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What are physics constant in Kerr metric?

by zerop
Tags: general relativity
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zerop
#1
Jan10-11, 07:45 AM
P: 1
1. What are the value of physics constant in Kerr metric, including G, M, c, a, r, or others?
I expect to simplify Gamma

2. why g_compts[1,4] has element and not [4,1]?

3. Some book assume G = c = 1, what is the meaning of this setting?

4. Different material have different metric, are there a metric table for element table?

5. What is theta in Kerr metric?

************** Kerr metric *****************
t r theta phi
t
r
theta
phi

with(tensor):
coord := [t, r, theta, Phi]:

g_compts:=array(sparse,1..4,1..4):

G := 6.67*10^(-11)

triangle := r^2 - 2*G*M*r/c^2 + a^2:
p2 := r^2 + ((cos(theta))^2)*a^2:
A := (r^2+a^2)^2 - (a^2)*triangle*(sin(theta))^2:

g_compts[1,1]:= (triangle - (a^2)*(sin(theta))^2)*(c^2)/p2:
g_compts[1,4]:= 4*G*M*a*r*(sin(theta))^2/(c*p2):
g_compts[2,2]:= -p2/triangle:
g_compts[3,3]:= -p2:
g_compts[4,4]:= -A*(sin(theta)^2)/(p2):

g1 := create([-1,-1], eval(g_compts)):
g1_inv := invert( g1, 'detg' ):

D1g := d1metric( g1, coord ):

Cf1_1 := Christoffel1(D1g):
Cf2_1 := Christoffel2(g1_inv, Cf1_1):
displayGR(Christoffel2,Cf2_1):
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stevebd1
#2
Jan19-11, 06:28 AM
P: 611
G is the gravitational constant in units m^3/(kg s^2) (1 in geometric units)

m is mass in kg where M is the geometric unit for mass (M=Gm/c^2) in metres

c is the speed of light in m/s (or 1 in geometric units)

a is the geometric units for angular momentum in metres (a=J/mc where J is angular momentum in SI units)

r is radius in metres

Delta (or triangle as you call it) is the radial parameter in m^2.

when writing delta, you have written delta=r^2-2*G*m*r/c^2+a^2. If geometric units are used, you can simply write delta=r^2-2M+a^2 where M=*G*m*r/c^2, the answers are the same.

g_compts[1,4] does include for [4,1], they've just substituted the 2*(2*.. with a 4*.., it can be rewritten-

g_compts[1,4]=2*(2*M*a*r*(sin(theta))^2/(p2)), [1,4] & [4,1] being the same, another way of writing it is 2*g_compts[1,4].

theta is the latitude approach, 90 degrees (or pi/2) at the equator and 0 at the poles.

You may also find this web page useful-
http://www.astro.ku.dk/~milvang/RelViz/000_node12.html
stevebd1
#3
Jan20-11, 02:01 AM
P: 611
The text above relating to delta should read-

'when writing delta, you have written delta=r^2-2*G*m*r/c^2+a^2. If geometric units are used, you can simply write delta=r^2-2Mr+a^2 where M=G*m/c^2, the answers are the same.'


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