0
$\begingroup$

The retarded fields for a moving charge are:

$$\mathbf{E}(\mathbf{r}, t) = \frac{1}{4 \pi \varepsilon_0} \left(\frac{q(\mathbf{n} - \boldsymbol{\beta})}{\gamma^2 (1 - \mathbf{n} \cdot \boldsymbol{\beta})^3 |\mathbf{r} - \mathbf{r}_s|^2} + \frac{q \mathbf{n} \times \big((\mathbf{n} - \boldsymbol{\beta}) \times \dot{\boldsymbol{\beta}}\big)}{c(1 - \mathbf{n} \cdot \boldsymbol{\beta})^3 |\mathbf{r} - \mathbf{r}_s|} \right)_{t_r}$$

and

$$\mathbf{B}(\mathbf{r}, t) = \frac{\mu_0}{4 \pi} \left(\frac{q c(\boldsymbol{\beta} \times \mathbf{n})}{\gamma^2 (1-\mathbf{n} \cdot \boldsymbol{\beta})^3 |\mathbf{r} - \mathbf{r}_s|^2} + \frac{q \mathbf{n} \times \Big(\mathbf{n} \times \big((\mathbf{n} - \boldsymbol{\beta}) \times \dot{\boldsymbol{\beta}}\big) \Big)}{(1 - \mathbf{n} \cdot \boldsymbol{\beta})^3 |\mathbf{r} - \mathbf{r}_s|} \right)_{t_r} = \frac{\mathbf{n}(t_r)}{c} \times \mathbf{E}(\mathbf{r}, t)$$

What are the corresponding advanced fields?

$\endgroup$
  • $\begingroup$ I've got to imagine that there is some information on this on the internet, or in standard E&M textbooks. Did you try checking any resources? Or did you make any attempt at a calculation? $\endgroup$ – David Z Mar 29 '18 at 22:11
  • $\begingroup$ @DavidZ the advanced solution is mentioned, but generally dismissed as non-causal and not pursued further so only the retarded potentials are calculated. I've had a go myself, but I doubt it's correct: the directions of the fields opposite to the retarded case and quantities evaluated at advanced rather than retarded time. $\endgroup$ – Physiks lover Mar 29 '18 at 22:17
  • $\begingroup$ It would still be helpful to mention that you've checked a few sources, and which ones. And if you can include an overview of the calculation you tried (without making the question excessively long), even better. That helps avoid duplication of effort, and it might point to a conceptual misunderstanding or something that would really help future readers. $\endgroup$ – David Z Mar 29 '18 at 22:29
1
$\begingroup$

The 4-potential of a moving charge is given by$^1$

$$ A^{\alpha}(x) = \frac {eV^{\alpha}(\tau)}{V\cdot[x - r(\tau)]}|_{\tau = \tau_0}$$

where $\tau_0$ is defined by the light-cone condition; $[x - r(\tau_0)]^2$ = 0, together with the retarded or advanced solution: $x_0 < r(\tau_0)$ or $x_0 > r(\tau_0)$.

Therefore the expressions for the retarded and advanced electric and magnetic fields only differ in either the retarded or advanced point being selected on the world-line of the moving charge, which is just replacing $t_r$ with $t_a$:

$$\mathbf{E}(\mathbf{r}, t) = \frac{1}{4 \pi \varepsilon_0} \left(\frac{q(\mathbf{n} - \boldsymbol{\beta})}{\gamma^2 (1 - \mathbf{n} \cdot \boldsymbol{\beta})^3 |\mathbf{r} - \mathbf{r}_s|^2} + \frac{q \mathbf{n} \times \big((\mathbf{n} - \boldsymbol{\beta}) \times \dot{\boldsymbol{\beta}}\big)}{c(1 - \mathbf{n} \cdot \boldsymbol{\beta})^3 |\mathbf{r} - \mathbf{r}_s|} \right)_{t_a}$$

$$\mathbf{B}(\mathbf{r}, t) = \frac{\mu_0}{4 \pi} \left(\frac{q c(\boldsymbol{\beta} \times \mathbf{n})}{\gamma^2 (1-\mathbf{n} \cdot \boldsymbol{\beta})^3 |\mathbf{r} - \mathbf{r}_s|^2} + \frac{q \mathbf{n} \times \Big(\mathbf{n} \times \big((\mathbf{n} - \boldsymbol{\beta}) \times \dot{\boldsymbol{\beta}}\big) \Big)}{(1 - \mathbf{n} \cdot \boldsymbol{\beta})^3 |\mathbf{r} - \mathbf{r}_s|} \right)_{t_a} = \frac{\mathbf{n}(t_a)}{c} \times \mathbf{E}(\mathbf{r}, t)$$


1 Classical Electrodynamics, Jackson, page 662

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.