- #1
JD_PM
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- Homework Statement
- I will not use summation sign: repeated pair of (upper and lower) indices are summed over: [itex]\sum_{a} A_{a} B^{a} \equiv A_{a}B^{a} = A_{c}B^{c}[/itex] (summed over indices are dummy indices so you can rename them as you like).
Let ##V(q) = 0##. Then the Hamiltonian (##H = T + V##) and the Lagrangian (##L = T - V##) are equal:
$$H = L = 1/2 G_{ab} \dot q^a \dot q^b \ \ \ \ (1)$$
Show that you get ##(1)## either using Hamilton's equation ##\dot{p}_k=\frac{\partial L}{\partial q^k}## or Euler-Lagrange equation ##\frac{d}{dt} \Big(\frac{\partial L}{\partial \dot q^k }\Big) = \frac{\partial L}{\partial q^k}##
- Relevant Equations
- $$H = L = 1/2 G_{ab} \dot q^a \dot q^b$$
$$\dot{p}_k=\frac{\partial L}{\partial q^k}$$
$$\frac{d}{dt} \Big(\frac{\partial L}{\partial \dot q^k }\Big) = \frac{\partial L}{\partial q^k}$$
We know that the momentum for such a lagrangian is:
$$p_k = \frac{\partial L}{\partial \dot q^k} = 1/2 \Big( G_{ab} \delta_k^a \dot q^b + G_{ab} \delta_k^b \dot q^a\Big) = 1/2 \Big( G_{kb} \dot q^b + G_{ak} \dot q^a\Big) = G_{ak} \dot q^a$$
OK so using Hamilton's equation we get (and recalling that ##L = H## for this particular problem):
$$\dot p_a = -\frac{\partial H}{\partial q^a} = -\partial_a H = -1/2\partial_a G_{cb} \dot q^c \dot q^b \ \ \ \ (2)$$
By using ##\frac{d}{dt} \Big(\frac{\partial L}{\partial \dot q^k }\Big) = \frac{\partial L}{\partial q^k}## we get:
$$\frac{d}{dt} \Big(\frac{\partial L}{\partial \dot q^a }\Big) = \frac{d}{dt}\Big( G_{ab} \dot q^b\Big) = \frac{\partial L}{\partial q^a} = 1/2\partial_a G_{cb} \dot q^c \dot q^b \ \ \ \ (3)$$
Mmm so there's a sign difference between ##(2)## and ##(3)##. They should be equal. Why am I wrong?
$$p_k = \frac{\partial L}{\partial \dot q^k} = 1/2 \Big( G_{ab} \delta_k^a \dot q^b + G_{ab} \delta_k^b \dot q^a\Big) = 1/2 \Big( G_{kb} \dot q^b + G_{ak} \dot q^a\Big) = G_{ak} \dot q^a$$
OK so using Hamilton's equation we get (and recalling that ##L = H## for this particular problem):
$$\dot p_a = -\frac{\partial H}{\partial q^a} = -\partial_a H = -1/2\partial_a G_{cb} \dot q^c \dot q^b \ \ \ \ (2)$$
By using ##\frac{d}{dt} \Big(\frac{\partial L}{\partial \dot q^k }\Big) = \frac{\partial L}{\partial q^k}## we get:
$$\frac{d}{dt} \Big(\frac{\partial L}{\partial \dot q^a }\Big) = \frac{d}{dt}\Big( G_{ab} \dot q^b\Big) = \frac{\partial L}{\partial q^a} = 1/2\partial_a G_{cb} \dot q^c \dot q^b \ \ \ \ (3)$$
Mmm so there's a sign difference between ##(2)## and ##(3)##. They should be equal. Why am I wrong?