[Math] Gradient in arbitrary coordinate system

[Deleted]

Ooookay, my question is: Why is the gradient operator different in non-orthonormal coordinate systems?

According to wikipedia (as well as Wolfram Alpha), the gradient in, say cylindrical coordinates is

grad(f)=(df/dr)*e_r + (1/r)*(df/d(phi))*e_(phi) + (df/dz)*e_z

Where there is a 1/r factor (which, as far as I remember is one of the non-vanishing Christoffel-symbols in cylindrical, but I dunno if that is a coincidence or not) that is not present in cartesian.

However, assuming an arbitrary inner product space, the gradient of a scalar function f, is basically the covariant derivative of f (resulting in a (0;1) tensor field).
But, the covariant derivative of a scalar field is the same as its partial derivative, as for scalars, there is no need for a metric connection, so no extra factors would need to appear anywhere.

I assume this probably has something to do with the fact that a f(x,y,z) scalar field in {u,v,w} coordinates will become f(x(u,v,w), y(u,v,w), z(u,v,w)) with composite functions, however I have no idea how to formulate this properly in a general case.

If someone could deduce this for me, I'd be terribly appreciative.

[Payday]

When I couldn't figure out a math question in school I just wrote undefined.

[Blueobelisk]

Rofl at this show off. It's not just the gradient, if you change coordinate systems you have to change your equations and variables. What's so hard to understand about that?

[videotape]

"If you change coordinate systems you have to change your equations and variables" is vague to the point of either nonsense or fallacy.

I think the answer to the question as stated (in short form, not the long-winded elaboration) is pretty straightforward. Think about what gradient represents conceptually, then think about what it means to have a space which is not orthonormal.

In fact the real answer is: It isn't different. It's just that the additional complexity of the general case reduces to identity in the orthonormal case.

[Garnier Fructis]

Rofl at this show off. It's not just the gradient, if you change coordinate systems you have to change your equations and variables. What's so hard to understand about that?

Counterexample: The Euler-Lagrange equations have the exact same form no matter what coordinate system you choose.

[Deleted]

Rofl at this show off. It's not just the gradient, if you change coordinate systems you have to change your equations and variables. What's so hard to understand about that?

Yes, it MUST be showoff, cause, you know, I spent about an hour yesterday trying to calculate a laplacian in general for reasons important to my studies (no this is NOT a "homework") and when I was done I applied it for cylindrical and got something wrong, because, as it turns out, I assumed the gradient was unchanged.

However what you just said really made no sense, as it had no actual substance and also, when you change coordinate systems, SCALARS are not changed because scalars are not expressed with a coordinate basis. Of course variables can be expressed with the new coordinate system's variables but that is not the same as vector or tensor transformations.

"If you change coordinate systems you have to change your equations and variables" is vague to the point of either nonsense or fallacy.

I think the answer to the question as stated (in short form, not the long-winded elaboration) is pretty straightforward. Think about what gradient represents conceptually, then think about what it means to have a space which is not orthonormal.

In fact the real answer is: It isn't different. It's just that the additional complexity of the general case reduces to identity in the orthonormal case.

I'm sorry but I can't do much with this answer either. I was taught that the covariant derivative of a scalar field is the same as its partial derivative.
If I take a scalar field c(x,y,z) in {x,y,z}, transform to {u,v,w}, for which c will simply become c(x(u,v,w), y(u,v,w), z(u,v,w)).
If I now calculate the covariant derivative of c, that is (D_i)c=((D_u)c; (D_v)c; (D_w)c), that is simply a covector valued expression whose components are the partial derivatives of c.

However, apparantly the bolded part is somewhere incorrect, I would like to know what is incorrect about that.
I am interested in the exact mathematical deduction of the 'gradient' operator in ANY kind of coordinate system, assuming I know the transformation fuctions and the metric exists.

[paralleluniverse]

Use the chain rule: http://physics.stackexchange.com/que...rst-principles

Alternatively, you can use the Jacobian.

[Deleted]

Use the chain rule: http://physics.stackexchange.com/que...rst-principles

Alternatively, you can use the Jacobian.

Hmm, that site is actually useful so thanks. They used a bit different approach compared to mine though.

Anyways, I managed to solve the problem, my Laplacian was correct, just the form I got was a bit different compared to the form stated in wolfram alpha, and their equivalence was not trivial for me initially.
My gradient was also correct, but I failed to notice the gradient in cylindrical expressed in wolfram and wikipedia was in a normed cylindrical coordinate system so that they can use basic vector calculus with it, while I used tensor calculus in a dual space, so in essence I expressed it in two different coordinate bases, hence why it seemed incorrect.