Some Idle musings whilst flicking through the Usenet Physics FAQ

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SCW - 30 Jan 2007 05:47 GMT

Whilst idly thumbing through the Usenet Physics FAQ (http://
www.weburbia1.demon.co.uk/physics (v. Nov2006) and the and Charter
(http://www.weburbia1.demon.co.uk/physics/Administrivia/Charters/
relativity.txt) I came up with the following paradox which is closely
related to "If you go too fast do you become a black hole?" question.

Having read through the "Does mass change with velocity?" section and
seeing the explanation of relativist mass (m-r) and invariant mass
(m-0; sorry, these should contain subscripts as in the FAQ, but do not
transfer to plain text), it appears to me, please correct me if I
misinterpreted this, that any increase in m-r is counter-acted by a
slowing of time, so that F = d/dt(mv) (cf. Planck, Lewis and Tolman) -
am I correct in this thinking?

Anyway, here's the question, which as mentioned, is perilously close
to "If you go too fast do you become a black hole?":

Imagine two system in at motion wrt each other at a constant
relativistic velocity. If each system had a device similar to a
Newton's cradle but with a gap between the suspended balls, would
either system observe the balls in the opposing system change in mass
with a corresponding increase in gravitational attraction that then
manifests itself by moving the balls? Or would this effect be
cancelled out by F = d/dt(mv)? Or am I completely off-track with this
one?

I assume in this that there must be an agreement between any
corresponding solutions for Minkowski and Newtonian/Euclidean
geometries?

Comments, and no doubt abuse, welcome.

SCW

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Dumbledore_ - 30 Jan 2007 05:59 GMT

> Whilst idly thumbing through the Usenet Physics FAQ (http://
> www.weburbia1.demon.co.uk/physics (v. Nov2006) and the and Charter
[quoted text clipped - 9 lines]
> slowing of time, so that F = d/dt(mv) (cf. Planck, Lewis and Tolman) -
> am I correct in this thinking?

No.

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dlzc - 30 Jan 2007 15:14 GMT

Dear SCW:

> Whilst idly thumbing through the Usenet Physics FAQ (http://www.weburbia1.demon.co.uk/physics(v. Nov2006) and
> the and Charter
[quoted text clipped - 7 lines]
> and invariant mass (m-0; sorry, these should contain
> subscripts as in the FAQ, but do not transfer to plain text),

Probably m_r and m_0 (or just m) would be clearer.

> it appears to me, please correct me if I misinterpreted this,
> that any increase in m-r is counter-acted by a slowing of
> time, so that F = d/dt(mv) (cf. Planck, Lewis and Tolman) -
> am I correct in this thinking?

... "measured slowing of time in the other frame" yes.

> Anyway, here's the question, which as mentioned, is
> perilously close to "If you go too fast do you become a
[quoted text clipped - 4 lines]
> similar to a Newton's cradle but with a gap between the
> suspended balls,

The balls would appear oblate, and their common centers would appear
to adjust corresponding to this axis's orientation to the motion axis.

> would either system observe the balls in the opposing
> system change in mass with a corresponding increase
> in gravitational attraction that then manifests itself by
> moving the balls? Or would this effect be cancelled out
> by F = d/dt(mv)? Or am I completely off-track with this
> one?

There would be no change. The balls would not appear to "sway"
towards each other. Relativistic mass is not gravitational /
inertial / rest / invariant mass. m_r is a whore's child, with
different values depending on how you want to accelerate the mass in
question, along the line of motion or transverse to it.

> I assume in this that there must be an agreement between any
> corresponding solutions for Minkowski and Newtonian/Euclidean
> geometries?

Minkowski doesn't do mass. Mass implies curvature, remember?

> Comments, and no doubt abuse, welcome.

The two passing systems energy describes their individual contribution
to spacetime. If you double the energy that one system measures /
infers the other to have, you change the effect of one system on the
other... not within each system itself. So as they sail by one
another, you alter the final trajectories (think impulse, since the
interaction duration will become smaller with increasing speed).

Don't know if that helps.

David A. Smith

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Sue... - 30 Jan 2007 15:48 GMT

> Dear SCW:
>
[quoted text clipped - 35 lines]
> different values depending on how you want to accelerate the mass in
> question, along the line of motion or transverse to it.

A whore's child? That might be a bit severe just because it
would fit better in the GR section. Having no better way
to equate gravitational mass to the energy density of a
volume of space-time it is an effective heurism.

A more realistic example is the bucket of water whose
temperature rises the same amount if a projectile hitting
it becomes hotter, heavier or hastier.

If my bucket of water was heating up faster than your
bucket of water we don't know if I am catching heavier
bullets, faster bullets or hotter bullets.

http://en.wikipedia.org/wiki/James_Prescott_Joule

Sue...

> > I assume in this that there must be an agreement between any
> > corresponding solutions for Minkowski and Newtonian/Euclidean
[quoted text clipped - 10 lines]
>
> David A. Smith

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Dumbledore_ - 30 Jan 2007 16:55 GMT

> Dear SCW:
>
[quoted text clipped - 60 lines]
>
> Don't know

Of course not. You are totally fuckin' ignorant.

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sal - 30 Jan 2007 16:38 GMT

> Whilst idly thumbing through the Usenet Physics FAQ (http://
> www.weburbia1.demon.co.uk/physics (v. Nov2006) and the and Charter
[quoted text clipped - 9 lines]
> that F = d/dt(mv) (cf. Planck, Lewis and Tolman) - am I correct in this
> thinking?

I think so, but you need to be careful of the "t", "m", "v", and
"F" in the equation. If F=4-force, m=rest mass, v=4-velocity, and
t=proper time in a momentarily comoving reference frame, then the equation
is correct.

In particular beware 4-force, it's not so simply related to 3-force as you
might expect. Not like 4-momentum, where the spacelike components _are_
the 3-momentum.

> Anyway, here's the question, which as mentioned, is perilously close to
> "If you go too fast do you become a black hole?":
[quoted text clipped - 6 lines]
> itself by moving the balls? Or would this effect be cancelled out by F =
> d/dt(mv)?

I think that's not too far off: the "relativistic mass
increase" cancels out and inertial and gravitational mass remain
equivalent. _BUT_ you can't go very far with this -- as David observed,
SR doesn't do gravity, and doesn't do black holes, and so you can't turn
into a black hole in the SR universe. You need GR for that.

You also can't transform gravity from one frame to another in SR,
AFAIK; simple Newtonian gravity isn't described with a 4-tensor.

> Or am I completely off-track with this one?

Actually try transforming it from one frame to another and see what you
get, if you want a check on your intuition.

> I assume in this that there must be an agreement between any
> corresponding solutions for Minkowski and Newtonian/Euclidean
> geometries?

Not sure what you mean here. Newton does gravity, Minkowski doesn't, so
there aren't any corresponding solutions to things with gravity in them.

> Comments, and no doubt abuse, welcome.
>
> SCW

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John C. Polasek - 30 Jan 2007 21:42 GMT

>Whilst idly thumbing through the Usenet Physics FAQ (http://
>www.weburbia1.demon.co.uk/physics (v. Nov2006) and the and Charter
[quoted text clipped - 29 lines]
>
>SCW

I'll bet you read and re-read the weburbia article, and aside from
absorbing the idea that real physicists want nothing to do with
relativistic mass, regrettably, you failed to find a solution
therein.
My motto is "If you have something to say, write an equation; if you
have nothing to say, write an essay". The weburbia essay only
obfuscates.
Generally it comes down to E = gamma mc2, where gamma x m is
mistakenly taken to be relativistic mass.
But no, no, no. Well, then what?
We are talking about energy and the effect that velocity v has,
whether it would increase mass or not. It is simple to show that with
constant mass you have instead gamma c, which I call Big C.

The total energy equation, a pillar of relativity, is at the heart of
this:
TE = sqrt(p2c2 + m2c4) with p = mv
Factor it into
TE = mc*sqrt(v2 + c2) = mc*BigC,
where BigC exceeds c and is gamma c.
The total energy thus is given (my version; I haven't seen it thus
analyzed, yet there it is; there seems no way out). as the momentum
times BigC. BigC is just so out of place, not?
It also has a false ring and goes against rule number 1.
Actually, to make this less painful, the total energy is bogus. The
treatment of energy scalars as components of a vector is algebraically
odious.
John Polasek

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Natural Science Forum / Physics / Relativity / January 2007

Some Idle musings whilst flicking through the Usenet Physics FAQ