What is 4-Vector Potential Transformation under Gauge Fixing?

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4-vector potential transformation under gauge fixing involves changing the 4-vector potential without altering the electric (E) and magnetic (B) fields. This is achieved by modifying the 3-vector part of the potential, as changes that add a 3-vector with zero curl do not affect the B field. Gauge fixing simplifies calculations by imposing specific conditions on the 4-vector potential, allowing for easier problem-solving. Common gauge fixing conditions exist to streamline these calculations while maintaining the gauge-invariant nature of the equations. Ultimately, this ensures that results remain consistent across different gauge choices, provided no errors occur.
Kulkarni Sourabh
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4- vector potential transformation under Gauge fixing.
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What is 4- vector potential transformation under Gauge fixing ?
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A gauge transform is a change in the 4-vector potential that leaves the ##E## and ##B## fields unchanged. So, for example, consider the 3-vector part of ##A##. Since the ##B## field is the curl of the 3-vector part of ##A##, then changing ##A## by adding a 3-vector with zero curl does not change ##B##.

Gauge fixing means to select particular conditions on ##A## to simplify the calculations you are currently doing. You do this by adding the corresponding things to ##A## such that the required condition is true. There are a number of commonly used gauge fixing conditions.

They are useful because we usually express the original version of the equations in a gauge-invariant (or covariant) form. That means we write all the equations in such a way that they are still true regardless of the gauge conditions we apply. That means if we were to do the calculation in another gauge we would necessarily get the same answer. Assuming, of course, we didn't make a mistake.
 
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I want to find the solution to the integral ##\theta = \int_0^{\theta}\frac{du}{\sqrt{(c-u^2 +2u^3)}}## I can see that ##\frac{d^2u}{d\theta^2} = A +Bu+Cu^2## is a Weierstrass elliptic function, which can be generated from ##\Large(\normalsize\frac{du}{d\theta}\Large)\normalsize^2 = c-u^2 +2u^3## (A = 0, B=-1, C=3) So does this make my integral an elliptic integral? I haven't been able to find a table of integrals anywhere which contains an integral of this form so I'm a bit stuck. TerryW

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