Coordinate change in de Sitter spacetime

[Parvulus]

Hello folks. Just registered, first post (moved here from the Physics forum).

Let there be a de Sitter metric in static coordinates:

ds^2 = - [1 - (r/R)^2] c^2 dt^2 + dr^2 / [1 - (r/R)^2] + r^2 d(omega)^2

where:

r is the radial coordinate
R is the cosmological horizon
coordinate time t is as observed from r = 0, the "origin", which we will call point O.

Let rP be the radial coordinate (i.e. as observed from the origin O) of a static point P.

Let's change now to a static frame of reference centered in said point P, and call r' the radial coordinate in that frame of reference.

What is r'O, that is the distance from P to O as observed from the frame of reference centered in P?

If you prefer to give the general equation for r' as a function of r, rP and R, that's fine.

Thank you in advance for your help.

 

[Jhero]

Hello and welcome to the forum! It's great to have you here.

To answer your question, let's first define some terms:

- r: the radial coordinate as observed from the origin O
- R: the cosmological horizon
- t: coordinate time as observed from the origin O
- rP: the radial coordinate of a static point P as observed from the origin O
- r': the radial coordinate of a point observed from a static frame of reference centered in P

To find the distance r'O, we can use the Pythagorean theorem. From the de Sitter metric, we know that

ds^2 = - [1 - (r/R)^2] c^2 dt^2 + dr^2 / [1 - (r/R)^2] + r^2 d(omega)^2

Since we are looking at a static frame of reference centered in P, we can set dt = 0. This leaves us with

ds^2 = dr^2 / [1 - (r/R)^2] + r^2 d(omega)^2

Now, let's plug in our values for rP and r' to this equation. We know that rP is the radial coordinate of P as observed from the origin O, so we can say that

rP = r

Similarly, we can say that

r' = rO

where rO is the radial coordinate of point O as observed from the frame of reference centered in P.

Plugging these values into our equation, we get:

ds^2 = dr'^2 / [1 - (r'/R)^2] + r'^2 d(omega)^2

Now, we can use the Pythagorean theorem to solve for r'O:

r'O = sqrt(dr'^2 / [1 - (r'/R)^2] + r'^2 d(omega)^2)

To get the general equation, we can substitute r' with rP and rO with r'O:

r'O = sqrt(drP^2 / [1 - (rP/R)^2] + rO^2 d(omega)^2)

I hope this helps answer your question. Please let me know if you need any further clarification. Keep exploring and asking questions!