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.
Hi,
I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem.
Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$
Where ##b=1## with an orbit only in the equatorial plane.
We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$
Ultimately, I was tasked to find the initial...
The value of H equals ## 10^{3}## in natural units,
According to : https://en.wikipedia.org/wiki/Natural_units, ## t \sim 10^{-21} sec = 10^{21} Hz ##, and since ## \text{GeV} \sim 10^{24} \text{Hz } ##,
## GeV \sim 10^{24} \times 10^{-21} = 10^3 ## in natural units.
So is this conversion correct?
Also in the above formula, can I convert H to that natural units , since it’s a constant, while keeping k in Hz ?