Grad of a Scalar Field: Computing ∇T in Spherical Coordinates

[ConorDMK]

Homework Statement

Let T(r) be a scalar field. Show that, in spherical coordinates ∇T = (∂T/∂r) rˆ + (1/r)(∂T/∂θ) θˆ + (1/(r*sin(θ)))(∂T/∂φ) φˆ

Hint. Compute T(r+dl)−T(r) = T(r+dr, θ+dθ, φ+dφ)−T(r, θ, φ) in two different ways for the infinitesimal displacement dl = dr rˆ + rdθ θˆ + r*sin(θ)dφ φˆ and compare the two results.

Homework Equations

∇ = (∂/∂x)xˆ + (∂/∂y)yˆ + (∂/∂z)zˆ

The Attempt at a Solution

dT(r) ≡ T(r+dl)-T(r) = T(r+dr, θ+dθ, φ+dφ) - T(r,θ,φ) = (T(r,θ,φ) + (∂T(r)/∂r)dr + (∂T(r)/∂θ)dθ + (∂T(r)/∂φ)dφ) - T(r,θ,φ)

⇒ dT(r) = (∂T(r)/∂r)dr + (∂T(r)/∂θ)dθ + (∂T(r)/∂φ)dφ

But I don't know where I can go from here, and I don't think what I've done previously is correct (I rubbed out some of the work that continued form this, as I don't know what I can and can't use.)

 

[stevendaryl]

Homework Statement

Let T(r) be a scalar field. Show that, in spherical coordinates ∇T = (∂T/∂r) rˆ + (1/r)(∂T/∂θ) θˆ + (1/(r*sin(θ)))(∂T/∂φ) φˆ

Hint. Compute T(r+dl)−T(r) = T(r+dr, θ+dθ, φ+dφ)−T(r, θ, φ) in two different ways for the infinitesimal displacement dl = dr rˆ + rdθ θˆ + r*sin(θ)dφ φˆ and compare the two results.

Homework Equations

∇ = (∂/∂x)xˆ + (∂/∂y)yˆ + (∂/∂z)zˆ

The Attempt at a Solution

dT(r) ≡ T(r+dl)-T(r) = T(r+dr, θ+dθ, φ+dφ) - T(r,θ,φ) = (T(r,θ,φ) + (∂T(r)/∂r)dr + (∂T(r)/∂θ)dθ + (∂T(r)/∂φ)dφ) - T(r,θ,φ)

⇒ dT(r) = (∂T(r)/∂r)dr + (∂T(r)/∂θ)dθ + (∂T(r)/∂φ)dφ

But I don't know where I can go from here, and I don't think what I've done previously is correct (I rubbed out some of the work that continued form this, as I don't know what I can and can't use.)

I think they are defining the gradient \nabla T to be a vector such that (\nabla T) \cdot \vec{dl} = dT. You already have computed dT; it's just \frac{\partial T}{\partial r} dr + \frac{\partial T}{\partial \theta} d\theta + \frac{\partial T}{\partial \phi} d\phi. So you have the equation:

(\nabla T) \cdot \vec{dl} = \frac{\partial T}{\partial r} dr + \frac{\partial T}{\partial \theta} d\theta + \frac{\partial T}{\partial \phi} d\phi

The left-hand side of that equation can be written as: (\nabla T)_r (dl)_r + (\nabla T)_\theta (dl)_\theta +(\nabla T)_\phi (dl)_\phi, where (\nabla T)_r means the r-component of \nabla T, etc. and (dl)_r means the r-component of dl, etc.

So just expand (\nabla T) \cdot \vec{dl} in terms of components of \nabla T and \vec{dl}, and see what you get.

 

[Chestermiller]

What is the equation for $\vec{dl}$ in spherical coordinates (I.e., using the spherical coordinate unit vectors)?

 

[ConorDMK]

I think they are defining the gradient \nabla T to be a vector such that (\nabla T) \cdot \vec{dl} = dT. You already have computed dT; it's just \frac{\partial T}{\partial r} dr + \frac{\partial T}{\partial \theta} d\theta + \frac{\partial T}{\partial \phi} d\phi. So you have the equation:

(\nabla T) \cdot \vec{dl} = \frac{\partial T}{\partial r} dr + \frac{\partial T}{\partial \theta} d\theta + \frac{\partial T}{\partial \phi} d\phi

The left-hand side of that equation can be written as: (\nabla T)_r (dl)_r + (\nabla T)_\theta (dl)_\theta +(\nabla T)_\phi (dl)_\phi, where (\nabla T)_r means the r-component of \nabla T, etc. and (dl)_r means the r-component of dl, etc.

So just expand (\nabla T) \cdot \vec{dl} in terms of components of \nabla T and \vec{dl}, and see what you get.

(∇T)_r(dl)_r = (∂T/∂r)(dr/dr)rˆ = (∂T/∂r)rˆ

(∇T)_θ(dl)_θ = (∂T/∂r)(dθ/dθ)θˆ = (∂T/∂θ)(1/r)θˆ

(∇T)_φ(dl)_φ = (∂T/∂φ)(dφ/dφ)φˆ = (∂T/∂φ)(1/(r*sin(θ)))φˆ


∇T(r) = (∇T)_r(dl)_r + (∇T)_θ(dl)_θ + (∇T)_φ(dl)_φ = (∂T/∂r)rˆ + (∂T/∂θ)(1/r)θˆ + (∂T/∂φ)(1/(r*sin(θ)))φˆ

This is what I had before, but I didn't think this was right.
And I also, apparently, have to do it with another method.

 

[stevendaryl]

(∇T)_r(dl)_r = (∂T/∂r)(dr/dr)rˆ = (∂T/∂r)rˆ

No, there is no \hat{r}. When you take a dot-product, you just get a number:
(\nabla T)_r (dl)_r = \frac{\partial T}{\partial r} dr

So, you didn't actually take my hint. Let me spell it out for you more explicitly:

One way of calculating (\nabla T) \cdot \vec{dl}:
(\nabla T) \cdot \vec{dl} = (\nabla T)_r (dl)_r + (\nabla T)_\theta (dl)_\theta + (\nabla T)_\phi (dl)_\phi

We have: (dl)_r = dr, (dl)_\theta = r d\theta, (dl)_\phi = r sin(\theta) d\phi. So we have:

(\nabla T) \cdot \vec{dl} = (\nabla T)_r dr + (\nabla T)_\theta r d\theta + (\nabla T)_\phi r sin(\theta) d\phi

Second way of calculating (\nabla T) \cdot \vec{dl}:

(\nabla T) \cdot \vec{dl} = dT = \frac{\partial T}{\partial r} dr + \frac{\partial T}{\partial \theta} d\theta + \frac{\partial T}{\partial \phi} d\phi

So putting those two together gives you:

(\nabla T)_r dr + (\nabla T)_\theta r d\theta + (\nabla T)_\phi r sin(\theta) d\phi = \frac{\partial T}{\partial r} dr + \frac{\partial T}{\partial \theta} d\theta + \frac{\partial T}{\partial \phi} d\phi

So, what do you think (\nabla T)_r must be? What is (\nabla T)_\theta? What is (\nabla T)_\phi?

 

[ConorDMK]

No, there is no \hat{r}. When you take a dot-product, you just get a number:
(\nabla T)_r (dl)_r = \frac{\partial T}{\partial r} dr

So, you didn't actually take my hint. Let me spell it out for you more explicitly:

One way of calculating (\nabla T) \cdot \vec{dl}:
(\nabla T) \cdot \vec{dl} = (\nabla T)_r (dl)_r + (\nabla T)_\theta (dl)_\theta + (\nabla T)_\phi (dl)_\phi

We have: (dl)_r = dr, (dl)_\theta = r d\theta, (dl)_\phi = r sin(\theta) d\phi. So we have:

(\nabla T) \cdot \vec{dl} = (\nabla T)_r dr + (\nabla T)_\theta r d\theta + (\nabla T)_\phi r sin(\theta) d\phi

Second way of calculating (\nabla T) \cdot \vec{dl}:

(\nabla T) \cdot \vec{dl} = dT = \frac{\partial T}{\partial r} dr + \frac{\partial T}{\partial \theta} d\theta + \frac{\partial T}{\partial \phi} d\phi

So putting those two together gives you:

(\nabla T)_r dr + (\nabla T)_\theta r d\theta + (\nabla T)_\phi r sin(\theta) d\phi = \frac{\partial T}{\partial r} dr + \frac{\partial T}{\partial \theta} d\theta + \frac{\partial T}{\partial \phi} d\phi

So, what do you think (\nabla T)_r must be? What is (\nabla T)_\theta? What is (\nabla T)_\phi?

Sorry, I kept thinking ∇ had to be a vector.

(∇T)_r = (∂T/∂r)

(∇T)_θ = (∂T/∂θ)(1/r)

(∇T)_φ = (∂T/∂φ)(1/(r*sin(θ)))

 

[Chestermiller]

Sorry, I kept thinking ∇ had to be a vector.

(∇T)_r = (∂T/∂r)

(∇T)_θ = (∂T/∂θ)(1/r)

(∇T)_φ = (∂T/∂φ)(1/(r*sin(θ)))

Your answer is correct. But your implication that del is not a vector is not quite correct. Del is a vector operator.

 

[stevendaryl]

Sorry, I kept thinking ∇ had to be a vector.

(∇T)_r = (∂T/∂r)

(∇T)_θ = (∂T/∂θ)(1/r)

(∇T)_φ = (∂T/∂φ)(1/(r*sin(θ)))

It is a vector (in the way you're being taught--it actually should be a covector, but that's kind of an advanced topic). A vector has three components.