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http://physics.stackexchange.com/questions/93390/field-of-moving-charge-lorentzlienard-wiechert

The Lorentz contraction of the electric field of a charge with uniform velocity is supposed to be symmetric across the plane pi/2 radians from the velocity vector of the charge. Now if that is true, then I should expect that the electric field of the charge should have the same magnitude on both sides, in the sense of a mirror. In other words, if I were to reverse the component of the relative position vectorr-rthat is co-linear with the uniform velocity of charge q, then the co-linear component of the electric field derived from the Liénard–Wiechert potentials should likewise reverse, due to the obvious symmetry. However, upon examining the equation for the electric field according to the Liénard–Wiechert potentials:_{S}

https://wikimedia.org/api/rest_v1/media/math/render/svg/1956a7a54e74bcfee533510e33d219646a81e75f

....this appears to not be the case.nin this example is a unit vector based onr-r. In the simple case that the velocity is constant, the second term in the parentheses drops out. Let's make things simple by setting_{S}beta= 0.5 times theunit velocity vector. Then in the case that the unit radial vector is aligned with the unit velocity vector, then the contents contained within the parenthesis of the numerator of the first term has the magnitude (1-0.5) or 0.5, while the parentheses in the denominator would have the value (1-1*0.5) which has to be cubed. The magnitude of the ratio of former and the cube of the latter is (1-0.5)/(1-1*0.5)^3 = 1/(1-0.5)^2 = 1/0.5^2 = 4. What happens then if I reverse the direction ofn? In that case, the result is (-1-0.5)/(1-(-1)*0.5)^3 = (-1.5)/(1.5)^3 = -1/(1.5)^2 = -4/9.

How can this result be reconciled with the claim that the electric field is symmetric across pi/2?

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# I The Symmetry of the Liénard–Wiechert potentials

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