Units in GR

[shifty89]

Hello, I'm currently studying general relatively and am trying to plot orbits of planets around the sun using the schwarzchild metric. I've worked out the geodesic equations, working with c=1 to simplify things, and written a matlab script to plot trajectories, but I'm struggling to work out what magnitudes or units to use.

Basically i need a vector of initial conditions for (r, \phi, t, r',\phi', t'), where r and \phi are the polar coordinates, t is time, r' is the speed in the r direction, etc. Obviously i can set \phi and t = 0 initially. I think t' = 1, i.e. initially time is travelling at the regular speed. This just leaves me with r, r', and \phi' to deal with.

i know that for mercury:
mass of sun = 2x10^30 kg
distance from sun = 5.8x10^10 m
orbital velocity = 4.8x10^3 m/s (would this be phi' ? or would i need to break it up into r and phi components... somehow?!)

but what units should i be working with in the relativistic units where c=1?
apologies if this kind of thing has been asked before or if it seems a little silly, but I am a maths student, we don't generally worry about units!

Any help would be greatly appreciated!

[LAHLH]

OK so c not only has value 1 it is also dimensionless in the units we would like to use. In order to achieve this we adopt a new unit for time, the meter! One meter of time is the time it takes light to travel 1m (of distance) (this is the opposite approach to the more familiar 'light year', whereby the measure distance in units usually reserved for time, you get the idea...)

c=(distance light travels in given time interval)/(the time interval)
c=1m/(time to travel 1m)
c=1m/1m=1

To convert from these weird units to the usual you can use 3x10^8 ms^-1=1, so 1s=3x10^8m and hence 1m of time =(1/(3x10^8))s

For less fundamental units than metres,seconds, kg and so on,such a joules and Newtons. For example 10J=10kgm^2s^-2, and since 1s=3x10^8m, s^-2=(9x10^16)^-1 m^-2, so we obtain 10J=1.1x10^-16kg

This is lifted from Schutz Ch1.

Another convention of seen in GR, Geometrized units, is when we set c=1 and G=1 (if we set hbar=1 we then get Planck units), in these units mass gets measured in length too, so as you have probably seen in the Schwarzschild solution 2m/r is dimensionless. Meaning the interval [ds^2]=L^2 (again because [dt^2]=L^2 too as we measure time in length)

The conversion factor is G/c^2 so the Sun's mass of 2x10^30kg is equal to 1.5km.

To see why:

[G]=L^3 M^-1 T^-2

we already had from [c]=1 that L=T, so we have:

[G]=L M^-1

and if now G is to be dimensionless it must also be that L=M so therefore mass is length. If G is also to be not just dimensionless but also 1, then it must be that:

1=(6.673x10^-11) m^3 kg^-1 s^-2
1=(6.673x10^-11) m^3 kg^-1 (9x10^16)^-1 m^-2
1=G/c^2 m/kg

Thus if this is to be 1, then 1kg=G/c^2 m, so 1kg is this fraction of 1 metre.

For a velocity such as the one you quoted, v=4.8x10^3 m/s, well again 1s=3x10^8m, so you need to divide through by a c to get v=1.6x10^-5 where [v]=1 (dimensionless now)

[bcrowell]

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

You can also simply put factors of c and G back in at the end. For example, if you obtained an expression like 1-v, the 1 is unitless, and the v is unitless in geometrized units. To convert it back to SI units, you'd have to insert a 1/c, making it 1-v/c.

[LAHLH]

I think t' = 1, i.e. initially time is travelling at the regular speed. This just leaves me with r, r', and LaTeX Code: \\phi ' to deal with.

Because of various symmetries (See Carroll ch5 available free online), you have various conserved quantities in the Schwarzschild solution, one of them is energy E, and it happens that \dot{t}=\frac{E}{1-\frac{2M}{r}} , you have have axial symmetry leading to the analogue of Kepler: \dot{\phi}=\frac{L}{r^2}

You are right to say that initial you can set t=0, phi=0 and this is because of these symmetries (you can't do the same with r- things depend on where you start in radial coordinate). Also because motion occurs in the plane, you can without loss of generality set theta=pi/2, which can massively simplifies your geodesic equations. At the end of all this you should find geodesic equation for timelike particles (in geometrized natural units):

\frac{1}{2}\dot{r}^2+\left[\frac{1}{2}-\frac{M}{r}+\frac{L^2}{2r^2}-\frac{ML^2}{r^3}\right]=\frac{E^2}{2}

You can get a good idea how these look and where the circular orbits are etc by plotting this effective potential.