# How long is a “day” in intergalactic space?

Imagine a human colony with the concept of an Earth day being transported somehow into the deep, deep space between galaxies.

As we know, gravity wells affect the relative passage of time. So, in this case, Earth time would run slower than the colony.

But by how much? Do the other gravity wells, such as the Sun or galaxy mass, make much of a difference?

A day in deep space is the same length as on Earth.

About 86,400 seconds. It looks the same to someone in either place. The only difference shows up when deep space clocks and Earth clocks check against each other.

Gravitational time dilation on Earth goes like this:

$$T = {{T_0}\over {\sqrt{1-{{2gR}\over{c^2}}}}}$$

Where,

• $T_0$ is the proper time between events A and B for a slow-ticking observer within the gravitational field (on Earth)
• $T$ is the coordinate time between events A and B for a fast-ticking observer at an arbitrarily large distance from the massive object (deep space)
• $R$ is the radial coordinate of the observer (which is analogous to the classical distance from the center of the object, but is actually a Schwarzschild coordinate)
• $g$ is the surface gravity of Earth
• $c$ is the speed of light

For Earth vs deep space, it's about a billionth of a second per second difference.

That is, on the deep space colonies, after one billion seconds ($\approx$ 31.68 years), the clocks on Earth would be one second behind.

• In nearly all situations, the difference is negligible by all but very precise uses. However, at very high fractions of $c$ or when approaching an event horizon. – Jim2B Jun 11 '15 at 1:23
• I just discovered Barycentric Coordinated Time. When adjusting for the sun's gravitational well, it's off by 490ms/yr. Would this mean that the galaxy's gravity well makes an even bigger time difference? en.wikipedia.org/wiki/Barycentric_Coordinate_Time – Sam Washburn Jun 11 '15 at 2:42
• @SamWashburn I haven't seen that before! Yes, you're correct. The gravitational time dilation is a continuum between centers of mass. Outside the solar system will be different than intergalactic space, but the effect won't ever be very large (likely still less than one second per year). – Samuel Jun 11 '15 at 3:11
• So how does 490ms reconcile with your computation of 1ns? The correction for interstellar space in our galaxy should be less than correcting out to the closest you can get to an infinitely distant observer beyond any gravitational sources. – JDługosz Jun 11 '15 at 23:20
• @JDługosz One nanosecond is for the Earth's gravity well alone. Over a year, it's about 32 milliseconds (which was the period for the 490 ms figure). The sun is over 330,000 times as massive as the Earth but we're much further away from it already, so we're getting a few hundred milliseconds of gravitational time dilation from the Sun's gravity well. On the surface we're experiencing the superposition of both and they total to about 490 ms. – Samuel Jun 12 '15 at 3:20

Here is something your colony may consider, though it has nothing to do with relativity. Human circadian rhythms don't exactly follow a 24 hour cycle. Here on planet Earth, we are stuck with the day/night cycle and we build our clock around it.

There have been experiments conducted in deep caves where people had no time pieces and no access to direct sunlight to find where their patterns settled. There was some variation, and the experiments I know about didn't last long enough to stabilize, but I found it surprising that people fell into a longer day cycle than 24 hours. The average is about 27. Here's one account with other measurements indicating internal chemical processes that relate to day/night biology.

Long story short: Once we are out in space and have to manufacture a day/night cycle through ship lighting, who's to say we don't stretch it out to the 27ish hours that seems to fit us better?

I know this isn't the info you wanted in a response, but I feel the discussion isn't complete without its consideration and Samuel already covered the relativity issue well.

This would actually be an interesting question, since most bodies in deep space would not have a rotational or orbital period resembling an Earth day or year at all.

For totally artificial colony structures, the inhabitants could set the "day" to whatever they like, but since they probably would have an earthly biosphere, it would be much easier set their clocks and systems to replicate Earth rather than try to tweak entire ecosystems to match some arbitrary system.

On planetary bodies, where they would have a difficult time resetting the planetary rotation and orbit unless they have some truly heroic megaengineering skills, they will probably still use the second as a basic unit of measurement. This will allow them to divide the days and years into convenient slices (kiloseconds and megaseconds are common SF tropes) that can be roughly matched up with the planet's own cycle.

Using seconds also means they can be on a universal calendar like the UNIX calendar (which by odd coincidence starts very close to when Man first set foot on the Moon: UNIX time starts Jan 01 1970. A few centuries from now, the difference between July 21 1969 and Jan 01 1970 will seem trivial). Once again, they may opt to set their time to match the standard Earth day and year in order to maintain biospheres and ecosystems with minimum fuss. If you are living in a bubble hundreds of metres below the ice surface of Europa, for example, you might not want or need to take reference to the cycles of Jovian orbital time anyway, except to calculate windows for spaceflight.

Since, as noted by Samuel, the effects of gravity will be minuscule, this should serve to allow for everyone to be on one universal time/calendar standard for simplified record keeping, timestamps on documents and transactions and so on. The only real issues will be for spacecraft travelling at a high fraction of c and near very deep gravity wells like event horizons of black holes.

There would be no significant difference. The key factor is, as already stated, that c/√(c²-gR) is very very close to 1 unless gR is a large fraction of c², and it's not. (I re-expressed the terms under the radical sign). The major difference would be the absence of the leap-second corrections which occur on Earth due to various geological and orbital perturbations. There's be no reason to impose them in the absence of a planet.

• Did you read Samuel's answer from last year before you posted? – JDługosz Jun 5 '16 at 10:33