Can we simplify the following surface integral in some way, or say anything about the properties it must have? $$ \oint_{\partial\mathbb{B}\left(\vec{0},R\right)} \left|\vec{U}\left(\vec{r}\right) \times \hat{r} \right|^2 \;ds\left(\vec{r}\right) $$ $\vec{U}$ is a continuous vector field the value of which depends on the position $\vec{r} \in \mathbb{R}^3$, while $\mathbb{B}\left(\vec{0},R\right)$ is a sphere centered on the origin ($\vec{0}$) with radius $R$. The unit vector $\hat{r}$ is the unit normal (pointing outward) to the sphere at position $\vec{r}$ (so $\vec{r}=R\;\hat{r}$).

Is it, specifically, possible to say that this integral will be proportional to $R^2$, for instance?


Update: It has been pointed out that I need to provide some more details about the field $\vec{U}$, so here it is. Actually, the $\vec{U}$ I'm interested in is time dependent, and is defined as: $$ \vec{U}\left(\vec{r}, t\right) = \iiint_{\mathbb{B}\left(\vec{0},R_s\right)} \vec{J}\left(\vec{a},t-\frac{\left|\vec{r}-\vec{a}\right|}{c} \right)\;dV\left(\vec{a}\right) $$ Here, $\mathbb{B}\left(\vec{0},R_s\right)$ is another sphere centered on the origin, with a radius $R_s < R$, while $c$ is a constant. $\vec{J}$ is another vector field which is not necessarily continuous, and has the following properties: $$ \forall t, \quad \forall \vec{a} \in \partial\mathbb{B}\left(\vec{0},R_s\right)\qquad \vec{J}\left( \vec{a}, t \right) = 0 $$ $$ \forall t, \quad \forall \vec{a} \notin \mathbb{B}\left(\vec{0},R_s\right)\qquad \vec{J}\left( \vec{a}, t \right) = 0 $$ That is, $\vec{J}$ is zero on the surface of and outside $\mathbb{B}\left(\vec{0},R_s\right)$.

Update 2: The following may or may not be true for my particular $\vec{U}$:

$$ \iiint_{\mathbb{B}\left(\vec{0},R_s\right)} \vec{J}\left(\vec{a},t\right)\;dV\left(\vec{a}\right) = 0 $$

[Due to a brain freeze on my part, I had previously thought that this must be the case, but it isn't really. Sorry for the confusion there...]

(This stuff comes from classical electrodynamics, as you can probably guess!)

  • 1
    $\begingroup$ Unless you provide more information about $U$, nothing is known about the integral aside from the obvious fact it is non-negative. At the least, you should tell us how you get your $U$. $\endgroup$ Aug 6 '13 at 6:42
  • $\begingroup$ Thanks @achillehui -- I have updated my question to add some information about my $\vec{U}$. $\endgroup$
    – Avijit
    Aug 6 '13 at 7:21
  • $\begingroup$ If $\nabla \cdot \vec{J} = 0$, you do have $\int_{\mathbb{B}(\vec{0},R_s)} \vec{J}(a,t) d V(\vec{a}) = \vec{0}$. If not but $\vec{J}$ satisfies some sort of conservation law $\frac{\partial \rho}{\partial t} + \nabla\cdot\vec{J} = 0$, then your integral will be equal to $\frac{\partial}{\partial t} \int_{\mathbb{B}(\vec{0},R_s)} \rho(\vec{a},t) \vec{a} dV(\vec{a})$. As long as your "charge" $\rho$ didn't accumulate at some points indefinitely, then over long enough time scale, your integral over $\vec{J}$ will time average to 0. $\endgroup$ Aug 8 '13 at 6:29
  • $\begingroup$ I had two things in mind -- [1] My $\vec{U}$ could refer to the volume integral for $\vec{J}$ as well as for $\frac{\partial\vec{J}}{\partial t}$, and [2] I was interested in the instantaneous values of $\vec{U}$ (i.e. not time averaged) $\endgroup$
    – Avijit
    Aug 8 '13 at 6:45

For simplicity of discussion, let us assume your $\vec{J}$ has only one relevant length scale $R_s$, i.e.

$$\left|\frac{\partial \vec{J}(\vec{a},t)}{\partial \vec{a}}\right| \sim~ O\left(\frac{|\vec{J}(\vec{a},t)|}{R_s}\right)$$ If that is the case, then $\vec{J}$ has only two relevant time scales $T_s$ and $T_c$. $T_s = \frac{R_s}{c}$ is the time scale for a light beam to travel across the inner sphere of radius $R_s$. $T_c$ is the time scale for temporal changes of $\vec{J}$. i.e. $$\left|\frac{\partial \vec{J}(\vec{a},t)}{\partial t}\right| \sim~ O\left(\frac{|\vec{J}(\vec{a},t)|}{T_s}\right)$$

If $T_c \gg T_s$, $\vec{J}$ is slowly varying in time and in the defining integral of $\vec{U}$, the $t$ dependence effectively goes away. In this limit,

$$ \vec{U}\left(\vec{r}, t\right) = \int_{\mathbb{B}\left(\vec{0},R_s\right)} \vec{J}\left(\vec{a},t-\frac{\left|\vec{r}-\vec{a}\right|}{c} \right)\;dV\left(\vec{a}\right) \rightarrow \int_{\mathbb{B}\left(\vec{0},R_s\right)} \vec{J}\left(\vec{a},t-\frac{\left|\vec{r}\right|}{c} \right)\;dV\left(\vec{a}\right) = 0 $$

If $T_c \ll T_s$, then for any two points inside $\mathbb{B}(0,R_s)$ separated by a distance $c T_c$, the spatial variation of $\vec{J}$ is significantly smaller than the temporal variation. In the defining integral of $U$, nothing stop these two points from destructively interfere with each other. One should expect $\vec{U}$ vanishes in this limit too.

This leaves us the case $T_c \sim O(T_s)$. Let's imagine we perform some sort of multipole expansion of the angular dependence of $\vec{J}$. Once again, the assumption of single length scale $R_s$ tells us in the multipole expansion, the lower term is more important.

If the multipole expansion is indeed dominated by the lowest term or $T_c$ is big (but not too big) compared to $T_s$, then for a point $\vec{r}$ at a distance $|\vec{r}| = R \gg R_s$, the leading contribution to $\vec{U}(\vec{r},t)$ should take the form:

$$ \begin{align} \vec{U}\left(\vec{r}, t\right) \sim & \int_{\mathbb{B}\left(\vec{0},R_s\right)} \frac{\partial\vec{J}}{\partial t}\left(\vec{a},t-\frac{R}{c} \right) \left(\frac{|\vec{r}|-|\vec{r}-\vec{a}|}{c}\right) \;dV\left(\vec{a}\right)\\ = & \int_{\mathbb{B}\left(\vec{0},R_s\right)} \frac{\partial\vec{J}}{\partial t}\left(\vec{a},t-\frac{R}{c} \right) \left(\frac{\vec{a}\cdot\hat{r}}{c} \right) \;dV\left(\vec{a}\right) + O\left(\frac{R_s}{R}\right)\\ \end{align} $$ where $\displaystyle \hat{r} = \frac{\vec{r}}{|\vec{r}|}$ is the unit vector in $\vec{r}$ direction.

Let $\displaystyle n_{\mu} = \frac{r_{\mu}}{R}$ be the components of the unit vector $\hat{r}$. Let $T_{\mu\nu}(t)$ be the $3 \times 3$ tensor defined by: $$T_{\mu\nu}(t) = \frac{1}{c}\int_{\mathbb{B}\left(\vec{0},R_s\right)} \frac{\partial\vec{J}_{\mu}(\vec{a},t)}{\partial t} a_{\nu} \;dV\left(\vec{a}\right) $$ Using Einstein summation convention, we have:

$$\left( \vec{U}(\vec{r},t) \times \hat{r} \right)_\mu = \epsilon_{\mu\nu\omega} U_{\nu}(\vec{r},t) n_{\omega} \sim \epsilon_{\mu\nu\omega} T_{\nu\rho}(t-\frac{R}{c}) n_\rho n_{\omega}$$

where $\epsilon_{\mu\nu\omega}$ is the Levi-Civita symbol. Notice $d S(\vec{r}) = R^2 d\hat{r}$, the surface integral becomes:

$$\begin{align} & \int_{\partial\mathbb{B}\left(\vec{0},R\right)} \left|\vec{U}\left(\vec{r}\right) \times \hat{r} \right|^2 \;dS\left(\vec{r}\right)\\ \sim & R^2 T_{\nu\rho} T_{\nu'\rho'} \left\{ \epsilon_{\mu\nu\omega} \epsilon_{\mu\nu'\omega'} \int n_{\rho} n_{\omega} n_{\rho'} n_{\omega'} d\hat{r}\right\}\\ = & R^2 T_{\nu\rho} T_{\nu'\rho'} \left\{ (\delta_{\nu\nu'}\delta_{\omega\omega'} - \delta_{\nu\omega'}\delta_{\nu'\omega}) \int n_{\rho} n_{\omega} n_{\rho'} n_{\omega'} d\hat{r}\right\}\\ = & R^2 \left\{ T_{\nu\rho} T_{\nu\rho'} \int n_{\rho} n_{\rho'} d\hat{r} - T_{\nu\rho} T_{\nu'\rho'} \int n_{\rho} n_{\nu'} n_{\rho'} n_{\nu} d\hat{r} \right\}\\ = & R^2 \left\{ \frac{4\pi}{3} T_{\nu\rho} T_{\nu\rho} - \frac{4\pi}{15} ( T_{\nu\nu} T_{\rho\rho} + T_{\nu\rho} T_{\nu\rho} + T_{\nu\rho}T_{\rho\nu} ) \right\}\\ = & \frac{4\pi R^2}{15}\left\{4 T_{\nu\rho} T_{\nu\rho} - T_{\nu\nu} T_{\rho\rho} - T_{\nu\rho} T_{\rho\nu} \right\}\tag{*} \end{align}$$ where $\delta_{\mu\nu}$ is the Kronecker delta and we have used the identities:

$$\int n_{\mu} n_{\nu} d\hat{r} = \frac{4\pi}{3}\delta_{\mu}{\nu} \quad\text{ and }\quad \int n_{\mu} n_{\nu} n_{\omega} n_{\rho} d\hat{r} = \frac{4\pi}{15} ( \delta_{\mu\nu} \delta_{\omega\rho} + \delta_{\mu\omega}\delta_{\nu\rho} + \delta_{\mu\rho}\delta_{\nu\omega} ) $$

For general $\vec{J}$, your surface integral is most likely to grow like $R^2$.


For the benefit of those not familiar with tensor, let me work out an important special case using vector. If $\nabla\cdot\vec{J} = 0$, then one can use the fact $\vec{J} = \vec{0}$ on $\partial\mathbb{B}(0,R_s)$ to show $T_{\mu\nu}$ is an antisymmetric tensor. In particular, this implies one can find two vectors $\vec{A}$, $\vec{B}$ such that

$$T_{\mu\nu} = A_{\mu}B_{\nu} - B_{\mu}A_{\nu}$$

If one substitute this into $(*)$, the surface integral becomes: $$\begin{align} & \frac{4\pi R^2}{15}\left( 4 ( A_{\nu}B_{\rho} - B_{\nu}A_{\rho})( A_{\nu}B_{\rho} - B_{\nu}A_{\rho} ) - ( A_{\nu}B_{\rho} - B_{\nu}A_{\rho})( A_{\rho}B_{\nu} - B_{\rho}A_{\nu} ) \right)\\ = & \frac{8\pi R^2}{3} \left( |\vec{A}|^2 |\vec{B}|^2 - (\vec{A}\cdot\vec{B})^2 \right) = \frac{8\pi R^2}{3} |\vec{A} \times \vec{B} |^2 \end{align}$$

In terms of vector $$U_{\mu} = T_{\mu\nu} n_{\nu} \quad\implies\quad \vec{U}(\vec{r}) = \vec{A} (\vec{B}\cdot\hat{r} ) - \vec{B} (\vec{A}\cdot\hat{r}) = \hat{r} \times ( \vec{A} \times \vec{B} ) $$ and we can evaluate the surface integral as: $$ \begin{align} & R^2 \int |( \hat{r} \times ( \vec{A} \times \vec{B} )) \times \hat{r}|^2 d\hat{r} = R^2 \int \left( |\vec{A} \times \vec{B}|^2 - ( ( \vec{A} \times \vec{B} ) \cdot \hat{r} )^2 \right) d\hat{r}\\ = & 2\pi R^2 |\vec{A} \times \vec{B}|^2 \int_{0}^{\pi} (1 - \cos^2\theta) \sin\theta d\theta = \frac{8\pi R^2}{3} |\vec{A} \times \vec{B}|^2 \end{align}$$ using suitable chosen polar coordinates. Of course, we get the same answer we obtained before using tensors.

For the more general case when $T_{\mu\nu}$ is not antisymmetric, computing the surface integral using vectors is actually more complicated and horrible than using tenors. I'll skip the mess here.

  • $\begingroup$ Hi @achillehui -- small question -- by $\hat{n}(\vec{r})$ I actually meant $\hat{r}$ [just in case I wasn't clear] so I'm really asking for $\oint |\vec{U}(\vec{r}) \times \hat{r}|^2 dS(\vec{r})$ -- does that change anything in your answer? (sorry I am not up to speed on tensors, so couldn't follow the last part of your answer completely -- would understand vectors in $\mathbb{R}^3$ though, could you please explain that way? Thanks...) $\endgroup$
    – Avijit
    Aug 6 '13 at 17:44
  • $\begingroup$ @Avijit In that case, the integral does grow like $R^2$, let me fix my answer. $\endgroup$ Aug 6 '13 at 17:48
  • $\begingroup$ Thanks @achillehui -- that was real helpful. $\endgroup$
    – Avijit
    Aug 6 '13 at 18:02

In addition to the full answer posted by @achillehui above, I'd like to add something that I should have thought of earlier (and in which case I wouldn't even have asked the question, as it turns out!)

We note that $\left|\vec{U}\left(\vec{r}\right) \times \hat{r}\right|^2$ is a scalar function or $\vec{r}$. So now if we choose a spherical coordinate system $\langle 0\le \mathrm{r} \lt \infty, \;0 \le \theta \le 2\pi,\; 0 \le \phi \le \pi\rangle$ with the origin at $\vec{0}$, we find that $$ds\left(\vec{r}\right) = \left(R\;\sin{\phi}\;d\phi\right)\left(R\;d\theta\right)$$ ... and since $R$ is a constant $\forall \vec{r} \in \partial\mathbb{B}\left(\vec{0},R\right)$, the the surface integral becomes:

$$ \oint_{\partial\mathbb{B}\left(\vec{0},R\right)} \left|\vec{U}\left(\vec{r}\right) \times \hat{r} \right|^2 \;ds\left(\vec{r}\right) = R^2 \int_0^{2\pi} \int_0^{\pi} \left|\vec{U}\left(\vec{r}\right) \times \hat{r} \right|^2 \sin \phi \;d\phi\;d\theta $$

So while the surface integral isn't proportional to $R^2$ (as the double integral is not a constant), it always contains $R^2$ as a factor. (This property was, in fact, all I needed for what I was doing.)


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