Potential energy of a sphere in the field of itself

AI Thread Summary
The discussion centers on calculating the gravitational potential energy of a sphere due to its own gravitational field, utilizing the shell theorem. The initial approach involves considering spherical shells and integrating their contributions, but issues arise regarding the factor of 1/2 to avoid double counting and the treatment of potential from outer shells. It is clarified that while outer shells do not contribute to the gravitational field inside, they still affect the potential, necessitating separate integrals for accurate results. The conversation also touches on the mathematical derivation of the potential energy, emphasizing the importance of correctly accounting for all contributions to avoid miscalculations. Ultimately, the dialogue highlights the complexities of gravitational interactions within spherically symmetric mass distributions.
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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?
 
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kuruman said:
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.
 
Delta2 said:
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.
 
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kuruman said:
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

kuruman said:
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...
 
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?
 
Delta2 said:
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.
 
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Delta2 said:
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.
 
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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?
 
  • #10
Delta2 said:
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.
 
  • #11
So both @Orodruin and @haruspex you think my approach gives the correct result but for the wrong reasons lol?
 
  • #12
Orodruin said:
##\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.
 
  • #13
Delta2 said:
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.

Delta2 said:
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.
 
  • #14
Orodruin said:
The particular form yes
Yes in what? That it holds only for spherically symmetrically distributions?
 
  • #15
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.
 
  • #16
Delta2 said:
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?
 
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  • #17
haruspex said:
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.
 
  • #18
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?
 
  • #19
Delta2 said:
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.
 
  • #20
haruspex said:
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 $$\phi_2(r)=2\pi\rho G(R^2-r^2)$$
 
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  • #21
Delta2 said:
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.
 
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  • #22
We want to calculate $$\begin {align*} \int_0^R \rho \phi_2(r) d^3r=\\ &=\int_0^R\rho 4\pi r^2 2\pi\rho G(R^2-r^2)dr\\&=8\pi^2\rho^2G\int_0^R (R^2r^2-r^4) dr\\&=8\pi^2\rho^2G(\frac{R^5}{3}-\frac{R^5}{5})\\&=\frac{16}{15}\pi^2\rho^2G R^5 \end {align*}$$
 
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  • #23
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.
 
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  • #24
@Orodruin thanks for the alternative approach, but do you mind answering my question at post #14.
 
  • #25
Delta2 said:
That it holds only for spherically symmetrically distributions?
Yes.
 
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  • #26
Delta2 said:
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.
 
  • #27
kuruman said:
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.
 
  • #28
Orodruin said:
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
 
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  • #29
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')##.
 
  • #30
Delta2 said:
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.
 
  • #31
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.
 
  • #32
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 🤷‍♂️
 
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  • #33
Orodruin said:
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).
 
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  • #34
Delta2 said:
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? ;-)).
 
  • #35
vanhees71 said:
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.
 
  • #36
kuruman said:
So you're saying that post #30 is nonsense because we are trying to solve Poisson's equation here? (I dread the answer).
No, I attempted "humor". I may have erroneously assumed that many people would be at least vaguely aware that those are among the contents of my textbook. Hence, saying that there is no good textbook on the subject would be along the lines of self-derogative humor.

To be on topic though ...

The Spherical harmonics ##Y_{\ell m}(\theta,\varphi)## are eigenfunctions of the angular part of the negative of the Laplace operator ##-\nabla^2##, i.e., in spherical coordinates
$$
-\nabla^2 = -\partial_r^2 - \frac 2r \partial_r + \frac 1{r^2} \hat \Lambda,
$$
where ##\hat \Lambda Y_{\ell m}(\theta,\varphi) = \ell(\ell+1) Y_{\ell m}(\theta,\varphi)##. Being constructed from individual Sturm-Liouville problems in the ##\theta## and ##\varphi## directions, it follows that any function on ##\mathbb R^3## may be written on the form
$$
f(\vec r) = \sum_{\ell = 0}^\infty \sum_{m = -\ell}^\ell f_{\ell m}(r) Y_{\ell m}(\theta, \varphi)
$$
and that the ##Y_{\ell m}## are linearly independent so the expansion is unique. (From now on, let me just write ##\sum_{\ell, m}## for the sums to save typing.)

So, armed with this knowledge, we can diagonalise any linear problem involving the Laplace operator, in particular on the form of Poisson's equation
$$
-\nabla^2 u(\vec r) = \rho(\vec r)
$$
where I have used convenient constant normalisation. As both ##u## and ##\rho## are functions of position, they can both be written in terms of spherical coordinates and therefore expanded in terms of the spherical harmonics as described above with expansion coefficients ##u_{\ell m}(r)## and ##\rho_{\ell m}(r)##, respectively. (Note that the spherical harmonics form the expansion basis for functions on a sphere. We are therefore expanding each function on the separate spheres of radius ##r##, which is why the expansion coefficients depend on ##r##.)

Inserting the expansion into Poisson's equation directly leads to
$$
\sum_{\ell, m} \left[ - u_{\ell m}''(r) - \frac 2r u_{\ell m}'(r) + \frac 1{r^2} \ell(\ell + 1) u_{\ell m}(r)\right] Y_{\ell m}(\theta, \varphi) = \sum_{\ell, m} \rho_{\ell m}(r) Y_{\ell m}(\theta, \varphi).
$$
Now, since the spherical harmonics are linearly independent, it directly follows that
$$
- u_{\ell m}''(r) - \frac 2r u_{\ell m}'(r) + \frac 1{r^2} \ell(\ell + 1) u_{\ell m}(r) - \rho_{\ell m}(r) = 0,
$$
which is an ordinary inhomogeneous differential equation of second order that can be solved. Boundary conditions are typically found from ##u|_{r = 0}## being regular (although in actuality this will follow from the differential equation itself in coordinates other than spherical) and ##\lim_{r \to \infty} u = 0##. Solving for each ##u_{\ell m}(r)## solves Poisson's equation for an arbitrary source term ##\rho(\vec r)##.

Note that the functions ##\rho_{\ell m}## may be found through the inner product (restricting to real functions for simplicity)
$$
\langle f, g \rangle = \int_{\theta = 0}^\pi \int_{\varphi = 0}^{2\pi} f(\theta,\varphi) g(\theta,\varphi) \sin(\theta) d\theta \, d\varphi
$$
as
$$
\rho_{\ell m}(r) = \frac{\langle Y_{\ell m},\rho\rangle }{\langle Y_{\ell m}, Y_{\ell m}\rangle},
$$
where I am leaving the normalisation term ##\langle Y_{\ell m}, Y_{\ell m}\rangle## as written since there are a relatively large number of different normalisation conventions for spherical harmonics and there will always be someone complaining if you pick a particular one ...

Regardless of normalisation, the numerator is on the form
$$
\langle Y_{\ell m},\rho\rangle = \int_{\theta = 0}^\pi \int_{\varphi = 0}^{2\pi} Y_{\ell m}(\theta,\varphi) \rho(r,\theta,\varphi) \sin(\theta) d\theta \, d\varphi,
$$
making it clear that ##\rho_{\ell m}## is indeed a function of ##r##.

Now, looking at a source distribution such that ##\rho(r) = 0## for all ##r > R## for some ##R##, the differential equations become homogeneous
$$
- u_{\ell m}''(r) - \frac 2r u_{\ell m}'(r) + \frac 1{r^2} \ell(\ell + 1) u_{\ell m}(r)= 0
$$
for ##r > R##. This has the general solution
$$
u_{\ell m}(r) = \frac{A_{\ell m}}{r^{\ell+1}} + B_{\ell m} r^{\ell}.
$$
With the boundary condition at infinity, this leads to ##B_{\ell m}= 0## and therefore ##u_{\ell m}(r) = A_{\ell m}/r^{\ell + 1}## outside of the source distribution. This leads to the multipole expansion
$$
u = \sum_{\ell, m} \frac{A_{\ell m}}{r^{\ell + 1}} Y_{\ell m}(\theta,\varphi)
$$
for ##r > R##.

(I had planned to go a bit further but it is getting late ...)

Edit: Minor typos.
 
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  • #37
Oh wait, this is the introductory physics homework forum … :sorry:
 
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  • #38
Orodruin said:
Oh wait, this is the introductory physics homework forum … :sorry:
Well this "self" potential energy is kind of a bit advanced concept since it is not cover in introductory physics book, where they usually talk about potential energy of a body in the field of another body. However I decided to put it here since the solution seemed to me a bit easy and straightforward, without the use of heavy math.

I didn't expect of course that the thread would evolve to spherical harmonics and multipole expansion of a field which are of course advanced physics theory.
 
  • #39
Really? I'd also say that spherical harmonics are at a higher level, but to understand the energy in the gravitational field (Newtonian theory of gravitation of course) of a mass distribution is not so difficult.

You start with a gedanken experiment, asking what energy it takes (or in this case how much energy you get out) when bringing a set of ##N## point masses from infinity into a given static configuration, i.e., particle number ##i## to position ##s\vec{x}_i## given Newton's law for the gravitational force between two point particles:
$$\vec{F}_{21}=-\gamma m_1 m_2 \frac{\vec{x}_2-\vec{x}_1}{|\vec{x}_2-\vec{x}_1|}$$
is the force on particle 2 at presence of particle 1.

For the following it's good to know that this interaction force has a potential, i.e., it's easy to show that
$$\vec{F}_{21}=-\vec{\nabla} \left [-\frac{\gamma m_1 m_2}{|\vec{x}_2-\vec{x}_1|} \right].$$

Now start with bringing particle 1 to ##\vec{x}_1##. Since there are no other particles present yet, there's no force acting on it and thus you neither need nor gain energy.

Now bring particle 2 at its position. This is very much easier, because there's the attractive gravitational force due to the presence of particle 1, i.e., you get energy out. Since we count the energy contained in the mass configuration that means this configuration looses energy, i.e., you count this energy as negative:
$$E_{2}=-\frac{\gamma m_1 m_2}{|\vec{x}_1-\vec{x}_2|}.$$
Now you bring particle 3 to its position. Now the attractive gravitational force from both particles 1 and 2 act on it, and according to Newton's theory there is no generic 3-body force involved in gravity, i.e., the total force is simply the sum of the two pair forces. This implies that for the three particles the total energy is
$$E_3=E_2 - \frac{\gamma m_1 m_3}{|\vec{x}_1-\vec{x}_3|} - \frac{\gamma m_2 m_3}{|\vec{x}_2-\vec{x}_3|} = -\gamma \sum_{j<k} \frac{m_j m_k}{|\vec{x}_j-\vec{x}_k|} = -\frac{\gamma}{2} \sum_{j \neq k} \frac{m_j m_k}{|\vec{x}_j-\vec{x}_k|},$$
and here you see the factor ##1/2## which compensates for counting the interaction energies between all possible pairs twice. Here ##j## and ##k## each run from ##1## to ##3##, but in the sum you leave out the divergent "self-energy contributions" by contraining the sum to ##j \neq k##.

Of course the same kind of sum you have for all ##N## particles, i.e., the ##j,k## each run from 1 to ##N## but leaving out the undefined contributions for ##j=k##.

Now go to the continuum limit and think of it as composed of very many infinitesimal volume elements ##\mathrm{d}^3 x##. Then instead of the sum you get the integral
$$E_{\text{grav}}=-\frac{\gamma}{2} \int_{\mathbb{R}^3} \mathrm{d}^3 x_1 \int_{\mathbb{R}^3} \mathrm{d}^3 x_2 \frac{\rho(\vec{x}_1) \rho(\vec{x}_2)}{|\vec{x}_1-\vec{x}_2|}.$$
Now tacitly I omitted the important assumption that I don't consider the self-energies, i.e., I tacitly didn't take into account that in the sum over the discrete point particles I've put ##j \neq k##. Of course in the latter model I'd get an infinite self-energy due to the singularities of the gravitational interaction potential. In the continuum limit these self-energy contributions are finite, because one integrates out the singularity. So in the continuum model the gravitational field-energy also contains these self-energy contributions.

Often it is simpler to bring this integral in other forms. You can introduce the gravitational field as
$$\vec{g}(\vec{x})=-\gamma \int_{\mathbb{R}^3} \mathrm{d}^3 x' \rho(\vec{x}') \frac{\vec{x}-\vec{x}'}{|\vec{x}-\vec{x}'|^3},$$
which you also can derive from a potential ##\Phi##,
$$\vec{g}(\vec{x})=-\vec{\nabla} \Phi \quad \text{with} \quad \Phi(\vec{x})=-\gamma \int_{\mathbb{R}^3} \mathrm{d}^3 x' \frac{\rho(\vec{x}')}{|\vec{x}-\vec{x}'|}.$$
Then the gravitational-field energy obviously can be written as
$$E_{\text{grav}}=\frac{1}{2} \int_{\mathbb{R}^3} \mathrm{d}^3 x \rho(\vec{x}) \Phi(\vec{x}).$$
So often it's simpler to first calculate the potential ##\Phi## and then evaluate the field energy using this formula.

It's a good exercise to calculate ##\Phi## and ##E_{\text{grav}}## for a homogeneous spherical mass distribution,
$$\rho(\vec{x})=\rho_0 \Theta(a-|\vec{x}|).$$
 
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  • #40
@Delta2

There is a very nice discussion from slides 6 to 9: https://home.iitk.ac.in/~akjha/PHY103_Notes_HW_Solutions/PHY103_Lec_6.pdf (albeit in an analogous discussion of electrostatics) That talks about the factor of ##\frac{1}{2}## and summing interaction terms while not double counting. It generalizes from the discrete case to the continuous case.

I'm going to try and derive an analogous "field squared" equation for gravitational potentials. Although their example is for a spherical shell I think we can apply it to a solid sphere. (Others feel free to correct me if I am wrong).

##W_{grav} = \frac{1}{2} \int_{all \text{ } space} \vec{g} \cdot \vec{g} \, d \tau = \frac{1}{2} \int_{inside} \vec{g}_{inside} \cdot \vec{g}_{inside} \, d \tau + \frac{1}{2} \int_{outside} \vec{g}_{outside} \cdot \vec{g}_{outside} \, d \tau ##

##\vec{g}_{inside} = -\frac{GMr}{R^3}## where ##r \lt R## (Can be derived through the integral form of Gauss' Law of Gravitation)
##\vec{g}_{outside} = -\frac{GM}{r^2}## where ##r \gt R##

##\vec{g}_{inside} \cdot \vec{g}_{inside} = \frac{G^2 M^2 r^2 }{R^6}##

##\vec{g}_{outside} \cdot \vec{g}_{outside} = \frac{G^2 M^2}{r^4}##

Dividing the integral into 2 parts one for the inside and one for the outside while realizing that this is a spherically symmetric distribution/field so ##d \tau = 4 \pi r^2 dr##. We integrate ##\vec{g}_{inside} \cdot \vec{g}_{inside}## from ##0## to ##R## and ##\vec{g}_{outside} \cdot \vec{g}_{outside}## from ##R## to ##\infty##

Here we go

##W_1 = \frac{4 \pi G^2 M^2}{R^6} \int_{r=0}^{4} r^4 \,dr = \frac{4 \pi G^2 M^2}{5 R} ##

##W_2 = 4 \pi G^2 M^2 \int_{r=R}^{\infty} \frac{1}{r^2} \,dr = \frac{4 \pi G^2 M^2}{R} ##

Adding them together ##W_{total} = W_{1} + W_{2} = \frac{24 \pi G^2 M^2}{5 R}##

Or something like that. I might be completely wrong.
 
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  • #41
Just realized @Orodruin used a similar approach before me. I must have skimmed past it. Seems like we have slightly different answers. Gonna look at mine more closely.
 
  • #42
PhDeezNutz said:
Just realized @Orodruin used a similar approach before me. I must have skimmed past it. Seems like we have slightly different answers. Gonna look at mine more closely.
You are missing the crucial differences between the gravitational and electrostatic cases in terms of the sign and the constants appearing in Gauss’ law for each scenario when deriving the energy density. I discussed this towards the end of my post.
Orodruin said:
Edit:
*: Compare this with the electrostatic potential energy stored in an electric field
ue(x→)=12ε0E→(x→)2.
The constants are a bit different because of how ε0 and constants appear in Gauss' law, but the essential argument is exactly the same. (Just replace g→→E→ and −1/4πG→ε0.)
 
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  • #43
Let's check it with the other method using the potential. It's easier to work it out with differential equations but you can also do it by using the integrals given in my previous posting. The equation for the gravitational potential is
$$\Delta \Phi=4 \pi \gamma \rho.$$
With ##\rho=\rho_0 \Theta(a-|\vec{x}|)## and in spherical coordinates, using the spherical symmetry of the problem by making the ansatz ##\Phi(\vec{x}) \equiv \Phi(r)##, where ##r=|\vec{x}|##, you get
$$\Delta \Phi(r)=\frac{1}{r} \partial_r^2 (r \Phi)=4 \pi \gamma \rho(r).$$
For ##r<a## we have ##\rho(r)=\rho_0## and thus by integrating you get
$$\partial_r (r \Phi)=2 \pi \gamma \rho_0 r^2 + A_< \; \Rightarrow \; r \Phi=\frac{2 \pi}{3} \gamma \rho_0 r^3 + A_< r + B_> \; \Rightarrow \; \Phi=\frac{2 \pi}{3} \gamma \rho_0 r^2 + A_<+\frac{B_<}{r}.$$
Since there's no singularity at the origin, we must have ##B_<=0##, while ##A_<## is an arbitrary integration constant.

For ##r>a## you get in the same way, because of ##\rho=0##,
$$\Phi(r)=\frac{B_>}{r},$$
where I've set the other constant to ##0##, because we define the potential such that it goes to 0 at infinity (that's just choosing an arbitrary physically irrelevant additive constant to the total energy).

Now both the potential as well as the gravitational field must be continuous at ##r=a##, i.e.,
$$\Phi(a-0^+)=\Phi(a+0^+) \; \Rightarrow \; \frac{2 \pi}{3} \gamma \rho_0 a^2 + A_<=\frac{B_>}{a}$$
and
$$\Phi'(a-0^+)=\Phi'(a-0^+) \; \Rightarrow \; \frac{4 \pi}{3} \gamma \rho_0 a = -\frac{B_>}{a^2}.$$
So we have
$$B_>=-\frac{4 \pi}{3} \gamma \rho_0 a^3=-M \gamma,$$
where ##M## is the total mass of the sphere and from this
$$A_<=\frac{B_>}{a}-\frac{2 \pi}{3} \gamma \rho^0 a^2 = -2 \pi \gamma \rho_0^2 a^2.$$
So we have
$$\Phi(r)=\begin{cases} \frac{2 \pi}{3} \gamma \rho_0 (r^2-3 a^2) & \text{for} \quad r<a \\
-\frac{4 \pi \gamma \rho_0 a^3}{3 r}=-\frac{M}{r} &\text{for} \quad r>a. \end{cases}$$
The total energy of the gravitational field thus is
$$E_{\text{grav}}=\frac{1}{2} \int_{\mathbb{R}^3} \mathrm{d}^3 x \rho(r) \Phi(r) = \frac{1}{2} 4 \pi \int_{0}^a \mathrm{d} r r^2 \rho_0^2 (r^2-3a^2)=-\frac{16}{15} \gamma \pi^2 \rho_0 a^5=-\frac{3}{5a} \gamma M^2.$$
That's different from the result above.

Now let's see how to express the field energy in terms of ##\vec{g}=-\vec{\nabla} \Phi##. Note that you have
$$\vec{\nabla} \cdot \vec{g}=-4 \pi \rho$$
and thus
$$E=\frac{1}{2} \int_{\mathbb{R}^3} \mathrm{d}^3 x \rho \Phi=-\frac{1}{8 \pi \gamma} \int_{\mathbb{R}} \mathrm{d}^3 x (\vec{\nabla} \cdot \vec{g}) \Phi = +\frac{1}{8 \pi \gamma} \int_{\mathbb{R}} \mathrm{d}^3 x \vec{g} \cdot \vec{\nabla} \Phi = -\frac{1}{8 \pi \gamma} \int_{\mathbb{R}} \vec{g}^2.$$
For our case, obviously you have ##\vec{g}=g(r) \vec{e}_r## with
$$g(r)=-\Phi'(r) = \begin{cases} -\frac{4 \pi}{3} \gamma \rho_0 r &\text{for} \quad r<a \\
-\frac{4 \pi a^3}{3r} \gamma \rho_0 & \text{for} \quad r>a. \end{cases}$$
Plugging this into the said integral you get the same as above with the potential.
 
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  • #44
Well @vanhees71 I am Greek and you are German and I like it that you use the Greek letter "gamma" ##\gamma## for the gravitational constant G , however it looks like it has something to do with the gamma of special relativity...
 
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