Re: Mass was Re: Questions (Space)

In article <146rcu32wmyji$.5txij1ijamjf.dlg@xxxxxxxxxx>,
Ric Locke <warlocke@xxxxxxxxxxxx> wrote:

On Tue, 11 Sep 2007 11:07:23 -0700, David Friedman wrote:

In article <1eyeyjjo8wi2d.ogtpt7eg4s2d$.dlg@xxxxxxxxxx>,
Ric Locke <warlocke@xxxxxxxxxxxx> wrote:

On Tue, 11 Sep 2007 09:40:39 -0700, David Harmon wrote:

On Tue, 11 Sep 2007 00:43:04 -0500 in rec.arts.sf.composition, Ric Locke
<warlocke@xxxxxxxxxxxx> wrote,
Quibble: it's much easier to redefine "acceleration" as "add kinetic
energy", or (for short) "add energy".

Is an object in orbit undergoing constant acceleration?

Yes. So are you. It's called "gravity".

Nothing can change the magnitude /or/ direction of its velocity vector
without acceleration; that's the difference between Special and General
Relativity.

Is it constantly having kinetic energy added to it?

Yep. Of course it's always giving up kinetic energy at the same time, so
the total KE (which is a scalar, with no direction to it) remains
constant.

No. It is not having kinetic energy added to it, because acceleration at
right angles to its velocity does not increase its speed and so does not
increase its kinetic energy.

Are you next going to tell us that an object which is decelerating is
having kinetic energy added to it--but giving up more kinetic energy?

You have just re-created one of the "thought experiments" that leads to
the Theor(y|ies) of Relativity.

The root of Special Relativity is that there is /no/ preferred
"reference frame", that the only thing that can be measured is qualities
(velocity, etc.) relative to something else. When the "something else"
is a force from outside, that gets you into General Relativity.

You are in a spaceship [fiction, first person present tense]. It is not
being accelerated. An outside observer may see that the ship has a
velocity, but from your point of view it isn't moving at all. There is,
from your point of view inside the ship, no difference between moving at
cee minus epsilon and being stationary.

Along comes a force which changes the velocity of the ship, that is,
accelerates it. From your point of view, inside the ship, the direction
of that acceleration vector is immaterial as regards energy required and
imparted. An outside observer sees your velocity vector, relative to
him, change from cee minus epsilon to cee minus (2*epsilon) and calls
the result "deceleration" (and watch out for reading that statement as
privileging that observer's reference frame). You, inside, see only an
acceleration sufficient to change your velocity by epsilon -- and after
that, you're stationary again. Your buddy, in the ship next to you and
moving at the same speed, originally saw /both/ of you as stationary,
and now sees you moving at a velocity of epsilon -- you accelerated,
according to him.

And according to both of us, my velocity changed--which is what
acceleration means. According to one of us energy increased, according
to another observer it decreased. So acceleration equals the first
derivative of velocity wrt time is a reference frame independent
description, acceleration equals an increase in kinetic energy is a
reference frame dependent description. So it's the latter that doesn't
work from a relativistic standpoint, not the former.

....

In the case of an orbit, at time T the orbiting body has a velocity
vector relative to the planet of X=V, Y=0, Z=0. At time T+epsilon the
velocity in the Z direction (which was toward/away from the center of
the planet at time T, and is now parallel to that line, i.e. it no
longer passes through the center) is no longer zero. At the same
T+epsilon the velocity in the X direction is equal to V minus some small
amount, such that SQRT((new X velocity)^2 plus (new Z velocity)^2) = V,
the original velocity. The object was "decelerated" in X and
"accelerated" in Y. The total "speed", and therefore the kinetic energy,
is constant, but the vector direction has changed. (Setting that up
formally and integrating it gives the equation for an orbit.)

Yes. I know all that. So, I think, does the poster you were initially
disagreeing with.

You then rotate the axes, such that X is in the direction of V and Z is
again toward the center of the planet, and once again have velocities of
X=V, Y=0, Z=0. What you're allowing yourself to do is rotate the axes
without cost, and you can't do that. If you could, you could set an
object up so that it was moving in a circle without a planet in the
center to make it an "orbit", and it would continue to move in a circle;
and that Does Not Occur.

I have no idea what your "without cost" means, or what it has to do with
the argument. In the reference frame of the planet energy has not
changed, velocity has, and acceleration has occurred.

--
http://www.daviddfriedman.com/ http://daviddfriedman.blogspot.com/
Author of _Harald_, a fantasy without magic.
Published by Baen, in bookstores now
.

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