Re: mass increase due to speed

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.

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

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...)


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? 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". 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.

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, 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? 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.

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.


Think about the situation...

The Oh My God particle was travelling at such a speed that it would cross
the Milky Way in 1 week proper time (0.999999+c). Did it cause the Earth
and Moon to become black holes, or deform their orbits in some
catastrophic way? Is it plausible that there was only one of such
particles *ever*, or even that this was the fastest particle to have
interacted with the Earth?

Yes you can duck into the math, or you can just look at what you are
expecting of Nature.

As to the first question, it is only slightly more complex...

Realtivistic mass =/= gravitational mass. And gravitational mass is about
the only kind of mass you are going to measure at high speeds.

David A. Smith

--
Nospam becomes physicsinsights to fix the email

.

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