How is time dilated in a gravitational field?

Question

I picture a box that contains a particle that travels back and forth at the speed of light. One round trip is a unit of time. We place this box on the surface of a large planet. The particles run upwards and down, perpendicular to the surface.

If the speed of light is constant and it is predicted that the rate of time in a gravitational field will decrease, how is this accomplished?

Speed is constant, length of time is increased, so that only leaves an increase in particle travel length. Has the box been elongated?

Answer 1

If the speed of light is constant and it is predicted that the rate of time in a gravitational field will decrease, how is this accomplished? Speed is constant, length of time is increased, so that only leaves an increase in particle travel length. Has the box been elongated?

Not exactly. The box isn’t simply elongated, the spacetime is curved. It is common enough to say that gravity is due to curved spacetime, but sometimes it is easy to forget what curved geometry is like.

Now, suppose that we have this box at rest in a gravitational field, and suppose we have two observers, one at the top of the box and one at the bottom. Each observer measures the height of the box by radar. Due to time dilation the clock at the bottom ticks slower, so the observer at the bottom measures the height to be shorter than the observer at the top. This may seem impossible, but it happens in curved spaces or with curved axes.

For example, consider the surface of the earth. Suppose that you have a plot of land whose sides are due north, south, east and west. If you measure the east-west distance then you will find that it is longer on the south side than on the north side (in the northern hemisphere). This happens even though the angles are all right angles. So in a curved space we can have a figure with four right angles and one side shorter than the other three.

That is what is happening here. The clock on the bottom is like the length of the north side of the plot of land, and vice versa for the other clock. The curvature of the time axis allows this to happen.

Answer 2

According to the equivalence principle, for a small box compared with the planet, the situation is like being inside an uniformed accelerated rocket. The time for the EM radiation "goes up" is longer than for "going down".

The light speed in the vacuum is a constant everywhere in a Minkowski spacetime.

Answer 3

The first kind of time dilation Einstein inferred was in special relativity, and due to relative motion. His light clock thought experiment shows frame shifts relate up-down and up-down-but-also-sideways motions: since $c$ is invariant, time dilation is equivalent to the difference in the paths' lengths.

We can explain gravitational time dilation with a thought experiment too.

Switch to an accelerated frame where objects in your box experience no gravity because it's cancelled by the acceleration (think Einstein's famous falling-in-a-lift thought experiment). In that frame special relativity appears to be applicable, so light moves in apparent straight lines. Therefore, in your original frame, its path curves (if you neglect $g$ variations, pretend the planet's surface is flat etc., it's one of those SUVAT parabolas you've studied before). Obviously, this makes the distance travelled greater.

Answer 4

Time dilation is widely misconceived, even in special relativity. You shouldn't think of it as time slowing down- instead, it is a consequence of the geometry of spacetime, in which different paths between two events can have different durations.

If you and I take two different paths between two events, such that my path is twelve minutes long, say, and yours is only ten minutes long, I could conclude that time has slowed down for you, but in fact your watch has continued to tick at the same rate- it is the duration of the path which has caused the elapsed time in your case to be less than in mine.

In SR, the differences in path lengths arise from the relativity of simultaneity. If you and I are moving relative to each other, our planes of simultaneity are tilted- for you there is a plane of simultaneity over which it is 'now' everywhere. Your plane is sloping in my reference frame, rising upwards through time in your direction of travel, so where it is 'now' ahead of you, it is later than 'now' in my reference frame. As a result, when you move from point A to point B in my frame, the elapsed time from my perspective is not just the time on your watch- it is increased by the fact that your baseline in time is sloping upwards in my frame.

In GR the effect arises not from sloping planes of simultaneity but from curvature of spacetime. It is analogous to curved paths through space- if you and I walk around a circular track, and I walk on the outside lane while you walk on the more tightly curved inside lane, you have less distance to walk than I do. If we both wear pedometers, your pedometer will 'tick' fewer times on a lap than mine will- it is not caused by 'pace dilation' but by the fact that the geometry of the track means that your path requires fewer paces than mine.

Answer 5

"I picture a box that contains a particle that travels back and forth at the speed of light."

Very good!

"[...] We place this box on the surface of a large planet. The particles run upwards and down, perpendicular to the surface."

As far as the relations between constituents of a box being held rigidly wrt. a mass can be approximated by relations between constituents of a box being uniformly (hyperbolically) and rigidly accelerated in a flat region, which can be calculated explicitly, the signal front roundtrip durations (ping durations) "top-to-bottom-to-top" and "bottom-to-top-to-bottom" are found as

$$\tau_{\text{TBT}} = \tau_{\text{BTB}} \, \text{Exp}\left[ \, \frac{a_B \, \tau_{\text{BTB}}}{2\, c^2} \, \right] = \tau_{\text{BTB}} \, \text{Exp}\left[ \, \frac{a_T \, \tau_{\text{TBT}}}{2\, c^2} \, \right],$$

where $a_B$ is the acceleration magnitude of the bottom, and $a_T$ is the acceleration magnitude of the top constituents of the box.

"One round trip is a unit of time."

The ping durations are separately constant (that's what we mean by constituents remaining rigid wrt. each other), and therefore separately useful "units". But these "units" are not necessary equal; especially, in consequence of the above equation:

$$\tau_{\text{TBT}} \gt \tau_{\text{BTB}}.$$

"it is predicted that the rate of time in a gravitational field will decrease"

At best, apparently that's a very cryptic and very much depreciated attempt to express, that between top and bottom constituents which are held rigid wrt. each other the (constant) ping durations "top-to-bottom-to-top" are larger than the (constant) ping durations "bottom-to-top-to-bottom",

$$\tau_{\text{TBT}} \gt \tau_{\text{BTB}}.$$

"Has the box been elongated?"

As far as "length" is understood in the sense of chronometric distance between two ends which are and remain at rest wrt. each other in a flat region, as

$$c/2 \, \text{Ping duration between the two ends},$$

any pair of bottom and top constituents of the box cannot be attributed a particular (mutually equal, symmetric) distance value, but two unequal values of quasidistance.

Of course, instead of "length", the term "height" is used when referring to spatial extension between rigidly held constituents ("directly above each other"), along the direction of acceleration. Nevertheless, it is not trivial to attribute exactly one "height" value to such a pair.

"If the speed of light is constant"

By definition, the speed of light (signal front) is constant between any pair of constituents (signal source and receiver) to whom a particular distance value can be attributed. Since this doesn't apply to any pair of one top and one bottom constituent of the box, therefore there cannot be any particular "speed" value be attributed to whatever would be exchanged between top and bottom, or vice versa.

Answer 6

A simple and reasonably accurate approach: Consider a photon rising in a gravitational field. It gains gravitational energy (with m = hf/$c^2$) and loses kinetic energy (hΔf). A distant observer would say that the emitting atom was vibrating atom was vibrating at a lower rate.