Metric for a surface of a cone

[gboff21]

Homework Statement

The metric for this surface is $ds^2 = dr^2 + r^2\omega^2d\phi^2$, where $\omega = sin(\theta_0)$.
Solve the Euler-Lagrange equation for phi to show that $\dot{\phi} = \frac{k}{\omega^2r^2}$. Then sub back into the metric to get $\dot{r}$

Homework Equations

$L = 1/2 g_{ab} \dot{x}^a \dot{x}^b$

The Attempt at a Solution

I've solved it to get $\ddot{\phi} + 2\frac{\dot{r}}{r}\dot{\phi} = 0$
and
$\ddot{r} - r\omega^2\dot{\phi}^2 = 0$

So how on Earth do you get that answer?

 

[TSny]

I've solved it to get $\ddot{\phi} + 2\frac{\dot{r}}{r}\dot{\phi} = 0$

Note how you got this equation. Back up a step where you must have had $\frac{d}{dt} (\rm { expression}) = 0$

What can you conclude about the expression?

 

[gboff21]

Ok I get it! $d/dt (\dot{\phi} r^2 \omega^2)=0$. So $\dot{\phi} = k/(r^2\omega^2)$
Thanks!

 

[TSny]

That's it. Good!

 

[paranormal]

I would first clarify what the problem is asking for. Is it asking for the metric of the surface of a cone or is it asking to solve the Euler-Lagrange equation for phi? These are two separate tasks and it is important to have a clear understanding of what is being asked in order to provide an accurate response.

Assuming the problem is asking for the metric of the surface of a cone, the given metric ds^2 = dr^2 + r^2\omega^2d\phi^2 is correct. This metric represents the infinitesimal distance on the surface of a cone in terms of the coordinates r and phi. The parameter \omega is a constant that depends on the angle \theta_0, which is the angle between the cone's axis and its generating line.

If the problem is asking to solve the Euler-Lagrange equation for phi, then the given solution is also correct. The Euler-Lagrange equation is used to find the equations of motion for a system described by a Lagrangian, which is given by L = 1/2 g_{ab} \dot{x}^a \dot{x}^b. In this case, the Lagrangian is the metric ds^2 and the equations of motion are \ddot{\phi} + 2\frac{\dot{r}}{r}\dot{\phi} = 0 and \ddot{r} - r\omega^2\dot{\phi}^2 = 0. Solving these equations gives the solution \dot{\phi} = \frac{k}{\omega^2r^2}, where k is a constant, and \dot{r} can be obtained by substituting this solution back into the second equation.

In conclusion, the given metric for the surface of a cone is correct and the solution for the Euler-Lagrange equation for phi is also correct. However, it is important to have a clear understanding of what is being asked in order to provide a relevant response.