How to find the metric given the interval

In summary, the conversation discusses a Lagrangian and the corresponding metric, and how to determine the metric from the interval. The Lagrangian is shown to be equal to a specific form, and the topic is addressed in the subject area of classical mechanics.
  • #1
vertices
62
0
If we have a Lagrangian which looks like this:

[tex]L=\frac{m}{2}g_{ij}(x)\dot{x}^i\dot{x}^j[/tex]

where:

[tex]ds^2=g_{ij}(x)dx^idx^j[/tex]

If we are told that:

[tex]ds^2=d\phi^2 +(sin^2 \phi) d\theta^2[/tex]

How can we show that the Lagrangian is:

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

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

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

If none of this is obvious, write out the summation [itex]g_{ij}dx^{i}dx^{j}[/itex] and convince yourself of it.
 
  • #3
nicksauce said:
It is easy. The term in the interval proportional to [itex]d\phi^2[/itex] comes from the term [itex]g_{11}[/itex]. Therefore you can read off g_{11} = 1. Similarly you can read off [itex]g_{22} = sin^2(\phi)[/itex]. There are no terms in the interval proportional to [itex]d\phi d\theta[/itex], so you can conclude the off-diagonal terms are 0 in the metric.

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

thanks nicksauce:)
 
  • #4
In what subject area is this topic addressed? It sounds very interesting.
 
  • #5
Prologue said:
In what subject area is this topic addressed? It sounds very interesting.

Classical Mechanics...
 
  • #6
vertices said:
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.)
 

1. How do you find the metric given the interval?

To find the metric given the interval, you need to know the formula for the specific metric you are looking for. For example, if you are looking for the metric of distance, the formula is distance = rate x time. Once you have the correct formula, you can plug in the values from the interval to find the metric.

2. What is the difference between interval and metric?

An interval is a range or distance between two points, while a metric is a measurement of a specific quantity within that interval. An interval can be thought of as a container, while a metric is the content within that container.

3. Can you give an example of finding a metric from an interval?

Sure, let's say we have an interval of 0-100 degrees Celsius and we want to find the metric of temperature in Fahrenheit. The formula for converting from Celsius to Fahrenheit is F = (C x 1.8) + 32, where F is the temperature in Fahrenheit and C is the temperature in Celsius. So, if we plug in the values from our interval (0-100), we get the metric of temperature in Fahrenheit to be 32-212 degrees.

4. How important is it to have the correct interval when finding a metric?

Having the correct interval is crucial when finding a metric because it determines the range of values that can be used in the calculation. If the interval is incorrect or too narrow, the calculated metric may not be accurate or may not make sense in the given context.

5. Are there any common mistakes to avoid when finding a metric from an interval?

One common mistake to avoid is using the wrong formula for the given metric. It's important to double-check the formula and make sure it corresponds to the metric you are trying to find. Another mistake is not paying attention to units and making sure they are consistent throughout the calculation.

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