No idea how to word this. Finding the gradient with vector?

[Flucky]

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$\begin{pmatrix} x+y\\ x-y\\ 0 \end{pmatrix}$

The Attempt at a Solution

Is this at all possible?

 

[stevendaryl]

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$\begin{pmatrix} x+y\\ x-y\\ 0 \end{pmatrix}$

The Attempt at a Solution

Is this at all possible?

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

 

[Flucky]

I'm not sure about your notation, is $A$ a matrix, or a vector? Do the three components $x+y, x-y, 0$ 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.

 

[stevendaryl]

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

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

$\nabla \Lambda = -A = -B(x+y, x-y, 0)$

That means that there are three equations for $\Lambda$:

1. $\frac{\partial}{\partial x} \Lambda = -B (x + y)$
2. $\frac{\partial}{\partial y} \Lambda = -B (x - y)$
3. $\frac{\partial}{\partial z} \Lambda = 0$

 

[stevendaryl]

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

$\nabla \Lambda = -A = -B(x+y, x-y, 0)$

That means that there are three equations for $\Lambda$:

1. $\frac{\partial}{\partial x} \Lambda = -B (x + y)$
2. $\frac{\partial}{\partial y} \Lambda = -B (x - y)$
3. $\frac{\partial}{\partial z} \Lambda = 0$

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

$\frac{d}{d x} \Lambda = -B x + c$

where $c$ is a constant? What's the most general solution?

 

[Flucky]

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

$\nabla \Lambda = -A = -B(x+y, x-y, 0)$

That means that there are three equations for $\Lambda$:

1. $\frac{\partial}{\partial x} \Lambda = -B (x + y)$
2. $\frac{\partial}{\partial y} \Lambda = -B (x - y)$
3. $\frac{\partial}{\partial z} \Lambda = 0$

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.

 

[Orodruin]

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.

You should always include the full problem as well as your attempts to solve it.