- #1

- 95

- 1

## 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?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- #1

- 95

- 1

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

A [/B]+

where

Is this at all possible?

- #2

- 8,938

- 2,926

## 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

- 8,938

- 2,926

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

- 8,938

- 2,926

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

- 20,004

- 10,663

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.

Share:

- Replies
- 4

- Views
- 793

- Replies
- 4

- Views
- 286

- Replies
- 8

- Views
- 1K

- Replies
- 13

- Views
- 665

- Replies
- 13

- Views
- 1K

- Replies
- 1

- Views
- 324

- Replies
- 2

- Views
- 573

- Replies
- 4

- Views
- 664

- Replies
- 1

- Views
- 56

- Replies
- 7

- Views
- 1K