# Lagrangian depend upon upon my choice of generalized coordinates?

1. Aug 30, 2007

### pardesi

does the lagrangian depend upon upon my choice of generalized coordinates

2. Sep 4, 2007

### fantispug

The Lagrangian does depend on your generalised coordinates.

Consider a particle mass m in one dimension, in a potential V=V(x).
Then
$$L=T-V= \frac{1}{2}m\dot{x}^{2} - V(x)$$

If instead of using x we used some shifted coordinate y=x+c, where c is some constant, then V=V(y-c)

$$L=T-V=\frac{1}{2}m\dot{y}^{2} - V(y-c)$$

Now notice that in general V(x) and V(y-c) are different (for example if V(x)=x).

However the equations of motion you derive from them will be equivalent. (Often choosing a clever set of coordinates makes the equations of motion easier to solve).

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