Re: mass increase due to speed

Dear sal:

"sal" <pragmatist@xxxxxxxxxx> wrote in message
news:pan.2006.10.22.03.43.57.899600@xxxxxxxxxxxxx
Note one thing, before we go on:

The ratio of gravitational to inertial mass is fixed, and
the same for every material which has ever been tested,
to the limits of our ability to test it. To all intents and
purposes they are apparently the same thing, and if
"inertial mass" changes then so does gravitational mass.

.... at least until the equivalence principle is broken. Still,
there are so many examples of "inertial" being equal to
"gravitational"...

So if a moving body becomes more difficult to
accelerate (gains "inertia"), it must also gravitate
more. Right?

But it *doesn't* get more difficult to accelerate transverse to
the path. That resistance to acceleration stays exactly equal to
the rest mass.

Now, let's get on with the comments... (you
don't get any brownie points for observing that this
post is a mess, I already know that...)

I'm not keeping score.

On Fri, 20 Oct 2006 22:52:41 -0700, N:dlzc D:aol T:com (dlzc)
wrote:

Dear sal:

"sal" <pragmatist@xxxxxxxxxx> wrote in message
news:pan.2006.10.21.03.53.49.698638@xxxxxxxxxxxxx
On Fri, 20 Oct 2006 18:42:04 -0700, N:dlzc D:aol T:com (dlzc)
wrote:
...
Say you were in your ship and measured the mass
of a basketball. You were then accelerated (say to
99% speed of light). If you now measured the mass
of the basketball would you find it 'heavier' or would
it appear the same?

All mass measurements in your frame are with respect
to other masses in your frame. The basketball in your
ship will weigh the same as a basketball, no matter
how fast you are moving.

Now, you observe a basketball in the other ship, on
on the Earth. How do you propose to measure its
mass from your frame?

Let's make this easier. Let's say you are observing
the Moon orbiting the Earth as you fly by at 0.99c.
Will the Earth-Moon system curve your path more or
less than it would at 0.0001c? Will the Moon
suddenly fall into the Earth, will its period suddenly
alter beyond what your gamma accounts for?

You call this making it "easier"??? These are
questions that regularly confuse people who actually
know something of relativity, and you're throwing
them at a newbie here...

The second one is actually a slam dunk, Sal.

Sure, sure, you can't turn the Earth into a black hole
by changing your frame of reference ... but what about
the **first** one, about the effect of the Earth-Moon
system on someone flying by?

If you don't turn a massive object into a black hole by changing
frames of reference, then how can you alter the mass of a more
complex system by merely making a frame of reference change?
Sometimes you have to read the whole "test" through before you
can answer the first question.

Read my flounderings below to see why I claim the
question of how "massive" a moving object "appears to
be" may not be totally obvious.

To rephrase the question, if a planet flies by us going
"real fast", does it appear to have a stronger gravitational
field than it would if it went by going very slowly? After
considering the case where we shrink the planet and put
it in a box, so it can bounce back and forth, and then add
some more tiny planets to the box, until we have a box of
gas, I think the answer must be "yes".

Any normal process observed in another frame will be affected by
the gamma of the frame. Periodic motion or "kinetic motion of
gasses" included.

If we heat the gas (add energy to it) it must gravitate more,
and the case of a the free planet is just the simplest
possible reduction of that case.

This can be tested in various ways and none of them
turned out to be as trivial as I expected. (OTOH I didn't
try looking it up on Google, which might be the most
sensible approach, come to think of it.) The most
obvious approach is probably to start with the
Schwarzchild metric, transform it to a moving frame, and
solve the equation of motion of a particle in that frame.
But that's more than I'm going to tackle tonight. Using
the Newtonian metric instead of the Schwarzchild metric
would most likely simplify that approach a bit but it still
seemed like more work than I was willing to put into this.

Since "robert" is satisfied... it will be for satisfaction /
posterity.

A simpler approach, which certainly only works for
weak fields (where we can pretend we've got a
Newtonian field overlaid on special relativity), is to
observe that an object traveling past a (non-rotating)
planet, at the moment when it is traveling perpendicular
to its radius vector, ought to "fall" toward the planet with
the same (instantaneous) acceleration as an object
which is stationary at the same point in space, as
measured by an observer who is stationary on the
planet. If this weren't true, then hot things would fall at
a different rate from cold things,

Disproven by an eotvos experiment, I believe, for "terrestrial"
temperature differences.

and they apparently don't.

An obvious objection to the "hot-things-would-fall-
differently" argument is that the particles in "hot" things
(of the sort we can actually experiment on!) are moving
at non-relativistic velocities, so perhaps the claim is
only true in the low-speed limit. That can be tested by
solving the geodesic equation for a particle in a
Newtonian gravitational field
(metric diag[-(1+2phi),1-2phi,1-2phi,1-2phi]) and
seeing if, when it is moving "perpendicular" to the field,
its acceleration depends on its velocity. That turned
out to be messier than I expected; still working on it
but it looks like the answer is _NO_ it doesn't
necessarily accelerate at a rate independent of
velocity, even when it's moving "horizontally", which is
too bad but which may explain the bogus result I
exhibit a few paragraphs farther along.

Let's ignore that and barge ahead, assuming for the
moment that the acceleration toward the planet of a
passing body is a constant, "a", in the frame of an
observer stationary with respect to the planet. Call
the moving body "B".

Now, the question is: How fast does someone in
B's instantaneous rest frame see the _planet_
accelerating?

Different rates at different places in its orbit. It might be
better / simpler to concentrate on *period*, because the orbiter
ends up at the same place wrt the orbitee periodically. Maybe
not arriving at an "instantaneous" "continuous" resolution is not
necessary.

B's proper clock ticks at 1/gamma relative to the
planet's clock. B's line of motion is perpendicular to the
line along which he/she is accelerating, which means
there's no Fitzgerald contraction to worry about: distances
along the line of acceleration are the same in both frames.
So, we only need to worry about the clock rate difference.

After t seconds, in the planet's frame, the distance B
moves toward the planet along the line of acceleration
would be:

(1/2)at^2

In B's instantaneous rest frame, the distance the planet
moves is the same, since there's no length contraction
along the line perpendicular to the line of flight. But the
time is different, by a factor of 1/gamma. So we have

(1/2)at^2 = (1/2)a(gamma*tau)^2

or, casting it in terms of the acceleration observed in
the moving frame, which we will call a',

(1/2)at^2 = (1/2)a'tau^2

or

(1/2)at^2 = (1/2)a'(t/gamma)^2

Solving for a',

a' = a * gamma^2

Which, unfortunately, looks totally bogus -- a sensible
answer should look more like

a' = a * gamma

but the assumption that B's acceleration doesn't depend
on B's velocity as viewed from the planet's frame seems
to lead inevitably to this result, as verified by computing
it a couple different ways, so it appears that getting a
correct answer here really does require solving the
geodesic equation for that case.

I think the orbitting body has:
* a *slightly* different gamma with some variation based on
position in the orbit (very minor),
* the orbit will be seen to be an ellipse, if classically
circular,
* the orbit will be even more elliptical, if already elliptical,
and
* the body being orbitted will move from a focus of the ellipse
towards the midpoint between the focii, as a function of gamma.

Or alternatively I messed up the transformation of time
going between frames, because the planet's moving in
B's frame just as B is moving in the planet's frame so
the clock issues are subtler than they appear at first.

Sigh... Wish I had more time to spend on this but I
don't, which is why I'm posting this with all its obvious
warts still intact. References to correct explanations of
this (trivial?) issue would be appreciated.

Google
relativity flyby "orbital period" site:.edu
.... only 78 hits.

Someone that can follow your derivation and assumptions might be
able to help you. I'm just a duffer.

David A. Smith


.

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