I was trying to read up on the effects of the Lorentz transforms on electromagnetic fields in the stationary and moving frames and came across the following equation for the E-field:(adsbygoogle = window.adsbygoogle || []).push({});

[1] [tex] E= \frac {Q}{4 \pi \epsilon_0 r^2}* \frac {1- \beta^2}{(1- \beta^2 sin^2 \theta) ^{3/2}} [/tex]

The first part of the equation appears to be the standard electrostatic expression derived from Coulomb’s law, while the second expression encompasses a relativistic component based on velocity [tex]( \beta = v/c}) [/tex] and the angle [tex]( \theta )[/tex] formed between the position, charge and angle of motion along the x-axis. As such, I am assuming [1] returns the value of (E) as measured in the moving frame. If so, I would have thought you could determine the relative strength of [Ex] and [Ey] to compare against the value predicted by the Lorentz transforms in [3]. As the value of [Ex] aligns to the direction of charge motion, (sin0=0), while [Ey] is perpendicular to the motion (sin90=1). If so, the relativistic component of [1] would reduce to:

[2] [tex]E_X \Rightarrow 1- \beta ^2; E_Y \Rightarrow \frac {1}{(1- \beta^2) ^{1/2}} [/tex]

By way of orientation, the following Lorentz transforms define the prime (*) frame being stationary, while in the moving frame, the velocity (v) of the source charge [Q] is again along the x-axis:

[3] [tex] E_X^* = E_X; E_Y^* = \frac {E_Y - vB_Z}{(1- \beta^2) ^{1/2}}; E_Z^* = \frac {E_Z + vB_Y}{(1- \beta^2)^{1/2}} [/tex]

Now there appears to be some discrepancy in the implications of [2] and [3]. We know that the moving frame will have an associated magnetic (B) field, which will not exist in the stationary (prime*) frame. However, I am assuming that [3] can be reversed by simply changing the sign of [v], since the moving frame moves with a negative velocity with respect to the stationary frame:

[4] [tex] E_X = E_X^*; E_Y = \frac {E_Y^* + vB_Z^*}{(1- \beta^2) ^{1/2}}; E_Z = \frac {E_Z^* - vB_Y^*}{(1- \beta^2) ^{1/2}} [/tex]

However, we also know that there can be no magnetic fields in the stationary (prime*) frame; therefore [4] can presumably reduce to the form:

[5] [tex] E_X = E_X^*; E_Y = \frac {E_Y^*}{(1- \beta^2) ^{1/2}}; E_Z = \frac {E_Z^*}{(1- \beta^2) ^{1/2}} [/tex]

As such, I would have thought that the relativistic components in [2] and [5] should be directly comparable. However, this is only true for the E-fields perpendicular to the motion, i.e. Ey & Ez, but not (Ex).

Any thoughts?

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# Relativistic Effects on the E-field

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