2
$\begingroup$

Under a Lorentz transformation, a vector field transforms as: $A'_{\mu}(x')=\Lambda^{\nu}_{\mu}A_{\nu}(x)$

My question is, why is the Lorentz transformed vector field evaluated at $x'=\Lambda x$, rather than at $x$ only. One way I can make sense of it is if I take the example of a scalar field under a rotation, my transformed field evaluated at the rotated coordinates is same as my original field evaluated at the original coordinates, but somehow, I am not able to find a convincing argument for the vector field case.

$\endgroup$
3
$\begingroup$

Precisely because of the reason you have put forward,

the Lorentz transformation for a scalar field $\phi(x)$ is usually written as:

$$ \phi(x) \rightarrow \phi(x)' = \phi(\Lambda^{-1}x), $$

and for a vector field $V^{\mu}(x)$ is usually written as:

$$V^\mu(x) \rightarrow V^\mu(x)' = \Lambda^{\mu}_{\nu}\,V^{\nu}(\Lambda^{-1}x).$$

So that you leave your coordinates $x$ fixed.

Hence, your expression is correct, but it’s in terms of the rotated coordinate system $x’$. You can just write $x = \Lambda^{-1} \, x’$ in the RHS and get to my expression.

$\endgroup$
0
$\begingroup$

It's actually two separate transformations that you do. The one is a change of basis in the tangential space (under which scalar fields remain entirely unchanged by definition), the other one is that you re-express the coordinates that these fields may depend on. Normally you do both together because otherwise it's pretty impossible to interpret the quantities that you have.

However, in GR people sometimes don't do the second transformation because while the first is linear, there are cases where you cannot explicitly solve a coordinate transformation for the new coordinates. So you may well end up having, eg, a metric tensor in one coordinate basis as an implicit function of other coordinates.

$\endgroup$
-1
$\begingroup$

Sure. That's because vectors do not transform in the same way as scalars. A good example is the electric field.

Imagine an infinite possitively charged plane along the $y,z$-directions (a vertical plane).

Charged plane electric field

It creates an $\vec{E}$ field in the $x$ direction, as in the picture above.

Now suppose you turn 90 degrees around the z axis. Then you expect

Rotating charged plane

Because physics are invariant under rotations. However, if you try to apply the same rule as with scalars:

$\vec{E}_{after}(new\ point)=\vec{E}_{before}(previous\ point)$

You would get the same "$\rightarrow$" vector, above the plane, whereas you know that now the vector is upwards, "$\uparrow$".

Mathematically, the original $E$ field was $E\cdot\hat{x}$. That would be saying that the new electric field is the same $E\hat{x}$, but dispolaced to the new point. That means translating the $\rightarrow$, but you actually want the $\uparrow$.

So you see that this method fails. That's interesting, but why? The reason is simple:

Physics are invariant under rotations, but full rotations. What you're doing is rotating everything except one thing: the reference frame.

Indeed, if you also rotated the axes, the result would still be $E\hat{x}$, but now $\hat{x}$ would be upwards as well, so you'd get what you're looking for.

That would be a "full rotation", everything has rotated, so physics should not be different. But of course, if you only rotate part of it, it changes it all.

So what do we do? We use infinitesimal generators, because we know that it works.

You can also deduce the roattion matrix by other means, but that's just mathematics.

$\endgroup$
  • $\begingroup$ Downvoters: please, explain if I said something wrong. $\endgroup$ – FGSUZ Jan 21 '19 at 21:15

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.