After waiting for long, Blogphysica gets \LaTeX . (See the announcement at the wordpress.com Blog )

The Basic syntax is $latex <LaTeX Equation> $. You can find a list of \LaTeX symbols here(pdf) . In particular, if you find the formulae are too small try $latex {\displaystyle <LaTeX Equation>}$ . It’s great !

Update(18/02/07) : See this FAQ for some more options.

Now, to illustrate LaTeX, I’ll take up a particular problem. Consider two equal masses falling towards each other, as shown below, starting from rest.

m—>—o—<—m

The question is this – How much power does this system lose as the two masses fall towards each other ?

Let r be the distance from the centre of mass (which I’ve denoted by an ‘o’ above and I choose it to be the origin). Take the common axis to be z-axis.

Hence the position of two masses are respectively (x,y,z)=(0,0,r) and (0,0,-r)

To the zeroth approximation, Newtonian mechanics tells you that

{\displaystyle \frac{d^2 r}{dt^2} = - \frac{G_Nm}{(2r)^2}}

(We neglect the effect of gravitational wave on the masses)

The quadrupole moment Q_{ij} of a mass distribution is defined by

{\displaystyle Q_{ij}(t) = \iiint d\forall\ \ \ \rho(x,t) \left[ x_i x_j - \frac{1}{3} \delta_{ij} r^2 \right]}

Where the integral is done over the whole source(\rho is the mass density of the source). It is basically negative of the traceless part of the moment of inertia.

The Einstein formula for the power emitted by the source (in the form of Gravitational waves) is

{\displaystyle P = \frac{G_N}{5c^5} \sum\ \frac{d^3 Q_{ij}}{dt^3}\  \frac{d^3 Q_{ij}}{dt^3} }

where the symbol \sum denotes a sum over i,j=1,2,3.

Assuming that the masses are small in size, the components of quadrupole moment in this case are

{\displaystyle Q_{zz}= m\left(r^2 - \frac{1}{3}r^2\right)+ m\left((-r)^2 - \frac{1}{3}r^2\right)= \frac{4}{3}mr^2}
{\displaystyle Q_{xx}=Q_{yy} = m(0^2 -\frac{1}{3}r^2)+ m(0^2 -\frac{1}{3}r^2)= -\frac{2}{3}mr^2 }
{\displaystyle Q_{xy}=Q_{yx}=Q_{zx} =\ldots = 0}

Now to calculate the third time derivative, we first use chain rule to get

{\displaystyle \frac{d^3}{dt^3} r^2 = 6 \frac{dr}{dt}\ \frac{d^2 r}{dt^2} + 2r\ \frac{d^3 r}{dt^3} }

We can now employ Newton’s Law to get

{\displaystyle \frac{d^3}{dt^3} r^2  = - \frac{G_Nm}{2r^2}\ \frac{dr}{dt}}

This leads to

{\displaystyle \frac{d^3 Q_{zz}}{dt^3} = - \frac{2G_Nm^2}{3r^2}\ \frac{dr}{dt}}
{\displaystyle \frac{d^3 Q_{xx}}{dt^3}=\frac{d^3 Q_{yy}}{dt^3} =  \frac{G_Nm^2}{3r^2}\ \frac{dr}{dt}}

All the other components are zero. Substituting this into the Einstein’s formula which I quoted above, the total power radiated comes out to be

{\displaystyle \mbox{Power emitted} = \frac{G_N^3m^2}{15c^5 r^4}\ \left(\frac{dr}{dt}\right)^2 }

which is terribly small in most cases.

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