|
Time in a Centrifuge |
|
|
|
It is well known in the context of special relativity that time dilation depends only on speed, not on acceleration. To be precise, let τ denote the proper time showing on a given ideal clock moving with speed v in terms of inertial coordinates x,y,z,t in flat spacetime (or more generally any circumstance in which changes in gravitational potential are negligible). Then, regardless of the clock’s acceleration, we have |
|
|
|
|
|
|
|
Consequently, to the lowest order, the moving clock loses (1/2)(v/c)2 seconds per second compared with the coordinate time t. Again, this is true for an ideal clock regardless of the acceleration. Einstein’s original justification for this independence was that “it is at once apparent that this result holds good if the clock moves from A to B in any polygonal line, and also when the points A and B coincide”, and then “we assume the result proved for a polygonal line is also valid for a continuously curved line”. Following this mode of expression, some authors (e.g., Rindler) have presented the independence from acceleration as a physical hypothesis, which they call the “clock hypothesis”. However, it’s debatable whether this is really a separate assumption, because, by definition, an instantaneous impulse acceleration occurs in an instant, i.e., zero elapsed time (as Einstein said, “when the points A and B coincide”), so it doesn’t contribute any elapsed time to the process. |
|
|
|
In practical terms, the independence from acceleration is part of the definition of an ideal clock, i.e., a clock that has been corrected for any locally sensible effects such as temperature, humidity, acceleration, etc. (Note that acceleration is absolute and locally sensible, whereas position and velocity are not, so an ideal clock can be intrinsically corrected if necessary for acceleration but not for position or velocity.) In any case, all experiments have confirmed that the temporal characteristics of elementary quantum phenomena are indeed independent of acceleration, up to extremely high values of acceleration. This is an essential aspect of local Lorentz invariance, and for the validity of the Minkowski spacetime metric, according to which |
|
|
|
|
|
|
|
from which it follows that |
|
|
|
|
|
|
|
People sometimes wonder about whether the independence of time dilation from acceleration violates Einstein’s equivalence principle, which relates acceleration to gravity. We know that clocks located at different points in a gravitational field exhibit gravitational time dilation. For example, the characteristic spectral lines from very dense stars are re-shifted. Likewise the clocks in GPS satellites exhibit time dilation relative to clocks on the ground, due both to their motion (per special relativity) and to their elevation in the Earth’s gravitational field. There have even been terrestrial experiments involving clocks flown on airplanes, and the different elevations compared with clocks on the ground resulted in the predicted gravitational time dilation, along with the effects of motion. The Rebka-Pound experiment showed gravitational time dilation between the top and bottom of a tower, and with modern clocks it has become possible to detect time dilation for differences of just 1 foot of elevation. |
|
|
|
Given all this, and in view of the equivalence principle, why don’t we expect the pseudo-gravity experienced on a centrifuge to produce time dilation? The answer is that it does, but it is not an additional effect, it is the same effect, merely described in different terms. First, in terms of any system of inertial coordinates there is no gravitational field, we simply have a clock moving with speed v, and the time dilation is simply given by special relativity as (1/2)v2/c2. As discussed above, the fact that the clock is undergoing acceleration (by moving in a circle) doesn’t affect this result. |
|
|
|
Now, consider the same situation in terms of a rotating coordinate system, in terms of which the arm of the centrifuge is stationary. In these coordinates the clock is not moving, but there is a pseudo-gravitational field due to the acceleration a = (v/R)2r for each location on the arm at a distance r from the hub, where R is the length of the centrifuge arm and v is the speed at the end of the arm. In a real gravitational field the time dilation, Δt/t, between two locations is equal to the change in energy, ΔE/E. (This follows, for example, from the fundamental quantum relation E=hν.) For a unit mass m the loss in energy equals the work needed to move the mass from the end of the arm (r=R) to the center (r=0), where there is no time dilation. To the lowest order we simply integrate Fdr = m(v2/R2)r from r = R to 0. This gives ΔE = (1/2)mv2. The total energy is E=mc2, so we have ΔE/E = (1/2)(v/c)2. This is equal to Δt/t, which shows that the time dilation between the hub and edge of the centrifuge is the same, regardless of whether we compute it as the effect of motion in an inertial frame or the effect of a pseudo-gravitational field in a rotating frame. |
|
|
|
It’s worth emphasizing that the amount of time dilation is not proportional to the gravitational force or acceleration, it is proportional to the difference in potential energy. For example, the time dilation in a large centrifuge that produces 20g acceleration at a radius of 30 feet would result in time dilation proportional to 600 g-feet between the hub and the rim, whereas the time dilation between the Earth’s surface with just 1g and a location at 30,000 feet (the height of an airplane cruising) would result in time dilation proportional to 30,000 g-feet, which is 50 times greater. |
|
|
|
Another illustration of this concerns the often-asked question about what the time dilation would be at the center of the Earth (or any other large gravitating body), where the force of gravity is essentially zero. If time dilation was dependent on the force (or acceleration) of gravity, we would expect the time dilation to be zero, but in fact it is greater than at the Earth’s surface. Qualitatively this is because it takes work to rise from the center to the surface. Quantitatively a simple first-order analysis is sufficient. The force of gravity on any test particle of unit mass at a distance r from the center is Gm/r2 in geometrical units, where m is the mass of the gravitating body inside the radius r. The mass outside that radius does not contribute any force. Integrating this from r = infinity to r = R (the surface of the Earth) gives GM/R where M is the total mass of the Earth. Converting this from unit mass to unit energy of a test particle, the time dilation on the Earth’s surface compared with time at infinity is GM/(Rc2). |
|
|
|
Now, to determine the time dilation at the center of the Earth, strictly speaking we need to know how the mass of the Earth is distributed. It is actually much less dense near the surface than at the core, but to illustrate the concept, let us suppose the density is constant. In that case the “interior” mass as a function of radial distance from the center is (4/3)πr3ρ where ρ = 3M/(4πR3) is the density. Thus we have m(r) = M(r/R)3, which implies the force is (GM/R3)r. Integrating this from the surface to the center gives GM/(2R), and hence the time dilation at the center of this spherical mass of constant density is (3/2)GM/(Rc2), which is 3/2 times greater than the time dilation at the surface. |
|
|
|
In general, the time dilation between any two states, whether due to motion or gravitational potential, is proportional to the amount of specific work needed to change from one state to the other. In the case of gravitation this is not symmetrical, because we have a definite difference between the potential at two locations (provided we have an asymptotically flat background as a reference).On the other hand, differences between the states of inertial motion of objects in flat spacetime are reciprocal, meaning for two relatively moving particles we can accelerate either one to match the state of motion of the other. |
|
|
