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

  1. Feb 1, 2015 #1
    1. The problem statement, all variables and given/known data
    I need to find Λ using the equation below (I think).

    2. Relevant equations
    A
    + Λ = 0

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

    3. The attempt at a solution
    Is this at all possible?
     
  2. jcsd
  3. Feb 1, 2015 #2

    stevendaryl

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    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?
     
  4. Feb 1, 2015 #3
    Sorry it is a vector. I was trying to mimic exactly how it is written in front of me.
     
  5. Feb 1, 2015 #4

    stevendaryl

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    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]:
    1. [itex] \frac{\partial}{\partial x} \Lambda = -B (x + y)[/itex]
    2. [itex]\frac{\partial}{\partial y} \Lambda = -B (x - y)[/itex]
    3. [itex]\frac{\partial}{\partial z} \Lambda = 0[/itex]
     
  6. Feb 1, 2015 #5

    stevendaryl

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    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?
     
  7. Feb 1, 2015 #6
    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.
     
  8. Feb 1, 2015 #7

    Orodruin

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