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Grad of a scalar field

  1. Oct 2, 2016 #1
    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. jcsd
  3. Oct 2, 2016 #2

    stevendaryl

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    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.
     
  4. Oct 2, 2016 #3
    What is the equation for ##\vec{dl}## in spherical coordinates (I.e., using the spherical coordinate unit vectors)?
     
  5. Oct 3, 2016 #4
    (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.
     
  6. Oct 3, 2016 #5

    stevendaryl

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    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]?
     
  7. Oct 3, 2016 #6

    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)))
     
  8. Oct 3, 2016 #7
    Your answer is correct. But your implication that del is not a vector is not quite correct. Del is a vector operator.
     
  9. Oct 3, 2016 #8

    stevendaryl

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    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.
     
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