Re: C-ship

[T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden)]

>Yes, and that's why I've spent all my time on that so far, with a
>relatively brief digression into astrophysics to develop a simple
>but adequate model of starlight and a 3-d star database from
>catalogs.

Do you have a "DOS" computer? If so, you should try the program Circumspace.
That program allows you to fly around between the local stars, you can use
warp-speed but it doesn't show relativistic effects. It is very difficult to
find your way back to Sol after a trip of 100 ly. (If you like I could put
it on the web for you.)

>Check out the exercises in Chapter L (that's what they call it)
>of _Spacetime Physics_.  Aberration and the "headlight effect"
>both follow directly from the geometry of the Lorentz
>transformation applied to light rays.

I see, it has to do with velocity addition, I thought only of
lenght-contraction. But what did you mean by Lorenz transforming the world
lines? Does that also mean that velocity addition is done at the same time?

>Remember that a displacement along a worldline of a light ray is
>different than a plain spacial displacement, so applying a
>Lorentz transform to it produces a different result.  Try out the
>numbers and I think you'll be surprised.  I just ran an example
>for myself with a Lorentz transform matrix for a boost of 0.9 c
>in the x direction, and light rays coming from a perpendicular
>direction (say from along the y or z axes).  They really do end
>up looking like they're in front of you when transformed into the
>"moving" frame.

Yes, I had already seen that in the c-ship chapter about abberation. The
name "headlight" is quite applicable because most of the light is
concentrated in the front and thus that part will be much brighter then the
back.

>Correct.  If your optics all take into account the finite speed
>of light, though, you will not necessarily see the amount of
>Lorentz contraction you'd think.  Farther portions of the object
>are seen farther back in time when they were farther away.

That's right, but did you know that a moving sphere always has a circular
outline to all observers even after LT?

>Which is why you should be suspicious.  In the case of a
>constant-velocity observer, it is not possible to decide whether
>the observer or its surroundings are moving.  (An accelerated
>observer experiences something that the rest of the universe does
>not; an accelerated observer knows that he is accelerating.)

I realized the next morning that it indeed could not be true, but still
wasn't sure where I went wrong. Now I think I know why: If the observer is
"at rest" and the objects move, then the curvature is caused by the finite
travel of light which causes to show objects at places where they WERE.
But when the observer moves and the objects are "at rest", then the
curvature is caused by abberation.
Do you understand this subtle difference?

So again nature has effectively removed all possibilities to find a absolute
rest frame.

>If you are at a point S = [ t x y z ], and you want to view an
>object whose worldline is described by P(t') = P0 + V * t' =
>[ t0 x0 y0 z0 ] + [ v0 v1 v2 v3 ] * t', then the path of a light ray
>between S and P satisfies the equation (S - P(t'))^2 = 0, The
>solution is:
>
>t' = ((S - P0)|V - sqrt(((S - P0)|V)^2 - (S - P0)^2 * V^2)) / V^2

Shouldn't it be:

 t' = ((S - P0)|V - sqrt(((S - P0)|V)^2 - 4(S - P0)^2 * V^2)) / 2 V^2

>This is pretty much simple application of the quadratic formula,
>choosing the smallest solution to get the t' that corresponds to
>light leaving the object at an earlier time than the observer's.
>
>Things are substantially more complicated when dealing with
>accelerated worldlines.  I've got a preliminary solution stated
>in similar terms as the above discussion; perhaps you'd like to
>check my math :-).

I tried checking it, but I don't get the same answer. At the non-substituted
result I already have a different answer. How sure are you about the
calculation? Did you do it by hand or did you use a mathematical program?

To be short I also have some negative exponent terms: ...exp(-a * t')^2 and
....exp(-a * t')

>I have yet to program this expression into my calculator and play
>with some solutions to see if they have the predicted results.
>In particular, there should be regions of spacetime where you
>cannot see the accelerated object (see Chapter 6 of
>_Gravitation_).
>
>You will also note that this is a somewhat less general statement
>of the problem; it uses a frame in which the accelerated object
>was at rest at the origin of the frame.  For a general solution
>you would need to transform the view point into the appropriate
>frame, apply this solution, and transform it back out afterwards.

A while ago I studied the accelerate formulas, it took quite a time before I
figured out that the acceleration was measured relative to the "momentary"
restframe. But I still don't know how the acceleration formulas are derived.


Timothy