# I Verify a Formula for Poynting Vector?

1. Sep 29, 2016

### Twigg

Hi all,

I derived a formula last night for the Poynting vector I have never seen before, and wanted some verification and perhaps insight on.

I started with the definition for the Poynting vector in free space: $$\vec{S} = \frac{1}{\mu_{0}} \vec{E} \times \vec{B}$$ and substituted the potential formalism for radiation in free space: $$\vec{E} = -\frac{\partial \vec{A}}{\partial t}$$ $$\vec{B} = \nabla \times \vec{A}$$

Much algebra later, I wind up with the following formula in index notation using the standard Cartesian basis:

$$S_{i} = \frac{1}{\mu_{0}} \frac{\partial A^{j}}{\partial t}(\frac{\partial A_{i}}{\partial x^{j}} - \frac{\partial A_{j}}{\partial x^{i}})$$

As far as verification, I don't see any obvious signs that it's false. The units appear to add up (A has units of $T*m$ and $\mu_{0}$ has units of $T^{2} * m^{3} / J$, so the whole thing has units of $J / m^{2} * s$, which is the units of S). Also, the formula is independent of $\nabla \cdot \vec{A}$, since the $\frac{\partial A_{i}}{\partial x^{j}} - \frac{\partial A_{j}}{\partial x^{i}}$ term is traceless, which makes the formula gauge-invariant. All I can tell is that it doesn't break any of the core rules, although it's not a covariant formula (since Lorentz transformations would transform some of the Poynting vector into momentum density, IIRC). I'm still not 100% convinced though. Does anyone have a reference to confirm/disprove this expression?

2. Oct 4, 2016

### Jonathan Scott

As a general vector cross-product identity, I think the following general scheme applies (at least my handwritten notes say so - it should be easy enough to work out) and if you apply that to the above case with your expressions for the field components (and taking account of the minus sign in E), I think it gives the answer which you quote.

$$P \times (Q \times R) = P_j Q_i R_j - P_j Q_j R_i = P_j \, (Q_i R_j - Q_j R_i)$$

3. Oct 4, 2016

### Twigg

I completely forgot about that identity. Would've saved a lot of time. Thanks!

4. Oct 5, 2016

### Jonathan Scott

The identity is usually written using scalar products as something like $Q (P.R) - (P.Q) R$ but that form is not useful when the terms do not commute, as in this case where one of the terms is the del operator. I see that the Wikipedia entry for Triple product shows this identity (on the last line of the main text, just before the Notes) in both forms.