 #1
 432
 35
Homework Statement:

I will not use summation notation; repeated pair of (upper and lower) indices are summed over (Einstein's notation).
Given the following action (note there's no potential term):
$$S = \int dt \Big( G_{ab} \dot q^a\dot q^b \Big) \ \ \ \ (1)$$
Let us define a vector ##v^a## (scarily called killing vector) which makes the following equation true:
$$\Big( \partial_a G_{bc}v^a + G_{ba}\partial_c v^a + G_{ca}\partial_b v^a \Big) = 0 \ \ \ \ (2)$$
Prove that ##Q_v = v^a \dot q^b G_{a b}## is a constant of motion.
HINT: You can solve the problem using Noether's Theorem
Relevant Equations:

$$S = \int dt \Big( \sum_{ab} G_{ab} \dot q^a\dot q^b \Big) \ \ \ \ (1)$$
$$\sum_a \Big( \partial_a G_{bc}v^a + G_{ba}\partial_c v^a + G_{ca}\partial_b v^a \Big) = 0 \ \ \ \ (2)$$
Noether's theorem tells us that an invariance of the Lagrangian yields a constant of motion. In this problem, that constant is:
$$Q_v = p^a \Big( \frac{\partial q_a^{\lambda}}{\partial \lambda}\Big)_{\lambda = 0} + p^b \Big( \frac{\partial q_b^{\lambda}}{\partial \lambda}\Big)_{\lambda = 0}= v^a \dot q^b G_{a b} \ \ \ \ (3)$$
So the approach is simple: calculate the 4 terms you see in the middle part of equation ##(3)## and you should get ##v^a \dot q^b G_{a b}##
OK so I will first calculate ##p^a## and ##p^b##
For ##p^k## we have
$$p^k = \frac{\partial L}{\partial \dot q_k} = G_{ab} \delta_k^a \dot q^b + G_{ab} \delta_k^b \dot q^a = G_{kb} \dot q^b + G_{ak} \dot q^a = 2 G_{ak} \dot q^a$$
Thus:
$$p^a = \frac{\partial L}{\partial \dot q_a} = 2 G_{aa} \dot q^a$$
$$p^b = \frac{\partial L}{\partial \dot q_b} = 2 G_{ab} \dot q^a$$
Then:
$$Q_v = 2 G_{aa} \dot q^a \Big( \frac{\partial q_a^{\lambda}}{\partial \lambda}\Big)_{\lambda = 0} + 2 G_{ab} \dot q^a \Big( \frac{\partial q_b^{\lambda}}{\partial \lambda}\Big)_{\lambda = 0} \ \ \ \ (4)$$
But how to proceed with ##\frac{\partial q_a^{\lambda}}{\partial \lambda}## and ##\frac{\partial q_b^{\lambda}}{\partial \lambda}## terms?
I think that the way I used Noether's theorem is OK though (AKA I think that equation (4) is OK).
$$Q_v = p^a \Big( \frac{\partial q_a^{\lambda}}{\partial \lambda}\Big)_{\lambda = 0} + p^b \Big( \frac{\partial q_b^{\lambda}}{\partial \lambda}\Big)_{\lambda = 0}= v^a \dot q^b G_{a b} \ \ \ \ (3)$$
So the approach is simple: calculate the 4 terms you see in the middle part of equation ##(3)## and you should get ##v^a \dot q^b G_{a b}##
OK so I will first calculate ##p^a## and ##p^b##
For ##p^k## we have
$$p^k = \frac{\partial L}{\partial \dot q_k} = G_{ab} \delta_k^a \dot q^b + G_{ab} \delta_k^b \dot q^a = G_{kb} \dot q^b + G_{ak} \dot q^a = 2 G_{ak} \dot q^a$$
Thus:
$$p^a = \frac{\partial L}{\partial \dot q_a} = 2 G_{aa} \dot q^a$$
$$p^b = \frac{\partial L}{\partial \dot q_b} = 2 G_{ab} \dot q^a$$
Then:
$$Q_v = 2 G_{aa} \dot q^a \Big( \frac{\partial q_a^{\lambda}}{\partial \lambda}\Big)_{\lambda = 0} + 2 G_{ab} \dot q^a \Big( \frac{\partial q_b^{\lambda}}{\partial \lambda}\Big)_{\lambda = 0} \ \ \ \ (4)$$
But how to proceed with ##\frac{\partial q_a^{\lambda}}{\partial \lambda}## and ##\frac{\partial q_b^{\lambda}}{\partial \lambda}## terms?
I think that the way I used Noether's theorem is OK though (AKA I think that equation (4) is OK).