Question of three clocks

[Genady]

57 minutes ago, joigus said:

3 hours ago, Genady said:

(v/c)² = (1-r_s/r)²(r_s/r)

Where do you get this from?

From here, among many other similar sources: general relativity - Will an object always fall at an infinite speed in a black hole? - Physics Stack Exchange

The link is to the Wikipedia: Schwarzschild geodesics - Wikipedia

[Mordred]

Continuing from here

 

for the orbiting Observer to Observer at CoM for simplicity. Once again will be using the Lewis Ryder reference in the previous post above.

were going to set the orbital rotation along the z axis in cylindrical coordinates

\[ds^2=-c^2dt^2+dr^2+rd\phi^2+dz^2\]

\[\acute{t}=t, \acute{r}=r, \acute{\phi}=\phi-\omega t, \acute{z}=z\]

\[\acute{ds}^2=-ct^2d\acute{t}^2+\acute{r}^2(d\acute{\phi}+\omega d\acute{t})^2+d\acute{z}^2\]

\[=-(c-\omega^2\acute{r}^2)d\acute{t}^2+2\omega\acute{r}^2d\acute{\phi}^2 d\acute{t}^2+d\acute{r}^2+\acute{r}^2d\acute{\phi}^2+d\acute{z}2\]

dropping the primes we have the invariant spacetime in a rotating frame

\[ds^2=-(c^2\omega^2 r^2)dt^2+2\omega r^2d\phi^2 dt+dr^2+r^2d\phi^2+dz^2\]

this becomes

\[ds^2=g_{\mu\nu}dx^\mu d^\nu=g_{00}(dx^0)^2+g_{0i}dx^0dx^j+g_{ik}dx^i dx^k\]

ijk sums over 1-3 where

\[x^\mu=(x^0,x^1,x^2,x^3)=(ct,r,\phi,z)g_{\mu\nu}\]

\[g_{\mu\nu}=\begin{pmatrix}-(1-\frac{\omega^2 r^2}{c^2}&0&\frac{\omega r^2}{c}&0\\0&1&0&0\\\frac{\omega r^2}{c}&0&r^2&0\\0&0&0&1\end{pmatrix}\]

time interval between events

\[ds^2=-c^2d\tau^2\] as world time not proper time

relation between world time and proper time is given by

\[d\tau^2\sqrt{-g_{00}}dt\]

giving in the above case

\[d\tau=\sqrt{1-\frac{\omega^2 r^2}{c^2}}dt\]

lengthy process but hope that helps better understand the 3 clock scenario described by the OP... all the material as mentioned is from Lewis Ryder "Introductory to Relativity"

Sagnac effect as shown above further detail here

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

I'm also considering applying the Kerr metric to the rotating case to place observer at sea level instead of CoM have to think about that though

 

 

 

Edited December 26, 2022 by Mordred

[Lorentz Jr]

On 12/24/2022 at 3:31 PM, mistermack said:

will clock C be running at the same speed as clock A, or clock B, or will it be slower than both?

C will start out at the same rate as B, because both clocks are moving slowly and far away from Earth.

From B's perspective, C will slow down as it falls, according to the Schwarzschild metric:

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

(with thanks to @joigus for the TeX 🙂)

dτ is called an object's "proper time". That's the amount of time that passes for A or C (from B's perspective) when a time dt passes for B (assuming B is moving slowly and is very far away).

As you can see, the dr term is negative (because [math]r>r_s[/math]), so C will be running more slowly than A as it falls past A (because r is changing for C).

The [math]d\phi[/math] term is also negative, so a clock D orbiting at the same altitude as A will also run more slowly (because [math]\phi[/math] is changing for D).

C will be slower than D because (a) it's moving faster (more total energy - same potential = more kinetic, and the distance D travels is [math]ds = r d\phi[/math] < dr) and (b) the dr term has a larger coefficient than the [math](rd\phi)[/math] term.

Edited December 26, 2022 by Lorentz Jr

[Genady]

50 minutes ago, Lorentz Jr said:

C will start out at the same rate as B, because both clocks are moving slowly and far away from Earth.

From B's perspective, C will slow down as it falls, according to the Schwarzschild metric:

c2dτ2=c2(1−rsr)dt2−(1−rsr)−1dr2−r2dϕ2

(with thanks to @joigus for the TeX 🙂)

dτ is called an object's "proper time". That's the amount of time that passes for A or C (from B's perspective) when a time dt passes for B (assuming B is moving slowly and is very far away).

As you can see, the dr term is negative (because r>rs ), so C will be running more slowly than A as it falls past A (because r is changing).

The dϕ term is also negative, so a clock D orbiting at the same altitude as A will also run more slowly (because ϕ is changing).

C will be slower than D because it's moving faster (more total energy - same potential = more kinetic, and the distance D travels is ds=rdϕ < dr) and the dr term has a larger coefficient.

Right. This gives us comparison of "tick rates", i.e., the dτ's, from the B's perspective, i.e., per one "tick" of the clock B. 

I think, if we write this equation for, say, dτ_A, solve it fordt , and substitute into the equations for dτ_B etc., then we can compare "tick rates" of the clocks from the A's perspective. And so on.   

[Mordred]

25 minutes ago, Genady said:

Right. This gives us comparison of "tick rates", i.e., the dτ's, from the B's perspective, i.e., per one "tick" of the clock B. 

I think, if we write this equation for, say, dτ_A, solve it fordt , and substitute into the equations for dτ_B etc., then we can compare "tick rates" of the clocks from the A's perspective. And so on.   

hopefully the proper acceleration example I provided aids you both in this GL I'm looking into finding a method to handle the elliptical orbit to have an observer at sea level while an observer is in orbit. I have found a couple of Kerr metric examples in elliptical orbit still examining them to see if they will be useful.

currently examining this one

https://arxiv.org/pdf/0903.3684.pdf

Edited December 26, 2022 by Mordred

[Genady]

9 minutes ago, Mordred said:

I'm looking into finding a method to handle the elliptical orbit to have an observer at sea level while an observer is in orbit.

Why did you pick this particular situation?

[Mordred]

I

4 minutes ago, Genady said:

Why did you pick this particular situation?

I like a good challenge lol and as the Earth also orbits the observer at sea level has numerous interesting effects though most are negligible in the case of the earth I am curious how to handle situations not so trivial

Edited December 26, 2022 by Mordred

[Genady]

6 minutes ago, Mordred said:

I

I like a good challenge lol and as the Earth also orbits the observer at sea level has numerous interesting effects though most are negligible in the case of the earth I am curious how to handle situations not so trivial

I see.

[Mordred]

lol its how I test myself for improvement. If I can help  others also learn then bonus. One thing about any physics theory you never know enough 

[Genady]

4 minutes ago, Mordred said:

lol its how I test myself for improvement. If I can help  others also learn then bonus. One thing about any physics theory you never know enough 

I understand and "approve" the first two points. I don't think I understand the third one. "Enough"?

[Mordred]

It simply means there is always something new to learn and if your dedicated in your studies you never stop trying to learn new aspects of a given theory.

from a preliminary quick research it seems to me applying the Carter Constant for the elliptical orbit may give me a methodology. I wonder if Markus or Jadus agrees with that approach

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

there we go I think I have an applicable method

https://en.wikipedia.org/wiki/Boyer–Lindquist_coordinates#Spin_connection

 

edit speaking of challenges I wonder at the steps to get get proper time for an observer at infinity to a clock orbiting Earth with the subsequent transverse blue/redshifts.

Edited December 26, 2022 by Mordred

[joigus]

6 hours ago, Genady said:

From here, among many other similar sources: general relativity - Will an object always fall at an infinite speed in a black hole? - Physics Stack Exchange

The link is to the Wikipedia: Schwarzschild geodesics - Wikipedia

OK. We're not talking about black holes. We're talking about the Earth. We're talking satellites orbiting the Earth. Schwarzschild's solution is not just about BH's. It's what any static, spherical distribution of mass looks like far enough away from it. We're talking more like somewhere between 6000 and 7000 Km from a horizon that, BTW, doesn't exist for the Earth. Initial velocities can, only too obviously, take any value we want, not related to r. The formula that you're using clearly comes from some assumption of initial conditions that's not compatible with the case we're discussing, and in particular implies radial approach. It's only too obviously not valid in general. Certainly not valid for rockets moving in an arbitrary way, for example. Make some substitutions and it will be transparent to you.

You can't take any velocity-radius dependence, or whatever other similar dependence used for a very particular problem from an article or book --in the case you took it from it's clearly a radial free fall--, and plug it into a completely differently motivated physical problem.

The method I used is described in such classics as Lightman's Problem Book in Relativity and Gravitation. It's clearly thought out and reasoned from the start. You may doubt its validity only because it doesn't allow you to separate in your mind the "kinematic" effects and the 1/r effect. But it's known to work, because it's derived from first principles and the approximations I have justified at every step. Essentially: Weak fields (but not so weak that all GR effects disappear) and slow velocities, which is an essential part of the discussion. I understand that @mistermack's clocks are moving much more slowly than the speed of light.

Edited December 27, 2022 by joigus
minor correction

[Genady]

Yes, I assume a radial free fall of C from B to A.

No, the formula is not specific for black holes. It is general formula for the Schwarzschild's metric.  

[joigus]

3 hours ago, Mordred said:

lol its how I test myself for improvement. If I can help  others also learn then bonus. One thing about any physics theory you never know enough 

I'm with you a 100 %.

8 minutes ago, Genady said:

It is general formula for the Schwarzschild's metric.  

No, it's not. Or the word "general" means something completely different from what I thought.

Edited December 27, 2022 by joigus
minor correction

[Genady]

39 minutes ago, joigus said:

Or the word "general" means something completely different from what I thought.

I mean that this formula describes velocity of a radially free falling body that starts at rest at infinity, in a Schwarzschild metric with the given r_s , at distance r from the center, in the coordinates of the far away observer. 

I don't know why you say it does not.

Edited December 27, 2022 by Genady

[Lorentz Jr]

It sounds like @Genady just means the solution applies for a central body of any mass. Planet, star, black hole, whatever. Not general in terms of initial conditions, but it's what the OP was asking about (clock C).

Edited December 27, 2022 by Lorentz Jr

[joigus]

Yes, I had to go back to @Genady's comment, and then further back to OP. He meant clock C. If you use the radial escape velocity relation you do get twice the r_s/r correction. Interesting...

Thanks both.

More discussion tomorrow. I'm tired and it's very late here.

[mistermack]

Say all three clocks are synchronised to an agreed event in the oldest galaxy visible. And all three are set to count down 1000 years and then explode. 

If C passes A at the instant A explodes, will C already be in bits, or still be in one piece, or explode at the same instant as A? 

(ignoring complications like orbits etc.)

[joigus]

15 minutes ago, mistermack said:

or explode at the same instant as A? 

 

According to whom? "Explode at the same instant" is a frame-dependent occurrence. As @MigL and @Markus Hanke pointed out, it's important to distinguish between frame-dependent and frame-independent.

[Genady]

49 minutes ago, mistermack said:

Say all three clocks are synchronised to an agreed event in the oldest galaxy visible. And all three are set to count down 1000 years and then explode. 

If C passes A at the instant A explodes, will C already be in bits, or still be in one piece, or explode at the same instant as A? 

(ignoring complications like orbits etc.)

C will be already in bits.

[joigus]

On 12/26/2022 at 9:15 PM, Mordred said:

Continuing from here

 

for the orbiting Observer to Observer at CoM for simplicity. Once again will be using the Lewis Ryder reference in the previous post above.

were going to set the orbital rotation along the z axis in cylindrical coordinates

 

ds2=−c2dt2+dr2+rdϕ2+dz2

 

 

t´=t,r´=r,ϕ´=ϕ−ωt,z´=z

 

 

ds´2=−ct2dt´2+r´2(dϕ´+ωdt´)2+dz´2

 

 

=−(c−ω2r´2)dt´2+2ωr´2dϕ´2dt´2+dr´2+r´2dϕ´2+dz´2

 

dropping the primes we have the invariant spacetime in a rotating frame

 

ds2=−(c2ω2r2)dt2+2ωr2dϕ2dt+dr2+r2dϕ2+dz2

 

this becomes

 

ds2=gμνdxμdν=g00(dx0)2+g0idx0dxj+gikdxidxk

 

ijk sums over 1-3 where

 

xμ=(x0,x1,x2,x3)=(ct,r,ϕ,z)gμν

 

 

gμν=⎛⎝⎜⎜⎜⎜⎜−(1−ω2r2c20ωr2c00100ωr2c0r200001⎞⎠⎟⎟⎟⎟⎟

 

time interval between events

 

ds2=−c2dτ2

as world time not proper time

 

relation between world time and proper time is given by

 

dτ2−g00−−−−√dt

 

giving in the above case

 

dτ=1−ω2r2c2−−−−−−−−√dt

 

lengthy process but hope that helps better understand the 3 clock scenario described by the OP... all the material as mentioned is from Lewis Ryder "Introductory to Relativity"

Sagnac effect as shown above further detail here

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

I'm also considering applying the Kerr metric to the rotating case to place observer at sea level instead of CoM have to think about that though

 

 

 

Thanks for your comments. Yes, I'm getting similar formulae for rotating satellites in free fall. I haven't thought of "proper" vs "coordinate" accelerations, but I think that's because the generality of the proper-time expression already accounts for that. Because the line element is 1st-order, quadratic in coordinate differences, when substituting general trajectories, the acceleration terms relevant are only quadratic in velocities, rather than 2nd-order time derivatives...

I think you can pull off a precise-enough discussion without entering into effects of rotating Earth --Kerr metric--, as the angular-momentum term for the Kerr metric is J/Mc --with J spin angular momentum of Earth, really small--, but it would be interesting to generalise to that case for further precision.

The thing that got me intrigued is @Genady's comment that consideration of exact expression for radial escape velocity falling from spatial infinity gives twice the correction that I was thinking about, corresponding to clock falling from finite --even if very far away-- distance. I think I'm on my way to understanding this discrepancy. Maybe I can comment on it later if anyone's interested. 

[Genady]

1 hour ago, joigus said:

exact expression for radial escape velocity falling from spatial infinity gives twice the correction that I was thinking about, corresponding to clock falling from finite --even if very far away-- distance. I think I'm on my way to understanding this discrepancy. Maybe I can comment on it later if anyone's interested. 

I am interested. I think this article is directly related although I didn't read it: Free_fall_in_the_Schwarzschild_field.pdf

[Mordred]

6 hours ago, joigus said:

Thanks for your comments. Yes, I'm getting similar formulae for rotating satellites in free fall. I haven't thought of "proper" vs "coordinate" accelerations, but I think that's because the generality of the proper-time expression already accounts for that. Because the line element is 1st-order, quadratic in coordinate differences, when substituting general trajectories, the acceleration terms relevant are only quadratic in velocities, rather than 2nd-order time derivatives...

I think you can pull off a precise-enough discussion without entering into effects of rotating Earth --Kerr metric--, as the angular-momentum term for the Kerr metric is J/Mc --with J spin angular momentum of Earth, really small--, but it would be interesting to generalise to that case for further precision.

 

I agree the earth spins slowly enough that it wouldn't make any significant differences. For the first part if your applying the line elements already derived such as he Schwartzchild metric etc  the coordinates vs proper time are already incorporated. You still have to be aware that those same line elements will return the proper time. Same goes for expressions for proper time already accounting for coordinate time

The commonly used  proper time to coordinate time expression is as follows

\[\Delta \tau =\int \sqrt{1- \frac{1}{c^2} \left ( \left (\frac{dx}{dt}\right)^2+ \left(\frac{dy}{dt}\right)^2+ \left (\frac{dz}{dt}\right)^2\right )}dt\]

which works fine until you have other factors such as frame dragging, rotation or acceleration. For rotation as above

\[d\tau=\sqrt{1-\frac{\omega^2 r^2}{c^2}}dt\]

 

 

Edited December 28, 2022 by Mordred

[Markus Hanke]

22 hours ago, joigus said:

The thing that got me intrigued is @Genady's comment that consideration of exact expression for radial escape velocity falling from spatial infinity gives twice the correction that I was thinking about, corresponding to clock falling from finite --even if very far away-- distance. I think I'm on my way to understanding this discrepancy.

To be honest, I’ve been struggling a little to follow the approach you took in your initial post here - I’m not saying anything is wrong there, it’s just that I don’t fully get it. 

The way I would work out radial free-fall times is by starting at the geodesic equation, in order to obtain an expression for \(\frac{dr}{d\tau}\), and then either integrate this up over an appropriate path, or solve directly for \(\tau\) - this is doable so long as one assumes a purely radial in-fall, so that the pesky angular momentum terms all vanish. Already the geodesic expression automatically leaves me with an additional factor of 2, as compared to your approach here (unless I’m missing something in your formalism, which is possible).

Curious to see how you resolved this

[joigus]

On 12/29/2022 at 8:24 AM, Markus Hanke said:

To be honest, I’ve been struggling a little to follow the approach you took in your initial post here - I’m not saying anything is wrong there, it’s just that I don’t fully get it. 

The way I would work out radial free-fall times is by starting at the geodesic equation, in order to obtain an expression for drdτ , and then either integrate this up over an appropriate path, or solve directly for τ - this is doable so long as one assumes a purely radial in-fall, so that the pesky angular momentum terms all vanish. Already the geodesic expression automatically leaves me with an additional factor of 2, as compared to your approach here (unless I’m missing something in your formalism, which is possible).

Curious to see how you resolved this

Ok. I totally understand what you mean. I've been trying to sketch the proof that you suggest and, yes, that would be the way to go. Even with the simplification of radial fall, the equation is not the simplest to solve, but it can be solved. In particular, it can be integrated to the expression that @Genady proposed --which is shown in many documents online-- with just one additional qualification about signs:

\[ \frac{\dot{r}}{c}=\pm\left(1-\frac{r_{S}}{r}\right)\sqrt{\frac{r_{S}}{r}} \]

This comes from integrating the geodesic equation, so it corresponds to bodies falling radially in Schwarzschild's metric, either approaching or escaping from the source. That it corresponds to very special initial conditions can be seen --by first picking out the minus sign--, and by analysing these limiting cases:

Radial velocity at spatial infinity --dotted expressions meaning "derivative with respect to coordinate time," not with respect to proper time:

\[ \dot{r}_{\infty}=\lim_{r\rightarrow\infty}-\left(1-\frac{r_{S}}{r}\right)\sqrt{\frac{r_{S}}{r}}c=0 \]

Radial velocity at Earth's Schwarzschild radius:

\[ \dot{r}_{r_{S}}=\lim_{r\rightarrow r_{S}}-\left(1-\frac{r_{S}}{r}\right)\sqrt{\frac{r_{S}}{r}}c=-c \]

Radial velocity at Earth's radius:

\[ \dot{r}_{r_{\oplus}}=-\left(1-\frac{r_{S}}{r_{\oplus}}\right)\sqrt{\frac{r_{S}}{r_{\oplus}}}c\simeq-11.144\times10^{3}\textrm{m}\textrm{s}^{-1} \]

which approximately matches Earth's radial escape velocity with a minus sign corresponding to the motion being an approach.

Now, one thing that I think should not be overlooked is the fact that such a trajectory --interpreted as "letting go" of an object with initial zero radial velocity--, is only consistent with doing so from spatial infinity, not from a finite distance. The observation might seem trivial, as far enough away we can disregard the difference to the effect of estimating velocities, but not so clearly --at least not to me-- to the effect of inferring relations between times. There is the possibility that it isn't, as a particle freely falling from infinity would make so with an infinite total proper time, as well as infinite total coordinate time. In any case, if we use the last relation \( \dot{r}\left(r\right) \) for the trajectory, and directly substitute in the expression for the proper time, we get,

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

\[ =\left(1-\frac{r_{S}}{r}\right)^{1/2}\left[1-\frac{r_{S}}{r}+\left(\frac{r_{S}}{r}\right)^{3}\right]^{1/2}dt\simeq\left(1-\frac{r_{S}}{r}\right)dt \]

If, on the other hand, we go directly to the expression of a body falling freely from a radius --no matter how big but finite-- with initial zero velocity, we can go directly to the expression for the proper time in terms of a general radial trajectory \( r\left(t\right)=r_{C}\left(t\right) \), and substitute,

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

and already disregard the term in \( \dot{r}/c \) from the start, because it's never going to reach a significant value in comparison with \( r_{S}/r \). But this produces half the value that's inferred from using the Schwarzschild escape velocity functional relation.

I think my argument that the Shapiro delay between the orbiting object and the object on the Earth surface would be intermediate between both still holds. We want the initial velocity to be zero at a finite distance, not coming from spatial infinity. The difference might look like small potatoes, but it could be non-trivial to the effect of inferring relations between proper times.