- #1

- 552

- 18

We have a retarded magnetic vector potential ##\mathbf{A}(\mathbf{r},t) = \dfrac{\mu_0}{4\pi} \int \dfrac{\mathbf{J}(\mathbf{r}',t_r)}{|\mathbf{r}-\mathbf{r}'|} \mathrm{d}^3 \mathbf{r}'##

And its curl, ##\mathbf{B}(\mathbf{r}, t) = \frac{\mu_0}{4 \pi} \int \left[\frac{\mathbf{J}(\mathbf{r}', t_r)}{|\mathbf{r}-\mathbf{r}'|^3} + \frac{1}{|\mathbf{r}-\mathbf{r}'|^2 c}\frac{\partial \mathbf{J}(\mathbf{r}', t_r)}{\partial t} \right] \times (\mathbf{r}-\mathbf{r}') \,\mathrm{d}^3 \mathbf{r}'##

I would like to express

In this paper, equation 35 looks like a promising step on the way there:

$$\mathbf{A}=\frac{1}{4 \pi c^{2}} \frac{\partial}{\partial t} \int \frac{\left[\boldsymbol{\nabla}^{\prime} \Phi+\partial \mathbf{A} / \partial t\right]}{R} d^{3} r^{\prime}+\boldsymbol{\nabla} \times \int \frac{[\mathbf{B}]}{4 \pi R} d^{3} r^{\prime}$$

Unfortunately, it has ∂

I'm not sure if the correct answer involves

If its easier to express ∂

And its curl, ##\mathbf{B}(\mathbf{r}, t) = \frac{\mu_0}{4 \pi} \int \left[\frac{\mathbf{J}(\mathbf{r}', t_r)}{|\mathbf{r}-\mathbf{r}'|^3} + \frac{1}{|\mathbf{r}-\mathbf{r}'|^2 c}\frac{\partial \mathbf{J}(\mathbf{r}', t_r)}{\partial t} \right] \times (\mathbf{r}-\mathbf{r}') \,\mathrm{d}^3 \mathbf{r}'##

I would like to express

**A**in terms of**B**In this paper, equation 35 looks like a promising step on the way there:

$$\mathbf{A}=\frac{1}{4 \pi c^{2}} \frac{\partial}{\partial t} \int \frac{\left[\boldsymbol{\nabla}^{\prime} \Phi+\partial \mathbf{A} / \partial t\right]}{R} d^{3} r^{\prime}+\boldsymbol{\nabla} \times \int \frac{[\mathbf{B}]}{4 \pi R} d^{3} r^{\prime}$$

Unfortunately, it has ∂

**A**/∂t on the RHS, and it also has a**∇'Φ**.I'm not sure if the correct answer involves

**Φ**If its easier to express ∂

**A**/∂t in terms of**B**, rather than**A**in terms of**B**, that'll do just fine as well.
Last edited: