Time Dilation On An Angularly Accelerating Earth

The spacetime interval, $ds^2$ in general relativity can be given in terms of the metric $g_{\mu\nu}$ as:

$ds^2=g_{\mu\nu}dx^{\mu}dx^{\nu}$ It can be shown that the spacetime interval expressed in co-ordinates such that the particle is at instantaneous rest is equal to $c^2d\tau^2$ where $\tau$ is the proper time (time measured by the particle).So:

$c^2d\tau^2=g_{\mu\nu}dx^{\mu}dx^{\nu}$ Working in spherical polar co-ordinates, the metric for minkowski flat space can be written as a diagonal metric with components:

$g_{tt},g_{rr},g_{\theta\theta},g_{\phi\phi}$

However as the Earth is rotating about it's equator, the $\phi$ coordinate will be time dependent and hence the metric fully describing a rapidly rotating Earth includes $g_{t\phi}$ and $g_{\phi t}$ , however the metric can be made diagonal WLOG, so $g_{t\phi}=g_{\phi t}$

Therefore, following Einstein summation notation:

$c^2d\tau^2=g_{\mu\nu}dx^{\mu}dx^{\nu} = g_{tt}dx^{t}dx^{t}+g_{rr}dx^{r}dx^{r}+g_{\theta\theta}dx^{\theta}dx^{\theta}+,g_{\phi\phi}dx^{\phi}dx^{\phi} + g_{t\phi}dx^{t}dx^{\phi}+g_{\phi t}dx^{\phi}dx^{t}$

Since the last two components are equal as the metric is diagonal $c^2d\tau^2= g_{tt}dx^{t}dx^{t}+g_{rr}dx^{r}dx^{r}+g_{\theta\theta}dx^{\theta}dx^{\theta}+,g_{\phi\phi}dx^{\phi}dx^{\phi} + 2g_{t\phi}dx^{t}dx^{\phi}$

$c^2d\tau^2= g_{tt}(dx^{t})^2+g_{rr}(dx^{r})^2+g_{\theta\theta}(dx^{\theta})^2+,g_{\phi\phi}(dx^{\phi})^2 + 2g_{t\phi}dx^{t}dx^{\phi}$

In spherical polars, it can be shown: $dx^{t}=cdt,dx^{r}=dr,dx^{\theta}=d\theta,dx^{\phi}=d\phi$

Hence: $c^2d\tau^2= g_{tt}(cdt)^2+g_{rr}(dr)^2+g_{\theta\theta}(d\theta)^2+,g_{\phi\phi}(d\phi)^2 + 2g_{t\phi}(cdtd\phi)$

Rewriting this as a sum multiplied by $(dt)^2$

$c^2d\tau^2= \left(c^2g_{tt}+g_{rr}\left(\frac{dr}{dt}\right)^2+g_{\theta\theta}\left(\frac{d\theta}{dt}\right)^2+g_{\phi\phi}\left(\frac{d\phi}{dt}\right)^2 + 2cg_{t\phi}\left(\frac{d\phi}{dt}\right)\right)(dt)^2$

$\frac{dr}{dt}$ is radial velocity $=v$ , $\frac{d\theta}{dt}$ is angular velocity about the colatitude $=\omega_\theta$ , $\frac{d\phi}{dt}$ is the angular velocity about the longitude $=\omega_\phi$ . So substituting this into the spacetime interval:

$c^2d\tau^2= \left(c^2g_{tt}+g_{rr}v^2+g_{\theta\theta}\omega_\theta^2+g_{\phi\phi}\omega_\phi^2 + 2cg_{t\phi}\omega_\phi\right)(dt)^2$

As the orbit is circular, radial velocity is 0. As the orbit is about the equator, $\omega_\theta=0$ . Therefore the spacetime interval can be rewritten as

$c^2d\tau^2= \left(c^2g_{tt}+g_{\phi\phi}\omega_\phi^2 + 2cg_{t\phi}\omega_\phi\right)(dt)^2$

It can now be seen that for Alice and Bob, the ratio of the time's they measure can be found by taking a ratio of their spacetime intervals

$\frac{c^2d\tau^2_{A}}{c^2d\tau^2_{B}} =\frac{d\tau^2_{A}}{d\tau^2_{B}} = \frac{\left(c^2g_{ttA}+g_{\phi\phi A}\omega_{\phi A}^2 + 2cg_{t\phi A}\omega_{\phi A}\right)(dt)^2}{\left(c^2g_{ttB}+g_{\phi\phi B}\omega_{\phi B}^2 + 2cg_{t\phi B}\omega_{\phi B}\right)(dt)^2}$

$\frac{d\tau^2_{A}}{d\tau^2_{B}} = \frac{c^2g_{ttA}+g_{\phi\phi A}\omega_{\phi A}^2 + 2cg_{t\phi A}\omega_{\phi A}}{c^2g_{ttB}+g_{\phi\phi B}\omega_{\phi B}^2 + 2cg_{t\phi B}\omega_{\phi B}}$

Now we need to figure out what $\omega_{\phi}$ is, note that Bob is sat on the surface of the Earth, hence $\omega_{\phi B}$ is just the angular velocity of the Earth's rotation, hence $\omega_{\phi B} = \frac{2\pi}{T}$ where T is the time period of the Earth. $\omega_{\phi A}$ is slightly more tricky. As Alice is orbiting, she is in free-fall and hence travelling along a geodesic.

The geometric Lagrangian is defined as: $L=g_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu}$ Where $\dot{x} = \frac{dx}{ds}$

Therefore, following Einstein summation notation:

$L = g_{tt}\dot{x}^{t}\dot{x}^{t}+g_{rr}\dot{x}^{r}\dot{x}^{r}+g_{\theta\theta}\dot{x}^{\theta}\dot{x}^{\theta}+g_{\phi\phi}\dot{x}^{\phi}\dot{x}^{\phi}+g_{t\phi}\dot{x}^{t}\dot{x}^{\phi}+g_{\phi t}\dot{x}^{\phi}\dot{x}^{t}$

$L = g_{tt}(\dot{x}^{t})^2+g_{rr}(\dot{x}^{r})^2+g_{\theta\theta}(\dot{x}^{\theta})^2+g_{\phi\phi}(\dot{x}^{\phi})^2+2g_{t\phi}\dot{x}^{t}\dot{x}^{\phi}$

Since $dx^{t}=cdt,dx^{r}=dr,dx^{\theta}=d\theta,dx^{\phi}=d\phi$

$\dot{x}^{t}=c\dot{t},\dot{x}^{r}=\dot{r},\dot{x}^{\theta}=\dot{\theta},\dot{x}^{\phi}=\dot{\phi}$

Hence

$L = c^2g_{tt}\dot{t}^2+g_{rr}\dot{r}^2+g_{\theta\theta}\dot{\theta}^2+g_{\phi\phi}\dot{\phi}^2+2cg_{t\phi}\dot{t}\dot{\phi}$

Using the Euler-Lagrange equation for r and the fact that the metric components don't depended on the derivatives of the coordinates:

$\frac{d}{ds}\frac{dL}{d\dot{r}}=\frac{dL}{dr}$ $2\frac{d}{ds}(g_{rr}\dot{r})=\frac{dL}{dr}$ $2(\dot{r}\frac{d}{ds}g_{rr}+g_{rr}\frac{d}{ds}\dot{r})=\frac{dL}{dr}$ $2(\dot{r}\frac{d}{ds}g_{rr}+g_{rr}\ddot{r})=\frac{dL}{dr}$

For a circular orbit r is constant, hence $\dot{r}=\ddot{r}=0$

$0=\frac{dL}{dr}$ $0=\frac{dL}{dr}=c^2\dot{t}^2\frac{d}{dr}g_{tt}+\dot{r}^2\frac{d}{dr}g_{rr}+\dot{\theta}^2\frac{d}{dr}g_{\theta\theta}+\dot{\phi}^2\frac{d}{dr}g_{\phi\phi}+2c\dot{t}\dot{\phi}\frac{d}{dr}g_{t\phi}$

As $\dot{r}=0$

$0=c^2\dot{t}^2\frac{d}{dr}g_{tt}+\dot{\theta}^2\frac{d}{dr}g_{\theta\theta}+\dot{\phi}^2\frac{d}{dr}g_{\phi\phi}+2c\dot{t}\dot{\phi}\frac{d}{dr}g_{t\phi}$

Rewrite as a sum multiplied by $\dot{t}^2$

$0=\left(c^2\frac{d}{dr}g_{tt}+\left(\frac{\dot{\theta}}{\dot{t}}\right)^2\frac{d}{dr}g_{\theta\theta}+\left(\frac{\dot{\phi}}{\dot{t}}\right)^2\frac{d}{dr}g_{\phi\phi}+2c\frac{\dot{\phi}}{\dot{t}}\frac{d}{dr}g_{t\phi}\right)\dot{t}^2$

$0=\left(c^2\frac{d}{dr}g_{tt}+\omega_{\phi}^2\frac{d}{dr}g_{\phi\phi}+2c\omega_{\phi}\frac{d}{dr}g_{t\phi}\right)\dot{t}^2$

As the rate the Earth is being angularly accelerated is very low and the orbit is equatorial, due to symmetry conditions $g_{tt}$ and $g_{\theta\theta}$ are the same as the schwarzschild metric ( The symmetry condition for $g_{\theta\theta}$ due to Alice orbiting along the equator, hence there is no variation in $\theta$ over her journey, and the symmetry condition for $g_{tt}$ being the acceleration of the Earth is very low, therefore after a particle completes an orbit a reference frame transformation can be made to make the metric appear identical to the previous orbit )

So $g_{tt}=1-\frac{R_S}{r}, g_{\theta\theta} = -r^2$ (Schwarzschild metric)

Therefore:

$0=\left(c^2\frac{d}{dr}(1-\frac{R_S}{r})+\omega_{\phi}^2\frac{d}{dr}g_{\phi\phi}+2c\omega_{\phi}\frac{d}{dr}g_{t\phi}\right)\dot{t}^2$

$0=\left(c^2\frac{R_S}{r^2}+\omega_{\phi}^2\frac{d}{dr}g_{\phi\phi}+2c\omega_{\phi}\frac{d}{dr}g_{t\phi}\right)\dot{t}^2$

$\dot{t}$ is never 0, hence

$0=c^2\frac{R_S}{r^2}+\omega_{\phi}^2\frac{d}{dr}g_{\phi\phi}+2c\omega_{\phi}\frac{d}{dr}g_{t\phi}$

Attempt to solve the problem in the Schwarzschild metric, then check if answer involves an Earth rotating at a sufficient fraction of the speed of light to require the use of the Kerr metric, in the Schwarzschild metric $g_{\phi\phi}=-r^2,g_{t\phi}=0$ , to save time if the Kerr metric needs to be used later, rewrite $g_{\phi\phi}=-(r^2+K_1),g_{t\phi}=K_2$ , where $K_1$ and $K_2$ are the Kerr terms that differ from the Schwarzschild metric, where $R_S = \frac{2GM}{c^2}$ is the Schwarzschild radius

$0=c^2\frac{R_S}{r^2}-\omega_{\phi}^2\frac{d}{dr}(r^2+K_1)+2c\omega_{\phi}\frac{d}{dr}K_2$

$0=c^2\frac{R_S}{r^2}-\omega_{\phi}^2(2r+\frac{d}{dr}K_1)+2c\omega_{\phi}\frac{d}{dr}K_2$

$0=\omega_{\phi}^2\left(2r+\frac{d}{dr}K_1\right)-\omega_{\phi}\left(2c\frac{d}{dr}K_2\right)-c^2\frac{R_S}{r^2}$

Solve for $\omega_{\phi}$

$\omega_{\phi} = \frac{\left(2c\frac{d}{dr}K_2\right)\pm\sqrt{\left(2c\frac{d}{dr}K_2\right)^2+4\left(2r+\frac{d}{dr}K_1\right)c^2\frac{R_S}{r^2}}}{2\left(2r+\frac{d}{dr}K_1\right)}$

$\omega_{\phi} = \frac{\left(2c\frac{d}{dr}K_2\right)\pm\sqrt{4c^2\left(\frac{d}{dr}K_2\right)^2+4\left(2r+\frac{d}{dr}K_1\right)c^2\frac{R_S}{r^2}}}{2\left(2r+\frac{d}{dr}K_1\right)}$

$\omega_{\phi} = \frac{\left(2c\frac{d}{dr}K_2\right)\pm\sqrt{4c^2\left(\left(\frac{d}{dr}K_2\right)^2+\left(2r+\frac{d}{dr}K_1\right)\frac{R_S}{r^2}\right)}}{2\left(2r+\frac{d}{dr}K_1\right)}$

$\omega_{\phi} = c\left(\frac{\left(\frac{d}{dr}K_2\right)\pm\sqrt{\left(\left(\frac{d}{dr}K_2\right)^2+\left(2r+\frac{d}{dr}K_1\right)\frac{R_S}{r^2}\right)}}{\left(2r+\frac{d}{dr}K_1\right)}\right)$

As we are comparing Bob on the surface and Alice in orbit, it may be more convenient to redefine Alice's radial distance, r, as $nR$ where R is the the radius of the Earth, and hence n is the number of Earth radii Alice is from the centre of the Earth. Hence for Bob $n=1$ .

$\omega_{\phi} = c\left(\frac{\left(\frac{1}{R}\frac{d}{dn}K_2\right)\pm\sqrt{\left(\left(\frac{1}{R}\frac{d}{dn}K_2\right)^2+\left(2nR+\frac{1}{R}\frac{d}{dn}K_1\right)\frac{R_S}{n^2R^2}\right)}}{\left(2nR+\frac{1}{R}\frac{d}{dn}K_1\right)}\right)$

(For the Schwarzschild case $K_1=K_2=0$ , so $\omega_{\phi}= c\frac{\sqrt{2\frac{R_S}{nR}}}{2nR} = \sqrt{\frac{GM}{n^3R^3}}$ , the same as the Newtonian case)

So going back to our ratio of times measured by Alice and Bob, and substituting in our metric:

$\frac{d\tau^2_{A}}{d\tau^2_{B}} = \frac{c^2g_{ttA}+g_{\phi\phi A}\omega_{\phi A}^2 + 2cg_{t\phi A}\omega_{\phi A}}{c^2g_{ttB}+g_{\phi\phi B}\omega_{\phi B}^2 + 2cg_{t\phi B}\omega_{\phi B}}$

$\frac{d\tau^2_{A}}{d\tau^2_{B}} = \frac{c^2\left(1-\frac{R_S}{nR}\right)-(n^2R^2+K_{1A})\omega_{\phi A}^2 + 2cK_{2A}\omega_{\phi A}}{c^2\left(1-\frac{R_S}{R}\right)-(R^2+K_{1B})\omega_{\phi B}^2 + 2cK_{2B}\omega_{\phi B}}$

So now we can just re-arrange and square root to get

$d\tau_{A} = d\tau_{B}\sqrt{\frac{c^2\left(1-\frac{R_S}{nR}\right)-(n^2R^2+K_{1A})\omega_{\phi A}^2 + 2cK_{2A}\omega_{\phi A}}{c^2\left(1-\frac{R_S}{R}\right)-(R^2+K_{1B})\omega_{\phi B}^2 + 2cK_{2B}\omega_{\phi B}}}$

$\Delta\tau_{A} = \Delta\tau_{B}\sqrt{\frac{c^2\left(1-\frac{R_S}{nR}\right)-(n^2R^2+K_{1A})\omega_{\phi A}^2 + 2cK_{2A}\omega_{\phi A}}{c^2\left(1-\frac{R_S}{R}\right)-(R^2+K_{1B})\omega_{\phi B}^2 + 2cK_{2B}\omega_{\phi B}}}$

As $\Delta\tau$ is the change in time measured by the observer, we can see for every second Bob measures, Alice measures that many seconds multiplied by the square root term. So to work out the gain of microseconds Alice measures over Bob every 24 hours, simply set $\Delta\tau_{B}=86400$ seconds (24 hours) to get the number of seconds measured by Alice, then subtract 86400 from that to see how many seconds were gained, then multiply by $10^{-6}$ to convert to microseconds.

$\Delta\tau_{A} = \left(86400\sqrt{\frac{c^2\left(1-\frac{R_S}{nR}\right)-(n^2R^2+K_{1A})\omega_{\phi A}^2 + 2cK_{2A}\omega_{\phi A}}{c^2\left(1-\frac{R_S}{R}\right)-(R^2+K_{1B})\omega_{\phi B}^2 + 2cK_{2B}\omega_{\phi B}}}-86400\right)10^{-6}$

The rest of the solution is just tedious algebra, substitute in $\omega_{\phi A}$ and $\omega_{\phi B}$ as derived above and solve for n in the case $\Delta\tau_{A}=30*10^{-6}$ to figure out the height of the satellite (You'll find it's n = 3, just slightly below GPS satellites)

With this you can then solve the case when the gain is 1 second per 24 hours when n=3. $\omega_{\phi A}$ will stay the same as it does not depend on the angular velocity of the Earth, but $\omega_{\phi B}$ does, since Bob is stood on the surface of the Earth his angular velocity is the same as the Earth's so $\omega_{\phi B} = \frac{2\pi}{T}$ . You can then use this to solve for T in this case, to get the time period the Earth will have to be accelerated to. You'll find that T<28 seconds, meaning that the surface of the Earth is travelling at 0.4\% of the speed of light. This is hence low enough that the effect due to the Kerr metric is negligible.

To work out the number of years Charles would measure, simply note that T initially is 86400, since it decreases by 1 per year and it needs to decrease to less than 28 as after 28 years it won't quite of reached it, but after 27 years it will of, so just subtract 27 from 86400 = 86373 to get number of years Charles measures.

To solve this without the tedious algebra, I found it simpler to just use the programming software MATLAB, to make an animation of the solution (just by plotting the time gained function against orbital radius and iterating over T from 86400 to 0), which is shown here

The MATLAB code to make this is shown below:

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clear

writerObj = VideoWriter('C:\Users\Jack\Pictures\Brilliantorg.avi');
writerObj.Quality = 100;
writerObj.FrameRate = 60;
open(writerObj); 

c = 299792458;
Days = 1;
End = 3;
dv = End/100;
v1 = [-End:dv:End];
v1 = 0;
format long
nstart = 1;
nend=10;
dn = (nend-nstart)/50;
T = 86400;
G = 6.67408E-11;
M = 5.9722E28;
c = 299792458;
Rs=2*G*M/c^2;
r = 6378137 ;
Tstart= 86400;
T = Tstart;
Step = 0;
while T> 0.170038470439427
j=0;
Step = Step + 1;
T = T - T/100
Period(Step) = T ;
if T < 0.170038470439427
    break
end
for n=nstart:dn:nend
    j=j+1;
sigma = (1 - Rs/r);
sigmaOrb = (1 - Rs./(n.*r));
Kgoo = -(r.^2+((4*pi^2/c^2).*r.^4./(T.^2)).*(1+Rs./(r)));
KgooOrb = -(n.^2.*r.^2+((4*pi^2/c^2).*r.^4./(T.^2)).*(1+Rs./(n.*r)));
Sgoo = - r^2;
SgooOrb = - (n.*r).^2;

w1 = ((2*pi./T) + v1/r);
worb =((2*pi*Rs./(T.*n.^2))+sqrt(2*c^2*Rs./(n.*r)))./((4*pi^2*Rs)./(c^2.*T.^2.*r.^2) -2.*n.*r);
WschwarzOrb = (c*sqrt(2*Rs./(n.*r)))/(-2.*n.*r);

v1T = (2*pi./T)*r + v1;
vOrbT = worb.*n.*r;

gtphi1 = +(4*pi./T).*r.*Rs;
gtphiOrb = +(4*pi./T).*(r./n)*Rs;

TKerrSurf1 = sigma.*c^2 + Kgoo .*w1.^2 + gtphi1.*w1;
TKerrSurfOrb = sigmaOrb.*c^2 + KgooOrb .*worb.^2 + gtphiOrb.*worb;

TSchwarzSurf1 = sigma.*c^2 + Sgoo .*w1.^2;
TSchwarzSurfOrb = sigmaOrb.*c^2 + SgooOrb .*WschwarzOrb.^2;

TMinkSurf1 = c^2 - v1T.^2;
TMinkSurfOrb = c^2 - vOrbT.^2;

AKerrOrbDilation = ((sqrt(TKerrSurfOrb./TKerrSurf1)*86400*Days)-(86400*Days))*10^6;
ASchwarzOrbDilation = ((sqrt(TSchwarzSurfOrb./TSchwarzSurf1)*86400*Days)-(86400*Days))*10^6;
AMinkowskiOrbDilation = ((sqrt(TMinkSurfOrb./TMinkSurf1)*86400*Days)-(86400*Days))*10^6;

y1(j) = AKerrOrbDilation;
y2(j) = ASchwarzOrbDilation;
y3(j) = AMinkowskiOrbDilation;
end
f = figure('visible','off');
hold on
n = [nstart:dn:nend];
axis([nstart,nend,y1(1),y1(end)])
plot(n,y1)
plot(n,y2)
plot(n,y3)
GPS =3.163933293;
line([GPS GPS], get(gca, 'ylim'));
x1=[-1000:1:-999]; 
plot(x1,x1)

 TimePeriod = num2str(T);
s1 = 'A day is ';
s2 = ' seconds long';
s = strcat(s1,TimePeriod,s2);
xlabel('Orbital Radius,Earth Radii')
ylabel('TimeDilation,Microseconds Per Day')
legend('Kerr','Schwarzschild','Minkowski','GPS',s)
frame = getframe(gcf); 
writeVideo(writerObj, frame);
end
close(writerObj);