Laplacian calculation

  • #1

Homework Statement


Consider the scalar field φ=x2+y2-z2-1. Let H be the scalar field defined by
H = -0.5∇.(∇φ/ abs(∇φ)), where abs(∇φ) is the magnitude of ∇φ. Which makes that some sort of unit quantity. When H is evaluated for φ=0 it is the mean curvature of the level surface φ=0.
Calculate H. Write your answer as a function of z only.
Hint: Work in Cartesian coordinates x, y, z throughout the whole of part (c) and make sure that that you work out all of the derivatives before imposing the constraint Φ = 0. You will find it useful to use the abbreviation r = √(x2+y2+z2. I didn't use any of this, so I'm doing it wrong!

Homework Equations




The Attempt at a Solution


I got (∇φ/ abs(∇φ)) to be 1/√3 x i + 1/√3 y j +1/√3 z k where i, j and k are unit vectors. From this, ∇.(∇φ/ abs(∇φ))=1/√3. This is clearly wrong, wolfram alpha disagrees, I didn't use r, I don't know how to impose the constraint and I think it's something pretty fundamental that I'm missing! I have looked at examples online but they really haven't helped at all!
 

Answers and Replies

  • #2
Delta2
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You have made some serious mistake in calculating ##\nabla\phi/|\nabla\phi|## cause the result you give doesnt have magnitude 1. From what i can see, it is something like ## \frac{x}{r} i+ \frac{y}{r} j- \frac{z}{r} k##.

To impose the constraint use that ##\phi=0\rightarrow x^2+y^2=z^2+1## to eliminate x and y in the expression you ll calculate for H.
 
  • #3
You have made some serious mistake in calculating ##\nabla\phi/|\nabla\phi|## cause the result you give doesnt have magnitude 1. From what i can see, it is something like ## \frac{x}{r} i+ \frac{y}{r} j- \frac{z}{r} k##.

To impose the constraint use that ##\phi=0\rightarrow x^2+y^2=z^2+1## to eliminate x and y in the expression you ll calculate for H.
Perfect person for the question, love the name :D I think I seriously misunderstood our lecture, because I thought ∇φ was obtained by ∂φ/∂x i + ∂φ/dy j + ∂φ/dz k? Which would make ∇φ = 2x i + 2y j -2z k. Actually, maybe my magnitude was wrong! Should it be 2√(x2+y2+z2)?
 
  • #4
Delta2
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yes that should be the correct magnitude. and yes i love that you love my name, it is inspired by the laplacian operator :D.
 
  • #5
I got to an answer! Can't believe I didn't see that mistake, the magnitude was clearly meant to be 1. Thank you! :) Vector calculus, not one of my strengths!
 

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