Question of three clocks

[mistermack]

Imagine three ideal clocks. 

Clock A is on the surface of the Earth. The other two, B and C, are side-by-side in space, remotely distant from the Earth's gravitional well. As I understand it, clock A on the Earth's surface will be running slower than the two out in space, which will both be running at the same rate. 

Now imagine that the two in space become separated. Clock B is tethered so stays the same distance from Earth. Clock C is not, and falls very slowly, picking up speed faster and faster, down to the Earth, where it is momentarily side-by-side with the original Earth clock A. It's now going to be falling at Earth's escape velocity, about 11.186 km/s. (forget the atmosphere for this scenario)

What I'm wondering is, will clock C be running at the same speed as clock A, or clock B, or will it be slower than both? 
(It's not a trick question, and I don't know the answer. I suspect B and C will still be running at the same speed, but that's just my guess)

[Genady]

For an observer with clock A, C will be slower. For an observer with B, C will be slower, too.

[Sensei]

3 hours ago, mistermack said:

As I understand it, clock A on the Earth's surface will be running slower than the two out in space, which will both be running at the same rate. 

It's not that simple. In order to stay in space, they must have enough velocity to avoid falling back to Earth, the Sun or another space object.

To calculate position on Earth, a GPS satellite (in orbit > 20,000 km) must have both gravitational time dilation and velocity time dilation taken into account.

https://upload.wikimedia.org/wikipedia/commons/b/b4/Comparison_satellite_navigation_orbits.svg

3 hours ago, mistermack said:

It's now going to be falling at Earth's escape velocity, about 11.186 km/s. (forget the atmosphere for this scenario)

Escape velocity is not constant, but variable and depends on the distance. What can be found on the web as the escape velocity of a space object is the velocity from the surface of that object.

 

[Lorentz Jr]

What I wonder about is a clock sitting just above the surface (with little or no rotation about the center of the planet) vs one in high-speed orbit just above the surface (in a vacuum).

In terms of SR, the orbiting clock should be slower because it "moves" and accelerates* around the planet, but I'm afraid to guess when gravity is involved, and in GR it's the "stationary" clock that's accelerating! 😶

I guess C/orbiting would have to be slower than A/fixed, because C/orbiting is moving faster as seen by B/distant, which is at least close to inertial in both SR terms and GR terms.

 

* Not literally because of the acceleration, but in SR, acceleration shows that the orbiting clock passes through multiple reference frames, as the astronaut and spaceship do in the twin paradox.

Edited December 25, 2022 by Lorentz Jr

[Genady]

13 hours ago, Genady said:

For an observer with clock A, C will be slower. For an observer with B, C will be slower, too.

In addition, for the observer C, A will be slower. And (assuming that B is extremely far, and that C is in free radial fall from extremely far) for C, B will be running at the same speed.

So, I think that

13 hours ago, mistermack said:

B and C will still be running at the same speed

is correct, but only from the C's perspective.

[mistermack]

The reason for the question is that I have a maybe flawed notion that a clock A on the Earth's surface is 'actually' running slower than clocks B and C, because of the effect of the Earth's gravity well. As opposed to 'from the perspective of'. So if they are reunited, they will disagree as to the elapsed time. 

But the clock C that falls to Earth is not undergoing acceleration according to GR, it's free floating in curved space time. So does it slow, as it falls, or not? 

I'm guessing not, as I said in the OP, I think it will remain in sync with clock B. But that's just my guess. I don't know enough to do any more than guess.

Edited December 25, 2022 by mistermack

[Genady]

@mistermack,

There are two time dilations involved. One is a gravitational time dilation. You could call it "actual" as it depends only on a change in gravitational field between events. The other is a relativistic time dilation, due to the observers' relative motion. This one differs "from the perspective of" different observers. (BTW, don't "reunite" the clocks, as the motion will cause the relativistic time dilation and the test will be ruined.)

In your scenario, the two effects add in different ways:

A and B don't move relative to each other, so there is only the "actual" gravitational time dilation, and it makes A to run slower than B.

A and C are in the same place at the same time, so there is no gravitational time dilation between them. But they move relative to each other, which causes a relativistic time dilation. Thus, A is slower than C from the C's perspective, and C is slower than A from the A's perspective.

B and C are in different positions in the gravitational field. This causes the gravitational time dilation which makes C to run slower than B. But they also move relative to each other causing the relativistic time dilation. For B, it makes the C to run even slower. However, for C, it makes the B to run slower, compensating for the effect of the gravitational time dilation. In case of B being in infinity and C free falling straight from a rest in infinity, this compensation will be exact, and thus for C, they will be running in sync.

Edited December 25, 2022 by Genady

[MigL]

All these types of questions have as many answers as available frames of reference, as timings are frame dependent.
Unless the question includes the observer's frame, it is an ill posed question.

[Lorentz Jr]

C vs A can sort of be verified (as long as C is slower). A clock falling from infinity moves faster in Earth's frame than one in circular orbit (more total energy + same potential = more kinetic), and they're both in free fall, so the falling clock must tick more slowly than the orbiting one when they're at the same altitude. And it might be possible to compare them directly by digging a tunnel through Earth so they cross paths again on the other side. Not very practical, of course....

And finally, the orbiting clock passes by the stationary clock periodically, so those two can be compared directly.

 

Edited December 25, 2022 by Lorentz Jr

[Markus Hanke]

15 hours ago, MigL said:

All these types of questions have as many answers as available frames of reference, as timings are frame dependent.
Unless the question includes the observer's frame, it is an ill posed question.

I tend to agree. The “tick rate” of a clock is an observer-dependent quantity, so this type of question really does very little to demonstrate the underlying physics. A more illuminating kind of scenario would consider the total elapsed time on clocks, which is equal to the geometric lengths of their world lines, and thus a quantity that all observers agree on. Of course, one would have to connect the same set of events to make a meaningful comparison between world lines, which we don’t have in the OP’s scenario.

As for the problem at hand, clock A on Earth is not inertial, but under constant acceleration; the same goes for distant B, unless it is far enough away that we can roughly consider it inertial. So you have two distant non-inertial frames, and one inertial frame in radial free fall. All of this is embedded in a curved spacetime, so one must also consider how to actually define simultaneity here, otherwise comparing distant clocks is meaningless.

[joigus]

For Earth's escape velocity the v/c parameter (the beta relativistic parameter) is about 1.4x10^-11. For the other velocities involved it's much smaller.

The r_s/r term in Schwarzschild metric though is of order 2.4x10^-9. r_s being the Schwarzschild radius of the Earth, and r the actual radius of the Earth.

In order to tackle these problems, you'd better not start thinking in terms of combined effects of r-dependent time dilation plus kinematical time dilation. Otherwise, you get confused, make a mess, and probably get the answer wrong. As @Markus Hanke often says, GR is highly non-linear, so it's not a matter of this effect plus that effect. I'm sure @Mordred agrees on this particular point.

What you do is write the Schwarzschild metric and do all your calculations of proper times from there for different trajectories --elapsed time from different POV's, as Markus suggests. The metric really gives you everything you need for small clocks either falling or accelerating, etc. in the "background" field. You just plug in the trajectories \( r\left( t \right) \), \( \theta \left( t \right) \), and \( \phi \left( t \right) \). It's for finer approximations that things might get hairy.

I've tried to do that for this case, and here's what I get:

\[ d\tau^{2}=\left(1-\frac{r_{\textrm{s}}}{r}\right)dt^{2}-\frac{1}{c^{2}}\left(1-\frac{r_{\textrm{s}}}{r}\right)^{-1}dr^{2}-\frac{1}{c^{2}}r^{2}\sin^{2}\theta d\theta^{2}-\frac{1}{c^{2}}r^{2}d\phi^{2} \]

You can further assume that all trajectories are equatorial, so,

\[ d\tau^{2}\simeq\left[\left(1-\frac{r_{\textrm{s}}}{r}\right)-\left(1+\frac{r_{\textrm{s}}}{r}\right)\left(\frac{\dot{r}}{c}\right)^{2}-\left(\frac{r\dot{\phi}}{c}\right)^{2}\right]dt^{2} \]

For small velocities \( \dot{r},r\dot{\theta},r\dot{\phi}\ll c \),and after Taylor-expanding the square roots, you get,

\[ d\tau\simeq\left(1-\frac{r_{\textrm{s}}}{2r}\right)dt \]

For clocks A, B, and C, we get proper times from 0 to a certain time standard assymptotic time T:

\[ \tau_{A}\simeq\int_{0}^{T}dt\left(1-\frac{r_{\textrm{s}}}{2r_{A}}\right)=\left(1-\frac{r_{\textrm{s}}}{2r_{A}}\right)T \]

\[ \tau_{B}\simeq\int_{0}^{T}dt\left(1-\frac{r_{\textrm{s}}}{2r_{B}}\right)=\left(1-\frac{r_{\textrm{s}}}{2r_{B}}\right)T \]

\[ \tau_{C}\simeq\int_{0}^{T}dt\left(1-\frac{r_{\textrm{s}}}{2r_{C}\left(t\right)}\right) \]

Remembering that,

\[ r_{A}<r_{B} \]

and r_C monotonically goes from r_B to r_A,

\[ r_{C}\left(0\right)=r_{B} \]

\[ r_{C}\left(T\right)=r_{A} \]

So that,

\[ \tau_{A}<\tau_{C}<\tau_{B} \]

So the proper time for observer A, who is at sea level on the Earth is shorter –proper, so from his own POV– than it appears to be from the POV of an observer at distance r_A, and both shorter still than time elapsed from POV of an asymptotic observer, which is T:

\[ \tau_{\infty}=\lim_{r\rightarrow\infty}\left(1-\frac{r_{\textrm{s}}}{2r}\right)T=T \]

Which means that those far-away inertial observers see B slowing down, C slowing down even more, and A running the slowest of all.

For clocks moving in any funny arbitrary way this would not be true. I think this approximation is good enough for our purposes, as the kinematic terms are at least a couple of orders of magnitude smaller than the r-dependent term.

That's what I get from my analysis, anyway. The method I know to be correct. Please do tell me if you think I overlooked something important.

[Genady]

Thank you, @joigus for the clarity. This calculation shows - both formally and intuitively - how the proper time behaves if we consider only the gravitational time dilation. I am just not convinced that we can ignore the kinematical time dilation of the free-falling from infinity observer C. That is because for such an observer, 

(v/c)² = (1-r_s/r)²(r_s/r)

Substituting it in the second equation above gives, I believe,

dτ_C= (1-r_s/r)dt

rather than

dτ_C= (1-r_s/(2r))dt

making τ_C smaller than in the calculation above. 

[joigus]

30 minutes ago, Genady said:

Thank you, @joigus for the clarity. This calculation shows - both formally and intuitively - how the proper time behaves if we consider only the gravitational time dilation. I am just not convinced that we can ignore the kinematical time dilation of the free-falling from infinity observer C. That is because for such an observer, 

(v/c)² = (1-r_s/r)²(r_s/r)

Substituting it in the second equation above gives, I believe,

dτ_C= (1-r_s/r)dt

rather than

dτ_C= (1-r_s/(2r))dt

making τ_C smaller than in the calculation above. 

I see how you would think that. It sounds very common-sensical. But that's not how it works. Keep in mind that any exact solution of Einstein's field equations already has any possible kinematical effects included. It's a package deal. Most of these solutions have a structure,

\[ f\left(r\right)dt^{2}-\frac{1}{f\left(r\right)}dr^{2}-r^{2}d\Omega^{2} \]

The \( f\left( r \right) \) and \( 1/f\left( r \right) \) play the role of "local contraction and dilation gamma factors" so to speak. Because GR is obtained from --among other things-- a demand that it satisfies SR locally, it is guaranteed to take care of that. \( \left(1-\frac{r_{\textrm{s}}}{r}\right)^{-1} \) plays the role of a gamma factor of sorts, while \( \left(1-\frac{r_{\textrm{s}}}{2r}\right) \) plays the role of an inverse gamma factor of sorts, if you will. It's true that the SR metric is different, but GR is under no obligation to follow SR's intuitions so closely. The equation that you wrote is off by a square in the differentials, so it should be,

\[ d\tau^{2}\simeq\left(1-\frac{r_{\textrm{s}}}{r}\right)dt^{2} \]

Which, after taking the square root becomes,

\[ d\tau\simeq\left(1-\frac{r_{\textrm{s}}}{r}\right)^{1/2}dt \]

And, only after Taylor expanding and keeping first-order terms,

\[ d\tau\simeq\left(1-\frac{r_{\textrm{s}}}{2r}\right)dt \]

which is the approximation I used.

[Mordred]

@joigusUsing the Schwartzchild solution as your basis is a good methodology for the sea level observer however be aware that to observe the clock on the orbital you will require the energy momentum terms in the T^{oi} and the T^{oj} directions. This will be further complicated in the Schwartzchild treatment as the radius to either the sea level clock and the observer at infinity will vary.

 I would further suggest using proper acceleration along with rapidity for the orbiting clock.

I may have an appropriate solution for the orbiting clock but will have to dig it up later today

Edited December 26, 2022 by Mordred

[Genady]

Sorry, @joigus for not making myself clear. I do not suggest adding gravitational and kinematical effects as I know that all is already there in the equation.

What I did was not something common-sensical but rather something algebraical, by simply substituting expression for (r-dot)/c = v/c in the second equation above, instead of approximating it by 0, as you did in your approximation. Is there a mistake in my algebra? I'd like to know.

[Mordred]

putting these here as a visual aid to fill the GR and stress energy tensor

\[\frac{dx^\alpha}{dy^{\mu}}=\frac{dx^\beta}{dy^{\nu}}=\begin{pmatrix}\frac{dx^0}{dy^0}&\frac{dx^1}{dy^0}&\frac{dx^2}{dy^0}&\frac{dx^3}{dy^0}\\\frac{dx^0}{dy^1}&\frac{dx^1}{dy^1}&\frac{dx^2}{dy^1}&\frac{dx^3}{dy^1}\\\frac{dx^0}{dy^2}&\frac{dx^1}{dy^2}&\frac{dx^2}{dy^2}&\frac{dx^3}{dy^2}\\\frac{dx^0}{dy^3}&\frac{dx^1}{dy^3}&\frac{dx^2}{dy^3}&\frac{dx^3}{dy^3}\end{pmatrix}\]

Schwartzchild metric

Vacuum solution [latex]T_{ab}=0[/latex] which corresponds to an unaccelerated freefall frame [latex]G_{ab}=dx^adx^b[/latex] if

[latex]ds^2> 0[/latex] =spacelike propertime= [latex]\sqrt{ds^2}[/latex]

[latex]ds^2<0[/latex] timelike =[latex]\sqrt{-ds^2}[/latex]

[latex] ds^2=0[/latex] null=lightcone

 

spherical polar coordinates [latex](x^0,x^1,x^2,x^3)=(\tau,r,\theta,\phi)[/latex]

[latex] G_{a,b} =\begin{pmatrix}-1+\frac{2M}{r}& 0 & 0& 0 \\ 0 &1+\frac{2M}{r}^{-1}& 0 & 0 \\0 & 0& r^2 & 0 \\0 & 0 &0& r^2sin^2\theta\end{pmatrix}[/latex]

line element

[latex]ds^2=-(1-\frac{2M}{r}dt)^2+(1-\frac{2M}{r})^{-1}+dr^2+r^2(d \phi^2 sin^2\phi d\theta^2)[/latex]

Dust solution no force acting upon particle

[latex] T^{\mu\nu}=\rho_0\mu^\mu\nu^\mu[/latex]

[latex]T^{\mu\nu}x=\rho_0(x)\mu^\mu(x)\mu^\nu(x)[/latex]

Rho is proper matter density

Four velocity 

[latex]\mu^\mu=\frac{1}{c}\frac{dx^\mu}{d\tau}[/latex]

Leads to

[latex]ds^2=-c^2d\tau^2=-c^2dt^2+dx^2+dy^2+dz^2=-c^2dt^2(1-\frac{v^2}{c^2})^\frac{1}{2}=\frac{1}{\gamma}[/latex]

[latex]T^{00}=\rho_0(\frac{dt}{d\tau})^2=\gamma^2\rho_0=\rho[/latex] [latex]\rho[/latex] is mass density in moving frame.

[latex]T^{0i}=\rho_0\mu^o\mu^i=\rho^o\frac{1}{c^2}\frac{dx^o}{d\tau}\frac{dx^2}{d\tau}=\gamma^2\rho_0\frac{\nu^i}{c}=\rho\frac{\nu^i}{c}[/latex]

[latex]\nu^i=\frac{dx^i}{dt}[/latex]

[latex]T^{ik}=\rho_0\frac{1}{c^2}\frac{dx^i}{d\tau}\frac{dx^k}{d\tau}=\gamma^2\rho\frac{\nu^i\nu^k}{c^2}=\rho\frac{\nu^i\nu^k}{c^2}[/latex]

Thus

[latex]T^{\mu\nu}=\begin{pmatrix}1 & \frac{\nu_x}{c}&\frac{\nu_y}{c} &\frac{\nu_z}{c} \\\frac{\nu_x}{c}& \frac{\nu_x^2}{c} & \frac{\nu_x\nu_y}{c^2}& \frac{\nu_x\nu_z}{c^2}\\ \frac{\nu_y}{c}& \frac{\nu_y\nu_z}{c^2} & \frac{\nu_y^2}{c^2}& \frac{\nu_y\nu_z}{c^2}\\ \frac{\nu_z}{c} &\frac{\nu_z\nu_x}{c^2}&\frac{\nu_z\nu_y}{c^2}&\frac{\nu_z}{c^2}\end{pmatrix}[/latex]

Schwartzchild metric

Vacuum solution [latex]T_{ab}=0[/latex] which corresponds to an unaccelerated freefall frame [latex]G_{ab}=dx^adx^b[/latex] if

[latex]ds^2> 0[/latex] =spacelike propertime= [latex]\sqrt{ds^2}[/latex]

[latex]ds^2<0[/latex] timelike =[latex]\sqrt{-ds^2}[/latex]

[latex] ds^2=0[/latex] null=lightcone

 

spherical polar coordinates [latex](x^0,x^1,x^2,x^3)=(\tau,r,\theta,\phi)[/latex]

[latex] G_{a,b} =\begin{pmatrix}-1+\frac{2M}{r}& 0 & 0& 0 \\ 0 &1+\frac{2M}{r}^{-1}& 0 & 0 \\0 & 0& r^2 & 0 \\0 & 0 &0& r^2sin^2\theta\end{pmatrix}[/latex]

line element

[latex]ds^2=-(1-\frac{2M}{r}dt)^2+(1-\frac{2M}{r})^{-1}+dr^2+r^2(d \phi^2 sin^2\phi d\theta^2)[/latex]

 

the above will not work directly but it is informative and will be helpful in a solution development. We will require additional terms I do know of one example that I have yet to add (the above has no acceleration for starters so this will involve rotations likely via rapidity.)

 

 

Edited December 26, 2022 by Mordred

[Genady]

PS. When I said,

1 hour ago, Genady said:

if we consider only the gravitational time dilation

and 

1 hour ago, Genady said:

ignore the kinematical time dilation of the free-falling from infinity observer

what I meant was removing the r-dot term of the equation for approximation.

(@joigus)

Edited December 26, 2022 by Genady

[joigus]

OK. I've got a little bit to digest for the time being. I do believe you can get away with the r_s/r term in 1st order and ignore kinematical corrections. Reason being that r_s/r is of order \( 2.4\times10^{-9} \), while \( v/c \) is of order \( 1.4\times10^{-11} \). First corrections to "kinematic" terms are square of the latter, so order \( 2\times10^{-22} \) while first corrections for the r-dependent term are just of order (1st order in r_s/r for sqrt of g₀₀ and g_rr) as said before.

Mmmm. @Mordred. I think I understand your main argument on rotation/acceleration, but... I wouldn't invoke Cartesian coordinates either. It's all more transparent in spherical coordinates. I'm confused by isotropic dust in the discussion too... What are you deploying on me man?  

I did make a rough estimation for rotating A and B clocks and didn't find that much of a difference if speeds are safely orders below c, but I could be wrong. I picked circular paths with fixed radii and used a 3rd-Kepler kind of approximation. GR corrections to Kepler's law don't seem to change scenario significantly, I think. They differ only by a sqrt(g₀₀)...

I'll keep an eye on this for further comments. I always try to keep as pedestrian as humanly possible.

So far I stand by my estimation.

Maybe later...

[Mordred]

hrrm seems to me we will need the rotating frame described by the Sagnac effect. This will take me a bit of time to get down so please be patient I'm going to apply the form given by Lewis Ryder from his textbook as its layout is one of the easiest to follow also going to have to break it down further for readers not familiar with the Lorentz transforms in the x, y and z direction. (its worth the additional effort as it is informative to all readers.)

 

20 minutes ago, joigus said:

OK. I've got a little bit to digest for the time being. I do believe you can get away with the r_s/r term in 1st order and ignore kinematical corrections. Reason being that r_s/r is of order 2.4×10−9 , while v/c is of order 1.4×10−11 . First corrections to "kinematic" terms are square of the latter, so order 2×10−22 while first corrections for the r-dependent term are just of order (1st order in r_s/r for sqrt of g₀₀ and g_rr) as said before.

Mmmm. @Mordred. I think I understand your main argument on rotation/acceleration, but... I wouldn't invoke Cartesian coordinates either. It's all more transparent in spherical coordinates. I'm confused by isotropic dust in the discussion too... What are you deploying on me man?  

I did make a rough estimation for rotating A and B clocks and didn't find that much of a difference if speeds are safely orders below c, but I could be wrong. I picked circular paths with fixed radii and used a 3rd-Kepler kind of approximation. GR corrections to Kepler's law don't seem to change scenario significantly, I think. They differ only by a sqrt(g₀₀)...

I'll keep an eye on this for further comments. I always try to keep as pedestrian as humanly possible.

So far I stand by my estimation.

Maybe later...

the dust solution I simply had handy as it describes an unaccelerated frame stress energy tensor treatment as mentioned it wont work in this case by itself but the details of how to fill the two tensors are included in differential form

Lorentz transformations list spherical coordinates (rotation along the z axis through an angle ) \[\theta\]

\[\acute{x}=x\cos\theta+y\sin\theta,,,\acute{y}=-x\sin\theta+y \cos\theta\]

 

Edited December 26, 2022 by Mordred

[joigus]

19 minutes ago, Mordred said:

hrrm seems to me we will need the rotating frame described by the Sagnac effect. This will take me a bit of time to get down so please be patient I'm going to apply the form given by Lewis Ryder from his textbook as its layout is one of the easiest to follow also going to have to break it down further for readers not familiar with the Lorentz transforms in the x, y and z direction. (its worth the additional effort as it is informative to all readers.)

 

the dust solution I simply had handy as it describes an unaccelerated frame stress energy tensor treatment as mentioned it wont work in this case by itself but the details of how to fill the two tensors are included in differential form

I see. Thanks.

[Mordred]

Lorentz transformations list spherical coordinates (rotation along the z axis through an angle ) \[\theta\]

\[(x^0,x^1,x^2,x^3)=(ct,r,\theta\\phi)\]

\[(x_0,x_1,x_2,x_3)=(-ct,r,r^2,\theta,[r^2\sin^2\theta]\phi)\]

 

\[\acute{x}=x\cos\theta+y\sin\theta,,,\acute{y}=-x\sin\theta+y \cos\theta\]

\[\Lambda^\mu_\nu=\begin{pmatrix}1&0&0&0\\0&\cos\theta&\sin\theta&0\\0&\sin\theta&\cos\theta&0\\0&0&0&1\end{pmatrix}\]

generator along z axis

\[k_z=\frac{1\partial\phi}{i\partial\phi}|_{\phi=0}\]

generator of boost along x axis::

\[k_x=\frac{1\partial\phi}{i\partial\phi}|_{\phi=0}=-i\begin{pmatrix}0&1&0&0\\1&0&0&0\\0&0&0&0\\0&0&0&0 \end{pmatrix}\]

boost along y axis\

\[k_y=-i\begin{pmatrix}0&0&1&0\\0&0&0&0\\1&0&0&0\\0&0&0&0 \end{pmatrix}\]

generator of boost along z direction

\[k_z=-i\begin{pmatrix}0&0&0&1\\0&0&0&0\\0&0&0&0\\1&0&0&0 \end{pmatrix}\]

the above is the generator of boosts below is the generator of rotations.

\[J_z=\frac{1\partial\Lambda}{i\partial\theta}|_{\theta=0}\]

\[J_x=-i\begin{pmatrix}0&0&0&0\\0&0&0&0\\0&0&0&1\\0&0&-1&0 \end{pmatrix}\]

\[J_y=-i\begin{pmatrix}0&0&0&0\\0&0&0&-1\\0&0&1&0\\0&0&0&0 \end{pmatrix}\]

\[J_z=-i\begin{pmatrix}0&0&0&0\\0&0&1&0\\0&-1&0&0\\0&0&0&0 \end{pmatrix}\]

there is the boosts and rotations we will need

and they obey commutations

\[[A,B]=AB-BA\]

 

 

 

 

 

Edited December 26, 2022 by Mordred

[joigus]

 

51 minutes ago, Genady said:

what I meant was removing the r-dot term of the equation for approximation.

That's essentially what I was proposing. That along with ignoring r times angle-dot for approximantion, as speeds are much smaller than c.

[Genady]

12 minutes ago, joigus said:

 

That's essentially what I was proposing. That along with ignoring r times angle-dot for approximantion, as speeds are much smaller than c.

Right. And that was the proposal I've disagreed and still disagree with as the substitution shows a significantly different outcome (if my algebra is correct, of course).

(Ignoring the angle-dot is fine. I am talking only about the free-falling r-dot.)

[joigus]

4 minutes ago, Genady said:

Right. And that was the proposal I've disagreed and still disagree with as the substitution shows a significantly different outcome (if my algebra is correct, of course).

Are you not happy with the fact that \( 2\times10^{-22} \) is much less than \( 2.4\times10^{-9} \)?

Your algebra might be correct, but inspection of Schwarzschild's metric tells you how small objects move somewhere near an approximately spherical, static source of gravitation. I know my approximation to be correct to 1st order in r_s/r, as well as being the usual one. So I think I'm right. Centrifugal terms are quadratic in velocities, so negligibly small in comparison with r-dependent term. c² is already accounted for in r_s. For Earth it gives about .88 cm which is much more sizeable than so-called "kinematic" corrections.

I'll keep thinking about it as soon as I can though.

2 hours ago, Genady said:

(v/c)² = (1-r_s/r)²(r_s/r)

Where do you get this from?

[Mordred]

With the rotations and boosts above of the Lorentz tranformations we can now examine the different clocks under proper acceleration which I will detail below for each case. (I will need to set one observer at CoM instead of sea level for simplicity)

four velocity

\[u^\mu=\frac{dx^\mu}{d\tau}=(c\frac{dx}{d\tau},\frac{dx}{d\tau},\frac{dy}{d\tau},\frac{dz}{d\tau})\]

\[A=\frac{du}{d\tau}\]

without going through all the relations a constant acceleration along the x direction with constant acceleration g gives the following

\[c\frac{dt}{d\tau}=u^0,\frac{dx^1}{d\tau}=u^1,\frac{du^0}{d\tau}=a^0,\frac{du^1}{d\tau}=a^1\]

\[a^\mu a_\nu=-(a^0)^2+(a^1)^2=g^2\]

gives two solutions

\[a^0=\frac{g}{c}u^1,,a^1=\frac{g}{c}u^0\]

from which

\[\frac{da^0}{d\tau}=\frac{g}{c}\frac{du^1}{d\tau}=\frac{g}{c}a^1=\frac{g^2}{c^2}u^0\]

gives solution between observer A and B (A set at CoM, B falling observer)

\[u^1=Ae^{g\tau/c}+Be^{g\tau/c}\]

hence

\[\frac{du^1}{d\tau}=\frac{g}{c}(Ae^{g\tau/c}-Be^{g\tau/c}\]

hence 

\[x=\frac{c^2}{g}cosh(\frac{g\tau}{c}),, ct=\frac{ct}{g}sinh(\frac{g\tau}{c})\]

space and time coordinates fall onto the hyperbola

\[x^2-c^2t^2=\frac{c^4}{g^2}\]

\[\frac{dx}{d\tau}=c \sinh(\frac{a\tau}{c})\]

\[\tau=\frac{c}{a}\sinh^{-1}\]

there is your proper time under constant acceleration for the falling clock

Lewis Ryder pages Introductory to General Relativity pages 23 to 28 even thoug its from the twin paradox solution the Hyperbolic rotation is identical in the linear acceleration case 

https://en.wikipedia.org/wiki/Lorentz_transformation

https://en.wikipedia.org/wiki/Hyperbolic_functions

 

 

 

Edited December 26, 2022 by Mordred