SR and the earth, sun, and galaxy.

[Sammywu]

My first try to resolve a SR/GR clock question with this modern SR theory.

The easiest question is the spaceship flying in a circle about a statis observer in a flat spacetime.

The most important issue here seems to choose a stable reference frame. It's very difficult to use the spaceship's reference frame because it actually change with time.

Using the flat spacetime as the reference frame, (t,x,y,z) for the world line of the spaceship, (t, R*cos(2*3.14*t/T), Rsin(2*3.14*t/T), 0) denotes the path using t as the parameter.

Replacing t by t', the proper time of the spaceship, d(t,x,y,z)/dt'= dt/dt'*(1, -2*3.14*R/T*sin(2*3.14*t/T), 2*3.14*R/T*cos(2*3.14*t/T), 0). dt/dt' can be assumed to be a constant as 1/sqrt(1-v^2/c^2) since the spaceship is kept in constant velocity.

When integrating this from t =0 thru T or t'= 0 thru T*sqrt(1-v^2/c^2), we will get T'=T*sqrt(1-v^2/c^2).

The whole process actually is redundent. When dt/dt' = sqrt*(1-v^2/c^2) was determined, the outcome is basically completely based on this.

From the spaceship's view, by SR, dt'/dt shall be also sqrt*(1-v^2-c^2).

The only reason this at last shows that spaceship has a slower clock is because the flatspace has a referencible coordinates.

 

[Sammywu]

By people's help from here, especially Arcon ( pmb_phy ), I read a few documents on tensor algebra and modern SR/GR, I got enought to come back to solve this porblem, even though I still do not understnd all the informations contained in thes documents.

1). We need a stable coordinates to start with. The assumed coordinate is (T(lab), X, Y Z ). T(lab) is also T(world) or T(coordinate). You can see Okun's paper about Gravity Redshift for a better description of it.

2). In this coordinate, g(00) , the 00 component of metric tensor, is -(1+2Phi). I perfer to write it as -(1-2|Phi|), since Phi is negative. In this formula, we can easier see 1-2|Phi| is smaller than 1.

3). For a standing clock in the gravitational potential Phi, Its time can be denoted as T(stand). We can establish another coordinate as (T(stand), X, Y Z ).

4). T(stand) is the same as a clock at thecomoving ( in this case, velocity as zero ) local inertial frame, denoted as T(loc1), to the order of the first differentials.
You can see Okun's paper for better description of this frame. I personally think T(stand) and T(loc1) does have difference when we integrate them over the world line, but to the order of second differentials, ignored by most treatment and for now.

T(loc1) is really the clock held at the same area and just released to free fall. Just at this moment dT(loc1) is the same as dT(stand).

When the object falled to a diffrent area and with aspeed not zero any more, its ticking rate might be different.

5). Based on ds^2=g(ij)dx(i)dx(j), dT(loc1) is the ds and dT(lab) is the dx(0). dx(1), dx(2) and dx(3) are all zero because the clock never move. -(dT(Loc1)^2)=-(1-2|Phi|)*dT(lab)^2. So dT(Loc1)=sqrt(1-2|Phi|)dT(lab).

6). dT(stand) is very close to dT(loc1). So dT(stand)=sqrt(1-2|Phi|)dT(lab). The deeper you are in the gravitational field, the higher is 2|Phi|. dT(stand) is smaller and it means T(stand) is slower.

Once we established T(stand)'s relationship with T(lab), we can now use a different coordinates (T(lab), X, Y, Z). This cordinates can be unchangely describing the orbit of an orbiter. Now let's see how to calculate the T(orbit), the time of an orbiter around the center mass.

7). First, we can establish another T(loc2) for T(orbit), T(loc2) is the free fall object going on the same speed as the orbiter. So, the orbiter is in the local inertal frame. T(loc2)=T(orbit), no doubt.

8). From T(loc1)'s view, T(loc2)=T(loc1)*gamma. Here beta is v/c and gamma is sqrt(1-beta^2).

Why don't we look from T(loc2)'s view? It's the most important question. The coordinates we can establish with clear relationship between time and space is the coordinate (T(stand), X, Y Z ). The rotating of the orbiter makes it diffecult to relate X,Y,Z to its T(orbit).

The selection of the coordinate again determines how the resulit will be. Anyway, I will think more on this.

9) T(orbit)=T(loc2)=gamma*T(loc1)=gamma*T(stand)=gamma*sqrt(1-2|Phi|)*T(lab). So, the orbiter is slower than a standing clock.

Note all I did above, I simplified that taking T as c*T. The coordinates established assuming c=1.


My question is:

How much confidence will we have about the existence of T(lab) and the relationship established between T(lab), X, Y and Z? T(lab) is the same as the far remote obserevr free of gravity I have imagined in the first place.

 

[Sammywu]

Using EQ (3.48) in the "Lecture Notes on General Relativity" written by Sean M. Carroll, with the same coordinate labeled (T(lab), X, Y, Z ) as before, g(00)=-(1+2Phi), g(11), g(22) and g(33) are close to 1. T(orbit)=integration from 0 to 2Pie of SQRT((1-2|Phi|)*(T/2Pie)^2-R^2), where T is the lab. time for the orbiter to make a circle. R is the raius of the orbit. The item R^2 comes from dX/dA^2+dY/dA^2, where dA is the angular velocity of the orbiter. X=R*cosA and Y=R*sinA, so dX/dA=R*(-sinA) and dY/dA=R*cosA.

Any way, 2Pie*R=T*beta, where beta=v/c as SR people know. R=T*beta/2Pie.

So, T(orbit)= integration from 0 to 2Pie of SQRT((1-2|Phi|)*(T/2Pie)^2-(T/2Pie)^2*beta^2).

SQRT((1-2|Phi|)*(T/2Pie)^2-(T/2Pie)^2*beta^2)= (T/2Pie)*SQRT(1-2|Phi|-bate^2).
T(orbit)=T*SQRT(1-2|Phi|-beta^2). Note T=T(lab), so
T(orbit)=T(lab)*SQRT(1-2|Phi|-beta^2).
We already know T(stand)=T(lab)*SQRT(1-2|Phi|).

This shows a match with what we know as GPS time formula.
Also this shall match the H&K experiment.
Note the formula can be applied to an airplane flying around the Earth with smaller amount of v than the so-called escape velocity that drives an orbiter, where the airplane will need to rely on air pressure beneath the wing to support it from not falling down.