Why does not acceleration affect time dilation in addition to the time dilation caused by velocity?

Question

In his book, Relativity: The Special and General Theory, Einstein claimed that the clocks located on a rotating disc run slower solely due to their tangential velocities as appear in the Lorentz factor, no matter how much acceleration they take. He then replaces potential per unit mass with velocity square $(r^2\omega^2)$:

If we represent the difference of potential of the centrifugal force between the position of the clock and the centre of the disc by $\phi$ , i.e. the work, considered negatively, which must be performed on the unit of mass against the centrifugal force in order to transport it from the position of the clock on the rotating disc to the centre of the disc, then we have

$$\phi=\frac{\omega^2r^2}{2}$$

However, I cannot really understand why the centrifugal acceleration does not affect clocks at all. Assume we have two concentric rings one with a large radius and the other with a very small one. If the rings rotate at the same tangential velocity, according to Einstein, the clocks run slower at the same rate as measured by an inertial observer at rest with respect to the plate's center. However, according to the centrifugal acceleration formula:

$$a=\frac{v^2}{r}\space,$$

the clock on the ring with a smaller radius experiences much more acceleration than one located on that with a larger radius. How can it be possible that such a large centrifugal force/acceleration, which can easily mash the nearer clock to the center of rotation (if the radius is small enough), is ineffective in altering time rates? (Forget about the viewpoint of the rotating observers.)

Remember that if the radius approaches zero, the centrifugal acceleration tends to infinity, yet the tangential velocity can remain unchanged. It is really hard for me to understand why an infinite acceleration/force cannot affect clock rates!

Answer 1

Your question has the same ring as, say, the following question:

"We know that everything slows down and stops, unless you keep pushing. Push an object across the floor, it keeps moving. Stop pushing: it stops. It is really hard for me to understand how satelites in orbit keep moving without anything pushing them around."

In your case what you are stuck on is a wrong assumption (a wrong assumption about time dilation.)

I do grant you that your wrong assumption is very tempting. Many go down that road.

To address the wrong assumption let me take the Twin scenario as example.

As is often done, I will call them Alice and Bob.
Alice goes on a journey, Bob stays put.
Alice makes her far away U-turn, and when Alice and Bob rejoin the amount of elapsed proper time for Alice is less than the amount of elapsed proper time for Bob.

Both Alice and Bob are scientists with full understanding of time dilation, so given the travel plan they know in advance how large the difference in elapsed proper time will be.

The way to calculate the difference in amount of elapsed proper time is to apply the Minkowski metric. Bob has traveled forward in time along the shortest path; he hasn't moved. Alice has not taken the shortest path: in addition to travelling forward in time she has traveled spatial distance. To calculate the difference in elapsed proper time you evaluate one thing and one thing only: the difference in spatial distance traveled.

The shape of the journey that Alice made does not matter. Several other siblings can make all kinds of journeys, if all those journeys are in the end the same spatial distance (compared to each other) then for all those travellers the same amount of proper time will have elapsed.

The fact that the shape of the journey doesn't matter follows from this: as you evaluate the total spatial distance traveled (applying the Minkowski metric) the shape of the journey drops out of the calculation.

Of course, the acceleration is necessary; Alice has to make that U-turn, that's the only way to rejoin Bob. But whether the U-turn is sharp (pulling a lot of G's) or gradual (low G-load), that doesn't matter. The journey that Alice makes can be a zigzag course, pulling G's all the time; it doesn't matter, only the difference in spatial distance travelled counts.

Yes, that is very counter-intuitive.
Superficially you might expect that the difference in spatial distance travelled is irrelevant, it looks so passive.
By contrast, the acceleration is violent, surely that's where it happens.

In actual fact the difference of elapsed proper time is described by the Minkowski metric. The Minkowski metric is necessary, and sufficient.

To learn the logical implications of the Minkowski metric is to learn Special Relativity.

Answer 2

I would not expect centripetal acceleration to have any effect. The motivation for SR was the principle that the speed of light should be the same for all observers. The Lorentz transform explains how one can account for the effects of relative motion between observers, ie the speed and direction of the motion. Acceleration is simply a change either in the speed or the direction of a motion, so it simply requires you to plug different values into the transformation equations. At any instant an accelerating body has a definite speed and a definite direction to its motion, so at that instant the Lorentz transform adequately quantifies all relativistic effects. The magnitude of those effects changes from instant to instant as the velocity of the body changes, but we have no reason to suppose that the nature of the effects would change.

Answer 3

In special relativity time dilation is not affected by body's acceleration, however in general relativity - it is, because of equivalence principle (accelerating reference frame is indistinguishable from non-accelerating frame in gravity field). Gravitational time dilation is defined through Schwarzschild metrics:

$$ t_{0}=t_{f}{\sqrt {1-{\frac {2GM}{rc^{2}}}}} $$

Using Newton's gravitation and second laws, one can re-write equation above as :

$$ t_0 =t_f\sqrt{1-2\frac{GMmr}{r^2c^2m}} \\=t_f\sqrt{1-2\frac{F_{_G}r}{mc^2}} \\=t_f\sqrt{1-2a\frac{r}{c^2}} \\ =t_f\sqrt{1 - \frac{a}{a_s}} $$

where $a_s$ is Schwarzschild acceleration and is defined as :

$$ a_s = \frac{c^2}{2r} $$

Answer 4

for a rotate coordinate system with the constant angular velocity $\omega$ the coordinates are:

$$x=x'\cos(\omega\,t')-y'\sin(\omega\,t')$$ $$y=x'\sin(\omega\,t')+y'\cos(\omega\,t')$$ $$z=z'$$ $$t=t'$$

with $ds^2=\eta_{\mu\nu}\,dx^\mu\,dx^\nu=c^2\,dt^2-dx^2-dy^2-dz^2$

we get the metric

$$G=\left[ \begin {array}{cccc} \left( -{x'}^{2}-{y'}^{2} \right) {\omega} ^{2}+{c}^{2}&\omega\,y'&-\omega\,x'&0\\ \omega\,y'&-1&0 &0\\-\omega\,x'&0&-1&0\\ 0&0&0&-1 \end {array} \right] $$

with:

$dx'=dy'=dz'=0$ we get

$$d\tau=\frac{ds_{\text{clock}}}{c}=\sqrt{G_{00}}\,dt'= \sqrt{1-\frac{\omega^2\,(x'^2+y'^2)}{c^2}}\,dt'$$

so $d\tau$ is a function of the clock distance from the center $x'^2+y'^2=r^2$ and the angular velocity. with $\omega=\frac{v}{r}=\frac{v}{\sqrt{x'^2+y'^2}}$ we get for $d\tau$

$$d\tau=\sqrt{1-\frac{v^2}{c^2}}\,dt'$$

Answer 5

In Special Relativity, it is assumed that only relative velocity has an effect on the ideal-clock rate, as read from another inertial observer. It is implicitely postulated that the clock's acceleration has no effect, or else you would need some universal acceleration to compare, like there's an universal constant $c$ to compare velocities. On dimensional grounds, you would need to introduce some acceleration parameter $g$ of some sort into the Lorentz transformation, a kind of universal constant (the maximal or minimal acceleration available in the universe?).

It is assumed that this $g = 0$ in SR.

In the past, some authors have already created extended theories which go above SR, but empirical data have ruled out most of these theories. For example, you may be interested in doubly special relativity:

https://en.wikipedia.org/wiki/Doubly_special_relativity

From dimensional analysis, these theories could introduce some acceleration constant that could have an effect on clock rates.