Re: [Phys-L] another DIY relativity experiment

[David Bowman <David_Bowman@georgetowncollege.edu>]

Regarding Bob S's question:

> ...
> But, right now, please clarify this ==>
>
> In the equation
> (c^2)*(dτ)^2 = ((c + g*z/c)^2)*(dt)^2 -((dx)^2 + (dy)^2 + (dz)^2),
> does the "z" in the first term of the right hand side refer to
> the z coordinate of the K origin as measured in K' ?
> Thanks

No.  It refers to the (nearly) common z-coordinate of any pair of infinitesimally separated space-time events as seen in frame K.  The infinitesimal difference in the coordinate values of the pair of events is (dx, dy, dz, c*dt) in the accelerated frame K.  It is the z-dependence of the time-like contribution to the space-time interval between the events that makes for the "gravitational" time dilation seen at different height z-values in frame K.  It is also responsible for the inertial forces seen acting on any freely falling body in that frame whose behavior is summarized below.

The Lagrangian for a freely falling body of mass m in frame K is given by:

L = -m*(c^2)*sqrt((1 + g*z/c^2)^2 -(1/c^2)*((dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2)
  = -m*(c^2)/γ

where γ ≡ 1/sqrt((1 + g*z/c^2)^2 -(1/c^2)*((v_x)^2 + (v_y)^2 + (v_z)^2).

The canonical momentum is:
p_x = m*γ*v_x
p_y = m*γ*v_y
p_z = m*γ*v_z .

The canonical (inertial) force is:
f_x = 0
f_y = 0
f_z = -m*g*γ*(1 + g*z/c^2) .

The conserved Hamiltonian is given by:

H = (1 + g*z/c^2)*sqrt((m*c^2)^2 + (c^2)*((p_x)^2 + (p_y)^2 + (p_z)^2))
  = m*(c^2)*γ*(1 + g*z/c^2)^2

where the γ factor can be also written in terms of the momentum (and coordinate z) as

γ = sqrt(1 + ((p_x)^2 + (p_y)^2 + (p_z)^2)/(m*c)^2)/(1 + g*z/c^2) .

Everything written here is all in non-inertial frame K.

David Bowman