Transformation properties of derivative of a scalar field

[Muphrid]

Frederik, we still disagree I think. Active = observable, passive = unobservable.

How would you distinguish the two?

If you only knew, say, the coordinate tuple that describes a vector, how would you know that it's with respect to the same basis (and hence describes some $u'$, the result of an active transformation) or with respect to a different basis (and hence describes the original vector $u$)?

See this post for an example of active nd passive transformations being observable and unobservable, respectively: https://www.physicsforums.com/showpost.php?p=4110601&postcount=5

Honestly, all I get from that is a failure of proper application of gauge invariance. All the transformations we've been talking about can be considered gauge transformations, and as such, the results should be gauge invariant. The "size" of the AB effect should be one such quantity, or else it is not meaningful.

 

[ianhoolihan]

How would you distinguish the two?

If you only knew, say, the coordinate tuple that describes a vector, how would you know that it's with respect to the same basis (and hence describes some $u'$, the result of an active transformation) or with respect to a different basis (and hence describes the original vector $u$)?



Honestly, all I get from that is a failure of proper application of gauge invariance. All the transformations we've been talking about can be considered gauge transformations, and as such, the results should be gauge invariant. The "size" of the AB effect should be one such quantity, or else it is not meaningful.

If I have a vector $u$ and a vector $v$ (maybe a basis vector) then doing a passive transformation on one, and an active on the other will change their relative displacements/orientation. 'Physically'. Anyway, in response to you question, if we have a coordinate tuple, we must know the basis it is in, or it makes no sense.

I admit the link I provided was a bit beyond me, but I thought it was kosher --- obviously not!

I will have to go look at some actual books such as Goldstein, but this link http://www.phy.duke.edu/courses/211/faqs/faq20/node2.html seems to indicate that I am wrong, and that the passive and octive are just inverses of each other.

*sigh*

Will look again ofter QFT.

 

[TrickyDicky]

In the context of QFT active and passive transformations are indistinguishable and as commented one is just inverse of the other. Every general coordinate transformation defines both transformations, depending on the POV, that is depending on whether you choose to fix the vector bases or the components.
The confusion for many people arises (i.e. the distinction passive/active is no longer trivial) only in the presence of curvature. So if you want to stick to QFT (as long as you keep away from Hawking radiation kind of stuff) you need not get confused about it.

 

[Muphrid]

If I have a vector $u$ and a vector $v$ (maybe a basis vector) then doing a passive transformation on one, and an active on the other will change their relative displacements/orientation. 'Physically'. Anyway, in response to you question, if we have a coordinate tuple, we must know the basis it is in, or it makes no sense.

I guess my question is more, if you have a system and you make two copies of it, one that you transform according to an active transformation and another according to a passive one, how could you tell which one was which only by comparing the copies to the original (not to each other)?

I will have to go look at some actual books such as Goldstein, but this link http://www.phy.duke.edu/courses/211/faqs/faq20/node2.html seems to indicate that I am wrong, and that the passive and octive are just inverses of each other.

*sigh*

Will look again ofter QFT.

Ultimately, I think this just goes back to how, in both passive and active transformations, you can express the original vector as $u = {u'}^1 \underline f^{-1}(e_1) + {u'}^2 \underline f^{_1}(e_2)$. That active transformations have the notion of transforming $u$ to $u'$ where passives don't necessarily have that doesn't make the above statement any less true. It's just that in passive transformations we tend to think of the above as a one step process, where in active transformations it seems like a two-step process.

 

[haushofer]

Maybe you also like this thread,

https://www.physicsforums.com/showthread.php?p=4142511&posted=1#post4142511

which is related to yours.

 

[ianhoolihan]

OK, after some time away, and some different perspectives, I think I have it sorted.

Firstly, the short answer is that the answer to my original question is that method A. is correct.

Secondly, active and passive transformations are equivalent --- there is no physical difference. (Sorry for getting that one wrong.)

In short, an active transformation involves moving the actual thing, while keeping the basis fixed, while a passive transformation is keeping the point fixed and moving the basis (in the opposite sense).

To reconcile with previous discussion, a nice way to think of it is in diffeomorphisms in GR. If $\varphi$ is a diffeomorphism between two manifolds $M$ and $N$, then we can move points in the manifold (and the vector spaces with pushforwards and pullbacks etc). When we do a transformation, we can think of it as an active one, in that we actually move a point in $p\in M$ to a point in $\varphi(p)\in N$. A point $p\in M$ simply takes the coordinates of the point $\varphi(p) \in N$. If it is a vector we are transforming, then the basis will change from that for $T_p(M)$ to $T_{\varphi(p)}(N)$. Then the components of the vector, and the basis, both change. (Usually we have $N=M$, i.e. a transformation from $M$ to itself, so that an active transformation means moving to a different point, in the same coordinates. The components and basis of a vector would still change, as $T_p(M) \neq T_{\varphi(p)}(M)$.) However, it is equivalent do define a coordinate system in a neighbourhood of $\varphi(p)\in N$ and then pull that coordinate patch back to $M$. Now take this pulled back coordinate system as a new coordinate system on $N$ (in a neigbourhood of $p$), and express $p$ in terms of these coordinates. This is seen as passive --- we didn't actually move the point. (Again, if we have $N=M$ as usual, then this is just a change of basis.)

To conclude, I think a lot of the confusion arises from notation. I believe the correct statement is
$$\phi(x) \to \phi'(x') =\phi(x)= \phi(\Lambda^{-1} x')$$
(i.e. $\phi' =\phi\circ \Lambda^{-1}$ etc). However, if we do a transformation, we then want to work in those coordinates, so we just relabel them $x'\to x$, and hence the above may be written
$$\phi(x)\to \phi'(x) = \phi(\Lambda^{-1}x)$$
(The arrow here means 'our representation goes to', as the underlying object doesn't change: $\phi(x) = \phi'(x)$ in the above, if we accept that the $x$ is actually an $x'$ on the right.)

I really hope I am not wrong on all of this, and haven't confused the issue even more! If someone wants, they can go and show how the diffeomorphism argument gives the correct transformation of vectors, but I'm not 100% sure at the moment, and need to do some actual QFT!