Re: (d/dt) ACCELERATION

[David Bowman <David_Bowman@GEORGETOWNCOLLEGE.EDU>]

Regarding John's comments about my formula for P:

> At 06:22 AM 10/12/00 -0400, David Bowman wrote:
> > This gravitational radiation formula for the radiant power P is:
> >
> > P = (1/5)*(G/c^5)* tr(d^3q_i_j/dt^3)                            [1]
>
> 1a) That doesn't pass the dimensional-analysis test.
> 1b) It doesn't express the following "product" operation:

Oops. I it appears I accidentally eliminated the square power ^2 in the
matrix product when I rearranged the trace formula to be readable in
ASCII using a cumbersome 3rd derivative notation. I had said properly in
words that the *square* of the 3rd time derivative of the traceless mass
quadrupole tensor is the relevant quantity, but I accidentally dropped
the square from the notation in the formula.

The formula should read:

P = (1/5)*(G/c^5)*tr((d^3q_i_j/dt^3)^2)

or if you prefer the Einstein summation convention on the repeated _i_j
indices the trace would be implicitly built in as:

P = (1/5)*(G/c^5)*(d^3q_i_j/dt^3)*(d^3q_i_j/dt^3)

> 2) I think you need to take the time derivatives before taking that product.

Definitely, yes. This is correct.  First the mass quadrupole tensor is
formed. It is the symmetric second tensor moment of the mass.  It has
dimension [mass]*[length]^2.  Next 1/3 of the identity times its
trace is subtracted from it to make it traceless.  Next this symmetric
2nd rank traceless tensor is differentiated thrice w.r.t. time.  Next
the resulting tensor is multiplied by itself as a matrix product.  This
contracts two common indices on the two matrix factors so the resulting
expression is a 2-index matrix.  The resulting matrix is then traced
over.  This square of a traceless matrix is *not* necessarily traceless.
The resulting scalar is multiplied by (1/5)*G/c^5).  The result is the
radiated power.

> > Here d^3q_i_j/dt^3 is the third time derivative of the traceless mass
> > quadrupole tensor q_i_j.
>
> 3) Uhhh, if it's traceless, why are we taking its trace in equation [1]?

See above.  The square of a traceless matrix is not in general
traceless.  We are taking the trace because the power emitted is a
(3-)scalar.  We need to contract all the tensor indices present.
Actually because of the symmetry of the mass quadrupole tensor (and its
3rd derivative) the trace of the matrix square of it is really just the
sum of the squares of all 9 of the individual matrix elements of the
3rd derivative tensor.

> How about this:
>
>         P = (1/5) (G/c^5) average([ (d/dt)^3 Itick ]^2)         [2]
>
> where Itick would be typeset as an I with an overstrike tick-mark, and
> represents the "reduced quadrupole moment":
>
>         Itick_j_k := I_j_k - (1/3) delta_j_k trace(I)
>
> and where I is just the ordinary second moment of the mass
> distribution.  Reference: Misner, Thorne, Wheeler equations 36.1 and 36.3.

Yes, this will do.  MTW's notation is a little weird (the I seems to
suggest an inertia tensor rather than a quadrupole tensor) but it is
fine, esp. after it is reduced to tracelessness (except for a sign change
between q & Itick, but even this sign error cancels out once
the square is taken).

> > The constants G and c are Newton's gravitational constant and the speed
> > limit of causation respectively.
>
> Speed limit of causation.  I like that.  I wonder if it will catch on.

I doubt it, unfortunately.

David Bowman
David_Bowman@georgetowncollege.edu