Hamiltonian Qn Homework: Find T+U is Not Equal

[derrickb]

Homework Statement

The simple form H=T+U is true only if your generalized coordinates are "natural"(relation between generalized and underlying Cartesian coordinates is independent of time). If the generalized coordinates are not natural, you must use the definition H=Ʃpq'-L. To illustrate this point, consider the following: Two children are playing catch inside a railroad car that is moving with varying speed V along a straight horizontal track. For generalized coordinates you can use the position(x,y,z) of the ball relative to a point fixed in the car, but in setting up the Hamiltionian you must use coordinates in an inertial frame-a frame fixed to the ground. Find the Hamiltonian for the ball and show that it is not equal to T+U (neither as measured in the car, nor as measured in the ground-based frame).

Homework Equations

L=T-U
H=Ʃpq'-L
P_x=$\frac{dL}{dx'}$
P_y=$\frac{dL}{dy'}$
P_z=$\frac{dL}{dz'}$

' is time derivative in this case

The Attempt at a Solution

T=.5m(x'²+y'²+z'²)
U=mgz; z is the vertical axis
L=.5m(x'²+y'²+z'²)-mgz

P_x=mx'
P_y=my'
P_z=mz'
P_v=mv

H=Ʃpq'-L
H=P_xx'+P_yy'+P_zz'+P_vx'-.5m(x'²+y'²+z'²)+mgz
H=P_x(P_x/m)+P_y(P_y/m)+P_z(P_z/m)+mv(P_x/m)-.5m(x'²+y'²+z'²)+mgz

If I continue this way, I end up with
H=(P_x²+P_y²+P_z²)/(2m)+P_xv+mgz
but the correct answer is
H=(P_x²+P_y²+P_z²)/(2m)-P_xv+mgz

Does anyone see where I went wrong, or is my answer correct, just with different sign convention?

 

[vela]

Shouldn't your expression for T depend on V?