How can the gradient of a scalar field be covarient?

[dEdt]

According to Carroll, $\nabla \phi$ is covariant under rotations. This really confuses me. For example, how could equations like $\vec{F}=-\nabla V$ be rotationally covariant if force is a contravariant vector?

I know this is strictly speaking more of a mathy question, but I still figured this was the best place to ask this question.

 

[WannabeNewton]

Hi! There was a very, very recent thread on this: https://www.physicsforums.com/showthread.php?t=666861. I have no idea what you mean by covariant under rotations so if you could explain that I would be much obliged thanks. $\omega = df$ is actually a one - form because if we take $(dx^{i})$ to be the dual basis for $T^{*}_{p}(\mathbb{R}^{n})$ then we can write $df(p) = \partial _{i}f(p)dx^{i}$ so that for all u,v in $T_{p}(\mathbb{R}^{n})$ holds
$$\begin{eqnarray*} (df(p))(av + bu) & = & \partial _{i}f(p)dx^{i}(av + bu)\\ & = & \partial _{i}f(p)dx^{i}(av) + \partial _{i}f(p)dx^{i}(bu)\\ & = & a\partial _{i}f(p)dx^{i}(v) + b\partial _{i}f(p)dx^{i}(u)\\ & = & av^{j}\partial _{i}f(p)dx^{i}(\partial _{j}) + bu^{j}\partial _{i}f(p)dx^{i}(\partial _{j})\\ & = & av^{i}\partial _{i}f(p) + bu^{i}\partial _{i}f(p)\\ & = & a(df(p))(v) + b(df(p))(u) \end{eqnarray*}$$ so the differential of f at p is actually a linear functional on the tangent space at p so it is actually a one - form.

 

[atyy]

Maybe he meant the old fashioned 3D vector calculus gradient which requires a metric to be defined, and he meant "covariant" analogous to Newton's equations written in component form in an inertial frame are "covariant" under Galilean transforms?

Maybe the OP can give a link so we can see Carroll's text?

 

[WannabeNewton]

Maybe he meant the old fashioned 3D vector calculus gradient which requires a metric to be defined, and he meant "covariant" analogous to Newton's equations written in component form in an inertial frame are "covariant" under Galilean transforms?

Maybe the OP can give a link so we can see Carroll's text?

Yes that is probably what is being said here; covariance in the sense of general covariance if I understood you correctly, right? I searched my copy of the book and couldn't find the phrase itself as stated by the OP anywhere so indeed if the OP could give what page / section he/she is talking about that would be splendid. G'day to you atyy =D.

 

[atyy]

Yes, that's what I was thinking.

 

[stevendaryl]

Hi! There was a very, very recent thread on this: https://www.physicsforums.com/showthread.php?t=666861. I have no idea what you mean by covariant under rotations so if you could explain that I would be much obliged thanks.

I've read some authors who talk about thingies with indices as being "covariant" or "contravariant" under coordinate changes. I can never remember which is which, but when changing from one set of coordinates $x^i$ to another set $x'^j$, one of them works like:

$(V^j)' = \dfrac{\partial x'^j}{\partial x^i} V^i$

and the other works like:

$(V_j)' = \dfrac{\partial x^i}{\partial x'_j} V_i$

Components of velocity vectors and accelerations work like the first, while components of gradients of scalars works like the second.

 

[stevendaryl]

I've read some authors who talk about thingies with indices as being "covariant" or "contravariant" under coordinate changes. I can never remember which is which, but when change from one set of coordinates $x^i$ to another set $x'^j$, one of them works like:

$(V^j)' = \dfrac{\partial x'^j}{\partial x^i} V^i$

and the other works like:

$(V_j)' = \dfrac{\partial x^i}{\partial x'_j} V_i$

Components of velocity vectors and accelerations work like the first, while components of gradients of scalars works like the second.

Of course, the convention for using raised or lowered indices automatically insures that you're going to use the correct transformation, because the wrong kind is syntactically invalid (a summation over indices requires one upper index and one lower index).

 

[Nugatory]

I've read some authors who talk about thingies with indices as being "covariant" or "contravariant" under coordinate changes.

"Co is Lo"

 

[WannabeNewton]

Components of velocity vectors and accelerations work like the first, while components of gradients of scalars works like the second.

Oh if that is what he means, and it might be if the OP could clarify, then the statement "covariant under rotations" seems very specific considering a one - form must transform in that way under any diffeomorphism.

 

[vanhees71]

It's easy to remember, if you just keep in mind that the coordinates $q^{\mu}$ have upper indices. Then the one-forms $\mathrm{d} q^{\mu}$ are the dual basis to the holonomous basis $\partial_{\mu}$ induced by the coordinates.

You can read these one-forms in a intuitive way as infinitesimal changes of the coordinates. Then switching to new coordinates $q'^{\mu}$ with a local diffeomorphism gives
$$\mathrm{d} q'^{\mu}=\frac{\partial q'^{\mu}}{\partial q^{\nu}} \mathrm{d} q^{\nu}.$$
This defines the contravariant transformation law from one dual basis to the other, i.e., you have objects with an upper index.

Now consider a scalar field $\Phi$. By definition, it's transformation property is
$$\Phi'(q')=\Phi(q).$$
Then
$$A_{\mu}=\frac{\partial}{\partial q^{\mu}} \Phi=\partial_{\mu} \Phi$$
obviously give covariant vector-field components. They transform covariantly (i.e., contragrediently wrt. to the co-basis vectors):
$$A'_{\mu}=\partial_{\mu}' \Phi'=\frac{\partial}{\partial q'^{\mu}} \Phi'(q')=\frac{\partial q^{\nu}}{\partial q'^{\mu}} \frac{\partial}{\partial q^{\nu}} \Phi(q).$$
In other words the holonomous basis vectors transform like
$$\partial_{\mu}'=\frac{\partial q^{\nu}}{\partial q'^{\mu}} \partial_{\nu},$$
i.e., contragrediently to the co-basis vectors, i.e., you have covariant objects with a lower index.

 

[Bill_K]

Newton had a clear advantage - he'd never heard of one-forms and diffeomorphisms.

 

[stevendaryl]

"Co is Lo"

Okay, that's a good way to remember it.

Someone explained the co- and contra- to me in terms of category theory, but I'm not sure it clarified anything to me.

 

[WannabeNewton]

Newton had a clear advantage - he'd never heard of one-forms and diffeomorphisms.

Or did he? Dun dun duuuuuun...

 

[dx]

As long as one thinks about the gradient as a vector ∇φ instead of the 1-form dφ, one is restricted to Euclidean space.

One can say that the representation of the gradient as a vector ∇φ is covariant under rotations and translations, but not under linear transformations or diffeomorphisms.

Note that the use of the word covariant here is different from its use in the word 'covariant vector.' Here we are talking about the covariance of a representation.

 

[dEdt]

To clarify some ambiguities:
By "rotationally covariant" I meant something analogous to generally covariant. Given the context of the question this was probably a stupid term to use, and I should have used something like rotationally invariant. My point, though, is that if the representation of $\vec {F}$ in one Cartesian coordinate system (CCS) is equal to the representation of $-\nabla V$ in the same CCS, then the representation of the two things cannot be equal in another CCS because they transform differently under a change of coordinate system.

At least, this was what I thought until about two minutes ago when I actually computed how the components of the gradient of a scalar change under a change of coordinate system. I found that they change in exactly the same way as the components of force do. So now I'm wondering what the difference between covariance and contravariance really is. Is seems like they're the same thing.

I think that dx may have addressed some of these questions, but I didn't really understand what he wrote.

 

[WannabeNewton]

If $X = \triangledown f$ in some coordinate system and you define a change of coordinates then the relation doesn't have to hold once you transform the components of $X$. Try this simple counter - example: let $f:\mathbb{R}^{2} \rightarrow \mathbb{R}, (x,y) \rightarrow x^{2}$ and $X = \triangledown f = 2x\partial _{x}$ and change to polar coordinates. This is why we instead use $df(p) = \partial _{i}f(p)dx^{i}$ because this is coordinate independent.

 

[dEdt]

That's kinda my point. There doesn't seem to be any reason why $\vec {F}=-\nabla V$ should hold in all coordinate systems, but it does...

 

[WannabeNewton]

Who said it does? Try the example I gave you; you will see it doesn't.

 

[atyy]

At least, this was what I thought until about two minutes ago when I actually computed how the components of the gradient of a scalar change under a change of coordinate system. I found that they change in exactly the same way as the components of force do. So now I'm wondering what the difference between covariance and contravariance really is. Is seems like they're the same thing.

I think that dx may have addressed some of these questions, but I didn't really understand what he wrote.

As dx said, the traditional vector calculus gradient is a vector - not a covector.

The gradient that is a covector is a different object from the traditional vector calculus gradient. For clarity dx wrote gradient with df for the former and ∇f for the latter. One can get the latter from the former in a space which has a Euclidean metric.

 

[TrickyDicky]

As long as one thinks about the gradient as a vector ∇φ instead of the 1-form dφ, one is restricted to Euclidean space.

The reason one can think of the gradient as a vector in Euclidean space, and in the case one uses orthogonal coordinates is that in this particular case due to the Euclidean metric form, vector and covectors have the same components and are therefore indistinguishable.

One can say that the representation of the gradient as a vector ∇φ is covariant under rotations and translations, but not under linear transformations or diffeomorphisms.

Well rotations are linear transformations so it seems to be covariant under linear transformations. And yes the representation is not diffeomorphism invariant due to what I mentioned about the requirement to use orthogonal coordinates.

If $X = \triangledown f$ in some coordinate system and you define a change of coordinates then the relation doesn't have to hold once you transform the components of $X$. Try this simple counter - example: let $f:\mathbb{R}^{2} \rightarrow \mathbb{R}, (x,y) \rightarrow x^{2}$ and $X = \triangledown f = 2x\partial _{x}$ and change to polar coordinates. This is why we instead use $df(p) = \partial _{i}f(p)dx^{i}$ because this is coordinate independent.

Who said it does? Try the example I gave you; you will see it doesn't.

Polar coordinates are orthogonal so it should be covariant under such transformation.

As dx said, the traditional vector calculus gradient is a vector - not a covector.

As commented above in euclidean space and as long as orthogonal coordinates are used the distinction is superfluous.

 

[WannabeNewton]

Polar coordinates are orthogonal so it should be covariant under such transformation.

What I posted is an exercise straight out of Lee's Smooth Manifolds book. Try it, it shouldn't take too long. I can refer you to the pages the discussion takes place in the text if you can get access to it so that this thread doesn't get derailed.

 

[dx]

The gradient of a function is simply the local linear behavior of the function and has nothing to do with any particular coordinate system, and therefore must be an object which is coordinate independent. This is not the case when one thinks of it as ∇f. One can illustrate this using a simple scaling of coordinates in one dimension. Let X and Y be two coordinate functions on the line, with X = 2Y, and let f be a function on the line with f = X = 2Y

Calculating the vector gradient in the X system, we have

∇f = (df/dX)e_X = e_X = (1/2)e_Y​

If we calculate it in the Y system we get

∇f = (df/dY)e_Y = = 2e_Y = 4e_X​

These are different vectors. So which one is the gradient? The representation is not even covariant with respect to scaling of the coordinate.

On the other hand, let's calculate the 1-form df in X:

df = (df/dX)dX = dX = 2dY​

Calculating in Y:

df = (df/dY)dY = 2dY = dX​

They are the same, as they should be. The gradient is a 1-form, not a vector.

Well rotations are linear transformations so it seems to be covariant under linear transformations.

It is not covariant under linear transformations, as the example above shows. When one says something is covariant under some group G, one means that it is covariant under all transformations belonging to the group. A rotation is a linear transformation, but a linear transformation is usually not a rotation.

 

[TrickyDicky]

The gradient of a function is simply the local linear behavior of the function and has nothing to do with any particular coordinate system, and therefore must be an object which is coordinate independent. This is not the case when one thinks of it as ∇f. One can illustrate this using a simple scaling of coordinates in one dimension. Let X and Y be two coordinate functions on the line, with X = 2Y, and let f be a function on the line with f = X = 2Y

Calculating the vector gradient in the X system, we have

∇f = (df/dX)e_X = e_X = (1/2)e_Y​

If we calculate it in the Y system we get

∇f = (df/dY)e_Y = = 2e_Y = 4e_X​

These are different vectors. So which one is the gradient? The representation is not even covariant with respect to scaling of the coordinate.

On the other hand, let's calculate the 1-form df in X:

df = (df/dX)dX = dX = 2dY​

Calculating in Y:

df = (df/dY)dY = 2dY = dX​

They are the same, as they should be. The gradient is a 1-form, not a vector.




It is not covariant under linear transformations, as the example above shows.

See below.

When one says something is covariant under some group G, one means that it is covariant under all transformations belonging to the group. A rotation is a linear transformation, but a linear transformation is usually not a rotation.

True but in your example you used a non-uniform scaling (you scaled differently one coordinate wrt the other) which is a non-linear transformation, only uniform scaling is linear. See http://en.wikipedia.org/wiki/Scaling_(geometry)
Can you give me a true example of a linear transformation that changes the components of a covector wrt a vector in Euclidean space?

 

[micromass]

True but in your example you used a non-uniform scaling (you scaled differently one coordinate wrt the other) which is a non-linear transformation, only uniform scaling is linear. See http://en.wikipedia.org/wiki/Scaling_(geometry)
Can you give me a true example of a linear transformation that changes the components of a covector wrt a vector in Euclidean space?

Huh? So you are saying that T(x,y)=(2x,y) is not a linear transformation?? You seem to have a very weird definition of linear transformations...

 

[TrickyDicky]

Huh? So you are saying that T(x,y)=(2x,y) is not a linear transformation?? You seem to have a very weird definition of linear transformations...

That is a linear transformation of course, but I believe that is not what dx is presenting, I think he is giving two functions with two arguments, and the transformation is linear in each argument separately but not for both. At least if it is truly a non-uniform scaling.
EDIT:
Admittedly I might also have misunderstood the Wikipedia article about scaling.

 

[micromass]

That is a linear transformation of course, but I believe that is not what dx is presenting, I think he is giving two functions with two arguments, and the transformation is linear in each argument separately but not for both. At least if it is truly a non-uniform scaling.
EDIT:
Admittedly I might also have misunderstood the Wikipedia article about scaling.

I think the point is the following. Assume we have an n-dimensional manifold M and take p in M. Assume we have a chart: $\varphi:U\rightarrow \mathbb{R}^n$ of p.
We define the gradient as a vector wrt the chart $\varphi$ as (the variables of $\varphi$ are called $(x^1,...,x^n)$)

$$(grad f)_p= \sum_i \frac{\partial f}{\partial x^i}(p) \frac{\partial}{\partial x^i}\vert_p$$

then this depends crucially on the chart $\varphi$. In particular, if $T:\mathbb{R}^n\rightarrow \mathbb{R}^n$ is any invertible linear map, then we can form the chart $T\circ \varphi: U\rightarrow \mathbb{R}^n$. The variables wrt this chart are $(y^1,...,y^n)$. We can write the gradient wrt this chart as

$$\sum_j \frac{\partial f}{\partial y^j}(p)\frac{\partial}{\partial y^j}\vert_p$$

Let us write $X^i = \frac{\partial f}{\partial x^i}(p)$. We want to express this in the coordinate system $T\circ \varphi$. If $\tilde{X}^j$ are the coordinates in the new coordinate system, then we have the following relation

$$\tilde{X}^j = \frac{\partial y^j}{\partial x^i}(p)X^i$$

We notice that the matrix of $\left(\frac{\partial y^j}{\partial x^i}(p)\right)$ coincides with T.

So if the two formulas for gradient coincide, then we would have that

$$\frac{\partial f}{\partial y^j}(p) = \sum_i \frac{\partial y^j}{\partial x^i}(p) \frac{\partial f}{\partial x^i}(p)$$

But by the chain rule, we have that

$$\frac{\partial f}{\partial y^j}(p) = \frac{\partial f}{\partial x^i}(p)\frac{\partial x^i}{\partial y^j}(p)$$

So the two formulas for the gradient coincide in the special case that the following matrices
equal

$$\left(\frac{\partial y^j}{\partial x^i}\right)=\left(\frac{\partial x^i}{\partial y^j}\right)$$

This is the same as demanding that that $T^t = T^{-1}$. Thus in the special case that $T\in O_n(\mathbb{R})$, we have that the gradients coincide.
If $T\notin O_n(\mathbb{R})$, then we can easily find counterexamples for the statement.

 

[DrGreg]

The gradient of a function is simply the local linear behavior of the function and has nothing to do with any particular coordinate system, and therefore must be an object which is coordinate independent. This is not the case when one thinks of it as ∇f. One can illustrate this using a simple scaling of coordinates in one dimension. Let X and Y be two coordinate functions on the line, with X = 2Y, and let f be a function on the line with f = X = 2Y

...but in your example you used a non-uniform scaling (you scaled differently one coordinate wrt the other) which is a non-linear transformation, only uniform scaling is linear. See http://en.wikipedia.org/wiki/Scaling_(geometry)
Can you give me a true example of a linear transformation that changes the components of a covector wrt a vector in Euclidean space?

In dx's example, it was not the case that (X,Y) was a single coordinate system for a 2D space, but rather that (X) and (Y) were two different coordinate systems on a 1D space.

 

[TrickyDicky]

In dx's example, it was not the case that (X,Y) was a single coordinate system for a 2D space, but rather that (X) and (Y) were two different coordinate systems on a 1D space.

Ok, but the coordinate system is 2 dimensional right? And a one dimensional scaling amounts to the trivial property of any element of a vector space that can be multiplied by a scalar, obtaining another vector space element.
In the case of a uniform scaling which is clearly a linear transformation I consider this kind of thing in 2D: f(x,y)=2x,2y or for a contraction f(x,y)=(1/3)x, (1/3)y
If you do this kind of transformation in Euclidean space and as long as you keep using orthogonal coordinate systems you still get the same components for ∇f and df.

 

[micromass]

Ok, but the coordinate system is 2 dimensional right? And a one dimensional scaling amounts to the trivial property of any element of a vector space that can be multiplied by a scalar, obtaining another vector space element.
In the case of a uniform scaling which is clearly a linear transformation I consider this kind of thing in 2D: f(x,y)=2x,2y or for a contraction f(x,y)=(1/3)x, (1/3)y
If you do this kind of transformation in Euclidean space and as long as you keep using orthogonal coordinate systems you still get the same components for ∇f and df.

There was nothing 2 dimensional about the example.

The example went as follows: consider a one-dimensional manifold M and a point p in M. Consider a chart $\varphi:U\rightarrow \mathbb{R}$ of p, where $p\in U\subseteq M$. Also consider the chart $\psi:U\rightarrow \mathbb{R}:q\rightarrow 2\varphi(q)$. The point of dx is simply that the gradient (considered as a vector) with respect to $\varphi$ and $\psi$ are different in general.

For example, consider the manifold $M=\mathbb{R}$. Consider p=1. Consider $f:M\rightarrow \mathbb{R}:x\rightarrow x$.
Let $\varphi:M\rightarrow \mathbb{R}:x\rightarrow x$ and $\psi:M\rightarrow \mathbb{R}:x\rightarrow 2x$.

Then the gradient (considered as a vector) wrt $\varphi$ would be $\frac{\partial}{\partial x}$.
The gradient wrt $\psi$ would be $\frac{1}{2}\frac{\partial}{\partial y} =\frac{1}{4}\frac{\partial}{\partial y}$.

 

[TrickyDicky]

I think the point is the following. Assume we have an n-dimensional manifold M and take p in M. Assume we have a chart: $\varphi:U\rightarrow \mathbb{R}^n$ of p.
We define the gradient as a vector wrt the chart $\varphi$ as (the variables of $\varphi$ are called $(x^1,...,x^n)$)

$$(grad f)_p= \sum_i \frac{\partial f}{\partial x^i}(p) \frac{\partial}{\partial x^i}\vert_p$$

then this depends crucially on the chart $\varphi$. In particular, if $T:\mathbb{R}^n\rightarrow \mathbb{R}^n$ is any invertible linear map, then we can form the chart $T\circ \varphi: U\rightarrow \mathbb{R}^n$. The variables wrt this chart are $(y^1,...,y^n)$. We can write the gradient wrt this chart as

$$\sum_j \frac{\partial f}{\partial y^j}(p)\frac{\partial}{\partial y^j}\vert_p$$

Let us write $X^i = \frac{\partial f}{\partial x^i}(p)$. We want to express this in the coordinate system $T\circ \varphi$. If $\tilde{X}^j$ are the coordinates in the new coordinate system, then we have the following relation

$$\tilde{X}^j = \frac{\partial y^j}{\partial x^i}(p)X^i$$

We notice that the matrix of $\left(\frac{\partial y^j}{\partial x^i}(p)\right)$ coincides with T.

So if the two formulas for gradient coincide, then we would have that

$$\frac{\partial f}{\partial y^j}(p) = \sum_i \frac{\partial y^j}{\partial x^i}(p) \frac{\partial f}{\partial x^i}(p)$$

But by the chain rule, we have that

$$\frac{\partial f}{\partial y^j}(p) = \frac{\partial f}{\partial x^i}(p)\frac{\partial x^i}{\partial y^j}(p)$$

So the two formulas for the gradient coincide in the special case that the following matrices
equal

$$\left(\frac{\partial y^j}{\partial x^i}\right)=\left(\frac{\partial x^i}{\partial y^j}\right)$$

This is the same as demanding that that $T^t = T^{-1}$. Thus in the special case that $T\in O_n(\mathbb{R})$, we have that the gradients coincide.
If $T\notin O_n(\mathbb{R})$, then we can easily find counterexamples for the statement.

This is a good way to show that only in the spatial case of orthogonal coordinate systems in Euclidean space the vector and covector coincide, this is what I was saying all along. But an isotropic scaling in Euclidean space doesn't change that orthogonality because it is a linear transformation invariant to rotations, an anisotropic scaling does, but it is a non-linear transformation.

 

[micromass]

This is a good way to show that only in the spatial case of orthogonal coordinate systems in Euclidean space the vector and covector coincide, this is what I was saying all along.

What do you mean with an "orthogonal coordinate system" anyway?

But an isotropic scaling in Euclidean space doesn't change that orthogonality because it is a linear transformation invariant to rotations, an anisotropic scaling does, but it is a non-linear transformation.

Huh? Any scaling is a linear transformation. I really don't understand where you're going.

 

[TrickyDicky]

What do you mean with an "orthogonal coordinate system" anyway?

http://en.wikipedia.org/wiki/Orthogonal_coordinates

Huh? Any scaling is a linear transformation. I really don't understand where you're going.

Ok, sorry about that. I take it back. I agree that the identification of vectors and covectors is not covariant under the whole group of linear transformations. In fact by insisting from the start on the orthogonal coordinates I was saying that in other words, how silly of me to confuse myself like that.

dx example is an anisotropic scaling, which is a linear transformation so it is the example I was asking of him.

Thanks everyone.

 

[micromass]

http://en.wikipedia.org/wiki/Orthogonal_coordinates

OK. Are you aware that if you define the gradient in polar coordinates as

$$\frac{\partial f}{\partial dr}\frac{\partial}{\partial r} + \frac{\partial f}{\partial \theta}\frac{\partial}{\partial \theta}$$

and if you define the gradient in rectangular coordinates as

$$\frac{\partial f}{\partial x}\frac{\partial}{\partial x}+\frac{\partial f}{\partial y}\frac{\partial}{\partial y}$$

that you will get different answers? So the gradient (as vector) is entirely dependent of the chart you use. However, both polar coordinates as rectangular coordinates are "orthogonal".

 

[TrickyDicky]

OK. Are you aware that if you define the gradient in polar coordinates as

$$\frac{\partial f}{\partial dr}\frac{\partial}{\partial r} + \frac{\partial f}{\partial \theta}\frac{\partial}{\partial \theta}$$

and if you define the gradient in rectangular coordinates as

$$\frac{\partial f}{\partial x}\frac{\partial}{\partial x}+\frac{\partial f}{\partial y}\frac{\partial}{\partial y}$$

that you will get different answers? So the gradient (as vector) is entirely dependent of the chart you use. However, both polar coordinates as rectangular coordinates are "orthogonal".

Hmmm, I think so but I'm not sure we're talking about the same thing. Maybe I was thinking about a slightly different thing

My point was that if we use polar coordinates in the Euclidean plane, the gradient of a function as a vector ∇f and the gradient of that function as a covector(df) have the same components. Now I'm not completely sure. Can you confirm?
This seems to me a different thing from the fact that ∇f in cartesian coordinates has different components in polar coordinates, which is kind of trivial since we have changed coordinates so the components can't remain the same.

 

[micromass]

Hmmm, I think so but I'm not sure we're talking about the same thing. Maybe I was thinking about a slightly different thing

My point was that if we use polar coordinates in the Euclidean plane, the gradient of a function as a vector ∇f and the gradient of that function as a covector(df) have the same components. Now I'm not completely sure. Can you confirm?

Sure, no problem. You just define it the same way. But what does this have to do with orthogonal coordinates?

This seems to me a different thing from the fact that ∇f in cartesian coordinates has different components in polar coordinates, which is kind of trivial since we have changed coordinates so the components can't remain the same.

The point is that if we look at the gradient as covector, then defining it in polar and in rectangular yields exactly the same thing. So the covector is coordinate invariant. The gradient as vector does change as we change the coordinates.