Variational Mechanics: Deriving Momentum & Hamiltonians

[stunner5000pt]

Consider systems described by Lagrangians
$$L_{1} = \frac{1}{2} m\dot{x}^2$$
$$L_{2} = \frac{1}{2} m\dot{x}^2 - a \dot{x} x$$
when $$\dot{x} = \frac{dx}{dt}$$ and a is a constnat

a) Derive the momenta conjugate to x for each system and write down the coressponding hamiltonians. Are the hamiltonians constants of motion and why?
Momenta conjugate to x (not familiar with the term conjugate)... does this mean
$$p_{x} = \frac{\partial L}{\partial x}$$
but isn't the momentum
$$p_{\dot{x}} = \frac{\partial L}{\partial \dot{x}}$$

b) Obtain the Lagrange equations of motion for each system, work out their solutions and explain what htey represent.
simply using the Euler Lagrange equation
for L1
let x = x(t)
$$\frac{\partial L}{\partial x} - \frac{d}{dt} \frac{\partial L}{\partial \dot{x}} = 0$$
but dL/dx = 0 so the second term is zero
$$\frac{d}{dt} \frac{\partial L}{\partial \dot{x}} = 0$$
dL/dx dot is mx dot so
$$\frac{d}{dt} m \dot{x} = 0$$
intengrate both sides wrt t
$$m x(t) = C_{1}t + C_{2}$$
so far so good?

for L2
$$\frac{\partial L}{\partial x} = -a\dot{x}$$
$$\frac{\partial L}{\partial \dot{x}} = m \dot{x} - ax$$
$$\frac{d}{dt} \frac{\partial L}{\partial \dot{x}} = m \ddot{x} - a \dot{x}$$
Into Euler Lagrange
$$-a\dot{x} - (m\ddot{x} - a\dot{x}) = 0$$
$$m\ddot{x} = 0$$
smae solution as L1 above
$$m x(t) = C_{1}t + C_{2}$$

i suspect i made some mistake in L2. In any case they are the same trajectories. They represent straight lines?

c) Show taht
$$L_{2} = L_{1} + \frac{d}{dt} F(x,\dot{x})$$
and find the function F. What does this imply above the actions
$$A_{i} = \int_{t_{1}}^{t_{2}} L_{l} dt$$
and how does this relate to your results of b)?
well that's easy
$$\frac{d}{dt} F(x,\dot{x}) = -\int ax \dot{x} dt = -\frac{a x^2}{2}$$

the action for L2 is
$$A_{i} = \int_{t_{1}}^{t_{2}} L_{2} dt = \int_{t_{1}}^{t_{2}} \frac{1}{2} m \dot{x}^2 + \int_{t_{1}}^{t_{2}} \frac{-ax^2}{2} dt$$
not quite sure how to proceed from here... WOuld i use the solutions from b to solve the action for each lagrangian?

 

[StatusX]

but isn't the momentum
$$p_{\dot{x}} = \frac{\partial L}{\partial \dot{x}}$$

Yes, that is what they mean by the momentum conjugate to x. And you can drop the dot above the x subscript.

b) Obtain the Lagrange equations of motion for each system, work out their solutions and explain what htey represent.
simply using the Euler Lagrange equation
for L1
let x = x(t)
$$\frac{\partial L}{\partial x} - \frac{d}{dt} \frac{\partial L}{\partial \dot{x}} = 0$$
but dL/dx = 0 so the second term is zero
$$\frac{d}{dt} \frac{\partial L}{\partial \dot{x}} = 0$$
dL/dx dot is mx dot so
$$\frac{d}{dt} m \dot{x} = 0$$
intengrate both sides wrt t
$$m x(t) = C_{1}t + C_{2}$$
so far so good?

for L2
$$\frac{\partial L}{\partial x} = -a\dot{x}$$
$$\frac{\partial L}{\partial \dot{x}} = m \dot{x} - ax$$
$$\frac{d}{dt} \frac{\partial L}{\partial \dot{x}} = m \ddot{x} - a \dot{x}$$
Into Euler Lagrange
$$-a\dot{x} - (m\ddot{x} - a\dot{x}) = 0$$
$$m\ddot{x} = 0$$
smae solution as L1 above
$$m x(t) = C_{1}t + C_{2}$$

i suspect i made some mistake in L2. In any case they are the same trajectories. They represent straight lines?

I don't see any mistakes. They represent staright lines in spacetime. What kind of motion is this?

c) Show taht
$$L_{2} = L_{1} + \frac{d}{dt} F(x,\dot{x})$$
and find the function F. What does this imply above the actions
$$A_{i} = \int_{t_{1}}^{t_{2}} L_{l} dt$$
and how does this relate to your results of b)?
well that's easy
$$\frac{d}{dt} F(x,\dot{x}) = -\int ax \dot{x} dt = -\frac{a x^2}{2}$$

the action for L2 is
$$A_{i} = \int_{t_{1}}^{t_{2}} L_{2} dt = \int_{t_{1}}^{t_{2}} \frac{1}{2} m \dot{x}^2 + \int_{t_{1}}^{t_{2}} \frac{-ax^2}{2} dt$$
not quite sure how to proceed from here... WOuld i use the solutions from b to solve the action for each lagrangian?

First off, d/dt(F) is equal to the difference in the Lagrangians, not the actions. The idea is that, since the Lagrangians differ by a perfect differential, what can you say about how the actions over some fixed path differ? How are the equations of motion affected by this (ie, what happens to F when you vary S)?

 

[stunner5000pt]

we ll i9 can say that from the momentum (with respect to x dot) the momenta only differ by ax.
Would it be path independent. So if S was varied we could get the same answer?
im not usre ho to interpret the actions, however

 

[StatusX]

Forget about the actual forms of these Lagrangians for the moment. If you can show:

$$L_{2} = L_{1} + \frac{d}{dt} F(x,\dot{x})$$

Then, just using this result, what does this give you for S₁ in terms of S₂, both evaluated along the same path? What can you then say about the extrema of these actions?

 

[stunner5000pt]

just one thing ... is S the path of a path integral that is
$$\nt_{S} L dt$$??

if this is the case
$$\int_{t_{1}}^{t_{2}} L_{1} dt$$
and $$\int_{t_{1}}^{t_{2}} \left(L_{1} + \frac{d}{dt} F(x,\dot{x}) dt = \int_{t_{1}}^{t_{2}} L_{1} dt + F(x,\dot{x})\right]_{t_{1}}^{t_{2}}$$

they only differ the F term...
not sure how to put this in terms of paths

 

[StatusX]

S is the action. So you see that the actions differ by the function f evaluated at the endpoints. But when you vary S to get the equations of motion, you keep the endpoints fixed, so...

 

[stunner5000pt]

so it makes no difference? that is the action remains the same regardless of the path? is This a consequence of the solutions found in part b??

 

[StatusX]

Well I would say the result of b is a consequence of this fact. And, the actions are different for a given path, but the extra term makes no difference as far as the equations of motion are concerned. Do you see the distinction, and why this is the case?

 

[stunner5000pt]

the actions are different for a given path? Why si that so?

dont hte action just differ by some cosntant? SO F is not affected to change in the path..
the concept isn't registering in my head!

 

[StatusX]

Yes, the actions differ by a constant, namely F evaluated at the endpoints. This constant drops out when you vary the path (infinitessimally, with endpoints fixed) to find the extrema, and so doesn't enter the equations of motion, which are just the conditions a path must satisfy to extremize the action.