2
$\begingroup$

Let's say I have a moving charged particle, with constant velocity.

Its electric field is given by (generally):

$$ \mathbf{E} = -\nabla\phi - \frac{\partial \mathbf{A}}{\partial t}. $$

  • If I perform a Lorentz transformation, boosting myself to a reference frame with the same velocity as the particle, then I see it stationary, hence $\partial/\partial t = 0$ and $\mathbf{E} = -\nabla \phi$.

  • If I decide to fix the gauge such that $\mathbf{A} = 0$, i.e. $A^{\mu} = (\phi/c, \mathbf{0})$, - the opposite Weyl gauge?? - I get that $\mathbf{E} = -\nabla \phi$.

So both actions leave the physics ($\mathbf{E}$) unchanged.

Are they two sides or the same coin?

$\endgroup$
  • 1
    $\begingroup$ $\uparrow $ No. $\endgroup$ – Qmechanic Sep 15 '18 at 14:02
  • $\begingroup$ Thanks. So how can I make sense that both method give the same physics? How can I distinguish them if I am just given the result? $\endgroup$ – SuperCiocia Sep 15 '18 at 14:03
  • $\begingroup$ @Qmechanic Following the title of the post, suppose we are considering gravity as a gauge theory. Amongst other gauge symmetries, we have those generated by the Lorentz generators $M_{ab}$ with gauge parameters $K^{ab}(x)$ and gauge field $\omega_m{}^{ab}(x)$ (the spin connection). Is fixing a gauge the same as fixing a frame of reference? Or is it slightly different because these are local Lorentz transformations in the tangent space? $\endgroup$ – NormalsNotFar Sep 15 '18 at 14:07
1
$\begingroup$

Lorentz transformations and electromagnetic gauge transformations are completely different things. The former changes the observer, the latter has no physical meaning because it corresponds to superfluous degrees of freedom. The former acts on all spacetime tensors, the latter only on electromagnetic quantities. But that's not the main issue here - your argument for why they "appear" the same doesn't work to begin with:

There is no such thing as a "gauge where $\vec A = 0$.". A gauge transformation acts on the four-vector $A$ with a smooth function $f$ as $$ A\mapsto A + \mathrm{d}f,$$ or, in components, $A^\mu \mapsto A^\mu + \partial^\mu f$. For an arbitrary $A$, it is not possible to force $A^1 = 0, A^2 = 0, A^3 = 0$ through such a transformation alone (try solving for $f$ in all three conditions!). "Fixing a gauge" does not mean "imposing arbitrary conditions on $A$". It means choosing conditions on $A$ that can actually be reached by a gauge transformation on any $A$.

Physically, it is easy to see that such a gauge transformation doesn't exist: A moving charge is a current, and if $A^i = 0$, then the magnetic field also vanishes. But a current always produces a magnetic field. You can only have $A^i = 0$ in a frame where the charge is not moving, i.e. only after a Lorentz transformation into the charge's rest frame.

$\endgroup$
  • $\begingroup$ So if I boost to a frame where the particle is stationary, I see no $B$ field. Can I show this mathematically? And what if I were boosting in an accelerated frame wrt a stationary charge? Would I then see radiation emitted? $\endgroup$ – SuperCiocia Sep 17 '18 at 13:17
  • $\begingroup$ @SuperCiocia 1. Of course you can show this mathematically: Just examine the transformation of $A$ - and hence $E$ and $B$ - under the Lorentz boost into a frame with the same velocity as the charge. 2. There is no such thing as "boosting in an accelerated frame". All Lorentz transformations lead to frames with constant velocity. $\endgroup$ – ACuriousMind Sep 17 '18 at 14:12
1
$\begingroup$

They are very different. Lorentz transformations act on $A$ as a vector, but gauge transformations act non-linearly on $A$: $$A \mapsto A + df.$$

The physical result depends on neither your choice of gauge nor your choice of reference frame. You get to choose them both and typically there is no interaction between them, unless you choose a gauge like temporal gauge which breaks the boost symmetry (which is what you're seeing happen). On the other hand a gauge like Lorenz gauge is nice because it's Lorentz invariant (watch out for them t's :).

$\endgroup$
1
$\begingroup$

There exists no gauge in which $\vec A =0$ unless there is no current in your particular reference frame. If there is no current, then in the right gauge $\vec A=0$.

$\endgroup$
1
$\begingroup$

Not really.

A Lorentz transformation transforms all the degrees of freedom.

Whilst a gauge fixing generally fixes some degrees of freedom.

Conceptually speaking they are not the same thing.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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