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

In summary: I'm not sure. I thought it was (∂T/∂φ)(1/(r*sin(θ))) but that doesn't seem to work out when I try and expand it.Yes, that's correct. So now you have the expression for each component of the gradient. Putting them together, you have:∇T = (∂T/∂r)rˆ + (∂T/∂θ)(1/r)θˆ + (∂T/∂φ)(1/(r*sin(θ)))φˆwhich is the same as what was given in the problem statement.
  • #1
ConorDMK
25
0

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.)
 
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  • #2
ConorDMK said:

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

[itex](\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[/itex]

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

So just expand [itex](\nabla T) \cdot \vec{dl}[/itex] in terms of components of [itex]\nabla T[/itex] and [itex]\vec{dl}[/itex], and see what you get.
 
  • #3
What is the equation for ##\vec{dl}## in spherical coordinates (I.e., using the spherical coordinate unit vectors)?
 
  • #4
stevendaryl said:
I think they are defining the gradient [itex]\nabla T[/itex] to be a vector such that [itex](\nabla T) \cdot \vec{dl} = dT[/itex]. You already have computed [itex]dT[/itex]; it's just [itex]\frac{\partial T}{\partial r} dr + \frac{\partial T}{\partial \theta} d\theta + \frac{\partial T}{\partial \phi} d\phi[/itex]. So you have the equation:

[itex](\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[/itex]

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

So just expand [itex](\nabla T) \cdot \vec{dl}[/itex] in terms of components of [itex]\nabla T[/itex] and [itex]\vec{dl}[/itex], 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.
 
  • #5
ConorDMK said:
(T)r(dl)r = (∂T/∂r)(dr/dr)rˆ = (∂T/∂r)rˆ

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

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

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

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

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

Second way of calculating [itex](\nabla T) \cdot \vec{dl}[/itex]:

[itex](\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[/itex]

So putting those two together gives you:

[itex](\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 [/itex]

So, what do you think [itex](\nabla T)_r[/itex] must be? What is [itex](\nabla T)_\theta[/itex]? What is [itex](\nabla T)_\phi[/itex]?
 
  • #6
stevendaryl said:
No, there is no [itex]\hat{r}[/itex]. When you take a dot-product, you just get a number:
[itex](\nabla T)_r (dl)_r = \frac{\partial T}{\partial r} dr[/itex]

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

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

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

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

Second way of calculating [itex](\nabla T) \cdot \vec{dl}[/itex]:

[itex](\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[/itex]

So putting those two together gives you:

[itex](\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 [/itex]

So, what do you think [itex](\nabla T)_r[/itex] must be? What is [itex](\nabla T)_\theta[/itex]? What is [itex](\nabla T)_\phi[/itex]?
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
ConorDMK said:
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.
 
  • #8
ConorDMK said:
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.
 

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

1. What is a scalar field?

A scalar field is a mathematical concept that assigns a scalar value (such as temperature, pressure, or density) to every point in a given space. It can be represented by a function that takes in coordinates as input and outputs a scalar value at that point.

2. What is the gradient of a scalar field?

The gradient of a scalar field is a vector that points in the direction of the steepest increase of the scalar field at a specific point. It is a mathematical operation that involves taking the partial derivatives of the scalar field with respect to each coordinate axis.

3. How is the gradient of a scalar field computed in spherical coordinates?

In spherical coordinates, the gradient of a scalar field is computed by taking the partial derivatives of the scalar field with respect to the radius, polar angle, and azimuthal angle. These derivatives are then multiplied by the corresponding unit vectors and added together to form a vector representing the gradient.

4. Why are spherical coordinates used to compute the gradient of a scalar field?

Spherical coordinates are commonly used to represent physical systems with spherical symmetry, such as planets, stars, and atoms. They are also useful for solving problems involving radial symmetry. In such cases, computing the gradient of a scalar field in spherical coordinates can simplify the calculations and provide a more intuitive understanding of the system.

5. What are some applications of computing the gradient of a scalar field in spherical coordinates?

Computing the gradient of a scalar field in spherical coordinates has numerous applications in physics, engineering, and mathematics. It is used to analyze and model physical systems with spherical symmetry, such as the gravitational field of a planet or the electric field of a charged sphere. It is also used in solving differential equations and in vector calculus to find the direction and rate of change of a function.

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