Derivation of curl of magnetic field in Griffiths

Physics Asked on May 8, 2021

Can someone please derive how $$frac{d}{dx} f(x-x’) = -frac{d}{dx’} f(x-x’)~?$$
In Griffiths electrodynamics, this is directly mentioned. I’m really confused, can someone elaborate!

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2 Answers

This follows directly from the chain rule:

$$frac{partial}{partial x}[f(x-x')] = f'(x-x')frac{partial}{partial x}[x-x'] = f'(x-x')$$ whereas $$frac{partial}{partial x'}[f(x-x')] = f'(x-x')frac{partial}{partial x'}[x-x'] = -f'(x-x').$$ (Here I take $f'(x-x')$ to mean that (total) derivative of $f$ with respect to its single independent variable.)

Thus, we see that the two expressions are simply the negations of each other. More complex versions of this can similarly be derived for other vector calculus operators, such as $$nabla_x f(x-x') = -nabla_{x'} f(x-x'),$$ where $nabla_x$ denotes the gradient with respect to $x$. I think this is also explained in Griffiths somewhere, but hopefully this explanation suffices.

Hope this helps.

Correct answer by Uyttendaele on May 8, 2021

I think that this can come from this argument: You can prove that $$frac{vec{x}-vec{x}'}{|vec{x}-vec{x}'|^3}=-nablaleft(frac{1}{|vec{x}-vec{x}'|}right)$$ If you take this change $$vec{x}rightarrowvec{x}'$$ and $$vec{x}'rightarrowvec{x}$$ the previous equality become $$frac{vec{x}'-vec{x}}{|vec{x}'-vec{x}|^3}=-nabla'left(frac{1}{|vec{x}'-vec{x}|}right)$$

From the two equations you can get that $$nablaleft(frac{1}{|vec{x}-vec{x}'|}right)=-frac{vec{x}-vec{x}'}{|vec{x}-vec{x}'|^3}=frac{vec{x}'-vec{x}}{|vec{x}'-vec{x}|^3}=-nabla'left(frac{1}{|vec{x}'-vec{x}|}right)$$

I hope it is useful, by the way, i check the book Classical electrodinamycs second edition of Jackson in the page 33. Bye!

Answered by Luca Javier Gomez Bachar on May 8, 2021

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