0

I'm trying to set up a functional that outputs the length of a curve on the paraboloid surface. I then want to write the Euler-Lagrange equation for the geodesic.

The surface is given in the cylindrical coordinate system $(r, \phi, z)$ with $z = \alpha r^2$.

$$ J[y] = L = \int_1^2 dl = \int_1^2 \sqrt{r^2(d\phi)^2 + (dr)^2 + (dz)^2}$$

$z = \alpha r^2 \Rightarrow dz = 2\alpha r dr.$ Therefore,

$$ J[y] = L = \int_1^2 \sqrt{r^2(d\phi)^2 + (dr)^2 + (2\alpha r)^2(dr)^2} $$ $$ = \int_1^2 \sqrt{r^2(d\phi)^2 + (1 + (2\alpha r)^2)(dr)^2} $$ $$ = \int_0^{2\pi} d\phi \underbrace{\sqrt{r^2 + (1 + (2\alpha r)^2) (\frac{dr}{d\phi})^2}}_\text{F}$$

The Euler-Lagrange equation: $$\frac{d}{d\phi}\frac{\partial F}{\partial(\frac{dr}{d\phi})} - \frac{\partial F}{\partial r} = 0$$

$$\frac{d}{d\phi}\frac{\partial F}{\partial(\frac{dr}{d\phi})} = \frac{d}{d\phi}\left[\frac{(1 + (2\alpha r)^2)\frac{dr}{d\phi}}{\sqrt{r^2 + (1 + (2\alpha r)^2) (\frac{dr}{d\phi})^2}} \right]$$

$$ \frac{\partial F}{\partial r} = \frac{r + (4\alpha^2 r)(\frac{dr}{d\phi})^2}{\sqrt{r^2 + (1 + (2\alpha r)^2) (\frac{dr}{d\phi})^2}} $$

I have the following questions:

  1. Are the functional and the Euler-Lagrange equation correct?
  2. Which conservation law can I use to obtain a simpler form of the equation for the geodesic?
Qmechanic
  • 201,751
Ilia
  • 11
  • Two notes : (1) By default we use the symbol $:\rho:$ for the cylinder radius instead $:r;$, see Figure 02 in my answer here Unit vectors in the cylindrical coordinate system as functions of position.............. – Frobenius Jul 20 '23 at 14:34
  • .......... (2) The general equation of an elliptic paraboloid is $$ z = \alpha x^2 + \beta y^2,, \quad \alpha\cdot\beta > 0 \tag{Frob-01} $$ so that the vertical cross sections are all parabolas, while the horizontal cross sections are ellipses. You post a special case $:z=\alpha \rho^2=\alpha x^2+\alpha y^2:$ within which the horizontal cross sections are circles. – Frobenius Jul 20 '23 at 14:34
  • @llia your functional is correct. the EL equation should be $~r \left( \phi \right) \left( 4, \left( r \left( \phi \right) \right) ^{3}{\alpha}^{2}+r \left( \phi \right) \right) {\frac {d^{2} }{d{\phi}^{2}}}r \left( \phi \right) +r \left( \phi \right) \left( -4 , \left( r \left( \phi \right) \right) ^{2}{\alpha}^{2}-2 \right) \left( {\frac {d}{d\phi}}r \left( \phi \right) \right) ^{2}- \left( r \left( \phi \right) \right) ^{3}=0 ~$ – Eli Jul 20 '23 at 19:09
  • Use conservation of energy (autonomous) and angular momentum (rotation invariance about $z$). This renders your problem integrable. – LPZ Jul 21 '23 at 09:34

0 Answers0