0
$\begingroup$

In the case of fields, it is clear what a local transformation in the internal space of the field is:

$$\phi(x) \to \phi'(x')= G(x) \phi(x),$$

as opposed to a global transformation, where $G$ would not depend on $x$.

But I don't understand the difference between local and global coordinate transformations.

It is said that general coordinate transformations, $$x^\mu \to x'^\mu =x^\mu+\epsilon^\mu(x)$$ are local, while rotations $$x^\mu \to x'^\mu =x^\mu+\sigma^\mu_\nu x^\nu$$ are global.

How can we distinguish between the two? In both cases, the variation depends on $x$.

$\endgroup$
  • $\begingroup$ Related: physics.stackexchange.com/q/48188/2451 , physics.stackexchange.com/q/179592/2451 $\endgroup$ – Qmechanic Nov 9 '16 at 22:19
  • $\begingroup$ A transformation is said to be global if it acts the same at every point; it is said to be local otherwise. In your example the former is local, as $\epsilon(x)$ is any function of the point, whereas the latter is global, since the $\sigma^{\mu}_{\nu}$ are constant and play the role of multiplicative coefficients. $\endgroup$ – gented Nov 9 '16 at 23:12
  • $\begingroup$ @GennaroTedesco But a rotation acts different at every point, since the variation is $\sigma^\mu_\nu x^\nu$. Points far from the origin are displaced more than those near the origin. I'm guessing that we restrict ourselves to affine transformations, so any terms that would involved higher powers of $x$ are absorbed into the coefficients. $\endgroup$ – jinawee Nov 10 '16 at 13:47
  • $\begingroup$ @jinawee In that example the rotation acts on the point (that reflects into the multiplication $\sigma\cdot x$) but the rotation itself doesn't depend on it (the $\sigma$). $\endgroup$ – gented Nov 10 '16 at 14:29

Your Answer

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

Browse other questions tagged or ask your own question.