Understanding the active/passive transformation of a scalar field

[etotheipi]

$\mathcal{P}$ is a point in Minkowski spacetime $M$, and $\varphi_1: U \in M \mapsto \mathbb{R}^4$ and $\varphi_2: U \in M \mapsto \mathbb{R}^4$ are two coordinate systems on the spacetime. A scalar field is a function $\Phi(\mathcal{P}): M \mapsto \mathbb{R}$, and we can define coordinate representations of this scalar field as the compositions $\phi = \Phi \circ \varphi_1^{-1}$ and $\phi' = \Phi \circ \varphi_2^{-1}$. Since the value of a scalar field at some point $\mathcal{P}$ is coordinate independent,$$\Phi(\mathcal{P}) = \phi(x) = \phi'(x')$$where $x \in \mathbb{R}^4$ and $x' \in \mathbb{R}^4$ are the coordinate vectors of $\mathcal{P}$ w.r.t. the two different coordinate systems. The Lorentz transformation between the coordinates of $\mathcal{P}$ in the two different frames is$$x' = \Lambda x \iff \Lambda^{-1} x' = x$$we can just substitute that into obtain$$\phi'(x) = \phi(\Lambda^{-1} x)$$Is that an active or a passive transformation? What would the other type look like? And most importantly, why is there any difference between them: don't we just always need to satisfy $\Phi(\mathcal{P}) = \phi(x) = \phi'(x')$?

Maybe for an active transformation you need to actually move the manifold, and by extension the point $\mathcal{P}$, underneath the coordinate systems? I have no idea, it's quite confusing.

 

[JD_PM]

Hey I'd recommend you read about active transformations on Tong's 1.2 Lorentz invariance section in the attached PDF.

 

[PeterDonis]

A scalar field is a function $\Phi(\mathcal{P}): M \mapsto \mathbb{R}^4$

Shouldn't it just be $\Phi(\mathcal{P}): M \mapsto \mathbb{R}$? A scalar is just a number, not a 4-tuple of numbers.

we can define coordinate representations of this scalar field as the compositions $\phi = \Phi \circ \varphi_1^{-1}$ and $\phi' = \Phi \circ \varphi_2^{-1}$.

These compositions define maps $\mathbb{R}^4 \mapsto \mathbb{R}$. See above.

Is that an active or a passive transformation?

Passive. You're holding the point $\mathcal{P}$ fixed and changing the coordinate chart.

Maybe for an active transformation you need to actually move the manifold, and by extension the point $\mathcal{P}$, underneath the coordinate systems?

Underneath the coordinate system, singular. An active transformation is a map $T: M \mapsto M$, so we have $\mathcal{P}^\prime = T(\mathcal{P})$, and if we have a coordinate chart $\varphi$, this naturally induces a map on the coordinates, $T: \mathbb{R}^4 \mapsto \mathbb{R}^4$ given by $T(\varphi(\mathcal{P})) = \varphi(\mathcal{P}^\prime)$. Note that we haven't changed coordinate charts.

What may be confusing you here is that, for an active transformation, the values of physical quantities like scalar fields will not, in general, be left invariant, because we are changing which point in the manifold we are referring to, whereas for a passive transformation, values of physical quantities must be left invariant, since we are holding the point in the manifold constant. However, active transformations usually come up in the context of discussing symmetries, i.e., active transformations that do leave some physical quantity invariant. For that case, mathematically, active transformations look like passive transformations, as long as you only look at quantities that are left invariant.

Note, however, that saying an active transformation leaves some physical quantity invariant is just another way of saying that there are multiple points in the manifold that have the same value for some physical quantity. So if "active transformation" language is confusing you, it might help to translate it into the latter kind of statement.

 

[etotheipi]

Shouldn't it just be $\Phi(\mathcal{P}): M \mapsto \mathbb{R}$?

Yes, of course . That's a typo.

Passive. You're holding the point P fixed and changing the coordinate chart.

That's what I suspected, since we haven't touched the point $\mathcal{P}$. Like you say, the coordinate chart is moved over the fixed manifold.

But I doubted it because it's not entirely clear from the literature. The David Tong notes that @JD_PM linked appear to call this transformation an active transformation, whilst calling a passive transformation one of the form $\phi'(x) = \phi(\Lambda x)$. That didn't make a lot of sense to me.

 

[PeterDonis]

I'd recommend you read about active transformations on Tong's 1.2 Lorentz invariance section

Here he is using "active transformation" to mean something somewhat different from what I described in my previous post. He is using it to mean a transformation where we not only move from one point of the manifold to another, we also "move" the physical quantities along as well. In other words, we make the "new" point in the manifold, after the transformation, look exactly the same, in terms of physical quantities, as the "old" point in the manifold, before the transformation.

The David Tong notes that @JD_PM linked appear to call this transformation an active transformation, whilst calling a passive transformation one of the form $\phi'(x) = \phi(\Lambda x)$. That didn't make a lot of sense to me.

It doesn't make a lot of sense to me either, because if you "transform" everything--the point in the manifold and all the physical quantities along with it--you haven't actually changed anything physical at all. You've done the same thing as a passive transformation does, but used much more confusing and obfuscating language to describe it.

However, unfortunately, this usage of "active transformation" is also common in the literature, so there's no help for it. Note that Tong's objective here is to define what a "Lorentz invariant" theory is, so he is looking at the case where the "active transformation" is a symmetry, and hence leaves physical quantities invariant. And, as I noted, for that case, the distinction between "active" and "passive" transformations doesn't really show up in the math; they both look the same mathematically (or if different mathematical expressions are used to describe them, as Tong does, they are trivially equivalent).

 

[etotheipi]

Thanks. In that case, maybe it's best not to worry about the terminology. Also, what does it mean when you said that the two different expressions are equivalent? As far as I can tell from the section 1.2, what he calls an active transformation takes the form$$\phi'(x) = \phi(\Lambda^{-1}x)$$whilst what he calls a passive transformation takes the form$$\phi'(x) = \phi(\Lambda x)$$to me these look different mathematically, but I think I'm mis-interpreting what they mean.

 

[PeterDonis]

As far as I can tell from the section 1.2, what he calls an active transformation takes the form$$\phi'(x) = \phi(\Lambda^{-1}x)$$whilst what he calls a passive transformation takes the form$$\phi'(x) = \phi(\Lambda x)$$to me these look different mathematically, but I think I'm mis-interpreting what they mean.

Strictly speaking, they are describing inverse operations, mathematically, but that's just a change of viewpoint. For a simple example, consider a 90 degree rotation in a plane:

The "active transformation" formula says "rotate the plane 90 degrees clockwise under the coordinate grid, so the 3 o'clock point on the grid refers to the same physical point/field values as the 12 o'clock point on the grid did before".

The "passive transformation" formula says "rotate the coordinate grid 90 degrees counterclockwise over the plane, so the 3 o'clock point on the grid refers to the same physical point/field values as the 12 o'clock point on the grid did before".

Notice how the result, physically, is the same in both cases, and the mathematical descriptions, while they are, as I said, inverse operations strictly speaking, both describe the same relative change between the grid and the plane, just from different viewpoints (the first holds the grid fixed, the second holds the plane fixed). That is the sense in which they are trivially equivalent.

 

[etotheipi]

That makes sense when written like that, but it bugs me that I could choose a tuple of real numbers, say $x = (a,b,c,d)$, and then those two equations appear to be saying that $\phi(\Lambda x)$ is equal to $\phi(\Lambda^{-1} x)$, which doesn't look right.

The first one seems to come from $\phi(x) = \phi'(x')$, which is what I expect the transformation properties of a scalar field to look like (given that scalars are coordinate invariant), whilst the second relation seems to come from $\phi(x') = \phi'(x)$, which seems odd.

 

[PeterDonis]

I could choose a tuple of real numbers, say $x = (a,b,c,d)$, and then those two equations appear to be saying that $\phi(\Lambda x)$ is equal to $\phi(\Lambda^{-1} x)$

No; you can't apply both equations at once. You have to pick one or the other. They're two different viewpoints on the same thing; trying to apply both of them at once doesn't make sense.

 

[PeroK]

Thanks. In that case, maybe it's best not to worry about the terminology. Also, what does it mean when you said that the two different expressions are equivalent? As far as I can tell from the section 1.2, what he calls an active transformation takes the form$$\phi'(x) = \phi(\Lambda^{-1}x)$$whilst what he calls a passive transformation takes the form$$\phi'(x) = \phi(\Lambda x)$$to me these look different mathematically, but I think I'm mis-interpreting what they mean.

Here's how I think about this. Imagine we have some point in spacetime where a physical scalar quantity has the value $X$. In our first coordinate system that point is denoted by coordinates $x_0$. And if $\phi$ is our scalar field that means that:
$$\phi(x_0) = X$$
Now we perform an active transformation. That means we move the thing that has value $X$ to a new point with coordinates $\Lambda x_0$. This means we change the function that defines our scalar field to $\phi'$ and we know that:
$$\phi'(\Lambda x_0) = X = \phi(x_0)$$
Hence, in general:
$$\phi'(\Lambda x) = \phi(x)$$ or $$\phi'(x') = \phi(\Lambda^{-1}x')$$ or $$\phi'(x) = \phi(\Lambda^{-1}x)$$
For a passive transformation, we change the coordinate system, so that the coordinate $x_0$ now represents what was at the old coordinate $\Lambda x_0$. The thing with value $X$ is now at the coordinate $x_0' = \Lambda^{-1}x_0$. This is the coordinate that now denotes that point in spacetime. And we have:
$$\phi'( \Lambda^{-1}x_0) = X = \phi(x_0)$$
And, in general, we have:
$$\phi'(\Lambda^{-1} x) = \phi(x)$$ or $$\phi'(x') = \phi(\Lambda x')$$ or $$\phi'(x) = \phi(\Lambda x)$$

 

[etotheipi]

Woah, thanks! Spelling it out like that helps massively. The key thing, as far as I can tell, seems to be the duality of the matrix $\Lambda$ as a change of basis matrix (for the passive case), and as a matrix representation of a linear transformation (for the active case). I won't lie, the manipulations are fiddly to me and it still trips me out, but you've restored my faith in Professor Tong!

It seems there are quite a few different ways that you could look at it. I had a look at the manifolds section in my methods textbook and one thing that looked relevant for active transformations was the theory of diffeomorphisms, i.e. a function $f$ that takes a point $\mathcal{P}$, $f: M \mapsto M$. But to be honest I don't think I should be mucking around too much with manifolds yet because I still have to finish linear algebra, so this can wait

Anyway, now I can sleep easy

 

[PeterDonis]

The key thing, as far as I can tell, seems to be the duality of the matrix $\Lambda$ as a change of basis matrix (for the passive case), and as a matrix representation of a linear transformation (for the active case).

Yes, that duality of notation is one of the things that can easily cause confusion.

 

[etotheipi]

By the way, I did come up with one more question. For context...

Suppose that we deal with a simpler case, rotation in two dimensions. Let the rotation operator be $R$. An active transformation would rotate the actual vector ($\vec{v} = v^i \vec{e}_i \rightarrow \vec{v}' = v'^i \vec{e}_i$), and would look like$$\vec{v}' = R\vec{v} \iff v'^i =R^i_{j} v^j$$The passive interpretation of the same transformation is to instead rotate the basis vectors by $R$, and to leave the actual vector unchanged ($v^i \vec{e}_i = \tilde{v}^i \tilde{e}_i$), in which case the components also transform oppositely to the basis vectors:$$\tilde{e}_i = R \vec{e}_i \iff \tilde{v}^{i} = (R^{-1})^i_{j}v^j$$It's clear that $v'$ and $v$ represent two different vectors, whilst $\tilde{v}$ and $v$ represent the same vector.

To make an analogy now with the Lorentz transformations and switch out $R \equiv \Lambda$ and $R^{-1} \equiv \Lambda^{-1}$, now for instance the Lorentz transformation that corresponds to a simple re-labelling of coordinates is $$\tilde{x} = \Lambda^{-1} x$$whilst $\Lambda$ is the transformation that rotates the basis vectors. And for the active transformation, $\Lambda$ is the transformation applied to the event.

However, often the change of coordinates is written like $\tilde{x} = \Lambda x$. I.e. they've switched out $\Lambda \leftrightarrow \Lambda^{-1}$ and are now using $\Lambda$ to refer to the passive transformation.

How do we know which convention is being used?

 

[PeroK]

How do we know which convention is being used?

You have to keep your wits about you!

 

[etotheipi]

You have to keep your wits about you!

Fair enough

 

[JD_PM]

Old thread but I wanted to share an analogy between active/passive transformations and Schrödinger/Heisenberg pictures.

I think we agree that the most intuitive definition of active transformation is as follows: take a 3D vector and rotate it (let's say over the azimuthal angle $\phi$) about certain axis (let's say the $z-$axis) while keeping the basis vectors fixed.

A completely equivalent way of performing such transformation is keeping the 3D vector fixed while rotating its basis vectors (over the azimuthal angle $-\phi$) about certain axis (the $z-$axis). This way is known as passive transformation.

The wave function (in the Schrödinger picture) $\Psi(x,t)$ can be understood as a vector ('living' in Hilbert space) that is time-dependent (i.e. it can change over time, just as our vector in the active transformation), while its basis vectors remain fixed (i.e. eigenstates are time-independent). This idea is known as Schrödinger picture and is strongly related to the concept of active transformation.

The equivalent way of thinking about this is letting the wave function be fixed (i.e. $\Psi(x)$) while allowing its basis vectors to move/rotate in the opposite direction of $\Psi(x,t)$ (i.e. the the wave function in the Schrödinger picture ). This idea is known as Heisenberg picture and is strongly related to the concept of passive transformation.