What about the Big Ben Paradox?

[Dale]

Then there seems a disconnect: With gamma of 2 and Big Ben running slow ,1/2 of the proper time of the traveling observer. Then how is the clock stated to be still running normally during the revolution around the sun ?

Big Ben is a clock. The orbit of the earth around the sun is a clock. All clocks are time dilated equally.

 

[PeroK]

Then there seems a disconnect: With gamma of 2 and Big Ben running slow ,1/2 of the proper time of the traveling observer. Then how is the clock stated to be still running normally during the revolution around the sun ?

That's the definition of a year in the local reference frame of the solar system.

 

[Nugatory]

This thread is closed.
As with all such thread closures, if there is something else to add you can PM any mentor to ask that it reopened for a new contribution.

 

[Dale]

I wanted to add the actual GR math. The outcome of this is exactly as everyone who has any experience in GR said. Indeed, from first principles it could be no other way. But I had time yesterday to play around with this. All equations are using geometrized units where $c=G=1$.

We start with the weak field metric in cylindrical coordinates: $$ds^2 = (1 - 2 U) dr^2 + (-1 - 2 U) dt^2 + (1 - 2 U) dz^2 + (r^2 - 2 r^2 U) d\phi^2 $$ with the standard gravitational potential in cylindrical coordinates $$ U=-\frac{M}{\sqrt{r^2+z^2}} $$

Now, an object in orbit is in free-fall, so the worldline of the planet is a geodesic. To calculate the orbit of the earth we therefore calculate the geodesic using the equations described here. When we do so, we get the following equations: $$0 = \left(
\begin{array}{c}
0 \\
\frac{2 r^2 \left(z^2-2 M^2\right)
\ddot r+r \left(-z^2 \left(z^2-4
M^2\right) \ \dot \phi ^2-M \dot r^2 \left(2 M+3
\sqrt{r^2+z^2}\right)+M \dot z^2
\left(\sqrt{r^2+z^2}-2 M\right)+M
\left(\sqrt{r^2+z^2}-2
M\right)\right)+z \left(z
\left(z^2-4 M^2\right) \ddot r-4 M
\sqrt{r^2+z^2} \dot r
\dot z\right)+r^3 \ \dot phi^2 \left(M
\left(2 M+\sqrt{r^2+z^2}\right)-2
z^2\right)+r^4 \ddot r+r^5
\left(-\dot \phi
^2\right)}{\left(r^2+z^2\right)
\left(-4 M^2+r^2+z^2\right)} \\
\frac{2 \dot r \dot \phi \left(-4 M^2+r^2
\left(1-\frac{2
M}{\sqrt{r^2+z^2}}\right)+z^2\right)
+r \left(\left(r^2-4 M^2\right) \ddot \phi
-\frac{4 M z \dot z \dot \phi
}{\sqrt{r^2+z^2}}+z^2 \ddot \phi
\right)}{r \left(-4
M^2+r^2+z^2\right)} \\
\frac{r \left(r \left(r^2-4 M^2\right)
\ddot z-4 M \sqrt{r^2+z^2} \dot r
\dot z\right)+2 z^2 \left(r^2-2
M^2\right) \ddot z+M z \left(\dot r^2
\left(\sqrt{r^2+z^2}-2 M\right)-\dot z^2
\left(2 M+3
\sqrt{r^2+z^2}\right)+\left(\sqrt{r^
2+z^2}-2 M\right) \left(r^2 \dot \phi
^2+1\right)\right)+z^4
\dot z}{\left(r^2+z^2\right) \left(-4
M^2+r^2+z^2\right)} \\
\end{array}
\right) $$

To specifically find a circular orbit we can set $z=0$ and $r=R$ and $\phi = d\phi \ t$. That simplifies the geodesic equation to: $$ 0=\left(
\begin{array}{c}
0 \\
\frac{-\text{d$\phi $}^2 M R^2-\text{d$\phi
$}^2 R^3+M}{2 M R+R^2} \\
0 \\
0 \\
\end{array}
\right) $$ so $$ {d\phi}=\frac{\sqrt{M}}{\sqrt{M
R^2+R^3}} $$

Solving for $\phi = 2\pi$ we get $$ t_{2\pi}=\frac{2 \pi \sqrt{R^2 (M+R)}}{\sqrt{M}} $$

Evaluating proper time along the worldline of the earth we get $$ \frac{{d\tau}}{{dt}}=\frac{\sqrt{-4
M^2-2 M R+R^2}}{\sqrt{M R+R^2}} $$ which we can integrate to get Big Ben's time over one year to get $$\tau_{2\pi} =\int_0^{t_{2\pi}} \frac{d\tau}{dt} dt = 2 \pi R \sqrt{-\frac{4 M}{R}+\frac{R}{M}-2} $$ Plugging in the mass of the sun $M=1480$ in geometrized units (meters) and the orbital radius of the Earth $R=1.496 \ 10^{11}$ we get the proper time $\tau_{2\pi} = 9.45 \ 10^{15}$ which is one year in geometrized units.

I will post the analysis in a moving frame in a separate post.

 

[Dale]

OK, now adding the math for the other frame that the OP was going on about. We will boost along the cylindrical axis with the Lorentz transform in that direction so $$t=\frac{T-v Z}{\sqrt{1-v^2}}$$ $$z=\frac{Z-v T}{\sqrt{1-v^2}}$$ Note that the $r$ and $\phi$ coordinates are unchanged. Transforming the line element we get $$ds^2= {d}\phi ^2 \left(r^2-2 r^2
U\right)+{{d}r}^2 (1-2
U)+{{d}T}^2 \left(\frac{2 U
\left(v^2+1\right)}{v^2-1}-1\right)-\frac{8
{dT} {dZ} U
v}{v^2-1}+{dZ}^2
\left(\frac{2 U
\left(v^2+1\right)}{v^2-1}+1\right) $$ And transforming the potential we get $$U=-\frac{M}{\sqrt{r^2 + \left( \frac{Z-vT}{\sqrt{1-v^2}} \right)^2}}$$

With the line element determined in the boosted frame we simply apply the same math as before to obtain the geodesic equations in this frame. We get the following monstrosity (thank you Mathematica) $$0=
\left(
\begin{array}{c}
0 \\
\frac{\left(v^2-1\right)^2 r \dot \phi^2
\left(-\frac{4 M^2
\left(v^2-1\right)}{(Z-T
v)^2-\left(v^2-1\right)
r^2}-1\right)+\frac{M \left(1-v^2\right)^3
r \dot r^2 \left(2 M \left(v^2-1\right)+4 v
\dot Z \sqrt{r^2-\frac{(Z-T
v)^2}{v^2-1}}-\left(v^2+3\right)
\sqrt{r^2-\frac{(Z-T
v)^2}{v^2-1}}\right)}{\left((Z-T
v)^2-\left(v^2-1\right)
r^2\right)^2}-\frac{M \dot r (Z-T v)
\left(-\frac{2 M v \left(\left(v^2+1\right)
\dot Z^2-4 v
\dot Z+v^2+1\right)}{\sqrt{r^2-\frac{(Z-
T v)^2}{v^2-1}}}+\dot Z \left(v
\left(v^2-3\right) \dot Z+4\right)+v
\left(v^2-3\right)\right)}{\left(r^2-\frac
{(Z-T v)^2}{v^2-1}\right)^{3/2}}+\frac{M
\left(v^2-1\right)^3 r^2\dot \phi^2
\left(\sqrt{r^2-\frac{(Z-T
v)^2}{v^2-1}}-2 M\right) \left(v \dot r
(Z-T v)+\left(v^2-1\right)
r\right)}{\left((Z-T
v)^2-\left(v^2-1\right)
r^2\right)^2}-\frac{M v
\left(v^2-1\right)^3 \dot r^3 (T v-Z)
\left(\sqrt{r^2-\frac{(Z-T
v)^2}{v^2-1}}-2 M\right)}{\left((Z-T
v)^2-\left(v^2-1\right)
r^2\right)^2}+\frac{M \left(1-v^2\right)^3
r \left(\left(v^2+1\right) \dot Z^2-4 v
\dot Z+v^2+1\right)
\left(\sqrt{r^2-\frac{(Z-T
v)^2}{v^2-1}}-2 M\right)}{\left((Z-T
v)^2-\left(v^2-1\right)
r^2\right)^2}}{\left(v^2-1\right)^2
\left(\frac{4 M^2 \left(v^2-1\right)}{(Z-T
v)^2-\left(v^2-1\right)
r^2}+1\right)}+\ddot r \\
\frac{\dot \phi \left(\frac{4 M
\left(v^2-1\right)^2 r \dot r \left(v
\dot Z-1\right)}{\left((Z-T
v)^2-\left(v^2-1\right)
r^2\right)^{3/2}}+\frac{M (Z-T v)
\left(\frac{2 M v \left(-\left(v^2-1\right)
\dot r^2+\left(v^2+1\right) \dot Z^2-4 v
\dot Z+v^2+1\right)}{\sqrt{r^2-\frac{(Z-
T v)^2}{v^2-1}}}+v \left(\left(v^2-1\right)
\dot r^2-v^2+3\right)-v \left(v^2-3\right)
\dot Z^2-4 \dot Z\right)}{\sqrt{1-v^2}
\left(r^2-\frac{(Z-T
v)^2}{v^2-1}\right)^{3/2}}\right)-\left(1-v^2
\right)^{3/2} \ddot \phi \left(-\frac{4 M^2
\left(v^2-1\right)}{(Z-T
v)^2-\left(v^2-1\right)
r^2}-1\right)}{\left(1-v^2\right)^{3/2}
\left(\frac{4 M^2 \left(v^2-1\right)}{(Z-T
v)^2-\left(v^2-1\right)
r^2}+1\right)}+\frac{M v r^2 (T v-Z)
\dot \phi^3}{\left((Z-T
v)^2-\left(v^2-1\right) r^2\right) \left(2
M+\sqrt{r^2-\frac{(Z-T
v)^2}{v^2-1}}\right)}+\frac{2 \dot r \dot \phi
}{r} \\
\frac{\left(v \dot Z-1\right) \left(\frac{M
(Z-T v) \left(\frac{2 M
\left(\left(v^2-1\right)
\dot r^2+\left(v^2-1\right) r^2 \dot \phi
^2-\left(v^2+1\right) \dot Z^2+4 v
\dot Z-v^2-1\right)}{\sqrt{r^2-\frac{(Z-
T v)^2}{v^2-1}}}-\left(v^2-1\right)
\dot r^2-\left(v^2-1\right) r^2 \dot \phi
^2+v^2 \dot Z^2+4 v \dot Z-3 \dot Z^2-3
v^2+1\right)}{\sqrt{1-v^2}
\left(r^2-\frac{(Z-T
v)^2}{v^2-1}\right)^{3/2}}+\frac{4 M
\left(v^2-1\right)^2 r \dot r
\left(v-\dot Z\right)}{\left((Z-T
v)^2-\left(v^2-1\right)
r^2\right)^{3/2}}\right)}{\left(1-v^2\right)^{3/2} \left(-\frac{4 M^2
\left(v^2-1\right)}{(Z-T
v)^2-\left(v^2-1\right)
r^2}-1\right)}+\ddot Z \\
\end{array}
\right)
$$

Now, as before we will simplify this substantially by considering only circular orbits which will wind up as helical orbits in this frame. We will use $r=R$ and $\phi = d\phi \ T$ as before, but this time we will have $Z=v T$. With these, the geodesic equation simplifies to $$ 0=\left(
\begin{array}{c}
0 \\
-\frac{M \left(\text{d$\phi $}^2
R^2+v^2-1\right)+\text{d$\phi $}^2 R^3}{R (2
M+R)} \\
0 \\
0 \\
\end{array}
\right) $$

Solving for $d\phi$ we get $$ \text{d$\phi $}=\frac{\sqrt{M-M v^2}}{\sqrt{M
R^2+R^3}}$$ Note that this is slower than $d\phi$ in the other frame by a factor of $1/\gamma=\sqrt{1-v^2}$. So in this frame the angular speed required to maintain a stable orbit (geodesic) is "dilated". This is the key fact that everyone with any experience in GR already knew.

We continue with the rest of the calculations. In this frame the time required to get to $\phi=2\pi$ is $$T_{2\pi}=\frac{2 \pi \sqrt{R^2 (M+R)}}{\sqrt{-M
\left(v^2-1\right)}}$$ which we see is also "dilated". Meaning that the year is longer relative to coordinate time $T$ in the boosted frame. And finally, we plug this back into the line element to find the proper time $$\frac{\text{d$\tau $}}{\text{dT}}=\sqrt{\frac{4
M^2 v^2}{R (M+R)}-\frac{4 M^2}{R
(M+R)}+\frac{2 M v^2}{M+R}-\frac{R
v^2}{M+R}-\frac{2 M}{M+R}+\frac{R}{M+R}}$$ and integrate it over the year to obtain $$\tau_{2\pi}= \int_0^{T_{2\pi}}\frac{d\tau}{dT} dT =2 \pi \sqrt{R \left(\frac{R^2}{M}-4 M-2
R\right)}$$ Which is the exact same expression for proper time as before, and substituting numbers gets the same numbers as before. So Big Ben measures the same amount of time in a year. Both the year and Big Ben are "dilated" the same, and the angular velocity required to maintain a stable orbit matches the actual angular velocity.

Edit: so what did we actually learn from this exercise? The conclusion was exactly as everyone but the OP said. Big Ben and the year both time dilate the same and the boosted velocity is the correct orbital velocity. Indeed, from first principles it could not be any other way. Any invariants must be the same in all frames. So the conclusion was guaranteed. What was actually tested by the above math was whether or not I can program Mathematica for GR calculations.