Precursor to brachistochrone problem

[stunner5000pt]

Another long question but not that hard. Most of the writing is my work/questions
According to my prof if i cna solve this... the resulting relation can be used to solve Bernoulli's problem

For \delta \int_{x_{1}}^{x_{2}} F(x,y(x),\dot{y}(x)) dx = 0
where \dot{y} = \frac{dy}{dx}
Show that if F = F(y,\dot{y}) only (not dependant of x explicitly) then the quantity
H = \dot{y} \frac{\partial F}{\partial \dot{y}} - F
is a constant. That is \frac{dH}{dx} = 0.

is this the total derivative (like it asks) of H wrt x? Since F does depend on y and y dot... which depend on x... then would hteir derivatives wrt x be included in the rightmost expression below?
\frac{dH}{dx} = \frac{d \dot{y}}{dx} \frac{\partial F}{\partial \dot{y}} + \dot{y} \frac{\partial F}{\partial \dot{y} dx} - \frac{dF}{dx} [/tex<br /> is this true &gt;&gt; \frac{d \dot{y}}{\partial \dot{y}} =1 ??<br /> in that case all we&#039;re left with (if the dH/dx is correct) is <br /> \frac{dH}{dx} = \dot{y} \frac{\partial F}{\partial \dot{y} dx}<br /> if i were to differentiate the expression for F <br /> \delta \int_{x_{1}}^{x_{2}} F(x,y(x),\dot{y}(x)) dx = 0<br /> wrt to y dot and x i would get zero, yes? <br /> SO dH/dx = 0?<br /> <br /> <br /> <b>Determine the expression for H if F = 1/2 mv^2 - V(x) and explain the physical significance of this quantity.</b><br /> is this correct? chain rule application everywhere<br /> \frac{\partial F}{\partial \dot{y}} = m \dot{x} \frac{d \dot{x}}{d \dot{y}} - \frac{\partial V}{\partial \dot{y}} \frac{\partial x}{\partial \dot{y}}<br /> <br /> then H = \dot{y} m \dot{x} \frac{d \dot{x}}{d \dot{y}} - \frac{\partial V}{\partial \dot{y}} \frac{\partial x}{\partial \dot{y}} - \frac{1}{2} m \dot{x}^2 + V(x)<br /> <br /> the physical significance... hmm ...<br /> H is a consnat quantity wrt x. But I am not sure how to interpret this past that part. <br /> <br /> Please help! Thank you!

 

[Gokul43201]

1. You are asked for the total derivative and you can show that it will reduce to the partial derivative (the two extra terms : \dot{y}\partial H/\partial y} + \ddot{y}\partial H/\partial{\dot{y} will cancel off) which is clearly = 0.

2. You've started off on the wrong foot...

First notice that F = F(x, dx/dt), since v=dx/dt

So, H = \dot{x} \frac{\partial F}{\partial \dot{x}} - F

When you do it correctly, the physical significance will pop right out !

 

[stunner5000pt]

where did you get \dot{y}\partial H/\partial y} + \ddot{y}\partial H/\partial{\dot{y} from?

also domnt u mena F(y,dy/dt)? Or do you actually mean it for x?
The former is in the question...

 

[Gokul43201]

where did you get \dot{y}\partial H/\partial y} + \ddot{y}\partial H/\partial{\dot{y} from?

dH/dx = \partial H/\partial x + (\partial H/\partial y)dy/dx + (\partial H/\partial\dot{y}) d\dot{y}/dx

also domnt u mena F(y,dy/dt)? Or do you actually mean it for x?
The former is in the question...

No, it's the latter which is in the question. What you are given in part 1 is the general form in terms of some general parameters. In part 2, you are given :

F(x,\dot{x}) = \frac{1}{2}m \dot{x}^2 + V(x)

where \dot{x} = dx/dt

 

[dextercioby]

That H is a constant ON the equations of motion !

That is

\frac{dH}{dx}= - \frac{dy}{dx}\frac{\delta F}{\delta y}

,where \frac{\delta F}{\delta y} is the Euler-Lagrange (variational) derivative of the lagrangian.


Daniel.

 

[stunner5000pt]

also in this question... what is the partial derivative of F wrt x zero? Is it because the action is zero?



for hte second one since F = \frac{1}{2} m \dot{x}^2 + V(x)
then
\frac{\partial F}{\partial \dot{x}} = m \dot{x} + \frac{\partial V}{\partail \dot{x}} \frac{\partial x}{\partial \dot{x}} ...(1 )

so far so good?
im just wondering if the partial of V wrt to x dot is zero becuase V would not depend on x dot because of its definition?
F(x,\dot{x}) = - \nabla V(x) ?
so then when i sub 1 into the expression for H
H = \frac{1}{2} m \dot{x}^2 - V(x) this is given that the partial of V wrt x dot is zero. THis means that H is the force exerted on the particle? Doesnt this mean that H is invariant of F, for a potential that does not depend on the time deriavitve of x?

 

[Gokul43201]

im just wondering if the partial of V wrt to x dot is zero becuase V would not depend on x dot because of its definition?

Correct. But there's a sign mistake in the solution. The sign of V in F(x,v) has been changed (looks like I made that mistake in my previous post - sorry !)

PS : No time to look at the rest now.

 

[stunner5000pt]

Correction to the mistake

so the expression for H (the hamiltonian, apparently)
H = \frac{1}{2} m \dot{x}^2 + V(x) = T + V = E

Thus H represents the total energy of the system.