How to prove gradients vectors are the same in polar and cartesian co.

[davidbenari]

Suppose T=T(r,θ)=G(x,y)
How do you prove ∇T(r,θ)=∇G(x,y)?

I can think of some arguments in favor of this equality, but I want an actual proof or a very good intuitive argument. My arguments in favor go something like this:

-Gradient vectors should be the same because if my directional derivative is taken parallel to the gradient vector then I get its maximum/minimum value and if these two gradient vectors are the same then everything will be consistent . However this proves that they should point in the same direction but it doesn't prove they're the same vector!


Also why can't ∇T(r,θ) can't just be = ∂T/∂r (unitvector-r) + ∂T/∂θ (unitvector-θ)

instead of ∇T(r,θ) = ∂T/∂r (unitvector-r) + (1/r)∂T/∂θ (unitvector-θ)... What does the (1/r) term imply? Aside from the fact that it will be useful whenever doing the dot product with some infinitesimally small displacement with a term rdθ (unitvector-θ).

Thanks.

 

[pat1enc3_17]

hey,

i did it in some theoretical physics exercise years past

google it, if you don't find any i can summarize it, i think, anyway the derivation is, as i remember straith forwards, based on line-elements and the chain rule... even its a 1-2 pages calculation

 

[HallsofIvy]

The gradient vector, of differentiable function f, is defined as "a vector pointing in the direction of fastest increase of f whose length is the rate of increase of f in that direction." Since that is completely independent of a coordinate system, your question does not make sense to me. Perhaps you have found the formulas for the gradient in Cartesian and Spherical coordinates and want to show they give the same vector?

 

[davidbenari]

HallsofIvy:
Yeah, I agree this question should not even arise at all based on definitions; it was just that my textbook presented the idea in a somewhat weird way. The only thing that keeps troubling me is the last part of my post, I'd like to hear what you think.

"" Also why can't ∇T(r,θ) can't just be = ∂T/∂r (unitvector-r) + ∂T/∂θ (unitvector-θ)

instead of ∇T(r,θ) = ∂T/∂r (unitvector-r) + (1/r)∂T/∂θ (unitvector-θ)... What does the (1/r) term imply? Aside from the fact that it will be useful whenever doing the dot product with some infinitesimally small displacement with a term rdθ (unitvector-θ). ""I can think of some answers to this, but I'd like to see what you thought about this. Thanks.