# Grad of a scalar field

1. Oct 2, 2016

### ConorDMK

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
= (∂/∂x)xˆ + (∂/∂y)yˆ + (∂/∂z)zˆ

3. 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.)

2. Oct 2, 2016

### stevendaryl

Staff Emeritus
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.

3. Oct 2, 2016

### Staff: Mentor

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

4. Oct 3, 2016

### ConorDMK

(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.

5. Oct 3, 2016

### stevendaryl

Staff Emeritus
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 $(\nabla T)_\phi$?

6. Oct 3, 2016

### ConorDMK

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)))

7. Oct 3, 2016

### Staff: Mentor

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

8. Oct 3, 2016

### stevendaryl

Staff Emeritus
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.