Gauge transformation has no effect on equation of motion

[KleZMeR]

I have a question about this classical invariance problem I'm working on. I'm almost done, and I understand the theory I think, so my question may seem a bit more math-oriented (it's been a few years since crunching equations). I have found that under a gauge transformation for a single particle in an electric (magnetic) field, the equation of motion is not affected. However, in my final step I have been left with these two extra terms in each component, i.e. these two terms are in my x component: [(1/c)*(d/dx)*(dψ/dt)]-[(d/dx)*(dψ/dt)] , which should equal zero?

So my result for the electric field x component is:

Ex = [-∇ø-(dAx/dt)]+[(1/c)*(d/dx)*(dψ/dt)]-[(d/dx)*(dψ/dt)]

Where I think it should just result in: Ex = [-∇ø-(dAx/dt)]

Is there something with the partial differential that I am not recognizing? Or I'm thinking the state change has no impact on the motion?

Any help would be greatly appreciated.

 

[vanhees71]

In Gauss or Heaviside-Lorentz units, there's always a factor 1/c in time derivatives. That's what makes these units so much better suited for electromagnetism than the SI units! So the correct expression for the electric field in terms of the scalar and vector potential (in any gauge) is
$$\vec{E}=-\frac{1}{c} \partial_t \vec{A} -\vec{\nabla} \phi,$$
and the magnetic field reads
$$\vec{B}=\vec{\nabla} \times \vec{A}.$$
Then it's easy to see that the electromagnetic field components don't change under the gauge transformation
$$\phi'=\phi+ \frac{1}{c} \partial_t \chi, \quad \vec{A}'=\vec{A}-\vec{\nabla} \chi,$$
for any scalar field $\chi$.

 

[KleZMeR]

Thanks vanhees71! Yes, it is surprising that the literature I am using does not indicate that information. So at this level of electromagnetism it is universal that the time derivatives get the 1/c factor?

 

[vanhees71]

It depends on the system of units you use. The Gaussian or even better the Heaviside Lorentz units are most natural and best suited for theoretical analysis, because it makes a relativistically covariant notation easy and quantities which belong together have the same dimension (units) as the electric and the the magnetic field $\vec{E}$ and $\vec{B}$ (where the naming of $\vec{B}$ as the "magnetic field" is already deviating from the traditional scheme, where it was called magnetic induction). In fact nowadays we know that $\vec{E}$ and $\vec{B}$ are just the six components of one antisymmetric Minkowski tensor in fourdimensional spacetime.

Also that in the system of units a "conversion factor" $c$ of the dimension of a speed occurs is just due to our choice of units. We decided to measure times and lengths in different units although with relativity we know they should be measured only in one unit. Indeed, if you look at the modern definition, the conversion factor $c$, the universal limit speed of Minkowski spacetime, is fixed by defining the meter (unit of length) in terms of the travel time of light, which is with a very high accuracy identical with this universal limit speed (in modern language photons are with a very high acccuracy measured to be massless).

In Heaviside-Lorentz units, you also get rid of factors $4 \pi$ in Maxwell's equations compared to the older Gaussian units. These factors then occur where they belong, namely in Coulomb's Law. In these units the microscopic Maxwell equations read
$$\vec{\nabla} \cdot \vec{B}=0, \quad \vec{\nabla} \times \vec{E} +\frac{1}{c} \partial_t \vec{B}=0,$$
$$\vec{\nabla} \cdot \vec{E}=\rho, \quad \vec{\nabla} \times \vec{B}-\frac{1}{c} \partial_t \vec{E} = \frac{1}{c} \vec{j}.$$
The factor $1/c$ occurs, because we still like to measure times in seconds and lengths in metres, because in everyday life light seconds are a bit inconvenient units of length.

In high-energy physics one is more lazy and sets the speed of light in a vacuum to 1 and also the modified Planck constant $\hbar$ to one. Then only one unit is left, namely an energy unit. Usually one uses GeV (Giga electron volts). Sometimes it's also convenient to measure lengths in fm (Fermi or femto metres=$10^{-15} \; \mathrm{m}$) and times in $\mathrm{fm}/c$. To convert between these units you just have to remember that $\hbar c \simeq 0.197 \; \mathrm{fm} \, \mathrm{GeV}$.