# Can I practically demonstrate Gravitational Time Dilation by spinning a wheel very fast? [duplicate]

In an attempt to demonstrate gravitational time dilation, I was curious if it were practical to mount a clock to a fast spinning wheel, with the centripetal acceleration of the wheel being equivalent to a gravitational field.

I wanted to calculate the velocity the wheel would need to rotate at in order to measure a time dilation of 1μs, when the clock is placed 1m from the center of the wheel. Since this isn't a common lab or high-school experiment, I didn't expect the velocity to be very practical.

Is it correct to use the following approximation to compute the required velocity within the same order of magnitude as the "correct" answer (from: http://en.wikipedia.org/wiki/Gravitational_time_dilation)? I substituted $v^2/r$ and $r$ for $g$ and $h$ respectively.

$$T = 1 + \frac{gh}{c^2}$$ $$T = 1 + \frac{gr}{c^2}$$ $$T = 1 + \frac{rv^2}{rc^2}$$ $$T = 1 + \frac{v^2}{c^2}$$

For $T = 1 + 10^-6$s, $v = 3x10^5 m/s$, which is clearly far too fast.

Are my calculations correct, given my assumptions and requirements?

## marked as duplicate by John Rennie, ACuriousMind♦, JamalS, Prahar, JimNov 30 '14 at 14:57

• – HDE 226868 Nov 29 '14 at 19:49
• A lot of good information in that question, @HDE226868. – Steve Guidi Nov 29 '14 at 19:58
• Note that the equivalence principle states there is no way to distinguish between a constant gravitational field and a constant acceleration. With a spinning disk, the magnitude of acceleration may not change but the direction changes sinusoidally. You could argue that on small scales, this looks like a constant acceleration, but in order to mimic a sufficient magnitude of gravity, you need to have a decent angular velocity. This makes the acceleration non-constant and makes the equivalence principle irrelevant – Jim Nov 30 '14 at 14:57

What you're missing is that time dilation is a multiplicative factor. If you want a time dilation of $1\mu s$, then what you mean is that your clock will differ from a stationary clock by $1\mu s$. This is an offset. In general, your moving clock will tick at a rate of $\frac{1}{\gamma}$ slower than a stationary one. So you could say, that after 1 hour of spinning, your clock will differ by $1\mu s$. You could write $\gamma T_\mathrm{moving} = T_\mathrm{stationary}$, and require that $T_\mathrm{moving} = 1.0 h$ and $T_\mathrm{stationary} = 1.0h + 1\mu s$. Then you can find $$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$ that solves this and obtain the required $v$.
The above calculation essentially treated the problem like straight line motion, with no acceleration. In general, the clock moves along some spacetime path and it's proper time is just the length of the path in Minkowski space: $$c d\tau = c dt^2 - d\vec{x}^2$$ In general, you would compute the arc length $$\tau = \int\! \sqrt{1-\frac{v^2}{c^2} } dt$$ where $v^2 = \frac{d\vec{x}}{dt} \cdot \frac{d\vec{x}}{dt}$. For our clock in circular motion, $v^2$ is a constant, so the integral just reduces to $$\Delta \tau = \sqrt{1-\frac{v^2}{c^2}} \Delta t$$
$$T_\mathrm{moving} = \sqrt{1-\frac{v^2}{c^2}} T_\mathrm{stationary}$$ Just to note, this effect is already taken into account by GPS systems. There it counter-acts the gravitational time dilation difference that arises because the curvature experienced by the satellites is weaker than that observed on the surface of the Earth. Both these effects need to be taken into account in order to achieve the sort of accuracy we often take for granted.
• Just bad units, I guess. I think that time dilation you quoted assumes that you travel some distance $h$. – lionelbrits Nov 29 '14 at 20:48