Griffiths Electrodynamics - Chap 5 clarification curl of B field

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SUMMARY

This discussion clarifies the application of vector calculus, specifically the use of Product Rule 5, in deriving the curl of the magnetic field B as presented in Griffith's Electrodynamics, page 224. The second term in the equation, –(J * ∇) (r^-3), requires careful integration to ensure it evaluates to zero. The participants emphasize the importance of correctly identifying terms in the product rule, particularly distinguishing between divergence and dot products. The discussion highlights the need for precise manipulation of vector fields and scalar functions in electrodynamics.

PREREQUISITES
  • Understanding of vector calculus, specifically the product rule for differentiation.
  • Familiarity with Griffith's Electrodynamics, particularly Chapter 5.
  • Knowledge of magnetic field concepts and the curl operator.
  • Proficiency in manipulating vector and scalar fields in mathematical physics.
NEXT STEPS
  • Review the application of the curl operator in electromagnetic theory.
  • Study the implications of the divergence theorem in vector calculus.
  • Learn advanced vector calculus identities relevant to electrodynamics.
  • Practice problems involving the derivation of magnetic fields from current distributions.
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Students of physics, particularly those studying electrodynamics, as well as educators and researchers looking to deepen their understanding of vector calculus applications in electromagnetic theory.

Sparky_
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Homework Statement



This question is regarding clarifying some reading in Griffith’s Electrodynamics, page 224.
“deriving the curl of B”
In particular it’s less on electrodynamics and more on some vectors or vector calculus.
The book states: we must check that the second term integrates to zero:

The second term that is referred to is:
–(J *\nabla)\frac{\hat{r}}{r^3}
• = “dot product”
The x-component is –(J *\nabla')\frac{x-x’}{r^3} = \nabla' * [\frac{x-x’}{r^3} J]- (\frac{x-x’}{r^3})(\nabla'*J)
“using product rule 5”
Product rule 5 states: \nabla*(fA) = f(\nabla*A)+ A*(\nabla f)
F is a scalar, A is a vector”


Homework Equations





The Attempt at a Solution



I am having some difficulty matching up the terms and applying “product rule 5” in this case”

If I let the f(\nabla*A) term = (J*\nabla')(\frac{x-x’}{r^3})

And \nabla*(f*A) = \nabla'*(\frac{x-x'}{r^3} J)

The last term does not match.

Meaning, I have A*(\nabla f) from the product rule.

The vector A "dot" a divergence.My remaining term does not have a divergence.

How does this product rule fit?

Thanks
-Sparky_
 
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You want to use the identity: ##\vec{\nabla} \cdot(f \vec{A}) = f(\vec{\nabla} \cdot\vec{A}) + \vec{A} \cdot (\vec{\nabla} f) = f(\vec{\nabla} \cdot\vec{A}) +( \vec{A} \cdot \vec{\nabla}) f## where I have rewritten the last term in a slightly different form.

Sparky_ said:
If I let the f(\nabla*A) term = (J*\nabla')(\frac{x-x’}{r^3})

Note that on the left you have a divergence of a vector: ##f(\vec{\nabla} \cdot \vec{A})## whereas on the right you do not have a divergence. You'll need to reconsider which term in the product rule should be identified with (\vec{J} \cdot \vec{\nabla '})(\frac{x-x’}{r^3})
 

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