- #1
Xyius
- 508
- 4
Hello, I am stuck on the following problem.
1. Homework Statement
Consider the continuous family of coordinate and time transformations (for small ##\epsilon##).
[tex]Q^{\alpha}=q^{\alpha}+\epsilon f^{\alpha}(q,t)[/tex]
[tex]T= t+\epsilon \tau (q,t)[/tex]
Show that if this transformation preserves the action
[tex]S=\int_{t_1}^{t_2}L(q,\dot{q},t)dt=\int_{T_1}^{T_2}L(Q,\dot{Q},T)dT[/tex],
then
[tex]\frac{\partial L}{\partial \dot{q}^{\alpha}}(\dot{q}^{\alpha}\tau-f^{\alpha})-L\tau[/tex]
is a constant of motion.
I am guessing as to which equations I may need[/B]
Extended Noether's Theorem:
[tex]\Gamma=\sum_{\alpha}\frac{\partial L}{\partial \dot{q}_{\alpha}}\left( \frac{\partial \psi_{\alpha}}{\partial \epsilon} \right)_{\epsilon = 0}-\Psi_{\epsilon = 0} = \text{Constant}[/tex]
Where ##\Psi## is a function in which it's time derivative is the difference between the Lagrangian before and after the transformation (Note that it is possible that the extended Noether's theorem doesn't apply and the normal Noether's theorem is sufficient, I do not know).
The Energy Function:
[tex]E=\sum_{\beta}\dot{x}_{\beta}\frac{\partial L}{\partial \dot{x}_{\beta}}-L[/tex]
So the first thing I think I should do is apply the transformation to the Action. First I need to find ##dT##.
[tex]T= t+\epsilon \tau (q,t) \rightarrow dT=(1+\epsilon \dot{\tau})dt \rightarrow dt=\frac{1}{1-(-\epsilon \dot{\tau})}dT \approx (1-\epsilon \dot{\tau})dT[/tex]
Thus the action is,
[tex]S'=\int_{t_1+\epsilon \tau(q,t_1)}^{t_2+\epsilon \tau(q,t_2)}L(Q,\dot{Q},T)dT-\epsilon \int_{t_1+\epsilon \tau(q,t_1)}^{t_2+\epsilon \tau(q,t_2)}\tau L(Q,\dot{Q},T)dT.[/tex]
[tex]S'=\int_{T_1}^{T_2}L(Q,\dot{Q},T)dT-\epsilon \tau L[/tex]
[tex]S'=S-\epsilon \tau L[/tex]
Thus when applying this transformation, the two actions differ by ##\epsilon \tau L##. This seems very similar to Noether's theorem except for that theorem we are talking about Lagrangians and not Actions. However, since the action is just the integral of the Lagrangian, can I assume the same principles apply? In Noether's theorem, we had
[tex]\frac{\partial L'}{\partial \epsilon}=\frac{d \Psi}{dt}[/tex]
In this case it appears that we have the following,
[tex]\frac{\partial S'}{\partial \epsilon}=\frac{d \Psi}{dT}=-L\tau[/tex]
Can I therefore make the conjecture that,
[tex]\Gamma=\sum_{\alpha}\frac{\partial S}{\partial \dot{q}_{\alpha}}\left( \frac{\partial \psi_{\alpha}}{\partial \epsilon} \right)_{\epsilon = 0}-\Psi_{\epsilon = 0} = \text{Constant}[/tex]
However evaluating this I do not get the required answer. :(
1. Homework Statement
Consider the continuous family of coordinate and time transformations (for small ##\epsilon##).
[tex]Q^{\alpha}=q^{\alpha}+\epsilon f^{\alpha}(q,t)[/tex]
[tex]T= t+\epsilon \tau (q,t)[/tex]
Show that if this transformation preserves the action
[tex]S=\int_{t_1}^{t_2}L(q,\dot{q},t)dt=\int_{T_1}^{T_2}L(Q,\dot{Q},T)dT[/tex],
then
[tex]\frac{\partial L}{\partial \dot{q}^{\alpha}}(\dot{q}^{\alpha}\tau-f^{\alpha})-L\tau[/tex]
is a constant of motion.
Homework Equations
I am guessing as to which equations I may need[/B]
Extended Noether's Theorem:
[tex]\Gamma=\sum_{\alpha}\frac{\partial L}{\partial \dot{q}_{\alpha}}\left( \frac{\partial \psi_{\alpha}}{\partial \epsilon} \right)_{\epsilon = 0}-\Psi_{\epsilon = 0} = \text{Constant}[/tex]
Where ##\Psi## is a function in which it's time derivative is the difference between the Lagrangian before and after the transformation (Note that it is possible that the extended Noether's theorem doesn't apply and the normal Noether's theorem is sufficient, I do not know).
The Energy Function:
[tex]E=\sum_{\beta}\dot{x}_{\beta}\frac{\partial L}{\partial \dot{x}_{\beta}}-L[/tex]
The Attempt at a Solution
So the first thing I think I should do is apply the transformation to the Action. First I need to find ##dT##.
[tex]T= t+\epsilon \tau (q,t) \rightarrow dT=(1+\epsilon \dot{\tau})dt \rightarrow dt=\frac{1}{1-(-\epsilon \dot{\tau})}dT \approx (1-\epsilon \dot{\tau})dT[/tex]
Thus the action is,
[tex]S'=\int_{t_1+\epsilon \tau(q,t_1)}^{t_2+\epsilon \tau(q,t_2)}L(Q,\dot{Q},T)dT-\epsilon \int_{t_1+\epsilon \tau(q,t_1)}^{t_2+\epsilon \tau(q,t_2)}\tau L(Q,\dot{Q},T)dT.[/tex]
[tex]S'=\int_{T_1}^{T_2}L(Q,\dot{Q},T)dT-\epsilon \tau L[/tex]
[tex]S'=S-\epsilon \tau L[/tex]
Thus when applying this transformation, the two actions differ by ##\epsilon \tau L##. This seems very similar to Noether's theorem except for that theorem we are talking about Lagrangians and not Actions. However, since the action is just the integral of the Lagrangian, can I assume the same principles apply? In Noether's theorem, we had
[tex]\frac{\partial L'}{\partial \epsilon}=\frac{d \Psi}{dt}[/tex]
In this case it appears that we have the following,
[tex]\frac{\partial S'}{\partial \epsilon}=\frac{d \Psi}{dT}=-L\tau[/tex]
Can I therefore make the conjecture that,
[tex]\Gamma=\sum_{\alpha}\frac{\partial S}{\partial \dot{q}_{\alpha}}\left( \frac{\partial \psi_{\alpha}}{\partial \epsilon} \right)_{\epsilon = 0}-\Psi_{\epsilon = 0} = \text{Constant}[/tex]
However evaluating this I do not get the required answer. :(