- #1

- 95

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## Homework Statement

I need to find Λ using the equation below (I think).

## Homework Equations

A [/B]+

**∇**Λ = 0

where

**A**(x,y,z,t) = B[itex]\begin{pmatrix} x+y\\ x-y\\ 0 \end{pmatrix}[/itex]

## The Attempt at a Solution

Is this at all possible?

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- Thread starter Flucky
- Start date

- #1

- 95

- 1

I need to find Λ using the equation below (I think).

A [/B]+

where

Is this at all possible?

- #2

- 8,590

- 2,685

## Homework Statement

I need to find Λ using the equation below (I think).

## Homework Equations

A [/B]+∇Λ = 0

whereA(x,y,z,t) = B[itex]\begin{pmatrix} x+y\\ x-y\\ 0 \end{pmatrix}[/itex]

## The Attempt at a Solution

Is this at all possible?

I'm not sure about your notation, is [itex]A[/itex] a matrix, or a vector? Do the three components [itex]x+y, x-y, 0[/itex] correspond to vector components, or matrix components?

- #3

- 95

- 1

I'm not sure about your notation, is [itex]A[/itex] a matrix, or a vector? Do the three components [itex]x+y, x-y, 0[/itex] correspond to vector components, or matrix components?

Sorry it is a vector. I was trying to mimic exactly how it is written in front of me.

- #4

- 8,590

- 2,685

I'm not sure about your notation, is [itex]A[/itex] a matrix, or a vector? Do the three components [itex]x+y, x-y, 0[/itex] correspond to vector components, or matrix components?

Assuming that you mean that [itex]A[/itex] is a vector, then your problem is to find a function [itex]\Lambda(x,y,z)[/itex] such that

[itex]\nabla \Lambda = -A = -B(x+y, x-y, 0)[/itex]

That means that there are three equations for [itex]\Lambda[/itex]:

- [itex] \frac{\partial}{\partial x} \Lambda = -B (x + y)[/itex]
- [itex]\frac{\partial}{\partial y} \Lambda = -B (x - y)[/itex]
- [itex]\frac{\partial}{\partial z} \Lambda = 0[/itex]

- #5

- 8,590

- 2,685

Assuming that you mean that [itex]A[/itex] is a vector, then your problem is to find a function [itex]\Lambda(x,y,z)[/itex] such that

[itex]\nabla \Lambda = -A = -B(x+y, x-y, 0)[/itex]

That means that there are three equations for [itex]\Lambda[/itex]:

- [itex] \frac{\partial}{\partial x} \Lambda = -B (x + y)[/itex]
- [itex]\frac{\partial}{\partial y} \Lambda = -B (x - y)[/itex]
- [itex]\frac{\partial}{\partial z} \Lambda = 0[/itex]

To solve the equation [itex] \frac{\partial}{\partial x} \Lambda = -B (x + y)[/itex], consider how you would solve the ordinary differential equation

[itex] \frac{d}{d x} \Lambda = -B x + c[/itex]

where [itex]c[/itex] is a constant? What's the most general solution?

- #6

- 95

- 1

Assuming that you mean that [itex]A[/itex] is a vector, then your problem is to find a function [itex]\Lambda(x,y,z)[/itex] such that

[itex]\nabla \Lambda = -A = -B(x+y, x-y, 0)[/itex]

That means that there are three equations for [itex]\Lambda[/itex]:

- [itex] \frac{\partial}{\partial x} \Lambda = -B (x + y)[/itex]
- [itex]\frac{\partial}{\partial y} \Lambda = -B (x - y)[/itex]
- [itex]\frac{\partial}{\partial z} \Lambda = 0[/itex]

Hmm I think I may have to post the question in full. My original post is just a small part of a bigger question (about Gauge transformations) so I am probably going about it wrong.

- #7

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You shouldHmm I think I may have to post the question in full. My original post is just a small part of a bigger question (about Gauge transformations) so I am probably going about it wrong.

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