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Imagine a stationary charge [itex]q[/itex] located on the positive [itex]y[/itex]-axis at a distance [itex]r[/itex] from a stationary observer at the origin.

Let us assume that the distance [itex]r[/itex] is large enough such that the electrostatic field due to the charge is negligible at the origin.

Now let us assume that the charge [itex]q[/itex] is given an acceleration in the positive [itex]x[/itex]-direction.

The electric field at the origin at time [itex]t[/itex] is directed along the [itex]x[/itex]-axis and is given by

[tex]

E_x(t) = \frac{-q}{4 \pi \varepsilon_0 c^2 r} a_x(t-\frac{r}{c}).

[/tex]

This is the standard retarded expression for the electric field due to an accelerated charge measured by a stationary observer at the origin. The electric field occurs at time [itex]t[/itex] after the acceleration of the charge at time [itex]t-r/c[/itex].

Now let us imagine the complementary situation.

We start again with a stationary charge [itex]q[/itex] located on the [itex]y[/itex]-axis at a distance [itex]r[/itex] from a stationary observer at the origin.

Now let us assume that the

I believe the electric field at the origin at time [itex]t[/itex]

[tex]

E_x(t) = \frac{q}{4 \pi \varepsilon_0 c^2 r} a_x(t+\frac{r}{c}).

[/tex]

This is the advanced expression for the electric field due to a stationary charge as measured by an accelerating observer. From the observer's point of view the charge has an apparent acceleration [itex]-a_x(t+r/c)[/itex]. The electric field occurs at time [itex]t[/itex] before the apparent acceleration of the charge at time [itex]t+r/c[/itex]. We would expect this result as the field should come into existance as soon as the observer accelerates and thus before the apparent acceleration of any charge.

Thus there is an equivalence between charge accelerating/stationary observer and stationary charge/accelerating observer. The former situation is described by a retarded field and the latter by an advanced field.

Is this correct?

Let us assume that the distance [itex]r[/itex] is large enough such that the electrostatic field due to the charge is negligible at the origin.

Now let us assume that the charge [itex]q[/itex] is given an acceleration in the positive [itex]x[/itex]-direction.

The electric field at the origin at time [itex]t[/itex] is directed along the [itex]x[/itex]-axis and is given by

[tex]

E_x(t) = \frac{-q}{4 \pi \varepsilon_0 c^2 r} a_x(t-\frac{r}{c}).

[/tex]

This is the standard retarded expression for the electric field due to an accelerated charge measured by a stationary observer at the origin. The electric field occurs at time [itex]t[/itex] after the acceleration of the charge at time [itex]t-r/c[/itex].

Now let us imagine the complementary situation.

We start again with a stationary charge [itex]q[/itex] located on the [itex]y[/itex]-axis at a distance [itex]r[/itex] from a stationary observer at the origin.

Now let us assume that the

*observer*is given an acceleration in the positive [itex]x[/itex]-direction while the charge [itex]q[/itex] remains stationary.I believe the electric field at the origin at time [itex]t[/itex]

*in the frame of the accelerated observer*is given by[tex]

E_x(t) = \frac{q}{4 \pi \varepsilon_0 c^2 r} a_x(t+\frac{r}{c}).

[/tex]

This is the advanced expression for the electric field due to a stationary charge as measured by an accelerating observer. From the observer's point of view the charge has an apparent acceleration [itex]-a_x(t+r/c)[/itex]. The electric field occurs at time [itex]t[/itex] before the apparent acceleration of the charge at time [itex]t+r/c[/itex]. We would expect this result as the field should come into existance as soon as the observer accelerates and thus before the apparent acceleration of any charge.

Thus there is an equivalence between charge accelerating/stationary observer and stationary charge/accelerating observer. The former situation is described by a retarded field and the latter by an advanced field.

Is this correct?

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