I Invariance of Action: Lagrangian Transformation

[kent davidge]

The Lagragian $\mathcal L_e = e(\lambda)^{-1} \mathcal L - \frac{1}{2}m^2 e(\lambda)$, with $\mathcal L$ not depending on $\lambda$, transform as $\delta L_e = \frac{d}{d\lambda} (\epsilon (\lambda) \mathcal L_e)$* under the reparametrization $\lambda \rightarrow \lambda + \epsilon(\lambda)$, according to a paper that I'm reading.

Above I reproduced the expressions exactly how they are given in the paper. However, I think there's a mistake in *. Shouldn't the right-hand-side be $\delta (\epsilon (\lambda) \mathcal L_e)$ instead?

Also, I was able to derive * myself only for the second term in $\mathcal L_e$. Does anyone know how to derive for the first term?

 

[samalkhaiat]

The Lagragian $\mathcal L_e = e(\lambda)^{-1} \mathcal L - \frac{1}{2}m^2 e(\lambda)$, with $\mathcal L$ not depending on $\lambda$, transform as $\delta L_e = \frac{d}{d\lambda} (\epsilon (\lambda) \mathcal L_e)$* under the reparametrization $\lambda \rightarrow \lambda + \epsilon(\lambda)$, according to a paper that I'm reading.

I will assuming that you know that under \tau \to \tau + \epsilon (\tau), we have \delta x (\tau) = - \epsilon (\tau) \ \dot{x} (\tau) , \delta e (\tau ) = - \frac{d}{d \tau} ( \epsilon (\tau ) \ e (\tau) ) and \delta \dot{x} (\tau) = - \frac{d}{d \tau} ( \epsilon (\tau) \ \dot{x}(\tau) ).
Now,
\delta \mathcal{L}_{e} = e^{-1} \ \delta \mathcal{L}(x , \dot{x}) - e^{-2} \ \mathcal{L} (x , \dot{x}) \ \delta e - \frac{1}{2}m^{2} \ \delta e , or
\delta \mathcal{L}_{e} = e^{-1} \ \frac{\partial \mathcal{L}}{\partial x} \ \delta x + e^{-1} \ \frac{\partial \mathcal{L}}{\partial \dot{x}} \ \delta \dot{x} - e^{-2} \ \mathcal{L} \ \delta e - \frac{1}{2}m^{2} \ \delta e . Next, if you substitute the above transformations, then with a bit of algebra you obtain
\delta \mathcal{L}_{e} = - \frac{d}{d \tau} \left( \epsilon \ ( e^{-1} \ \mathcal{L} - \frac{1}{2}m^{2} \ e ) \right) + e^{-1} \ \frac{d \epsilon}{d \tau} \left( 2 \mathcal{L} - \frac{\partial \mathcal{L}}{\partial \dot{x}} \ \dot{x} \right) . That is
\delta \mathcal{L}_{e} + \frac{d}{d \tau} \left( \epsilon \ \mathcal{L}_{e}\right) = e^{-1} \ \frac{d \epsilon}{d \tau}\left( 2 \mathcal{L} - \frac{\partial \mathcal{L}}{\partial \dot{x}} \ \dot{x} \right) . The RHS vanishes if and only if \mathcal{L}(x , \dot{x}) is a homogeneous function of degree two in \dot{x}. That is
\delta \mathcal{L}_{e} + \frac{d}{d \tau} (\epsilon \ \mathcal{L}_{e}) = 0 \ \ \Leftrightarrow \ \ \frac{\partial \mathcal{L}}{\partial \dot{x}} \ \dot{x} - 2 \mathcal{L} = 0. This is the case when \mathcal{L} = \frac{1}{2} \dot{x}^{2}.

 

[kent davidge]

Thanks @samalkhaiat, I understood your solution, but I have two questions:

1 - is $\delta e^{-1} = (\partial e^{-1}/ \partial e) (\partial e / \partial \tau) \delta \tau$?

2 - are the form of the variations you give in the beggining of your post obtained by requiring invariance of the action?

 

[samalkhaiat]

Thanks @samalkhaiat,
1 - is $\delta e^{-1} = (\partial e^{-1}/ \partial e) (\partial e / \partial \tau) \delta \tau$?

No, the variation symbol \delta is a Lie derivative, i.e., it is a derivation but not a differential. So \delta e^{-1} = \left( \frac{d}{d e} e^{-1} \right) \delta e , but (as we will see below), \delta e \neq \frac{d e}{d \tau} \delta \tau = \frac{d e}{d \tau} \epsilon (\tau) .

2 - are the form of the variations you give in the beggining of your post obtained by requiring invariance of the action?

Yes, you can show that the invariance of the action S[x , e] under the diffeomorphism \tau \to \bar{\tau} = \bar{\tau} (\tau) implies (and is implied by) the following:

1) The coordinates behave like scalar fields, i.e., x (\tau) \to \bar{x} (\bar{\tau}) = x (\tau). So, for the infinitesimal diffeomorphism \bar{\tau} = \tau + \epsilon (\tau), we have, to first order in \epsilon, \bar{x}(\tau) = x ( \tau - \epsilon ) = x (\tau ) - \epsilon (\tau) \ \frac{d}{d \tau}x(\tau) = x (\tau) - \epsilon (\tau) \dot{x} (\tau) . The variation (or the Lie derivative) \delta x (\tau) is defined by \delta x (\tau) \equiv \bar{x}(\tau) - x (\tau). This clearly leads to \delta x (\tau) = - \epsilon (\tau) \dot{x}(\tau).

2) The e (\tau) field behaves as scalar density[*]: e (\tau) \to \bar{e}(\bar{\tau}) = \left( \frac{d \bar{\tau}}{d \tau} \right)^{-1} e ( \tau ) . Again, infinitesimally, we have \bar{e} (\tau + \epsilon ) = \left( 1 + \frac{d \epsilon}{d \tau} \right)^{-1} e ( \tau ) . Expanding both sides to first order in \epsilon, we get \bar{e} (\tau) + \epsilon \ \frac{d}{d \tau} e ( \tau ) = \left( 1 - \frac{d \epsilon}{d \tau} \right) e ( \tau ) . Note that in the second term on the left hand side we replaced \bar{e}( \tau ) by e ( \tau ) which is allowed by our first order approximations. Indeed \epsilon \ \frac{d}{d \tau} \bar{e}( \tau ) = \epsilon \ \frac{d}{d \tau} e ( \tau ) + \mathcal{O} (\epsilon^{2}). So, as usual, the Lie derivative of e is given by \delta e (\tau) \equiv \bar{e}(\tau) - e (\tau) = - \epsilon (\tau) \ \frac{d}{d \tau} \bar{e} (\tau) - e (\tau) \ \frac{d}{d \tau} \epsilon (\tau) = - \frac{d}{d \tau} \left( \epsilon (\tau) e (\tau)\right) .

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[*] Imagine a one-dimensional manifold with symmetric metric “tensor” g_{\tau \tau}(\tau). The square of distances on such manifold is given by ds^{2} = g_{\tau \tau} (\tau) \ d \tau d \tau , \ \ \Rightarrow \ \ ds = \sqrt{g_{\tau \tau}(\tau)} \ d \tau \equiv e (\tau) \ d \tau , where we have defined the (frame) field e (\tau) \equiv \sqrt{g_{\tau \tau}(\tau)} Now, the invariance of the distance ds = d \bar{s} under arbitrary general coordinate transformations \tau \to \bar{\tau} = \bar{\tau}(\tau), implies that the frame field e (\tau) transforms as scalar density: \bar{e}( \bar{\tau} ) \ d \bar{\tau} = e ( \tau ) \ d \tau \ \ \Rightarrow \ \ \bar{e}(\bar{\tau}) = \left( \frac{d \bar{\tau}}{d \tau} \right)^{-1} e ( \tau ) .