How to find the metric given the interval

[vertices]

If we have a Lagrangian which looks like this:

$$L=\frac{m}{2}g_{ij}(x)\dot{x}^i\dot{x}^j$$

where:

$$ds^2=g_{ij}(x)dx^idx^j$$

If we are told that:

$$ds^2=d\phi^2 +(sin^2 \phi) d\theta^2$$

How can we show that the Lagrangian is:

$$L=\frac{m}{2}[\dot{\phi}^2 +(sin^2 \phi) \dot{\theta}^2]$$

Is there a general way of determing the metric from the interval?

Thanks.

 

[nicksauce]

It is easy. The term in the interval proportional to $d\phi^2$ comes from the term $g_{11}$. Therefore you can read off g_{11} = 1. Similarly you can read off $g_{22} = sin^2(\phi)$. There are no terms in the interval proportional to $d\phi d\theta$, so you can conclude the off-diagonal terms are 0 in the metric.

If none of this is obvious, write out the summation $g_{ij}dx^{i}dx^{j}$ and convince yourself of it.

 

[vertices]

It is easy. The term in the interval proportional to $d\phi^2$ comes from the term $g_{11}$. Therefore you can read off g_{11} = 1. Similarly you can read off $g_{22} = sin^2(\phi)$. There are no terms in the interval proportional to $d\phi d\theta$, so you can conclude the off-diagonal terms are 0 in the metric.

If none of this is obvious, write out the summation $g_{ij}dx^{i}dx^{j}$ and convince yourself of it.

thanks nicksauce:)

 

[Prologue]

In what subject area is this topic addressed? It sounds very interesting.

 

[vertices]

In what subject area is this topic addressed? It sounds very interesting.

Classical Mechanics...

 

[Prologue]

Classical Mechanics...

Bummer, in my junior mechanics class, we didn't do this.

(i'm not talking about lagrangian stuff but, rather, the metric stuff.)