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Transformation of the Line-Element

by Anamitra
Tags: lineelement, transformation
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Oct12-11, 01:38 AM
P: 621
An Analytical Endeavor:
[tex]{Cot}{(}{\theta}{)}{=}{[}\frac{{\partial}^{2}{\theta}}{{\partial x}^{2}}{+}\frac{{\partial}^{2}{\theta}}{{\partial y}^{2}}{]}[/tex] ----- (1)

We look for a solution of the form:
[tex]\theta{=}{A}{(}{-}{ln}{Sin}{x}{-}\frac{1}{2}{x}^{2}{)}{+}{B}{(}{-}{ln}{Sin}{y}{-}\frac{1}{2}{y}^{2}{)}{+}{\lambda}{x}^{2}[/tex]------------- (2)
A and B are constants Lambda is a parameter independent of x and y.

Using the above trial in relation (1) we have

[tex]{A}{Cot}^{2}{x}{+}{B}{cot}^{2}{y}{+}{2}{\lambda} {=}{Cot}\theta[/tex]--- (3)

We may eliminate lambda between (2) and (3) to find Particular Integrals[surfaces theta=f(x,y)]
[Even if one has to use numerical methods, the equation will be an ordinary one instead of a“differential Equation”]
Instead of (2) one may use a trial of the form below, to obtain a greater variety of particular integrals:
[tex]\theta{=}{A}{f1}{(}{x}{)}{+}{B}{f2}{(}{y}{)}{+}{\lambda}{x}^{2}[/tex]------------- (4)
f1 and f1 are known ,arbitrary functions--well behaved ones of course,in terms of continuity ,differentiability etc.

The homogeneous part of equation (1) is simply Laplace’s equation in two dimensions. We have the familiar solutions.

[This posting will be revised]
Oct12-11, 07:38 PM
P: 17,329
So plug your transformation equation back into the metric and calculate the curvature.

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