Poynting’s theorem -- Check the relation of the energy balance

[Lambda96]

Homework Statement

Check the relation of the energy balance $\partial_t \omega + \nabla \cdot \vec S=- \vec j \cdot \vec E$

Relevant Equations

none

Hi

I have a problem with task c)

For the Poynting vector $\vec S$ and the energy density $\omega$ I got the following:

$$\vec S= \frac{1}{\mu_0} \frac{4 I_0^2}{c \varrho} \frac{t}{c^2 t^2 - \varrho^2} \Theta(ct-\varrho)^2 \vec{e}_\varrho$$
$$\omega= \frac{1}{2} \epsilon_0 \frac{4 I_0^2 \Theta(ct- \varrho)(\varrho^2+c^4t^2)}{c^4\varrho^2t^2-\varrho^4 c^2}$$

Then I calculated the following

$$\partial_t \omega=\frac{4 I_0^2 \epsilon_0(c^4t^2+\varrho^2)\delta(ct-\varrho)\Theta(ct-\varrho)}{c^3\varrho^2t^2-c\varrho^4}-\frac{4 I_0^2 (c^2+1)t \epsilon_0 \Theta(ct-\varrho)^2}{(c^2t^2-\varrho^2)^2}$$

and the i used the hint $\nabla (f \vec v)=(\nabla f) \cdot \vec v + f \nabla \cdot \vec v$

$$(\nabla f) \cdot \vec v + f \nabla \cdot \vec v=\Biggl( - \frac{8 I_0 t \delta(ct-\varrho) \Theta(ct-\varrho)}{c^3 \mu t^2 \varrho-c \mu \varrho^3}-\frac{4 I_0 t (c^2t^2-3 \varrho^2)\Theta(ct-\varrho)^2}{c \mu (\varrho^3-c^2 t^2 \varrho)^2} \Biggr) \cdot \vec e_\varrho + \frac{1}{\mu_0} \frac{4 I_0^2}{c \varrho} \frac{t}{c^2 t^2 - \varrho^2} \Theta(ct-\varrho)^2 \cdot \vec e_\varphi$$

Unfortunately I don't know what I can do with it, the results are quite a mess or have I miscalculated and also in the task part b) ?

 

[anuttarasammyak]

The relation
\theta(x)^2=\theta(x)
might easen your calculation. In your calculation of

\omega= \frac{1}{2} \epsilon_0 \frac{4 I_0^2 \Theta(ct- \varrho)(\varrho^2+c^4t^2)}{c^4\varrho^2t^2-\varrho^4 c^2}

I find $\varrho^2+c^4t^2$ is strange in dimension.

 

[Lambda96]

Thank you anuttarasammyak for your help and the tip


With $\Theta(x)^2=\Theta(x)$ the term now looks like this:

$$\partial_t \omega=\frac{2 I_0^2 \epsilon_0(c^4t^2+\varrho^2)\delta(ct-\varrho)}{c^3\varrho^2t^2-c\varrho^4}-\frac{4 I_0^2 (c^2+1)t \epsilon_0 \Theta(ct-\varrho)}{(c^2t^2-\varrho^2)^2}$$

Unfortunately, I still can't get any further with this term and I have recalculated and again get the term $\varrho^2+c^4t^2$ for $\omega$

 

[renormalize]

I have recalculated and again get the term $\varrho^2+c^4t^2$ for $\omega$

You clearly have an algebra error. The term $\varrho^2+c^4t^2$ is dimensionally inconsistent because $\varrho\sim L,c\sim\frac{L}T{},t\sim T$.

 

[TSny]

The problem statement expresses the fields in Gaussian units. In these units the Poynting vector is $\mathbf S = \dfrac {c}{4 \pi} \mathbf E \times \mathbf B$ and the energy density is $w = \dfrac{1}{8\pi}(E^2 + B^2)$.

Using the expressions for the fields given in the problem I find $$\mathbf S = \left[ \frac{I_0^2}{\pi} \frac{t}{c^2t^2 - \rho^2} \theta(ct-\rho) \right] \left(\frac{\hat e_{\rho}}{\rho} \right).$$ This is the same as your result except for differences in the constants. I could have made mistakes, so you can see whether or not you can get the same expression. The reason for writing the last factor as $\dfrac{\hat e_{\rho}}{\rho}$ is to prepare for using the divergence identity given in the problem statement.

The problem statement says $\nabla \cdot \left( \dfrac{\hat e_{\rho}}{\rho} \right) = 2\pi \delta(\rho)$. Actually, this should be $\nabla \cdot \left( \dfrac{\hat e_{\rho}}{\rho} \right) = 2\pi \delta^{(2)} (\boldsymbol {\rho})$, where the right side is the two-dimensional delta function $\delta^{(2)} (\boldsymbol {\rho})$ and $\boldsymbol {\rho}$ is the two-dimensional position vector for points in a plane perpendicular to the z-axis. The dimension of $\delta(\rho)$ is inverse length; whereas, the dimension of $\delta^{(2)} (\boldsymbol {\rho})$ is inverse area.

$\delta^{(2)} (\boldsymbol {\rho})$ is useful for expressing the current density for this problem: $$\mathbf J(\boldsymbol {\rho}, t) = I(t)\delta^{(2)} (\boldsymbol {\rho}) \hat e_z$$

 

[graphking]

Very gross understanding towards poynting, what is j and E vector?

 

[TSny]

Very gross understanding towards poynting, what is j and E vector?

E is the electric field.

j is the current density vector. For current $I$ in a wire, j is a vector with magnitude equal to the current per unit cross-sectional area of the wire and j points in the direction of the current. For the limiting case of an infinitely thin wire, j can be expressed in terms of the current $I$ and a Dirac delta function. See the last equation in post #5.

 

[Lambda96]

Thanks for your help and explanation TSny

I have repeated the calculation again with Gaussian units and get the same result for the Poynting vector


You were right, instead of $\nabla \cdot \left( \dfrac{\hat e_{\rho}}{\rho} \right) = 2\pi \delta(\rho)$ it should be $\nabla \cdot \left( \dfrac{\hat e_{\rho}}{\rho} \right) = 2\pi \delta^{(2)} (\boldsymbol {\rho})$. But my lecturer wrote it as $\nabla \cdot \left( \dfrac{\hat e_{\rho}}{\rho} \right) = 2\pi \delta(x) \delta(y)$

 

[TSny]

my lecturer wrote it as $\nabla \cdot \left( \dfrac{\hat e_{\rho}}{\rho} \right) = 2\pi \delta(x) \delta(y)$

Good.