David Tong's passive transformation of the fields is wrong

Question

David Tong's definition of active transformation is clear. Under active transformation coordinates (basis vectors) are not changed but rather the field is. I denote the old and new fields as $\phi$ and $\phi'$. So by active transformation

$\phi'(x) = \phi (\lambda^{-1} x) $.

Notice that I have put the prime on the field and not the coordinate since the field is changed ( rotated) and not the coordinate system.

For passive transformation Tong writes

$\phi'(x) = \phi (\lambda x) $. This looks wrong to me. I change the coordinate system from $x$ to $x'$ and the field in the old coordinate system is $\phi(x)$ and in the new one it is $\phi'(x') $. Since it is a scalar field I have $\phi'(x') = \phi (x) $. Now $x'= \lambda x.$

So I get

$\phi(x) = \phi'(\lambda x) $. Which is clearly not the same as Tong's. Now I think Tong's equation is not right because I can't find any fault in my definition. Can someone please explain this.

Further could please some write the correct transformation law for a vector field both for active and passive rotation ( both for a covariant and contravariant field) keeping in mind the following conventions ( which I take from GR). While writing it please specify the matrix with the field changes as well as the matrix with which the coordinate changes like I have written above for a scalar field.

When A is contravariant vector it transforms as

$A= \lambda A$

When A is covariant vector it transforms as

$A'= \lambda^{-1}A.$

If I understand correctly the above transformations are passive ones.

Edit:

I know that if the basis vector transform as $e' = \lambda^{-1} e$ where ${e'}$ are new basis vectors and ${e}$ are old basis vectors, then coordinates transform as $x'= \lambda x$. And in GR we know that covectors transform as basis vector ( with $\lambda^{-1}$) while contravariant ones transform like coordinates (with $\lambda$). And this is a passive transformation ( like in Caroll)

(Please use this convention throughout)

So in GR we have $A^{\mu'} = \lambda^{\mu'}{ }_{\nu} A^{\nu}$ -> (1)

and $A_{\mu'} =( \lambda^{-1})_{\mu'}{ }^{\nu} A_{\nu}$

Now I have these specific questions-

1. In the above transformations the components of the vector ( or covector), i. e $A^{\nu}$ change. But none of the books (Caroll) mention the change in the arguments. Why is that so. Why don't they change the arguments as well. Whereas you have written the transformation for coordinates as well.

So should (1) really be

$A^{\mu'}(x') = \lambda^{\mu'}{ }_{\nu} A^{\nu}(x)$

-> $A^{\mu'}(\lambda x) = \lambda^{\mu'}{ }_{\nu} A^{\nu}(x)$

and $A_{\mu'}(x') = (\lambda^{-1})_{\mu'}{ }^{\nu} A_{\nu}(x)$

-> $A_{\mu'}(\lambda x) = (\lambda^{-1})_{\mu'}{ }^{\nu} A_{\nu}(x)$

because remember in my notation coordinates change as $x'= \lambda x$ ( so I have just replaces that. But you are not getting the same result as mine. Am I wrong in the calculation or the understanding. Should eq(1) be as above ( with change in coordinates depicted too or without that like GR books denote). All this what I have written is for passive transformation, because the coordinates have changed ( and the functional form of the field/vector). Please point out if what I have written is right or not. If wrong please point out which exact equation or understanding is wrong.

2. Now based on my knowledge of the above transformation of basis vectors (with the inverse matrix) , I try and form the Active transformation. Here the fields/vector change and not the coordinates. So I should be really using the inverse matrix $\lambda^{-1}$ (is this reasoning correct) when writing the change for contravariant ones ( because now their components have not changes rather they have been rotated. And I should use the direct matrix ($\lambda$) for covariant ones because their transformation is inverse of the contravariant one. So

$A^{\mu'}(x) =( \lambda^{-1})^{\mu'}{ }_{\nu} A^{\nu}(x)$ ( No change in arguments since they are not changed).

and $A_{\mu'}(x) = \lambda_{\mu'}{ }^{\nu}A_{\nu}(x)$ ( No change in arguments since they are not changed).

Again please point out if any thing is wrong here, precisely the exat equation or assumption

3. Lastly I know for a scalar field

Since it is a scalar field I have $\phi'(x') = \phi (x) $. Now $x'= \lambda x.$

So I get

$\phi(x) = \phi'(\lambda x) $.

What exactly is wrong here. I understand your Active Transformation but I can't figure out what I am doing wrong with the Passive Transformation. Please point out the error.

Answer 1

Consider the following picture.

We have a field which is large in the red rectangle and small elsewhere. The function which tells us the field value at some point at coordinates $\mathbf x$ is $\phi$; that is, $\phi(\mathbf x)$ is the value of the field at the point labeled by coordinates $\mathbf x=(x^1,x^2)$.

Now we perform an active transformation corresponding to a rotation of the field by $60^\circ$.

We're using the same coordinates, but after the transformation $\phi$ is no longer the function which gives us the field values. We must consider a new function $\psi$ which is related to the old one via

$$\psi(\mathbf x) = \phi(R^{-1}\mathbf x)$$

where $R$ is the $60^\circ$ rotation matrix.

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Now we consider a passive transformation instead.

This is a change of coordinates where the new coordinates $\mathbf y = (y^1,y^2)$ are related to the old ones via

$$\mathbf y = R\mathbf x$$

Once we've adopted the $y$-coordinate system, $\phi$ is once again the wrong function. The field values at a point $\mathbf y$ are given by the function $\sigma$, which is related to $\phi$ via

$$\sigma(\mathbf y) = \phi(R\mathbf y)$$

Use the pictures to convince yourself that if $\phi$ is large at, say, $\mathbf x=(1,0)$, then $\sigma$ will be large at $\mathbf y = \big(\frac{1}{2},-\frac{\sqrt{3}}{2}\big)$. In other words, $\sigma\big(\frac{1}{2},-\frac{\sqrt{3}}{2}\big) = \phi(1,0)$, which is consistent with $\sigma(\mathbf y) = \phi(R\mathbf y)$.

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To address your second question, under an active (linear) transformation a vector field $\mathbf V$ transforms as $\mathbf V\rightarrow \mathbf V'$ where $$\mathbf V'(\mathbf x) = R\mathbf V(R^{-1}\mathbf x)$$

In component notation, $$V'^\mu(\mathbf x) = R^\mu_{\ \ \nu} V^\nu(R^{-1}\mathbf x)$$

Under a passive (linear) transformation defined by a coordinate change $\mathbf y = R\mathbf x$,

$$\mathbf V'(\mathbf y) = R^{-1}\mathbf V(R\mathbf y)$$ $$V'^\mu(\mathbf y) = (R^{-1})^\mu_{\ \ \nu} V^\nu(R\mathbf y)$$

Both transformation rules can be seen immediately by replacing the rectangles in my drawings with arrows.

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In response to the edit, the mistake you are making throughout is in saying that $\phi'(x')=\phi(x)$ and then substituting $x'=Rx$. This is not correct. Please read the example I gave. If the coordinates are rotated by $60^\circ$, then $\mathbf x = (1,0)$ will have new new coordinates given by $\mathbf y = \big(\frac{1}{2},-\frac{\sqrt{3}}{2}\big)$. Therefore, the new field evaluated at the point $\big(\frac{1}{2},-\frac{\sqrt{3}}{2}\big)$ will be equal to the old field evaluated at the point $(1,0)$, i.e.

$$\phi'\big(\frac{1}{2},-\frac{\sqrt{3}}{2}\big) = \phi(1,0)$$

But $\big(\frac{1}{2},-\frac{\sqrt{3}}{2}\big) = R(1,0)$, not $R^{-1}(1,0)$. In general then, $\phi'(\mathbf x) = \phi(R\mathbf x)$.

In the above transformations the components of the vector ( or covector), i. e Aν change. But none of the books (Caroll) mention the change in the arguments. Why is that so. Why don't they change the arguments as well. Whereas you have written the transformation for coordinates as well.

Presumably Carroll did not want to add too much to the notation. But if you are transforming a vector field, which takes different values at different points in spacetime, then you certainly have to treat each component as a function and transform arguments accordingly.

Now based on my knowledge of the above transformation of basis vectors (with the inverse matrix) , I try and form the Active transformation. Here the fields/vector change and not the coordinates. So I should be really using the inverse matrix λ−1 (is this reasoning correct)

No, it is not correct. If the field configuration is rotated by $60^\circ$, then the direction of the vector should also be rotated by $60^\circ$, as in my diagram.

Lastly I know for a scalar field. Since it is a scalar field I have ϕ′(x′)=ϕ(x). Now x′=λx.

Again, this is wrong. Perhaps it would be simpler to consider a single coordinate $x$ and a scaled coordinate $x'=2x$.

When we say $x'=2x$, we do not mean that the new label of a point is twice the old label. Instead, we mean that the $x'$ "tick marks" are twice as far apart as the $x$ tick marks are.

What this means is that the $x'$ coordinate of a point is half the corresponding $x$ coordinate of that point. As you can see on the diagram, $x=3$ corresponds to $x'=1.5$, not $x'=6$.

Ultimately, you're making the same error as if you said "1 meter is equal to 100 centimeters, so the position of a point in meters is 100$\times$ the position of the point in centimeters". You have it backwards.