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

In summary, the conversation discusses the equality of the gradient vectors for differentiable functions in Cartesian and Spherical coordinates. The question arises if ∇T(r,θ) can just be equal to ∂T/∂r (unitvector-r) + ∂T/∂θ (unitvector-θ), without the (1/r) term. The conversation also mentions the usefulness of the (1/r) term in dot product calculations. The concept of the gradient vector is defined as a vector pointing in the direction of fastest increase of a function, independent of a coordinate system. The conversation ends with the suggestion to google for a derivation of the gradient vector in this context.
  • #1
davidbenari
466
18
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.
 
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  • #2
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
 
  • #3
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?
 
  • #4
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.
 
  • #5


To prove that ∇T(r,θ)=∇G(x,y), we can use the chain rule to relate the derivatives in polar coordinates to those in Cartesian coordinates.

First, we can express T(r,θ) in terms of x and y by using the relationships between polar and Cartesian coordinates: x=r*cos(θ) and y=r*sin(θ). This gives us T(r,θ)=G(r*cos(θ), r*sin(θ)).

Next, we can take the partial derivatives of T(r,θ) with respect to x and y, using the chain rule. This gives us:

∂T/∂x = ∂G/∂x * ∂x/∂x + ∂G/∂y * ∂y/∂x
∂T/∂y = ∂G/∂x * ∂x/∂y + ∂G/∂y * ∂y/∂y

Simplifying these expressions, we get:

∂T/∂x = ∂G/∂x
∂T/∂y = ∂G/∂y

These are the same as the partial derivatives of G(x,y), which means that ∇T(r,θ)=∇G(x,y).

Now, why can't we just use the simpler form ∂T/∂r (unitvector-r) + ∂T/∂θ (unitvector-θ)? The reason is that this form only works for functions that are radially symmetric, meaning that they do not change with respect to the angle θ. In other words, they are only dependent on the distance from the origin, r. However, many functions in polar coordinates, such as T(r,θ), do depend on the angle θ, so we need to include the (1/r) term to account for this dependence. This term represents the change in the function as we move along a constant radius, r, in the polar coordinate system.

In summary, to prove that ∇T(r,θ)=∇G(x,y), we use the chain rule to relate the derivatives in polar coordinates to those in Cartesian coordinates. And the (1/r) term in the gradient vector accounts for the dependence of the function on the angle θ.
 

1. What is the difference between polar and cartesian coordinates?

Polar coordinates use a distance and angle to represent a point in a 2D plane, while cartesian coordinates use x and y coordinates. Polar coordinates are often used in situations where there is a circular or radial pattern, while cartesian coordinates are used for more general purposes.

2. How do I convert between polar and cartesian coordinates?

To convert from polar coordinates to cartesian coordinates, you can use the following equations:
x = r cos θ
y = r sin θ
where r is the distance from the origin to the point, and θ is the angle from the positive x-axis to the point. To convert from cartesian coordinates to polar coordinates, you can use the following equations:
r = √(x^2 + y^2)
θ = tan^-1 (y/x)

3. How do I prove that gradients vectors are the same in polar and cartesian coordinates?

To prove that gradients vectors are the same in polar and cartesian coordinates, you can use the chain rule for partial derivatives. This will show that the gradient vector in polar coordinates has the same magnitude and direction as the gradient vector in cartesian coordinates.

4. What is the relationship between gradients and directional derivatives?

Gradients and directional derivatives are closely related, as the gradient vector at a point represents the direction and magnitude of the steepest increase in a scalar field at that point. The directional derivative is the rate at which the scalar field changes along a specific direction from that point. The gradient vector can be used to calculate the directional derivative in any direction.

5. Can I use polar coordinates to solve problems involving gradients and directional derivatives?

Yes, polar coordinates can be used to solve problems involving gradients and directional derivatives. However, it may be more convenient to use cartesian coordinates in some cases, depending on the specific problem. It is important to understand the relationship between polar and cartesian coordinates and how to convert between them to effectively solve problems involving gradients and directional derivatives.

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