Discussion in 'Alien Hub' started by waitedavid137, Jul 31, 2020 at 8:00 PM.

1. ### waitedavid137Honorable

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152
So lets say an observer outside of any warp uses coordinates [cT, X, Y, Z], define
$\small r_{s}=\sqrt{X^{2}+Y^{2}+\left ( Z-\int \beta dcT \right )^{2}}$
where $\small \beta$ will be a function of T for this will be a standard for the distance from somewhere to the center of a warp ship.
The spacetime given by
$\large ds^{2}=\left [ A\left ( cT,r_{s} \right ) \right ]^{2}dcT^{2}-\left \{ dX^{2}+dY^{2}+\left [ dZ-\beta \left ( cT \right )f\left ( r_{s} \right )dcT \right ]^{2} \right \}$
will be an Alcubierre warp drive spacetime given boundary conditions
$\large \begin{matrix} f \rightarrow 1 & as & r_{s}\rightarrow 0 \end{matrix}$
$\large \begin{matrix} f \rightarrow 0 & as & r_{s}\rightarrow \infty \end{matrix}$
$\large \begin{matrix} A\rightarrow 1 & as & r_{s}\rightarrow \infty \end{matrix}$
A is called the lapse function because its value at the ship's location determines time dilation for the ship during the trip. The ship will have a speed of
$\large v_{s}=\beta c$
For the case that
A=1 everywhere, time does not lapse for the ship with respect to the frame for which it was initially at rest when
$\large \beta =0$
So what I'm interested in is the energy density according to coordinates that are actually appropriate for the ship. After all it is the ship that we want to generate the warp. So what we start out doing is Let A=1 everywhere and then do the transformation
$\large \begin{matrix} ct=cT\\ x=X \\ y=Y \\ z=Z-\int \beta dcT \end{matrix}$
The expression for the spacetime becomes
$\large ds^{2}=dct^{2}-\left \{ dx^{2}+dy^{2}+\left [ dz+\beta \left ( 1-f \right )dct \right ]^{2} \right \}$
And so for simplicity of calculation we define a new function g
g=1-f
So it becomes
$\large ds^{2}=dct^{2}-\left \{ dx^{2}+dy^{2}+\left [ dz+\beta \left ( ct \right ) g \left ( x,y,z \right)dct \right ]^{2} \right \}$
defining
$\large r=\sqrt{x^{2}+y^{2}+z^{2}}$
Yields the boundary conditions to be
$\large \begin{matrix} g\rightarrow 0 & as & r\rightarrow 0\\ g\rightarrow 1 & as & r\rightarrow \infty \end{matrix}$
From the perspective of the ship observer, he floats at zero velocity with no pushes or pulls in a locally inertial frame while it is the stars that are at warp in the opposite direction.
Now here's the first trick. I am going to reintroduce what was called the lapse function term into this standard of coordinates with the modifying exception that I am going to change its boundary conditions.
The modified warp drive spacetime will be
$\large ds^{2}=\left [ A\left ( ct,x,y,x \right ) \right ]^{2}dct^{2}-\left \{ dx^{2}+dy^{2}+\left [ dz+\beta \left ( ct \right )g\left ( x,y,z \right )dct \right ]^{2} \right \}$
with
$\large \begin{matrix} A\rightarrow 1 & as & r\rightarrow 0,\infty \end{matrix}$
You calculate or compute from the metric $\large G^{00}$, and use $\large T^{00}=\frac{c^{4}}{8\pi G}G^{00}$to arrive at the ship frame energy density term to be
$\large T^{00}=-\frac{c^{4}}{32\pi G}\frac{\beta ^{2}}{A^{4}}\left [ \left ( \frac{\partial g}{\partial x} \right )^{2} +\left ( \frac{\partial g}{\partial y} \right )^{2} \right ]$
So from the perspective of the ship, which is what we want to produce the stress-energy tensor for the warp, the energy density can be arbitrarily reduced by allowing A to be large in the region of the warp matter where the negative energy density would have otherwise been large. Yet with its boundary conditions there need be no lapse in time between the ship observer, and the inertial frame observers he was initially at rest with respect to before going into warp.
More later.

Last edited: Jul 31, 2020 at 9:07 PM
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2. ### waitedavid137Honorable

Messages:
152
The warp itself tends to deflect geodesics it goes against away from it, so an observer off to the side would see light coming from anything in the direction it is going as coming from it.
I think star trek took that idea from Chris Van Den Broeck that showed that doing such a thing could reduce the negative energy density's magnitude.
I usually describe it from the perspective of the craft when in warp with respect to some remote observer who is at rest with respect to the ship when beta=0.
This version doesn't radiate, but when we get a fully practical version, I'm sure it would radiate both electromagnetic and gravitational radiation.
My version does, yes.

Last edited: Jul 31, 2020 at 8:49 PM
3. ### Dejan CorovicNoble

Messages:
941
Five more to settle UFO lore, and I am done:

1) Does the infamous lenticular shape, aka flying saucer shape, have any privileged status in your equations? Or even somewhat asymmetrical lenticular shape, with the top dome being more convex than the bottom one.

2) Under your equations will stationary hovering spacecraft, near Earth's surface, need to spontaneously rotate around its own major axis of symmetry at about 0.3Hz?

3) Would the spacecraft's hull be able to have any openings, aka windows? This one might be going too far into the engineering side.

4) I guess this is given: these spacecraft would be able to instantaneously accelerate to great speeds, from standing still.

5) What would be the stronger propulsive direction? A) along the major axis of symmetry? or B) perpendicular to the axis of symmetry?