Transformation properties of derivative of a scalar field

[Fredrik]

I'll reply to this quickly, as I've got to shoot off. The point is, I disagree, and that's what I've been trying to say. In an active transformation, the coordinate system does not change, only the function --- as before $y(p) \to y(p)$. In a passive one, $y(p)\to z(p)$ and the corresponding change in $\phi \to \phi'$ equates to a trivial change in coordinates.

You need to distinguish between the point y(p) and its coordinate 4-tuple with respect to a basis (even when they happen to have the same components due to a choice of basis). The active/passive terminology is only used about transformations of coordinate 4-tuples. It simply doesn't apply to transformations of points. However, a transformation of the points induces both an active and a passive transformation of the coordinate 4-tuples.

The coordinate change $y\to z$ obviously induces the change $y(p)\to z(p)$. And this induces an active transformation by $z\circ y^{-1}$ of the coordinate 4-tuple of y(p) with respect to the standard basis, and it induces a passive transformation by $y\circ z^{-1}$ of the coordinate 4-tuple of y(p) with respect to the standard basis.

When you say that in an active transformation, we have $y(p)\to y(p)$, I'm not sure I even understand what you're saying. The active/passive terminology simply doesn't apply to transformations of the point y(p), and both active and passive transformations of a corresponding coordinate 4-tuple will change that coordinate 4-tuple. An active transformation by $\Lambda$ is a passive transformation by $\Lambda^{-1}$ and vice versa.

I can't seem to odit it.

There's a time limit for odits . I think it's currently set to 11 hours and 40 minutes (=700 minutes).

 

[ianhoolihan]

An active transformation by $\Lambda$ is a passive transformation by $\Lambda^{-1}$ and vice versa.


There's a time limit for odits . I think it's currently set to 11 hours and 40 minutes (=700 minutes).

For now this: why then is a passive transformation not physically observable, while an active one is?

Now I need just about an odit of sleep !

 

[Fredrik]

Active transformations lead to physically observable effects. Passive ones do not

If you need a more intuitive way to think about these things, I suggest that you think of an active transformation by a rotation matrix R as a physical rotation by R (say a counterclockwise rotation by an angle of π/4) of the object on which we're going to do measurements, and the corresponding passive transformation by R as a physical rotation by R^-1 (a clockwise rotation by π/4) of the labratory around the object (while the object is held fixed relative to the Earth).

In both cases, there's a physical change. The point is that the changes are equivalent, a far as physics experiments are concerned (unless of course we're doing experiments with something like a compass needle; in those cases, you have to imagine these things taking place in intergalactic space or something).

In the passive case, the orientation of the object relative to the Earth (or some other fixed stuff outside the laboratory) doesn't change. But we would still change our description of its orientation, if we describe it relative to the walls of the laboratory (the new basis vectors). In the active case, our description of the orientation of the object relative to the walls changes in exactly the same way as in the passive case.

Edit: Note that this last bit is consistent with what I've been saying about active and passive transformations of a coordinate 4-tuple $(x_\mu)_{\mu=0}^3$:

Active transformation by $\Lambda$: $x^\mu\to\Lambda^\mu{}_\nu x^\nu$.
Passive transformation by $\Lambda$: $x^\mu\to(\Lambda^{-1})^\mu{}_\nu x^\nu$.
Passive transformation by $\Lambda^{-1}$: $x^\mu\to\Lambda^\mu{}_\nu x^\nu$. (This is the same as the active transformation by $\Lambda$, as suggested by the informal argument above).

 

[Muphrid]

I disagree. Active transformations lead to physically observable effects. Passive ones do not --- they are trivial changes of coordinates.

In a passive transformation, both component and basis are changed (inversely). In an active transformation, only one of those is changed --- either the component or the basis, yes. Maybe this is what Frederik stated, and I mistook him.? I thought he meant that an active transformation was the same as a passive one, in the sense I've just described.

Kane

I think I've discovered the problem. Consider a tangent vector $u = u^1 e_1 + u^2 e_2$. We can transform this vector into the primed space.

$$u' = {u^1}' e_1 + {u^2}' e_2 = u^1 {e_1}' + u^2 {e_2}'$$

(Incidentally, I think I see now why some authors prefer $\phi'(x) = \phi(x')$ now. It makes talking about the transformation laws hideous, but it would keep all the primes on one side of the above statement.)

At any rate, we can now transform back to get the original $u$:

$$u = u^1 e_1 + u^2 e_2 = {u^1}' \underline f^{-1}(e_1) + {u^2}' \underline f^{-1}(e_2)$$

There's a certain symmetry here, which I think can be expressed as follows: the vector $u'$ can be expressed in terms of either (a) new components, same basis vectors or (b) same components, new basis vectors. This is what I was saying earlier.

However, you've also been talking about the untransformed vector $u$, which clearly can be described in terms of either (a) same components, same basis vectors or (b) new components, new basis vectors. The latter is what you expect in a passive transformation, while the former may be what you expect in an active transformation, since you're generally not even interested in the untransformed vector at all (we don't tend to think about it, at least).

 

[ianhoolihan]

In both cases, there's a physical change. The point is that the changes are equivalent, a far as physics experiments are concerned (unless of course we're doing experiments with something like a compass needle; in those cases, you have to imagine these things taking place in intergalactic space or something).

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

There's a certain symmetry here, which I think can be expressed as follows: the vector $u'$ can be expressed in terms of either (a) new components, same basis vectors or (b) same components, new basis vectors. This is what I was saying earlier.

However, you've also been talking about the untransformed vector $u$, which clearly can be described in terms of either (a) same components, same basis vectors or (b) new components, new basis vectors. The latter is what you expect in a passive transformation, while the former may be what you expect in an active transformation, since you're generally not even interested in the untransformed vector at all (we don't tend to think about it, at least).

If you mean former, as in the former paragraph, then yes, that's what I mean. An active transformation is not the inverse of a passive one (which, I admit, is what is usually bandied around).

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

 

[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!