- #1

olgerm

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Is on of these pages wrong or I misunderstand it?

I am asking because I want to know what does Δ

^{2}equal to in this

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- #1

olgerm

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Is on of these pages wrong or I misunderstand it?

I am asking because I want to know what does Δ

- #2

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- #3

olgerm

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These are Cartesian coordinates.

- #4

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The Laplace operator applied to a vector in Cartesian coordinates is such that the ##x## component of ##\nabla^2 \vec A## is equal to ##\nabla^2 A_x##, where ##A_x## is the ##x## component of ##\vec A##.

- #5

olgerm

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A is vector-field. I am trying to find out what does ##\nabla^2 A##, on this Wikipedia page

https://en.wikipedia.org/wiki/Mathematical_descriptions_of_the_electromagnetic_field#Gauge_freedom , equal to.

Is it that ##(\nabla^2 A)_j=\sum_{i=1}^3(\frac{\partial^2 A_j}{\partial x_i^2})##?

Where ##A_i## is the ##i## component of ##\vec A##.

https://en.wikipedia.org/wiki/Mathematical_descriptions_of_the_electromagnetic_field#Gauge_freedom , equal to.

Is it that ##(\nabla^2 A)_j=\sum_{i=1}^3(\frac{\partial^2 A_j}{\partial x_i^2})##?

Where ##A_i## is the ##i## component of ##\vec A##.

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- #6

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Neither, see my previous post.A is vector-field. I am trying to find out what does ##\nabla^2 A##, on this Wikipedia page

https://en.wikipedia.org/wiki/Mathematical_descriptions_of_the_electromagnetic_field#Gauge_freedom , equal to.

Is it that ##(\nabla^2 A)_i=\frac{\partial^2 A_i}{\partial x_i^2}## or ##(\nabla^2 A)_j=\sum_{i=1}^3(\frac{\partial^2 A_i}{\partial x_i \cdot \partial x_j})## or ##\nabla^2 A=\sum_{i=1}^3(\frac{\partial^2 A_i}{\partial x_i^2})##?

Where ##A_i## is the ##i## component of ##\vec A##.

Your first option is just one term in the sum, your second is grad(div(A)) and not the Laplacian of A, your third is a scalar.

- #7

olgerm

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https://en.wikipedia.org/wiki/Mathematical_descriptions_of_the_electromagnetic_field#Gauge_freedom ##(\nabla^2 A)_j=\sum_{i=1}^3(\frac{\partial^2 A_j}{\partial x_i^2})##?

- #8

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Yes. Although this is only valid in Cartesian coordinates.

https://en.wikipedia.org/wiki/Mathematical_descriptions_of_the_electromagnetic_field#Gauge_freedom ##(\nabla^2 A)_j=\sum_{i=1}^3(\frac{\partial^2 A_j}{\partial x_i^2})##?

- #9

olgerm

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Ok, thanks.Yes. Although this is only valid in Cartesian coordinates.

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