3
$\begingroup$

$\Phi, \Lambda$ are both scalars dependent upon, and $\mathbf u$ is a vector independent of coordinates. I'm trying to express $\Lambda$ in terms from $\mathbf U \cdot \nabla\Lambda = \Phi$ and to start with, since I'm just familiar with the basics, it looks pretty hopless. So:

  1. Get help from people more expert in this area
  2. Trawl through the list of standard vector identitites involving $\nabla$
  3. Search for an online mathematics program to solve it.
  4. Make it simpler to start with by solving in one dimension, and progress from there

what strategy would/did you use in tackling this problem? Solving it would be a bonus ;)

Edit: I've added the context of the problem since some people think this will help:

I'm trying to work out the conserved canonical momentum for a static electric field, using Noether's theorem on the relativistic electromagnetic Lagrangian $L = \frac{m_0c^2}{\gamma} + \frac{e}{c}\mathbf{u \cdot A} - e\Phi(x,y,z)$ $L$ needs to be independent of coordinates which can be done by transforming \mathbf A\rightarrow \mathbf A'= \mathbf A + \nabla\Lambda The conserved canonical momentum $ P = \frac\partial{\partial \mathbf u} ( \frac{-m_0c^2}{\gamma} + \frac{e}{c}\mathbf{u \cdot (A+\nabla\Lambda}) - e\Phi)$ With no magnetic field $A=0$ $P = -m_0\mathbf u\gamma +\frac e c \frac\partial {\partial\mathbf u}\mathbf u \cdot\nabla\Lambda$ becomes $P = -m_0\mathbf u\gamma +\frac e c (\nabla\Lambda + \mathbf u \cdot\frac\partial {\partial\mathbf u}\mathbf\nabla\Lambda)$ To get any further, I need to know the form $\Lambda$ must take, which comes from making $L$ independent of the coordinates before, and so $\nabla( \frac{e}{c}\mathbf u \nabla\Lambda - e\Phi) = 0$

  • 0
    @Willie yes $\mathbf u$ is the velocity of a relativistic charged particle in a static electric field with no magnetic field. From Noether's theorem, an $L$ independent of time gives rise to a conserved quantity called the energy. Likewise when independent of coordinates, it gives rise to another conserved quantity called the canonical momentum. I'm using the arbitariness of the magnetic vector potential $A$ to the addition of $\nabla\Lambda$, to compensate for $\Phi$'s dependence on position to then make $L$ independent of position.2011-07-04

2 Answers 2

5

The equation $U\cdot\nabla \Lambda=\Phi$ says precisely that the derivative of scalar function $\Lambda$ in the direction of the vector field $U$ is $\Phi$. To solve it then, locally, near points where $U$ is not zero, you find the integral curves of the field $U$ and integrate $\Phi$ along them. (Solving globally, or near a zero of $U$ requires rather more elaborate analyses in this generality...)

If, for example, $U$ is the constant field $(1,0,0)$, then the equation is $\frac{\partial}{\partial x}\Lambda(x,y,z)=\Phi(x,y,z),$ and it obviously can be solved by integrating. The general case reduces to this one, because near a non-zero point of $U$ coordinates can be changed to that the equation becomes the above one.

If you provide concrete details about the equations you are trying to solve, one can give more concrete details about the solution...

  • 0
    Nice clear answer, thanks. Directional derivative is the key.2011-07-04
0

Usually when faced with something like that I look to Green's identities:

http://en.wikipedia.org/wiki/Green%27s_identities

and start looking for a Gaussian surface:

http://en.wikipedia.org/wiki/Gaussian_surface

to integrate over as is commonly done in electromagnetism. The proper choice of the surface is made to ensure the integration is easy to carry out.