ds$$ $$V(r) = W(|r|)$$ then $\frac{d}{d\lambda} W(\lambda) = W'(\lambda) = \lambda\, w(\lambda)$ and hence by the chain rule $$\nabla\, V(r) = \nabla \, W(|r|) = W'(|r|) \, \frac{r}{|r|} = |r|\, w(|r|) \, \frac{r}{|r|} = w(|r|)\, r = g(r)$$ i.e. the rotationally-invariant vector field $g(r)$ is potential.

$$ $$ $$ $$

Older Version. Assume that for an odd dimensional space $\mathbb{R}^n$ we have a vector field $g(r)$ with the property that for any $T \in \text{SO}(n)$ $$T\,g(r) = g(Tr)$$

Theorem. Let $n$ be odd. For every non-zero vector $r \in \mathbb{R}^n$ there exists an orthogonal transformation $T \in \text{SO}(n)$ such that $Tr = r$ and the space of all vectors such that $Tv = v$ is one dimensional, spanned by $r$. In other words, for any vector $r \in \mathbb{R}^n$ there exists an orthogonal transformation $T \in \text{SO}(n)$ such that $r$ is an egienvector of $T$ with eigenvalue $1$, where the multiplicity of the eigenvalue is $1$.

Proof: Take an arbitrary non-zero vector $r \in \mathbb{R}^n$ and define the unit vector $\hat{r} = \frac{r}{|r|}$. Denote by $L$ the $n-1$ dimensional vector subspace of $\mathbb{R}^n$ orthogonal to $\hat{r}$. Since $L$ is non-trivial, choose a unit vector $\hat{r}_2 \in L$. Then $\hat{r}$ and $\hat{r}_2$ are pairwise orthogonal. Continue this procedure. Assume you have selected $k$ unit vectors $\hat{r}, \, \hat{r}_2, \, ..., \, \hat{r}_k$, that are pairwise orthogonal. Then, there exists a unique $n-k$ dimensional vector subspace $L_k$ which is the orthogonal complement of the span of $\hat{r}, \, \hat{r}_2, \, ..., \, \hat{r}_k$. If $k < n$ then $L_k$ is non-trivial and you can find aunit vector $\hat{r}_{k+1} \in L_k$. Thus, you have found $k+1$ unit vectors $\hat{r}, \, \hat{r}_2, \, ..., \, \hat{r}_k, \, \hat{r}_{k+1}$, that are pairwise orthogonal. Continue like this until $k = n$. You have constructed an orthonormal basis $\hat{r}, \, \hat{r}_2, \, ..., \, \hat{r}_n$ of $\mathbb{R}^n$. Think of this new vectors as column-vectors. Define the orthogonal matrix $$U = \big[ \hat{r} \, \, \hat{r}_2 \,\, ... \, \hat{r}_n \big] \, \in \, \text{SO}(n)$$ If you denote by $e_1 = [1, \, 0, \, ..., \, 0]^T$ the first basis vector of $\mathbb{R}^n$, interpreted as a column-vector, then $$\hat{r} = U \, e_1 \,\, \text{ and therefore } \,\, e_1 = U^T \, \hat{r}$$
If $n = 2\,m + 1$, construct the block-diagonal matrix $$R = R(\theta_1,..., \theta_m) = \begin{bmatrix} 1 & & & &\\ & R_2(\theta_1) & & & &\\ & & R_2(\theta_2) & & \\& & & ... & \\ & & & & R_2(\theta_m) \end{bmatrix}$$ where for each $j=1,..,m$ $$R_2(\theta_j) = \begin{bmatrix} \cos(\theta_j) & -\,\sin(\theta_j)\\ \sin(\theta_j) & \cos(\theta_j) \end{bmatrix}$$ and $\theta_j \in (0, 2\pi)$. By construction, $$R(\theta_1,..., \theta_m)\, v = v \,\, \text{ if and only if } \,\, v = \lambda \,e_1$$, i.e. this matrix has eigenvalue $1$ with multiplicity $1$. The rest of the eigenvalues are true complex unitary numbers. Finally, define the orthogonal transformation $$T = U \, R \, U^T \in \text{SO}(n)$$ By construction, $$T \, r = T \, |r| \, \frac{r}{|r|} = |r|\, T \,\hat{r} = |r|\, U\, R \, U^T \,\hat{r} = |r|\, U\, R\, e_1 = |r|\, U\, e_1 = |r|\, \hat{r} = r$$ Moreover, since $T$ and $R$ are conjugate matrices, they have the same eigenvalues, including multiplicities. Therefore $T$ has eigenvalue $1$ of multiplicity $1$ and eigenvector $r$. End of proof

Now, you can apply this theorem to any $\text{SO}(n)$-invariant vector field on $\mathbb{R}^n \setminus \{0\}$, i.e. $$g : \mathbb{R}^n \setminus \{0\} \to \mathbb{R}^n$$ $$g(T\,r) = Tg(r) \,\,\text{ for any } \,\, T \in \text{SO}(n)$$ For an arbitrary non-zero vector $r \in \mathbb{R}^n$, pick one orthogonal transfromation $T_r \in \text{SO}(n)$ with the property described in the theorem above. Hence $T_r\, r = r$. Furthermore $$T_rg(r) = g(T_rr) = g(r)$$ which means that $g(r)$ is also an eigenvector of $T_r$ with egienvalue $1$. Since the eigenspace of $T_r$ that corresponds to the eigenvaule $1$ is one dimensional and spanned by $r$, the vector $g(r)$ must be a multiple of $r$, i.e. there exists a scalar $\lambda(r) \in \mathbb{R}$ such that $$g(r) = \lambda(r) \, r$$ As you can see, the vector field $g(r) = \lambda(r) \, r$ is radial.

What happens if you take $n=2$? Can you find a counterexample? (I think it is fairly straightforward)

Now, there is another approach to your question, in particular that if you define the potential function
$$V(r) = G \, \int_{\Sigma} \, \frac{\rho(s)}{|r - s|} \, ds$$ and the function $\rho(s)$ is nice enough so that we can differentiate with respect to $r$ under the integral, then $$g(r) = \nabla \,V(r) = - \, G \, \int_{\Sigma} \, \rho(s)\,\frac{(r-s)\,\,}{|r - s|^3} \, ds$$
So basically, your $\text{SO}(n)$-invariant vector field is in fact a potential field and comes from a potential function $V(r)$ which is also $\text{SO}(n)$-invariant, i.e. for any $T \in \text{SO}(n)$ $$V(Tr) = V(r)$$ Fix one unit vector $r_0 \in \mathbb{R}^n$ ($|r_0| = 1$). Now take any other non-zero vector $r \in \mathbb{R}^n$. Both vectors $\frac{r}{|r|}$ and $r_0$ lie on the unit sphere and since $\text{SO}(n)$ acts transitively on the unit sphere (in fact with stabilizers always isomorphic to $\text{SO}(n-1)\,$) there exists an orthogonal transformation $T \in \text{SO}(n)$ such that $$T \, \frac{r}{|r|} = r_0$$ Now do the following: \begin{align} V(r) =& V(T\,r) \\ V(r)\, =& \, V\left(\,T \left(|r| \, \frac{r}{|r|}\right)\,\right) \\ V(r) \, =&\, V\left(|r| \, T \, \frac{r}{|r|}\right) \\ V(r) \, =&\, V\big(\,|r| \, r_0\,\big) \end{align} It follows from the last identity that there exists a single-variable function $$W : \mathbb{R}_{+} \to \mathbb{R}$$ $$W(\lambda) = V(\lambda \, r_0)$$ such that $$V(r) = W(|r|)$$ Then, by the chain rule, the gradient of $V$ with respect to $r$ is $$\nabla \, V(r) = \nabla \, W(|r|) = \frac{dW}{d\lambda}(|r|)\, \frac{r}{|r|} = W'(|r|)\, \frac{r}{|r|}$$ i.e. the vector field $g(r) = \nabla \, V(r) = W'(|r|)\, \frac{r}{|r|}$ is radial.

  • $\begingroup$ You're right, using the potential makes this much clearer. $\endgroup$
    – Pedro
    Jul 10, 2020 at 2:43
  • $\begingroup$ @Pedro I have added one possible proof of the main theorem I use when sowing that rotation-invariant vector fields in odd dimensional spaces are radial. Of course, there could be other proofs of this theorem plus the proof shows a bit more than what is really needed for the rest of the arguments, but it gives some idea about the general structure of the orthogonal matrices with exactly one rotation axis. $\endgroup$ Jul 11, 2020 at 12:54
  • $\begingroup$ @Pedro I updated my answer and I included a proof of the general fact that any rotationally invariant vector field in any dimension greater than 2 is radial and potential. $\endgroup$ Jul 11, 2020 at 15:35

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