0
$\begingroup$

I'm looking for examples of the following descriptions:

  1. A gauge invariant scalar which is not Lorentz-invariant

  2. A Lorentz covariant scalar

For 1. I was thinking about the scalar potential $A$ (for the magnetic field). This is gauge invariant, but I don't know whether or not it is Lorentz-invariant. For 2. I was thinking about the proper time (when you have relativistic speeds). Can somebody help me with this?

Improve this question
$\endgroup$
2
  • $\begingroup$ Your second example works according to this: en.wikipedia.org/wiki/Lorentz_covariance $\endgroup$
    – safesphere
    Jun 5, 2018 at 17:07
  • $\begingroup$ Oké so my first example is wrong, but I'm sure it isn't a contradiction, but I can't find a good example either $\endgroup$
    – Belgium_Physics
    Jun 5, 2018 at 17:10

3 Answers 3

Reset to default
4
$\begingroup$

The potentials $\phi$ and $\vec{A}$ in electromagnetism are the canonical examples of quantities which aren't gauge-invariant. For instance, we are free to shift the electric potential $\phi$ by a constant without changing the physics; this shift is called a gauge transformation. It is a change in our description of the system, but not a change in any physical properties of the system. All physical quantities must be gauge-invariant, therefore. Examples include $E$ or $E \cdot B$.

Usually when people use the word 'scalar' in this context, they mean 'number which is invariant under Lorentz transformations'. So there is no such thing as a scalar which is not Lorentz invariant. If by 'scalar' one means 'number which is invariant under rotations', then $E\cdot E$ does the trick for your first question. This quantity changes under Lorentz transformations. Finding numbers which don't change under Lorentz transformations is easiest to do if we reformulate electromagnetism in relativistic terms. The canonical examples are $E \cdot B$ and $E^2-B^2$.

Improve this answer
$\endgroup$
1
  • $\begingroup$ A great answer. I only want to add that gauge, in addition to being our description of the system, also reflects a symmetry or relation that actually exists in the universe. To say that the phase of a wave function is not observable, but the phase difference is observable (the U(1) symmetry) is similar to saying that the (absolute) speed is not observable, but the speed difference (relative speed) is observable (Lorentz symmetry). In either case it is the actual existing symmetry that allows us to change our description (gauge) without changing any observables. $\endgroup$
    – safesphere
    Jun 5, 2018 at 20:52
0
$\begingroup$

The Lagrangian of a charged particle moving non-relativistically in an electromagnetic field is an example for the first description: $$ L = \frac{\,\vec{p}^2}{2m} - q \left( \frac{\vec{v}}{c} \cdot \vec{A} - \phi \right) $$ where the kinetic term is not Lorentz invariant at all, but it is gauge invariant up to a total derivative.

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ Interestingly, this form of $L$ is not gauge invariant. It does not obey the beloved principle of gauge invariance [1]. Never use it :-) [1]: amazon.com/Under-Spell-Gauge-Principle-Mathematics/dp/… $\endgroup$
    – my2cts
    Jun 5, 2018 at 21:33
  • $\begingroup$ Yup, I fixed it. Sleep time :) $\endgroup$
    – Oktay Doğangün
    Jun 5, 2018 at 21:49
  • $\begingroup$ I was joking , the lagrangian was fine . $\endgroup$
    – my2cts
    Jun 6, 2018 at 7:48
0
$\begingroup$

The question mixes Newtonian physics with Special Relativity.

In Relativity the EM field does not have a scalar potential $\phi$ and a 3-vector potential $A$; it has a 4-vector potential whose component are $\phi$, $A_1$, $A_2$, and $A_3$, and that potential is Lorentz invariant. Adding a 4-divergence is a gauge change and gives the same field. equivalent

Improve this answer
$\endgroup$
1
  • $\begingroup$ Your answer seems incomplete. Please consider the reader in your logical flow. Also, please use proper capitals and punctuation to indicate sentence start and end. $\endgroup$
    – Dlamini
    Jun 6, 2018 at 1:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.