How can the gradient of a scalar field be covarient?

[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.

 

[TrickyDicky]

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

Just that if you were using non-orthogonal coordinates, like having the x and y-axis in the Euclidean plane forming an acute or obtuse angle, df and ∇f would have different components, (just like they would have if we were in a curved surface) and we couldn't identify the gradient of a function with a vector as we usually do in regular vector calculus in Euclidean space with orthogonal coordinates, it would have to be a covector.

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.

Yes, sure. Differential forms are coordinate independent while vector fields representation is coordinate dependent, was this what you wanted to highlight?

 

[WannabeNewton]

Yes, sure. Differential forms are coordinate independent while vector fields representation is coordinate dependent, was this what you wanted to highlight?

Yes that was what my original example was trying to do. This was the OP's original question.

 

[stevendaryl]

Just that if you were using non-orthogonal coordinates, like having the x and y-axis in the Euclidean plane forming an acute or obtuse angle, df and ∇f would have different components, (just like they would have if we were in a curved surface) and we couldn't identify the gradient of a function with a vector as we usually do in regular vector calculus in Euclidean space with orthogonal coordinates, it would have to be a covector.

Being nonorthogonal is one way that vectors and covectors can be different, but they are different even for orthogonal coordinates in lots of circumstances.

Here's an intuitive way to get an idea of which mathematical objects should be thought of as vectors, and which should be thought of as covectors: Suppose you are given a mathematical description of a situation in good old Cartesian, Euclidean coordinates (x,y,z), but with one difference: Different units are used in measuring distances in the x-y plane and in measuring distances in the vertical direction (z-direction). An example of such a situation is the measurements used by sailors in the olden days. Distances along the surface of the ocean were measured using nautical miles, while vertical distances below the surface of the ocean were measured using fathoms. How long is a fathom, in terms of nautical miles? If you don't know the conversion factor, then you can't convert between vertical distances and horizontal distances.

But you can still do a lot of vector analysis. For instance, you can compute velocities as vectors with components V^x = \dfrac{dx}{dt}, V^y = \dfrac{dy}{dt}, V^z = \dfrac{dz}{dt}. If there is a scalar function, say the temperature of the ocean, T(x,y,z), you can compute a one-form dT with components dT_x = \dfrac{\partial T}{\partial x}, dT_y = \dfrac{\partial T}{\partial y}, dT_z = \dfrac{\partial T}{\partial z}. You can combine a vector with a one-form to get a scalar: If a fish has velocity V and the water has a temperature "gradient" dT, then the rate of change of temperature for the fish will be given by:

\dfrac{dT}{dt} = V^i (dT)_i

But what you can't do is compute any kind of "dot-product" between two vectors, or between two one-forms.

 

[TrickyDicky]

Yes that was what my original example was trying to do.

Ah, ok, thanks WN.

This was the OP's original question.

On rereading it was not so easy to see that just by the wording of post #1.
But I,m glad I get this now.
I thought by the title of the thread he was more interested in the reasons why in usual vector calculus the gradient could be considered a vector field.

 

[WannabeNewton]

I thought by the title of the thread he was more interested in the reasons why in usual vector calculus the gradient could be considered a vector field.

I don't blame you; it was ambiguous for me as well. I only realized that was what he was asking later on. I still couldn't find the pages in Carroll where this was mentioned so I gave up on that haha.

 

[atyy]

Maybe Tricky Dicky meant positive definite metric and orthonormal?

I believe the usual vector calculus gradient is g^ijf_,j with Euclidean metric and f_,j are the components of the gradient covector.