Re: mass increase due to speed

On Sun, 22 Oct 2006 09:46:03 -0700, N:dlzc D:aol T:com (dlzc) wrote:

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.

Momentum = p^i = m * gamma * v^i (for the spatial components)

Suppose a body is moving along the x axis at 3-velocity V, so

v^1 = V

v^j = 0 (when j <> 1)

Then, among other things,

d(V*V)/dv^j = 0 for j<>1

and so

d(gamma)/dv^j = 0 for j<>1.

Now suppose we apply a force along the Y axis (where Y is coordinate
#2). Then

dp^2/dt = -F = m * gamma * dv^2/dt

Hmmm -- Looks like a factor of gamma got in there somehow.

Conclusion: Accelerating a moving body transverse to its line of
motion _DOES_ get "harder" as the body gains energy. The 3-force
needed to achieve a particular transverse acceleration goes up as
gamma. The reason is that gamma appears in the formula for momentum,
which must be conserved. If you didn't need to "push" any harder to
turn an object which was moving really fast, then momentum would not
be conserved.

There is a reason some people use "mass" to mean "relativistic mass"
-- 'cause the mass-energy behaves an awful lot like 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.

So if a speeding particle doesn't gravitate more due to its speed when
it's free, does it gravitate more due to its speed if it's inside a
box? If so, why doesn't it when it's "free" -- how does it know it's
inside a box?

Do two particles in a box gravitate more when they're moving faster
than when they're moving slower? If so, why doesn't _one_ particle?

If two don't show any effect, how about 200? 2000? 6*10^23?

At some point you've got a box of gas, and hot gas gravitates more
than cool gas, else something's rotten in stress tensor town. But hot
gas is just gas where the particles are moving faster. If one
particle moving through space doesn't gravitate more when it moves
faster, but a box of gas _does_ gravitate more when its "particles"
are moving faster, then just where is the line? How many particles
must be in the system before you'll see such an effect?

And finally, here's the kicker:

I have a u-235 nucleus sitting in space. I measure its gravitational
mass by testing its "far field". Suddenly its fissions into two
pieces (and a couple neutrons IIRC). The total _rest_ _mass_ of all the
pieces is smaller than the _rest_ _mass_ of the original atom. But the
pieces are moving hell-bent-for-election, while the original nucleus was
stationary (and so the difference is made up). Does the far field of the
ensemble _DECREASE_ at the moment of fission due to the loss of rest mass?
Or does it remain fixed, because the _energy_ of the pieces still totaled
the same? I'd claim it's the latter -- but a lot of that energy now takes
the form of linear motion of free particles.



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.

Sounds right to me.



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.

The planet example rapidly turns into a mess, I agree.


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.

Moi aussi. Went for about 6 months without touching a physics book
and I'm so rusty it's disgusting.


--
Nospam becomes physicsinsights to fix the email


.

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