The discussion revolves around finding a function Λ using the equation involving the gradient and a vector A defined in terms of variables x, y, z, and t. The context appears to involve concepts from vector calculus and potentially gauge transformations.
Some participants have provided guidance on interpreting the problem and have suggested considering the equations derived from the gradient. Others have noted the need for a more complete statement of the problem, indicating that the original post may not fully capture the complexity of the issue.
There is mention of gauge transformations, suggesting that the problem may be part of a larger theoretical framework. Participants have expressed uncertainty about the notation and the completeness of the information provided in the original post.
Flucky said: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 said: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?
stevendaryl said: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?
stevendaryl said: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:
- \frac{\partial}{\partial x} \Lambda = -B (x + y)
- \frac{\partial}{\partial y} \Lambda = -B (x - y)
- \frac{\partial}{\partial z} \Lambda = 0
stevendaryl said: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:
- \frac{\partial}{\partial x} \Lambda = -B (x + y)
- \frac{\partial}{\partial y} \Lambda = -B (x - y)
- \frac{\partial}{\partial z} \Lambda = 0
You should always include the full problem as well as your attempts to solve it.Flucky said: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.