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

  • Thread starter Flucky
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  • #1
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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?
 

Answers and Replies

  • #2
stevendaryl
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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?
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
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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?
Sorry it is a vector. I was trying to mimic exactly how it is written in front of me.
 
  • #4
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?
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]
 
  • #5
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]
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
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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]
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
Orodruin
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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.
 

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