# Can an observer B move in such a way that his clock will run faster than that of a fixed observer A?

My understanding of Special and General Relativity comes from books which attempt to explain them to non-experts in these fields.

Let A and B be observers, which at some time t(0), are together at some location P fixed to the earth's surface. A and B are provided with "clocks", which at time t(0), are identical lumps of the same radioactive element-that has a very long half-life. A and A's "clock" remain fixed to the earth's surface. B and B's "clock" leave A and stay together, but eventually return to A after some interval of time T (as measured by A's "clock"). If, during this time T, B's velocity relative to A is sufficiently close to the velocity of light or B remains close to a sufficiently massive body, it seems that time can "pass more slowly" for B than for A. So that when B rejoins A, B's "clock" will contain more of the radioactive element than A's "clock"

My question is: Are there any scenarios that B together with B"s "clock" can undergo (during the time T) which will cause time to "pass more quickly" for B than for A? So that when B rejoins A, B's "clock" will contain less of the radioactive element than A's "clock". If there are no such scenarios, is there any fact or law in Relativity theory which rules them out?

• This question has certainly been here many times, I just cannot find an accurate enough duplicate. Anyways, see the wiki on Hafele-Keating experiment for a realization of such a scenario.
– Void
Jan 27, 2015 at 20:26

If you consider first time dilation due to relative velocity in flat spacetime, then relative velocity always makes the moving clock run slower. If you're interested, I explain why this in the first part of my answer to Is gravitational time dilation different from other forms of time dilation?. So in flat spacetime in $A$'s rest frame there is no way for $B$ to move that will make $B$'s clock run faster.

But the situation is different when the time dilation is due to a gravitational field. To calculate the time dilation exactly you need to know the metric, but there is a useful approximation that links the time dilation to the gravitational potential (per unit mass):

$$\frac{dt_B}{dt_A} = \sqrt{1 - \frac{2(\Phi_A - \Phi_B)}{c^2}} \tag{1}$$

The quantity $\Phi_A - \Phi_B$ is the difference in gravitational potential energy between $A$ and $B$, and $dt_B/dt_A$ is the time dilation of $B$'s clock relative to $A$'s clock i.e. if this is less than one $B$'s clock runs slower and if it's greater than one $B$'s clock runs faster.

To make this concrete let's take $B$ to be hovering 1m above $A$. You may have come across the expression for the change in potential energy in a uniform gravitational field with acceleration $g$:

$$\Delta\Phi = mgh$$

In this case $\Phi_B < \Phi_A$, i.e. it is less negative, and we need to divide by $m$ to get the energy per unit mass, so putting in $h = 1$m we get:

$$\Phi_A - \Phi_B = -g$$

and substituting in equation (1) and using Earth's gravitational acceleration, $g = 9.8$m/s$^2$, gives:

\begin{align} \frac{dt_B}{dt_A} &= \sqrt{1 + \frac{2g}{c^2}} \\ &\approx 1.0000000000000001 \end{align}

This is very small but it is greater than one i.e. $B$'s clock runs faster than $A$'s clock.

However we can't make $B$'s clock run arbitrarily fast. The best we can do is take $B$ to well away from the Earth so $\Phi_B \approx 0$. If $A$ is on the Earth's then $\Phi_A$ is given by the Newtonian expression:

$$\Phi_A = -\frac{GM_E}{r_E}$$

where $M_E$ is the mass of the Earth and $r_E$ is the radius of the Earth. This gives $\Phi_A \approx 6.25 \times 10^7$, and putting this into equation (1) we get:

$$\frac{dt_B}{dt_A} \approx 1.000000001$$

which is still pretty small, but it's large enough that the clocks in GPS satellites run measurably faster than clocks on the Earth's surface.

To make the time difference any greater you'd have to move $A$'s clock to somewhere with a lower (i.e. more negative) gravitational potential. For example you could move $A$ to near the event horizon of a black hole.

• Let's say you had two clocks in the same inertia frame without any gravity present. If one clock accelerates to a substantial speed, travels for a while, then turns around and comes at the same speed back, wouldn't the clock that accelerated definitively show less time than the other? Nov 7, 2016 at 21:09