On 25 Dec 2017 at 15:33, Robert O'Connor wrote:

> Leonard Erickson wrote:
>  > Slight problem. the ship's kinetic energy depends on which frame of
>  > reference you use... It's energy with respect to the star and the >
>  various other bodies in a system will be different for each.
>
> Yes. Is it numerically significant?

Damn straight it is!

> In special relativity, the relationship:
>
> (kinetic energy)^2 = (momentum x c)^2 + (rest mass x c^2)^2
>
> is true in all frames of reference.
>
> [Where the ship's kinetic energy is (gamma-1) x m x c^2, and gamma is
> the Lorentz factor for the velocity.
>
> Similarly, the momentum is (gamma) x m x v].
>
> At worst, you will get variances comparable to the Shapiro effect
> (delay in radar travel time between planets caused by solar gravity)
> of a few percent between observers.

I think you blew your calculations *badly* somewhere.

Consider a planet orbiting a star at velocity X. (where X is very
much less than the speed of light)

Now add a ship moving at X relative to the planet, and in the
direction the planet is orbiting. This means it has a negligble
radial velocity with respect to the star the planet is orbiting.

Relative to the planet it has a Ke of 0.5*m*X^2

Relative to the star it has a Ke of 0.5*m*(2X)^2

That's a factor of *4* difference in KE. And a factor of 2 difference
in momentum. Yet the rest mass difference is negligible.

If you want to make it even simpler, have two ships moving at
velocity X with respect to the planet, but in opposite directions.

A  planet  B

From the planet both ships have a KE of 0.5*m*X^2, and a momentum of
m*X (in opposite directions)

From Ship A the planet has a KE of 0.5*m*X^2 and a momentum of m*X,
but ship B has a KE of 0.5*m*2X^2 and a momentum of m*2X

Again, a factor of 4 difference in KE and a factor of 2 difference in
momentum.

--
Leonard Erickson (aka shadow)
shadow at shadowgard dot com