My question concerns generation of graviational waves.
A topic I am not familiar with so please excuse likely misconceptions.


In MWT chap. 36 p. 975, an argument is given for no mass dipole
radiation, namely that the total momentum is conserved.

But what if an external force is applied? It seems that assuming
momentum is conserved is tacitly assuming no radiation, so this is a
somewhat circular argument...

Can anybody suggest a better argument or reference for why a linearly
accelerated mass will not emit graviational radiation, i.e. no mass
dipole radiation?

Thanks - luke

funk420 wrote:

In MWT chap. 36 p. 975, an argument is given for no mass dipole
radiation, namely that the total momentum is conserved.

But what if an external force is applied?

Well, someting has to supply the force, and once you take the
source of the force into account, the total momentum is still
conserved. (Think about Newton's third law -- if something
pushes your system, it gets pushed back as well.) And since
gravity couples to *everything*, it also couples to whatever is
supplying the force, you still get no dipole radiation.

There are, of course, more technical derivations -- since you
have MTW, see section 36.10 for the details. But the underlying
reason for no dipole radiation is still conservation, eqn. (36.41),
as used to get from (36.40) to (36.43).

(Note that conservation of momentum and energy, eqn. (36.41),
is a consequence of the Einstein field equations, not something
that has to be added from the outside.)

Steve Carlip

funk420:

My question concerns generation of graviational waves.
A topic I am not familiar with so please excuse likely misconceptions.


In MWT chap. 36 p. 975, an argument is given for no mass dipole
radiation, namely that the total momentum is conserved.

But what if an external force is applied? It seems that assuming
momentum is conserved is tacitly assuming no radiation, so this is a
somewhat circular argument...

Can anybody suggest a better argument or reference for why a linearly
accelerated mass will not emit graviational radiation, i.e. no mass
dipole radiation?

Yes, but since I don't have a copy of MTW and don't know what's on page
975, I'll use a different argument. What is a dipole? Draw a + and -
charge and the field lines. The dipole moment is just p = qd and points in
the direction from +q to -q. Now draw two masses. (call one your mass and
the other your "external" field, if you like). We don't have any negatives
masses, so the field can't point from one mass to the other. There are no
gravitational dipoles. The lowest order multipole moment is a quadrupole.

(Bilge) wrote in message ...
funk420:

My question concerns generation of graviational waves.
A topic I am not familiar with so please excuse likely misconceptions.


In MWT chap. 36 p. 975, an argument is given for no mass dipole
radiation, namely that the total momentum is conserved.

But what if an external force is applied? It seems that assuming
momentum is conserved is tacitly assuming no radiation, so this is a
somewhat circular argument...

Can anybody suggest a better argument or reference for why a linearly
accelerated mass will not emit graviational radiation, i.e. no mass
dipole radiation?

Yes, but since I don't have a copy of MTW and don't know what's on page
975, I'll use a different argument. What is a dipole? Draw a + and -
charge and the field lines. The dipole moment is just p = qd and points in
the direction from +q to -q. Now draw two masses. (call one your mass and
the other your "external" field, if you like). We don't have any negatives
masses, so the field can't point from one mass to the other. There are no
gravitational dipoles. The lowest order multipole moment is a quadrupole.


Thanks Bilge, this helps..

I was confused because (correct me if I'm wrong) a single point charge
emits dipole radiation (bremsstrahlung) when accelerated..

But now I see that only quadrupole terms come out if the acceleration
is done by another particle of the same charge, and as you point out
there is only one "charge" in mass/gravity, hence only quadrupole and
higher.

As for no negative masses, this is certainly true in the F=ma sense,
but could perhaps antimatter act as a gravitational negative mass?

(Bilge)
wrote in message


What is a dipole? Draw a + and -
charge and the field lines. The dipole moment is just p = qd and points in
the direction from +q to -q. Now draw two masses. (call one your mass and
the other your "external" field, if you like). We don't have any negatives
masses, so the field can't point from one mass to the other. There are no
gravitational dipoles. The lowest order multipole moment is a quadrupole.

If in a uniformly distributed mass there is an empty place
(like e.g. an air bubble in a tank with water) it makes a
"hole" which is like gravitational negative mass with all
the same consequences that you might know from
physics of positive holes made by missing electrons.

As a matter of fact those "holes" are directly responsible
for Hubble's redshift in our universe but it is much more
complex matter so please don't start arguing about this
last remark just think about the mechanism and you
easily find one.

-- Jim

Bilge wrote:
What is a dipole? Draw a + and -
charge and the field lines. The dipole moment is just p = qd and points in
the direction from +q to -q. Now draw two masses. (call one your mass and
the other your "external" field, if you like). We don't have any negatives
masses, so the field can't point from one mass to the other. There are no
gravitational dipoles. The lowest order multipole moment is a quadrupole.

Consider a single pointlike mass. It is a monopole.

Consider two pointlike objects of different mass, separated by a small
distance looked at far away. The multipole expansion of this is a
monopole plus a dipole. While no pure gravitational dipole exists, it is
CERTAINLY possible for a system to have a nonzero gravitational dipole
moment.

Still, even though gravitational monopoles and dipoles exist, there is
no monopole or dipole gravitational radiation. As others have pointed out.


Tom Roberts

Tom Roberts:
Bilge wrote:
What is a dipole? Draw a + and -
charge and the field lines. The dipole moment is just p = qd and points in
the direction from +q to -q. Now draw two masses. (call one your mass and
the other your "external" field, if you like). We don't have any negatives
masses, so the field can't point from one mass to the other. There are no
gravitational dipoles. The lowest order multipole moment is a quadrupole.

Consider a single pointlike mass. It is a monopole.

Consider two pointlike objects of different mass, separated by a small
distance looked at far away. The multipole expansion of this is a
monopole plus a dipole. While no pure gravitational dipole exists, it is
CERTAINLY possible for a system to have a nonzero gravitational dipole
moment.

Still, even though gravitational monopoles and dipoles exist, there is
no monopole or dipole gravitational radiation. As others have pointed out.

That's only an artifact of the coordinates you choose to write the
multipole expansion. A single point mass (or charge) trivially has
a multipole expansion m\sum r^l Y_lm(\theta, \phi). However,
for a system of masses, you can choose coordinates in which the
dipole moment vanishes, i.e., the center of mass. That isn't true
for charges and the reason is the one I gave.

Steve Carlip wrote in message
...
funk420 wrote:

In MWT chap. 36 p. 975, an argument is given for no mass dipole
radiation, namely that the total momentum is conserved.

But what if an external force is applied?

Well, someting has to supply the force, and once you take the
source of the force into account, the total momentum is still
conserved. (Think about Newton's third law -- if something
pushes your system, it gets pushed back as well.)

Yes, however if we consider e.g. two charged particles in motion
(1&2), it is more complex. The total momentum is only conserved if
you include the momentum carried off in the radiation, so Newton's
third law is not really followed (F_12 != F_21).

As you point out below momentum conservation is a consequence of the
field equations.. Is that a total momentum, so that some can be
carried off by radiation?

And since
gravity couples to *everything*, it also couples to whatever is
supplying the force, you still get no dipole radiation.

There are, of course, more technical derivations -- since you
have MTW, see section 36.10 for the details. But the underlying
reason for no dipole radiation is still conservation, eqn. (36.41),
as used to get from (36.40) to (36.43).

(Note that conservation of momentum and energy, eqn. (36.41),
is a consequence of the Einstein field equations, not something
that has to be added from the outside.)

Steve Carlip

Thanks! My apologies for posting when I should be studying

I can't help but notice that this question is on similar lines to the
"radiation of accelerated elektron" post from around the same time
(thanks Bastian and responders).. So does the gravitationally
accelerated electron emit (dipole?) bremsstrahlung? And the
vice-versa: does a (charged) mass accelerated by a static electric
field emit gravitational radiation?

-luke

Starblade Darksquall:

What exactly is a quadrupole?

Literally, "four poles". However, you are confusing "quadrupole"
with "quadrupole moment". A single point charge or point mass can
have a quadrupole moment, but it isn't a quadrupole. The reason is
that the multipole moments come from a multipole expansion of the
potential 1/|r-r'| and the expansion of 1/|r-r'| for the same
potential depends entirely upon what you choose for your origin.
If you choose the origin to be at the point charge, the _only_
multipole moment is the monopole moment. If you choose the origin
to lie elsewhere, then the vector |r-r'| is different and other
terms in the expansion contribute.

A useful theorem says that the lowest non-vanishing multipole
moment is independent of the origin. In general, multipole moments
other than the lowes non-vanishing multipole moment will have
a dependence on the coordinates chosen. So, if you want to know
what a quadrupole is, then consider a charge distribution which
has no monopole moment and no dipole moment. The monopole moment
is the total charge, so a quadrupole has a total charge of zero.
Similarly, the dipole moment has to be zero, so that the charge
distribution can't look like:

|- d -|
-q +q

Which would be a dipole (note that here the monopole moment is
zero, so the dipole moment is independent of origin and is
just p = qd). So, a quadrupole would have a configuration in
which the dipole moments cancel:

-q --
+q -q d
or +2q --
-q +q d
-q --

in the limit that the distance d - 0 and q - \infty such that
the quadrupole moment remains constand (analogusly to the case
of a dipole in which d - 0 and q - \infty such that qd remains
constant). Another charge distribution with a monopole moment
and no dipole moment (assuming the origin is taken to be at the
midpoint of the charge distribution):

+q +q

Is that just a dipole except the arrow is pointing both ways,
or does it have to do with the movement of the masses, meaning
it also involves the 'gap' left when the mass moves away from
a point?

It's not just a dipole with arrows pointing both ways. In the
case of electric charges, it's easier to visualize, because you
can create a pure quadrupole due to the charges having opposite
signs so that the monopole moment is zero. For masses, you have
a monopole moment. However, a monopole doesn't radiate and the
only dipole moment is due to the choice of origin, so it vanishes
by simply choosing coordinates at the center of mass (you don't
get radiation by choosing a different coordinate system). The
next possible multipole moment that might not vanish for any choice
of coordinates is the qudrupole moment and a system of two masses
connected by a spring or orbiting each other is such a configuration.
A spring connecting two masses will oscillate about the center of
mass and because you have an oscillating quadrupole, you can have
quadrupole radiation.