Does F=d^2m/dt^2 x describe any known physical systems?

[ANS]

Hello,

I suspect that the answer to the above question may be no, but I would
like see what others think. Perhaps it is possible to make up a
situation that is contrived enough to be describable by it (I tried but
did not succeed). Also, does this equation violate Lorentz invariance?

Thank you.

[MathematicalPhysicist]

well, mathematically it's the same as d^2x/dt^2m, so yes.

[Dirk Bruere]

ANS wrote:
> Hello,
>
> I suspect that the answer to the above question may be no, but I would
> like see what others think. Perhaps it is possible to make up a
> situation that is contrived enough to be describable by it (I tried but
> did not succeed). Also, does this equation violate Lorentz invariance?
>
> Thank you.

The mass of a bag of sand that has a hole in it?
As the sand empties the hole gets bigger because of the abrasion.
AFAIK Lorenz invariance is safe:-)

Dirk

[Doug Goncz]

Well, ANS, if F means force, then

F=m'' is what I read in the title of your post. F=ma or F=ms'' in most
analyses where s is distance. What comes to mind is F = m's', that is,
a flow situation in which for some reason m=v, a unit violation, and
m/t is mass per time, such as rocket exhaust, and s/t is distance per
time (speed or velocity constrained to a line).

So the physical system described _to me_ by your title equation is some
kind of "unit rocket" or "calibrated rocket", that is a rocket having
some relationship between m/t and v/t. Like a rocket that's shooting
one pound per second at one mile per second, or a kilogram per second
at a meter per second. A "standard" rocket.

Doug Goncz
Replikon Research
Falls Church, VA 22044-0394

[ANS]

Dear Dirk,

Thank you for your reply. While I certainly agree that a hole of
changing size in a sandbag could cause a second order change in the
bag's mass, I do not believe that your example actually fits the
equation because the change in the hole size is not dependent on the
position of the bag (at least not by the mechanism you
mentioned-abrasion).This is what makes it, at least to me, tricky to
think of a natural system that can satisfy it. One might contrive a
situation in which the hole is replaced by a variable opening which
changes dependent upon the position of the sandbag (say, via some
positioning system), so that as the bag is pushed around its change in
mass is second order and dependent upon the bag's position. The
problems I have with this scenario are 1) it is really position that is
now a function of mass per time squared and Force, not Force that is a
function of the other two 2) this scenario is highly contrived. So, the
question still remains...
Armin

[Mark Fisher]

You can come up with many possible situations mathematically, by writing

F = m'' x
=> (mx')' = m'' x
=> m'x'' + mx'' = m'' x (1)

where the dash is a t-derivative. Now you can avoid having to try and find a
general relationship between m and x by just choosing how you would like the
particle to move and solving the resulting ODE for m.

What you have to be careful of when looking for a physical situation is that
you know what x is describing, i.e. if it is a rocket or a bag of sand with
a hole, is x the centre of mass or the coordinate describing the position of
the bag or the payload of the rocket, and is m the mass remaining in the bag
or the total mass.

Since you want a physical solution we have that total mass M is unchanging
and centre of mass X is unchanging.

so you can use M = m + e and X = 1/M (mx + (e)h) (m is the mass of the
isolated system, x is it's coordinate. e is the mass that has left the
system and h is it's centre of mass). Use mx''= - eh" (Newton III) and the
fact that M'=0 and X'=0 (conserved quantities) combined with (1) to try and
see what conditions on m and x and must be met. I think trying to solve your
problem differs from the standard rocket problem because you are allowing
the escaping mass to escape at a non-uniform rate.



"ANS" <armin@umich.edu> wrote in message
news:1146536752.679212.229830@e56g2000cwe.googlegr oups.com...
> Hello,
>
> I suspect that the answer to the above question may be no, but I would
> like see what others think. Perhaps it is possible to make up a
> situation that is contrived enough to be describable by it (I tried but
> did not succeed). Also, does this equation violate Lorentz invariance?
>
> Thank you.
>

[a student]

ANS wrote:
> Hello,
>
> I suspect that the answer to the above question may be no, but I would
> like see what others think. Perhaps it is possible to make up a
> situation that is contrived enough to be describable by it (I tried but
> did not succeed). Also, does this equation violate Lorentz invariance?

There is an ambiguity in the question - what do you mean by "force"? I
will assume here that you mean the rate of change of momentum, i.e.,
F := dp/dt,
as this implies Newton's third law is equivalent to conservation of
momentum; and is compatible fairly general Hamiltonian systems
including rocket equations. Your equation then becomes
dp/dt = m'' x, (1)
where a prime denotes a time derivative.

If we assume that the momentum is, as is typical in the absence of e-m
fields, given by
p := m dx/dt = mx', (2)
it follows that your system is equivalent to a particle moving in the
potential
V(x) = - (1/2) m'' x^2.
In particular, the Hamiltonian
H = p^2/(2m) + V(x)
generates equations of motion (1) and (2) above.

If m'' is always negative, then V(x) is always positive, and
corresponds to a time-dependent harmonic oscillator (eg, a particle on
spring), with mass m(t) and frequency
w(t) = sqrt[-m''/m] .
Interesting cases include m''=0 (describing a free particle with
linearly varying mass m(t)=m(0)+kt), and m''=-k (describing a particle
attached to a spring with fixed spring constant k, and variable mass
m(t)=m(0)+At-k(t^2)/2). Both of these cases can only be physical over
a limited time, to avoid arbitrarily increasing/decreasing mass.

If m'' is always positive, then the system will be unstable/unphysical
in general, as the particle can obtain arbitrarily negative energies by
moving towards infinity. It would be interesting to analyse stability
for the case where m'' changes sign periodically (reminiscent of
rotating frogs and ion traps!).

For a relativistic particle, let P and U denote the 4-momentum and
4-velocity relative to some frame, m denote the rest mass, and T denote
the time measured by a clock co-moving with the particle (i.e., T is a
proper time, and not the time as measured by a clock stationary
relative to the frame). Then U=dX/dT=X', and the above equations
generalise to the Lorentz-boost invariant form
F := dP/dT = m'' X, P = mU,
where m'' denotes d^2 m/dT^2 (to be fully Lorentz invariant, one has
to replace X by X-A for some 4-vector A, corresponding to an "origin"
for the force). Substituting the second equation in the first gives
m' U + m U' = m'' X. (3)
Note that not all solutions of this equation are allowed: since
U.U=c^2=1 in suitable units, this implies that U.U'=0. Hence, taking
the contraction of (3) with U gives
m' = m'' U.X = m'' X.X' = (1/2) m'' (X.X)' . (4)
This places strong constraints on allowable functions m(T) - for
example, the first equality implies that m(T)=m(0)+kT is possible if
and only if k=0. Note that equation may be integrated to give
X.X = \int dT 2m'/m'' .

I don't know whether there are any solutions to (3) consistent with the
constraint U.U=1 (other than m(T)=constant). Note that a time-varying
rest mass could be interpreted as due to heating or cooling of the
particle, eg, due the environment it is travelling through.