Pressure versus stress in uniformly charged sphere

[Rob2024]

Homework Statement

What's the pressure and stress at a distance r within a uniformly charged sphere?

Relevant Equations

$E (r) = \frac{Q r}{4 \pi \epsilon_0 R^3}$
$\sigma(r) = \half \epsilon_0 E^2$

Let the charge density be $\rho$, radius be $R$, total charge be $Q = \rho \frac{4}{3} \pi R^3$. We know
from Gauss's law, $E (r) = \frac{Q r}{4 \pi \epsilon_0 R^3}$.

We also know from Maxwell stress tensor $\sigma(r) = \half \epsilon_0 E^2$.

We can compute the pressure due to the electric field directly, it causes the charge pressing outward due to
repulsion. Take a spherical layer of charge at distance $r$ whose area is $A = 4 \pi r^2$, apply force balance,
$P(r+dr) A - P(r) A = dq E = \rho 4\pi r^2 dr E$
$dP = \rho E dr = \frac{\rho Q r}{4 \pi \epsilon_0 R^3} dr$
$P = \half \frac{k Q \rho r^2}{R^3}$
$P = 3 \half \e\spilon_0 E^2 = 3 \sigma$

Why is the pressure derived directly different from the one from Maxwell's stress tensor? Are they different in physical meaning?

Thanks!

 

[renormalize]

Please rewrite this post in readable LaTeX format. There is a handy LaTeX guide at the left below.

 

[haruspex]

On this forum you need to use a double hash (# #, without the space) instead of $.
And \half is not recognized.
—————————
Homework Statement: What's the pressure and stress at a distance r within a uniformly charged sphere?
Relevant Equations: $E (r) = \frac{Q r}{4 \pi \epsilon_0 R^3}$
$\sigma(r) = \frac 12 \epsilon_0 E^2$

Let the charge density be $\rho$, radius be $R$, total charge be $Q = \rho \frac{4}{3} \pi R^3$. We know
from Gauss's law, $E (r) = \frac{Q r}{4 \pi \epsilon_0 R^3}$

We also know from Maxwell stress tensor $\sigma(r) = \frac 12 \epsilon_0 E^2$

We can compute the pressure due to the electric field directly, it causes the charge pressing outward due to
repulsion. Take a spherical layer of charge at distance $r$ whose area is $A = 4 \pi r^2$, apply force balance,
$P(r+dr) A - P(r) A = dq E = \rho 4\pi r^2 dr E$
$dP = \rho E dr = \frac{\rho Q r}{4 \pi \epsilon_0 R^3} dr$
$P = \frac 12 \frac{k Q \rho r^2}{R^3}$
$P = 3 \frac 12 \epsilon_0 E^2 = 3 \sigma$

Why is the pressure derived directly different from the one from Maxwell's stress tensor? Are they different in physical meaning?

Thanks!

 

[Rob2024]

Hi, I am asking again to see if people have some ideas why the normal stress value is different from pressure? I had thought that they would be the same. Thanks,

 

[haruspex]

$P(r+dr) A - P(r) A = dq E = \rho 4\pi r^2 dr E$

I'm not sure about this. That layer is subject to electrostatic forces from outside as well. Doesn't that reduce the pressure difference required to balance the force from inside?

Edit: Ignore that, it is wrong.

 

[Rob2024]

Thanks for the correction. With the correction, the pressure is 3/2 of the normal stress. After discussing with a professor, I believe the use of the word stress in this case is not to be confused with stress in elastic material since the Maxwell stress is not 0 in vacuum outside of the sphere. Let me know if any additional enlightenment can be shed on this question. Thanks.

 

[TSny]

We can compute the pressure due to the electric field directly, it causes the charge pressing outward due to
repulsion. Take a spherical layer of charge at distance $r$ whose area is $A = 4 \pi r^2$, apply force balance,
$P(r+dr) A - P(r) A = dq E$

I don’t think this way of deriving $P(r)$ is correct.

The area at $(r+dr)$ is $4 \pi (r+dr)^2 = 4 \pi (r^2 + 2rdr + dr^2)$. The term $dr^2$ in the parentheses can be neglected, but not the term $2rdr$. So, I believe leaving out this term is a mistake.

I’m uncomfortable using an entire spherical shell to deduce $P(r)$. Instead, consider an infinitesimal circular disk of material inside the sphere as shown.

Let $a$ be the (infinitesimal) radius of the disk and $dz$ the thickness of the disk. Let $r$ be the distance from the center of the sphere, $O$, to the center of the bottom surface of the disk. The distance from $O$ to the center of the top of the disk is then $r+dz$. Let $A$ be the area of the bottom (or top) of the disk. So, the volume of the disk is $dV = Adz$

The electric force $F_z^{\rm elec}$ experienced by the disk will be in the $z$ direction: $F_z^{\rm elec} = \rho dV E(r)$.

The net pressure force $F_z^{\rm pr}$ on the disk must be equal and opposite to the electric force: $$F_z^{\rm pr} = P(r)A – P(r+dz)A = - \rho dV E(r).$$ Approximating $P(r+dz) \approx P(r) + P’(r)dz$, show $$P’(r) = \rho E(r) = \frac{\rho^2 r}{3\varepsilon_0}.$$ I think this is what you got. But, as I mentioned above, I don’t believe your derivation is correct.

When you integrate the expression for $P'(r)$, you need to allow for a constant of integration. If there is no externally applied pressure at the surface of the sphere, then $P(R) = 0$. Use this boundary condition to find the constant of integration.

The Maxwell stress tensor components $T_{ij}$ are not directly related to the material pressure $P(r)$. In electrostatics, the Maxwell tensor $\overleftrightarrow{T}$ has the following property. If $S$ is a closed surface, then $\large \oint_S \normalsize \overleftrightarrow{T} \cdot \vec{dA}$ equals the net electric force on the total charge enclosed by $S$. Here, $\vec {dA}$ represents an outward pointing element of area of $S$. For example, if you let $S$ be the surface of the circular disk used in the discussion above, then evaluation of $\large \oint_S \normalsize \overleftrightarrow{T} \cdot \vec{dA}$ gives the result $\rho dV E(r)$ as expected.

 

[Rob2024]

@TSny, thanks. The integral using the stress tensor is $$\lim\limits_{dV \to 0} \oint \epsilon d \vec A = \iiint \nabla \epsilon dV = \iiint \frac{1}{r^2} \frac{d r^2 \epsilon_{rr}}{dr} d V = \frac{1}{r^2} \frac{d r^2 \epsilon_{rr}}{dr} dV = \frac{2 k Q^2 r}{R^6} dV $$ This is not equal to $\rho E(r) dV$. If you don't mind, can you demonstrate how you evaulated the net force integral using the stress tensor? Thanks,

 

[TSny]

Evaluate the surface integral $\large \oint_S \normalsize \overleftrightarrow{T} \cdot \vec{dA}$ over the surface of the disk in the figure of post #7 without invoking the divergence theorem. From the symmetry of the problem, the net electric force on the disk will be in the z direction in the figure. So, concentrate on the z-component of the surface integral: $$ \oint_S \left(\overleftrightarrow{T} \cdot \vec{dA}\right)_z = \int_{\rm bottom} \left(\overleftrightarrow{T} \cdot \vec{dA}\right)_z + \int_{\rm top} \left(\overleftrightarrow{T} \cdot \vec{dA}\right)_z +\int_{\rm rim} \left(\overleftrightarrow{T} \cdot \vec{dA}\right)_z $$
Note that in Cartesian components,$$ \left(\overleftrightarrow{T} \cdot \vec{dA}\right)_z = T_{zx} dA_x + T_{zy} dA_y + T_{zz} dA_z.$$ For the bottom surface of the disk, $dA_x = dA_y = 0$, and $dA_z = -dA$. So, for the bottom we have $$\int_{\rm bottom} \left(\overleftrightarrow{T} \cdot \vec{dA}\right)_z = -T_{zz}A = -\frac{\varepsilon_0}{2} E^2A = -\frac{\rho^2r^2}{18 \varepsilon_0}A$$ where $A$ is the area of the bottom of the disk. In the same way we get for the top surface $$\int_{\rm top} \left(\overleftrightarrow{T} \cdot \vec{dA}\right)_z = +T_{zz}A = \frac{\varepsilon_0}{2} E^2A = \frac{\rho^2(r+dz)^2}{18 \varepsilon_0}A$$ Adding the top and bottom contributions and dropping the higher order term proportional to $dz^2$, $$\int_{\rm top + bottom} \left(\overleftrightarrow{T} \cdot \vec{dA}\right)_z = \frac{\rho^2r}{9 \varepsilon_0}Adz = \frac{\rho^2r}{9 \varepsilon_0}dV.$$
Finally, consider a small patch of area $dA_R$ of the rim of the disk:

Choose our x-axis in the direction of $\vec{dA_R}$. Then $dA_{R, y} = dA_{R,z} = 0$. So for this patch of area, $$ \left(\overleftrightarrow{T} \cdot \vec{dA}\right)_z = T_{zx}\, dA_x + T_{zy}\, dA_y + T_{zz} \,dA_z = T_{zx}\, dA_R = \varepsilon_0 E_zE_x dA_R.$$ We can approximate $E_z = E\cos \theta \approx E$ and $E_x = E\sin \theta \approx E \dfrac{a}{r}$. [Previous sentence has been edited to correct a typo.] Also, we may approximate the distance from $O$ to the patch $dA_R$ by the distance $r$ shown in the figure (the distance from $O$ to the center of the bottom of the disk). Thus, $$ \left(\overleftrightarrow{T} \cdot \vec{dA}\right)_z \approx \varepsilon_0 E^2 \frac{a}{r}dA_R = \frac{\rho^2 r a}{9 \varepsilon_0}dA_R.$$ Summing all of the patches on the rim, $$ \int_{rim}\left(\overleftrightarrow{T} \cdot \vec{dA}\right)_z =\frac{\rho^2 r a}{9 \varepsilon_0} \int dA_R = \frac{\rho^2 r a}{9 \varepsilon_0}(2\pi a dz) = \frac{2\rho^2 r }{9 \varepsilon_0}(\pi a^2 dz) = \frac{2\rho^2 r }{9 \varepsilon_0}dV.$$ Adding this to the result for the top and bottom surfaces gives the net electric force on the disk: $\dfrac{3\rho^2 r }{9 \varepsilon_0}dV = \dfrac{\rho^2 r } {3 \varepsilon_0}dV = \rho E(r)dV.$

 

[Rob2024]

We can approximate $E_x$ and $E_z$...

@TSny, there appears to be a typo here, the rest looks reasonable... what if you used a curved element with the edges facing tangential direction instead of $x$ direction. It cannot be that the result depends on what coordinate system or what type of element is chosen...I am still checking. Thanks.

 

[TSny]

@TSny, there appears to be a typo here, the rest looks reasonable... what if you used a curved element with the edges facing tangential direction instead of $x$ direction. It cannot be that the result depends on what coordinate system or what type of element is chosen...I am still checking. Thanks.

Thanks for catching the typo.

For the curved element, I still get $\rho E dV$ for the surface integral of the Maxwell stress tensor.

In the figure $\theta$ is considered small and the "rectangle" should be rotated around the z-axis to generate the volume element. I hope this is the geometry that you mentioned. The stress forces on the top and bottom surfaces (blue) are due to $T_{rr}$ while the stress force on the rim is due to $T_{\theta \theta}$ (brown).

For the force on the bottom I get $F_z = - \dfrac {\pi \theta^2\rho^2}{18 \varepsilon_0} r^4$.

For the top I get $F_z = \dfrac {\pi \theta^2\rho^2}{18 \varepsilon_0} (r+dr)^4 \approx \dfrac {\pi \theta^2\rho^2}{18 \varepsilon_0} (r^4 + 4r^3 dr)$.

Together, these give $F_{\rm top+bottom, z} = \frac 2 9 \frac{\rho^2 r}{\varepsilon_0}dV$, where $dV = \pi (r \theta)^2 dr$.

For the rim I get $F_{\rm rim, z} = \frac 1 9 \frac{\rho^2 r}{\varepsilon_0}dV$.

So, the net force due to Maxwell stresses is $F_z = \frac 1 3 \frac{\rho^2 r}{\varepsilon_0}dV = \rho E(r) dV$.

 

[Rob2024]

I realized what the issue was, I neglected the $\sigma_{\theta \theta}$ and $\sigma_{\phi \phi}$ terms in the stress tensor. The stress tensor has 3 non-zero diagonal entries and the divergence of the tensor does give the correct answer, $$(\nabla \sigma)_r = \frac{1}{r^2} \frac{ \partial r^2 \sigma_{rr}}{ \partial r} - \frac{1}{r} (\sigma_{\theta \theta} + \sigma_{\phi \phi})= \frac{\rho^2 r}{3 \epsilon_0}$$

TSny made it clear there must be contributions from terms other than $\sigma_{rr}$. The stress tensor still holds mechanically.