# Relativistic Time Difference

I thought that it might be interesting to calculate the time difference between a clock placed on Earth when I was born and a hypothetical clock placed at the centre of the sun at the same time.

I thought that I might use special relativity and the Lorentz factor. However, even if the speed of the Earth were constant (which it is not), the Earth still experiences acceleration as it traverses its (almost) elliptical orbit around the sun. Given there is acceleration, I suppose that I must use general relativity.

How might one calculate the time difference experienced between me and my barbecued brother?

• Could the person that down voted my question please do the right and proper thing and use the comments section to make suggestions as to how I might improve my post. It is unfair to down vote a post without offering advice and opportunity for improvement. – Fly by Night Nov 5 '13 at 21:41

You should look into this experiment, where they tested GR using atomic clocks flying around on airplanes. If you dig into the details on it, they had to account for both the (primarily) special relativistic effects due to the motion of the airplanes, as well as the general relativistic effects due to gravity.

It is not necessary to use general relativity if you are ignoring the effects of gravity -- it is possible to merely use the special relativistic time dialation effect, remembering that, for any observer, $\eta_{\mu\nu}{\dot x^{\mu}}{\dot x^{\nu}} = -1$.

So, for a circularly travelling observer, we have:

\begin{align}x &= r \cos(\omega\,t)\\ y&= r\sin(\omega\,t)\end{align}

Our timelike condition then tells us that $-\left(\frac{dt}{d\tau}\right)^{2} + r^{2}\omega^{2} = -1$, so we get $\tau = \left(\sqrt{1 + r^{2}\omega^{2}}\right)t$

Meanwhile, our central observer will have $x = y = 0$, which gives (Obviously) $t=\tau$, so that's how you compare clocks.

But what you will find is that there is also a time dialation effect due to general relativity (clocks move slower near gravitating bodies), which says that $\tau = \left(\sqrt{1 - 2GM/r}\right)t$. If you calculate this effect for the Earth and the sun, you will find that it is larger than the special relativistic effect for the motion.

• Hi Jerry. Thank you very much for a very interesting answer. However, I didn't say that I was ignoring gravity and I didn't say that the orbit was circular. Could you please add to your answer so that it correlates more with my question. Thank you very much. – Fly by Night Nov 7 '13 at 19:55
• @FlybyNight: I tell you how to factor in gravity. You merely need to interpret $M$ as $M(r)$, or more propertly, replace $1-2M/r$ with $1- \phi$, where $\phi$ is the newtonian potential. As for a general path, it's just as easy to use the method I used above with a general path. Most of the planets have very small eccentricity (i.e., Earth's orbital eccentricity is 0.0167), and you can use a circular orbit to first order. – Jerry Schirmer Nov 7 '13 at 20:27
• And you only mentioned motion in your question, and in particular, you only mentioned acceleration. And I wanted to make it explicit that special relativity is perfectly capable of dealing with acceleration, and time in accelerated reference frames. – Jerry Schirmer Nov 7 '13 at 20:44
• When one discusses planets, gravity is a tacit assumption. Moreover, in general relativity acceleration and gravity are in some sense similar: "An accelerated system is completely physically equivalent to a system inside a gravitational field". The acceleration of the Earth as it orbits the Sun is due to the Sun's gravity. No gravity, no acceleration. – Fly by Night Nov 7 '13 at 21:00
• @FlybyNight: please don't lecture me on when General relativity applies and when it doesn't, when I cite the domains of applicability of special relativity and general relatvity.. – Jerry Schirmer Nov 7 '13 at 21:57