Potential energy of a sphere in the field of itself

[Delta2]

Homework Statement

Find the gravitational potential energy of a sphere of radius R and uniform density $\rho$, due to the gravitational field of itself.

Relevant Equations

shell theorem, $U=-G\frac{Mm}{r}$

My attempt was to consider spherical shells of radius $r$ ($r\leq R$))and thickness $dr$ and then the potential energy of this shell would be in the field only of the "residual" sphere of radius $r$ (a result also known as shell theorem) $$U_{dr}=G\frac{\rho\frac{4}{3}\pi r^3 \rho 4\pi r^2}{r}dr$$ and the whole gravitational potential energy of the sphere within the field of itself would be $$U=\int_0^R U_{dr}$$.

What do you think?

 

[kuruman]

You might find the equation in this post
https://www.physicsforums.com/threa...niformly-charged-sphere.1007823/#post-6550731
useful.

 

[Delta2]

You might find the equation in this post
https://www.physicsforums.com/threa...niformly-charged-sphere.1007823/#post-6550731
useful.

Ok, I see there are two problems with my approach
1) The coefficient $\frac{1}{2}$. Don't know how this should come up, probably I am double counting ..
2) I take into account only the first term of the potential. According to my reasoning, because the outer shells don't contribute to the field, they shouldn't contribute to the potential too but I guess that's where I am wrong.

 

[kuruman]

Ok, I see there are two problems with my approach
1) The coefficient $\frac{1}{2}$. Don't know how this should come up, probably I am double counting ..
2) I take into account only the first term of the potential. According to my reasoning, because the outer shells don't contribute to the field, they shouldn't contribute to the potential too but I guess that's where I am wrong.

1. There are two integrals to be done, one to find the potential $\varphi(r)$ at a point where there is mass $dm_i$ inside the sphere. The contribution to the potential energy from that mass is $dU=\varphi(r)dm$. The second integral is adding all the $dU$s to find the total. The factor of $\frac{1}{2}$ is needed to avoid double counting as you say.

2. At a point $r'$ inside the distribution, the potential of a shells of radius $r<r'$ is a $1/r$ potential as if the entire mass of the shell were concentrated at the center. By contrast, the potential inside a shell of radius $r>r'$, the potential is the same everywhere and proportional to the radius $r$ of the shell. That is why you need to do two separate integrals. Although the outer shells do not contribute to the field, each one contributes its own constant (relative to infinity) to the potential.

 

[Delta2]

Although the outer shells do not contribute to the field, each one contributes its own constant (relative to infinity) to the potential.

Yes ok I understand I missed that , but what I don't understand is the following

The factor of 12 is needed to avoid double counting as you say.

I really don't understand where I do double counting with my approach...

 

[Delta2]

BTW @kuruman something else that is left implied but not explicitly stated, is that the potential brakes in that two integrals only for spherically symmetric mass (or charge) distributions right?

 

[haruspex]

Homework Statement:: Find the gravitational potential energy of a sphere of radius R and uniform density $\rho$, due to the gravitational field of itself.
Relevant Equations:: shell theorem, $U=-G\frac{Mm}{r}$

My attempt was to consider spherical shells of radius $r$ ($r\leq R$))and thickness $dr$ and then the potential energy of this shell would be in the field only of the "residual" sphere of radius $r$ (a result also known as shell theorem) $$U_{dr}=G\frac{\rho\frac{4}{3}\pi r^3 \rho 4\pi r^2}{r}dr$$ and the whole gravitational potential energy of the sphere within the field of itself would be $$U=\int_0^R U_{dr}$$.

What do you think?

Your expression is correct, e.g. see https://scienceworld.wolfram.com/physics/SphereGravitationalPotentialEnergy.html.
But from subsequent posts I am not sure you arrived at it validly.

Your $U_{dr}$ expresses the work needed to remove the outermost shell dr to infinity. Integrating that gives the work to disperse the entire sphere to infinity.

 

[Orodruin]

BTW @kuruman something else that is left implied but not explicitly stated, is that the potential brakes in that two integrals only for spherically symmetric mass (or charge) distributions right?

No, what @kuruman says is generally applicable. You can compute the total gravitational potential of a set of point masses by summing up the potential of each point mass in the field created by the other point masses. However, this is where double counting occurs because what you really want to do is to sum up the potential energy among distinct pairs of masses - that is where the factor of 1/2 comes from. With each point mass at position $\vec r_i$, the potential energy is given by
$$
U = -\sum_{i<j} \frac{Gm_i m_j}{|\vec r_i -\vec r_j|}
= -\frac 12 \sum_{i\neq j} \frac{Gm_i m_j}{|\vec r_i -\vec r_j|}
= \frac 12 \sum _{i\neq j} m_i \varphi_j(\vec r_i)
$$
where $\varphi_j(\vec r)$ is the gravitational potential from mass j at $\vec r$.
The integral formula is the continuous limit of this sum.

The approach described by @haruspex (and yourself?) is different and involves gradually building up the object by bringing mass in from infinity in a particular order. Hence, no factor of 1/2 in that case.

 

[Delta2]

Sorry @Orodruin, the equation for $\phi (r)$ (i am not referring to the equation for U at the same post) at the link of post #2 holds for any kind of charge/mass density?

 

[Orodruin]

Sorry @Orodruin, the equation for $\phi (r)$ (i am not referring to the equation for U at the same post) at the link of post #2 holds for any kind of charge/mass density?

$\varphi_j(\vec r)$ is the gravitational potential for a point mass $m_j$ at position j. The total potential at point i (excluding mass i itself), is given by the sum over j. Being more pedantic about the sum sets, the sums are $\sum_{j}\sum_{i<j}$ and $\sum_i \sum_{j\neq i}$, respectively.

 

[Delta2]

So both @Orodruin and @haruspex you think my approach gives the correct result but for the wrong reasons lol?

 

[Delta2]

$\varphi_j(\vec r)$ is the gravitational potential for a point mass $m_j$ at position j. The total potential at point i (excluding mass i itself), is given by the sum over j. Being more pedantic about the sum sets, the sums are $\sum_{j}\sum_{i<j}$ and $\sum_i \sum_{j\neq i}$, respectively.

I don't think that address my question, I ask about the particular formula for $\phi (r)$ in that link of post #2.

 

[Orodruin]

So both @Orodruin and @haruspex you think my approach gives the correct result but for the wrong reasons lol?

I did not say that, I just agreed with @haruspex that it was not clear from the subsequent posts in the thread that it was for the right reasons.

I don't think that address my question, I ask about the particular formula for $\phi (r)$ in that link of post #2.

The particular form yes. Generally you would have three-dimensional integrals for both.

 

[Delta2]

The particular form yes

Yes in what? That it holds only for spherically symmetrically distributions?

 

[Delta2]

That formula calculates the potential as a sum of potentials from concentric shells and it considers that the mass of each shell is concentrated at the center which holds only if each shell has the same $\rho$ over its infinitesimal volume.

 

[haruspex]

That formula calculates the potential as a sum of potentials from concentric shells and it considers that the mass of each shell is concentrated at the center which holds only if each shell has the same $\rho$ over its infinitesimal volume.

Yes, but your integral only considered the potential for a shell created by the mass inside the shell. The validity question is why you did that. Was it because you had in mind the model of blowing successive shells away to infinity or because you thought the mass outside a shell generated no relevant potential?

 

[Delta2]

Was it because you had in mind the model of blowing successive shells away to infinity or because you thought the mass outside a shell generated no relevant potential?

Because I thought the mass outside a shell doesn't generate potential (because it doesn't generate field inside). I didn't have in mind the model of blowing successive shells.

 

[Delta2]

So if I put the $1/2$ in front of my result I should add also the term relating to the potential term from the mass outside the shell in order to get the correct result, right?

 

[haruspex]

So if I put the $1/2$ in front of my result I should add also the term relating to the potential term from the mass outside the shell in order to get the correct result, right?

No, I don't think that works. The potential due to the mass outside the shell is a rather different function.

 

[Delta2]

No, I don't think that works. The potential due to the mass outside the shell is a rather different function.

I did it and works (though I admit it didn't seem to work, seems like some sort of math conspiracy that it turns out that it works). Give me a few minutes to Latex my derivation. The missing potential term is $$\phi_2(r)=2\pi\rho G(R^2-r^2)$$

 

[haruspex]

I did it and works (though I admit it didn't seem to work, seems like some sort of math conspiracy that it turns out that it works). Give me a few minutes to Latex my derivation. The missing potential term is $$\phi_2(r)=2\pi\rho G(R^2-r^2)$$

On reflection, it should work, precisely because it is counting every pairwise shell interaction twice.

 

[Delta2]

We want to calculate $$\begin {align*} \int_0^R \rho \phi_2(r) d^3r=\\ &=\int_0^R\rho 4\pi r^2 2\pi\rho G(R^2-r^2)dr\\&=8\pi^2\rho^2G\int_0^R (R^2r^2-r^4) dr\\&=8\pi^2\rho^2G(\frac{R^5}{3}-\frac{R^5}{5})\\&=\frac{16}{15}\pi^2\rho^2G R^5 \end {align*}$$

 

[Orodruin]

For completeness, let me just mention an alternative approach that makes it clearer that the energy is stored locally in the field rather than in individual particles. The analogue in electromagnetism forms part of the foundation for the electromagnetic stress energy tensor.

Let us start from the expression for the total potential energy
$$
U = \frac{1}{2} \int \rho(\vec x) \phi(\vec x) d^3x
$$
where $\rho(\vec x)$ is the mass density and $\phi(\vec x)$ the gravitational potential such that $\vec g = -\nabla \phi$. The Poisson equation for gravity ensures that the potential satisfies
$$
\nabla \cdot \vec g = - \nabla^2 \phi = - 4\pi G \rho
$$
or, equivalently,
$$
\rho = \frac{1}{4\pi G} \nabla^2 \phi.
$$
Inserting this into the expression for the potential energy leads to
$$
U = \frac{1}{8\pi G} \int [\nabla^2 \phi] \phi\, d^3x.
$$
Applying the divergence theorem and assuming the potential to vanish sufficiently fast as $r \to \infty$ now leads to
$$
U = - \frac{1}{8\pi G} \int (\nabla \phi)^2 d^3 x = -\frac{1}{8\pi G} \int \vec g(\vec x)^2 d^3x.
$$
We can thus associate a gravitational field $\vec g(\vec x)$ with a local potential gravitational energy density*
$$
u_g(\vec x) = -\frac{1}{8\pi G} \vec g(\vec x)^2.
$$

Knowing the functional form of the gravitational field $\vec g(r) = - \vec e_r GM(r)/r^2$ with $M(r) = 4\pi r^3 \rho/3$ for $r < R$ and $M(r) = 4\pi \rho R^3/3$ for $r > R$ for the homogeneous sphere then leads to
\begin{align*}
U &= - \frac{1}{8\pi G} \int_0^\infty \frac{G^2 M(r)^2}{r^4} 4\pi r^2 dr \\
&= -\frac{G}{2} \int_0^\infty \frac{M(r)^2}{r^2} dr \\
&= - \frac{G \rho^2}{2} \left[\int_0^R \frac{16 \pi^2 r^4}{9} dr + \int_R^\infty \frac{16\pi^2 R^6}{9r^2} dr\right] \\
&= - \frac{8 \pi^2 G\rho^2}{9} \left(\left[\frac{r^5}{5}\right]_0^R - \left[\frac{R^6}{r}\right]_R^\infty \right) \\
&= - \frac{8\pi^2 G\rho^2 R^5}{9} \frac{6}{5} = - \frac{16 \pi^2 G \rho^2 R^5}{15}.
\end{align*}
This should now look familiar.

Note: The minus sign is essential. The gravitational potential energy of the assembled system is less than zero because you would have to do work to disassemble it.

Edit:
*: Compare this with the electrostatic potential energy stored in an electric field
$$
u_e(\vec x) = \frac{1}{2} \varepsilon_0 \vec E(\vec x)^2.
$$
The constants are a bit different because of how $\varepsilon_0$ and constants appear in Gauss' law, but the essential argument is exactly the same. (Just replace $\vec g \to \vec E$ and $-1/4\pi G \to \varepsilon_0$.)

Edit 2:
Including also the magnetic field, the EM energy density is given by
$$
u_{em} = \frac{1}{2}\left( \varepsilon_0 \vec E^2 + \frac{1}{\mu_0} \vec B^2\right).
$$
This is the time-time-component of the electromagnetic stress energy tensor. The time-space-components are given by the Poynting vector $\vec S = \vec E \times \vec B/\mu_0$ and the space-space-components are the components of the Maxwell stress tensor.

 

[Delta2]

@Orodruin thanks for the alternative approach, but do you mind answering my question at post #14.

 

[Orodruin]

That it holds only for spherically symmetrically distributions?

Yes.

 

[kuruman]

BTW @kuruman something else that is left implied but not explicitly stated, is that the potential brakes in that two integrals only for spherically symmetric mass (or charge) distributions right?

I am not sure that splitting the radial integral in two parts is limited to spherical distributions only. Using spherical harmonics $Y_l^m(\theta,\phi)$ decouples the radial integral from the angular integrals. Say you have an asymmetric mass distribution $\rho(\mathbf{r'})=\rho(r',\theta',\phi')$. The contribution to the potential from element $dm=\rho(\mathbf{r'})dV'$ at observation point $\mathbf{r}$ will be $$d\varphi=\frac{\rho(\mathbf{r'})dV'}{|\mathbf{r}-\mathbf{r'}|}.$$To integrate this over primed coordinates, one uses the Laplace expansion for the potential $$\frac{1}{|\mathbf{r}-\mathbf{r'}|}=\sum_{l=0}^{\infty}\frac{4\pi}{2l+1}\sum_{m=-l}^l\left(-1\right)^m\frac{r_<^l}{r_>^{l+1}} Y_l^{-m}(\theta,\phi)Y_l^{m}(\theta',\phi').$$You can put the above in the expression for $d\varphi$ and integrate over primed coordinates. Note that you are looking for $\varphi(\mathbf{r})$ where $\mathbf{r}$ is a point of observation inside the distribution. The radial integral will always have to be done in two parts because the integration variable $r'$ is identified with $r_<$ when $r'<r$ and with $r_>$ when $r'>r.$

Doing the first integral will give you $\varphi(r,\theta,\phi)$ which you can use in the second integral $$U=\frac{1}{2}\int_V \rho(r,\theta,\phi)\varphi(r,\theta,\phi)dV$$to find the potential energy.

 

[Orodruin]

I am not sure that splitting the radial integral in two parts is limited to spherical distributions only. Using spherical harmonics $Y_l^m(\theta,\phi)$ decouples the radial integral from the angular integrals. Say you have an asymmetric mass distribution $\rho(\mathbf{r'})=\rho(r',\theta',\phi')$. The contribution to the potential from element $dm=\rho(\mathbf{r'})dV'$ at observation point $\mathbf{r}$ will be $$d\varphi=\frac{\rho(\mathbf{r'})dV'}{|\mathbf{r}-\mathbf{r'}|}.$$To integrate this over primed coordinates, one uses the Laplace expansion for the potential $$\frac{1}{|\mathbf{r}-\mathbf{r'}|}=\sum_{l=0}^{\infty}\frac{4\pi}{2l+1}\sum_{m=-l}^l\left(-1\right)^m\frac{r_<^l}{r_>^{l+1}} Y_l^{-m}(\theta,\phi)Y_l^{m}(\theta',\phi').$$You can put the above in the expression for $d\varphi$ and integrate over primed coordinates. Note that you are looking for $\varphi(\mathbf{r})$ where $\mathbf{r}$ is a point of observation inside the distribution. The radial integral will always have to be done in two parts because the integration variable $r'$ is identified with $r_<$ when $r'<r$ and with $r_>$ when $r'>r.$

Doing the first integral will give you $\varphi(r,\theta,\phi)$ which you can use in the second integral $$U=\frac{1}{2}\int_V \rho(r,\theta,\phi)\varphi(r,\theta,\phi)dV$$to find the potential energy.

There is no need to use spherical harmonics here. The result follows directly from the continuous limit of the sums in #8.

 

[vanhees71]

For completeness, let me just mention an alternative approach that makes it clearer that the energy is stored locally in the field rather than in individual particles. The analogue in electromagnetism forms part of the foundation for the electromagnetic stress energy tensor.

Let us start from the expression for the total potential energy
$$
U = \frac{1}{2} \int \rho(\vec x) \phi(\vec x) d^3x
$$
where $\rho(\vec x)$ is the mass density and $\phi(\vec x)$ the gravitational potential such that $\vec g = -\nabla \phi$. The Poisson equation for gravity ensures that the potential satisfies
$$
\nabla \cdot \vec g = - \nabla^2 \phi = - 4\pi G \rho
$$
or, equivalently,
$$
\rho = \frac{1}{4\pi G} \nabla^2 \phi.
$$
Inserting this into the expression for the potential energy leads to
$$
U = \frac{1}{8\pi G} \int [\nabla^2 \phi] \phi\, d^3x.
$$
Applying the divergence theorem and assuming the potential to vanish sufficiently fast as $r \to \infty$ now leads to
$$
U = - \frac{1}{8\pi G} \int (\nabla \phi)^2 d^3 x = -\frac{1}{8\pi G} \int \vec g(\vec x)^2 d^3x.
$$
We can thus associate a gravitational field $\vec g(\vec x)$ with a local potential gravitational energy density*
$$
u_g(\vec x) = -\frac{1}{8\pi G} \vec g(\vec x)^2.
$$

Knowing the functional form of the gravitational field $\vec g(r) = - \vec e_r GM(r)/r^2$ with $M(r) = 4\pi r^3 \rho/3$ for $r < R$ and $M(r) = 4\pi \rho R^3/3$ for $r > R$ for the homogeneous sphere then leads to
\begin{align*}
U &= - \frac{1}{8\pi G} \int_0^\infty \frac{G^2 M(r)^2}{r^4} 4\pi r^2 dr \\
&= -\frac{G}{2} \int_0^\infty \frac{M(r)^2}{r^2} dr \\
&= - \frac{G \rho^2}{2} \left[\int_0^R \frac{16 \pi^2 r^4}{9} dr + \int_R^\infty \frac{16\pi^2 R^6}{9r^2} dr\right] \\
&= - \frac{8 \pi^2 G\rho^2}{9} \left(\left[\frac{r^5}{5}\right]_0^R - \left[\frac{R^6}{r}\right]_R^\infty \right) \\
&= - \frac{8\pi^2 G\rho^2 R^5}{9} \frac{6}{5} = - \frac{16 \pi^2 G \rho^2 R^5}{15}.
\end{align*}
This should now look familiar.

Note: The minus sign is essential. The gravitational potential energy of the assembled system is less than zero because you would have to do work to disassemble it.

Edit:
*: Compare this with the electrostatic potential energy stored in an electric field
$$
u_e(\vec x) = \frac{1}{2} \varepsilon_0 \vec E(\vec x)^2.
$$
The constants are a bit different because of how $\varepsilon_0$ and constants appear in Gauss' law, but the essential argument is exactly the same. (Just replace $\vec g \to \vec E$ and $-1/4\pi G \to \varepsilon_0$.)

Edit 2:
Including also the magnetic field, the EM energy density is given by
$$
u_{em} = \frac{1}{2}\left( \varepsilon_0 \vec E^2 + \frac{1}{\mu_0} \vec B^2\right).
$$
This is the time-time-component of the electromagnetic stress energy tensor. The time-space-components are given by the Poynting vector $\vec S = \vec E \times \vec B/\mu_0$ and the space-space-components are the components of the Maxwell stress tensor.

In this connection, there's a very interesting paper about the conception of gravitational-field energy (of course only in the here discussed Newtonian context), which is usually not discussed in the literature (it's an open-access paper in AJP):

https://doi.org/10.1119/10.0009889

 

[Delta2]

Ehm , I would love to see the derivation where we have a spherical (not necessarily spherical symmetric) distribution of radius R , how we can pass from $$\phi(r)=\int_0^R\frac{\rho(r')}{|r-r'|}d^3r'$$ to $$\phi(r)=\int_0^R \frac{\rho(r')}{|r|}d^3r'+\int_{|r|}^R\frac{\rho(r')}{|r'|}d^3r'$$ without making any assumptions on $\rho(r')$.

 

[kuruman]

Ehm , I would love to see the derivation where we have a spherical (not necessarily spherical symmetric) distribution of radius R , how we can pass from $$\phi(r)=\int_0^R\frac{\rho(r')}{|r-r'|}d^3r'$$ to $$\phi(r)=\int_0^R \frac{\rho(r')}{|r|}d^3r'+\int_{|r|}^R\frac{\rho(r')}{|r'|}d^3r'$$ without making any assumptions on $\rho(r')$.

If the question is addressed to me, I would say $$\begin{align}\varphi(r,\theta,\phi)= & \sum_{l=0}^{\infty}\frac{4\pi}{2l+1}\sum_{m=-l}^l\left(-1\right)^m Y_l^{-m}(\theta,\phi)\int_0^{2\pi}d\phi '\int_0^{\pi}\sin\theta'd\theta 'Y_l^{m}(\theta',\phi ') \nonumber \\ & \times\left[\int_0^r \rho(r',\theta',\phi')\frac{r'^l}{r^{l+1}} r'^2dr'+\int_r^R \rho(r',\theta',\phi')\frac{r^l}{r'^{l+1}} r'^2dr'\right] \nonumber \end{align}$$The only assumption on $\rho(r',\theta',\phi')$ is that you have an expression for it, otherwise you will not be able to go very far. If you expand it in spherical harmonics, the angular integrals become much easier to find.

If this question is addressed to someone else, I cannot answer for them.

 

[Delta2]

Ok I got no clue what you write cause I am really bad on spherical harmonics, but I know they are the solutions to Laplace's equation, not to Poisson's equation.

 

[Orodruin]

If only there was a good book that covers spherical harmonics and their use to solve things like Poisson’s equation. Alas, no such luck

 

[kuruman]

If only there was a good book that covers spherical harmonics and their use to solve things like Poisson’s equation. Alas, no such luck

So you're saying that post #30 is nonsense because we are trying to solve Poisson's equation here? (I dread the answer).

 

[vanhees71]

Ok I got no clue what you write cause I am really bad on spherical harmonics, but I know they are the solutions to Laplace's equation, not to Poisson's equation.

Are you saying, there's no multipole expansion? ;-)).

 

[Delta2]

Are you saying, there's no multipole expansion? ;-)).

No, I am just not very familiar with multipole expansion and solutions to Laplace's equations so I can't really judge what @kuruman said.