The speed of light is absolute, but time is relative. So would a light-year for us on earth be a different distance from a light-year on a different uniformly moving object? Why or why not?


6 Answers 6


The distance light travels in a given period is the same for every observer. That's the whole point of relativity.

You can figure out for yourself almost all the effects predicted by relativity if you start with that assumption and think through its consequences.

Indeed, the reason why times have to be relative is to allow observers who are moving relative to each other to agree on the speed of light.

Take the classic set-up where you are on a railway carriage and I am on the platform of the station you are passing.

As you pass me I flash a laser along the platform.

After what seems to me to be 100 nanoseconds, light has traveled to a certain point 100 feet along my platform (a foot is a light-nanosecond, hence a shorter version of a light-year). To you, however, that point is not 100 feet away, as you have been travelling relative to the platform, so it is some other distance.

If we both agree that the speed of light is the same, ie the ratio of the distance it has covered to the time it has taken, when we each think it has covered a different distance, then we must disagree about the elapsed time, ie time is relative.

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    $\begingroup$ Thank you for the explanation, it's easier to visualise (after reading it a few times). However, what if the platform was 100 feet long and on the other end of the platform was a wall. If the person on the train started a clock at the start of the platform and timed 99ns then dropped a ball out of the window. Would that ball immediately hit the wall or would it appear before or after the wall? $\endgroup$ Commented Oct 4, 2019 at 6:58
  • $\begingroup$ Hi Elliot, it is impossible to answer that question unless you know what speed the train is travelling compared with the platform (and its direction!). Let's assume the train is travelling towards the wall at 100 feet per second. After 99ns it will hardly have budged. $\endgroup$ Commented Oct 4, 2019 at 8:07
  • $\begingroup$ However, I think I know where you are coming from. Imagine the speed of light was quite low, say a foot per second, and the train was travelling an inch per second. And imagine that all along the carriage you have mates sitting ten feet apart. The light leaves you and travels along the carriage, and your mates record the time it passed them. When you compare notes afterwards your pals see that there are ten second gaps between the successive times they have noted (assuming their watches are good)... $\endgroup$ Commented Oct 4, 2019 at 8:10
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    $\begingroup$ @KeiNagase Because the speed of light is the same in all reference frames. Velocity is a measure of distance over time. The two are linked; an observer moving at relativistic speed will observe length contraction of objects, but their time dilation from a stationary observer is the same factor, thus the two cancel out. $\endgroup$ Commented Oct 4, 2019 at 18:56
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    $\begingroup$ @KeiNagase "I am not sure why we should make such an assumption." This is the way science works: make a hypothesis,work out its implications, & see if that explains so far unexplained phenomena. If you're looking for some idea of how someone might have come up with such a hypothesis, perhaps in historical context, consider that electromagnetism had described light as electromagnetic waves. The question Einstein asked himself was, what would an electromagnetic wave look like in a reference frame moving as fast as the wave? $\endgroup$
    – Diagon
    Commented Oct 4, 2019 at 23:29

A light-year is exactly 9,460,730,472,580,800 meters, the distance light travels in one Julian Earth year. Time dilation is irrelevant to its definition.

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    $\begingroup$ Yes, a simple Google search returns that. My question is, will that distance differ for an observer is in a different uniformly moving body? So would the distance still be relevant if we gave this distance to someone on mars (even if the different was negligible)? $\endgroup$ Commented Oct 4, 2019 at 5:37
  • $\begingroup$ Every observer thinks that a meter stick, or a light-year stick, at rest relative to that observer is the length it is supposed to be. But observers measure moving meter sticks or light-year sticks as being length-contracted. $\endgroup$
    – G. Smith
    Commented Oct 4, 2019 at 5:41
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    $\begingroup$ Alright, well I guess don't understand the question well enough to ask it then. $\endgroup$ Commented Oct 4, 2019 at 5:46
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    $\begingroup$ This seems to me like a circular argument: "A light-year is constant because a light-(1/299729458 seconds) is constant" $\endgroup$
    – JiK
    Commented Oct 4, 2019 at 14:18
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    $\begingroup$ Ignore the words "light" and "year" and think of it in the same you think of "meters" or "kilometers" and any confusion should disappear $\endgroup$ Commented Oct 4, 2019 at 18:06

The answer depends on the context of the question. Are we measuring the distance between objects that may be outside our frame of reference depending on our speed? Or are we measuring the distance between objects in one frame of reference, then re-measuring after changing all objects to a new frame of reference?

The distance between external objects, such as stars in the galaxy, will appear to be different depending on our reference frame.

The distance between internal objects, such as the length of the spaceship we're measuring from, will appear to be the same no matter the reference frame.

Distance Between External Objects

The distance is different if we measure static objects from each frame of reference. Due to time dilation and the assumption that light travels at a constant speed relative to a given frame of reference, distances along the velocity axis between two frames of reference will be measured differently.

Take, for example, an interstellar spaceship traveling 0.995c relative to Earth. This gives a Lorentz factor, γ (gamma), of 10. Now, the spaceship is traveling in some direction in a straight line. Conveniently, there are a bunch of stars along that line at consistent intervals.

Suppose Earth is able to broadcast powerful radio signals towards the stars and record the reflected signals. The radio signals take 20 years to bounce off the first star, 40 years for the second, etc. Using the time delay between signals, Earth determines the first star is 10 light years away, and each subsequent star is 10 light years from the previous star.

Now, the spaceship passes by Earth at the exact moment one of the broadcasts goes out. Because the spaceship is moving towards the stars, it will take less time for the return signal to get to the spaceship. So we'd calculate the same time out, and a (much) lower time back. Then we divide this time by the Lorentz factor to get the perceived time.

Earth and spaceship looking at a series of stars with Earth perceiving ten light years between stars and the spaceship perceiving half a light year between stars.

Specifically, the times for star number i are:

$t_{out}(i) = i \cdot 10 \text{y}$

$t_{back}(i) = \frac{ d_{remaining} }{ v + c }$

$= \frac{i \cdot 10\text{ly} \cdot \left( 1 - \frac{v}{c} \right) }{ v + c }$

$= \frac{i \cdot 10\text{ly} \cdot 0.005 }{ 1.995 \frac{ly}{y} }$

$= i \cdot 0.02506 \text{y}$

$t_{total} = t_{out} + t_{back}$

$= i \cdot 10 \text{y} + i \cdot 0.02506 \text{y}$

$= i \cdot 10.025 \text{y}$

$t_{perceived} = \frac{ t_{total} }{ \gamma }$

$= \frac{ i \cdot 10.025 \text{y} }{ 10 }$

$\approx i \cdot 1 \text{y}$

Because the round-trip signal takes an apparent 1 year, the spaceship might measure the distance between stars as being half a light year. However, repeated measurements would also show that the stars are moving towards the spaceship at nearly the speed of light, meaning light only travels half the distance before bouncing back (the radio signal goes out at about the same speed as the stars are coming in, so they meet in the middle). Thus, the spaceship would conclude the stars are 1 light year apart and moving quite quickly.

I'm too lazy to do the full math right now, but it should turn out that for any given time dilation, the apparent distance between two objects will be lower by the exact same factor. In this case, a Lorentz factor of 10 means Earth will simultaneously perceive the time taken to be 10 times longer, and the distance traveled to be 10 times farther.

Distance Between Internal Objects

Next, let's look at measuring distances between objects in our frame of reference after changing the reference frame of both us and the objects we're measuring.

For example, let's say we park our spaceship on Earth and measure the distance from stem to stern. Then, we accelerate our spaceship to 0.995c and measure again from the spaceship.

Instead of using light years, let's bring that down to light microseconds, or about 300 meters. Conveniently, our spaceship is about 300 meters long. (Says this site and Wikipedia, not that it matters.)

So when our spaceship is sitting on Earth, we put a mirror on the front and a laser pointer and detector on the back, activate the laser, and time how long the laser takes to bounce off the mirror and return to the detector. It takes 2 μs for the signal to hit the mirror and return, so we measure the spaceship to be 1 Lμs long. Which is obvious, because that's how we set up the problem.

Spaceship sitting on Earth. Light takes one microsecond each direction.

Now, we accelerate the spaceship to 0.995c, which gives a Lorentz factor of 10, as noted above. The light is just barely going faster than the mirror up front, so it takes a long time to get there. Then the return trip is almost instantaneous since the detector and light are moving in opposite directions. After calculating the total, we modify it for time dilation to get the perceived time taken.

Spaceship flying through space appears a tenth the length. Light takes twenty microseconds one way, perceived as two microseconds, and almost nothing the other way.

$t_{out}=\frac{ d }{ c - v }$

$t_{back}=\frac{ d }{ c + v }$

$t_{total} = t_{out} + t_{back}$

$= \frac{ d }{ c - v } + \frac{ d }{ c + v }$

$= d \frac{ (c + v) + (c - v) }{ (c + v) (c - v) }$

$= d \frac{ 2 c }{ c^2 - v^2 }$

$= 1 \text{Lμs} \frac{ 2 \frac{\text{Lμs}}{\text{μs}} }{ 1 \left( \frac{\text{Lμs}}{\text{μs}} \right)^2 - 0.995^2 \left( \frac{\text{Lμs}}{\text{μs}} \right)^2 }$

$= 200.5 \text{μs}$

But now we have to take a step back. There's an error here that's non-obvious (to me, anyway). We presumed the distance between the front and back remains the same when the spaceship starts moving. But, for the same reason the stars appear to be closer together in the first example, the spaceship, as measured from Earth, will appear to have contracted along its velocity axis. So we divide the measured resting distance by the Lorentz factor so all the moving values are proper according to Earth's reference frame.

$\require{enclose} \enclose{horizontalstrike}{t_{total} = 200.5 \text{μs}}$

$t_{total} = \frac{ d }{ \gamma } \frac{ 2 c }{ c^2 - v^2 }$

$= \frac{ 1 \text{Lμs} }{ 10 } \frac{ 2 \frac{\text{Lμs}}{\text{μs}} }{ 1 \left( \frac{\text{Lμs}}{\text{μs}} \right)^2 - 0.995^2 \left( \frac{\text{Lμs}}{\text{μs}} \right)^2 }$

$= 20.5 \text{μs}$

$t_{perceived} = \frac{ t }{ \gamma }$

$= \frac{ 20.5 \text{μs} }{ 10 }$

$= 2.05 \text{μs}$

The distance will thus be measured as just over 1Lμs from the spaceship, which is what we expected based on our Earth measurements.

As above, I'm fairly certain there's some math I'm missing that makes the distance exactly equal no matter your reference frame, instead of being a couple percent off, but I'm not sure where the error is.

  • $\begingroup$ "Then the return trip is almost instantaneous since the detector and light are moving in opposite directions." This can not be true. The light returning from the mirror - according to the passengers abord - would travel much faster than lightspeed. $\endgroup$ Commented Oct 7, 2019 at 11:29
  • $\begingroup$ @MadsAggerholm: That's from Earth's frame of reference. $\endgroup$
    – MichaelS
    Commented Oct 7, 2019 at 11:44
  • $\begingroup$ That does not change the fact, that the speed of light would be seen different for the people abord when it goes forward, and when it's returned. Going forward, the speed of the lightbeam looks like it has lightspeed seen from abord, because of the time dilation. When going the other way, it would look even more instantaneous, exactly because of the same time dilation. There is something here that does not add up, but I can't put my finger on why. $\endgroup$ Commented Oct 7, 2019 at 11:54
  • $\begingroup$ @MadsAggerholm: You can't see light while it's traveling. All you can do is model where you think it is, or has been. The Earth-based observer will model it one way, while the spaceship observer will model it a different way. Most of the calculations are done from Earth's point of view, with the last couple showing how the spaceship would see things. A proper spaceship frame of reference would see light going out and assume it was at the speed of light, then see light coming back and also assume it was at the speed of light, and build the model from there. $\endgroup$
    – MichaelS
    Commented Oct 7, 2019 at 12:39
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    $\begingroup$ This answer gives a good explanation of the ambiguity in interpreting the question (at least the first paragraph, which I read). "Is a light year a different distance if measured from a moving object?" Answer: Yes and no, depending on interpretation. $\endgroup$ Commented Oct 9, 2019 at 0:37

In 1983 the General Conference on Weights and Measures determined the definition of a meter to be the distance light travels in a vacuum in 1/299,792,458 of a second, it's current definition. So since all reference frames see light speed the same, a light year cannot change, it is 9,460,730,472,580,800 meters, and the length of a meter is based on the speed of light, a constant, therefore a light year is a constant.

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    $\begingroup$ That makes sense in terms of a description, but if I wanted to travel one light-year and I was traveling at 0.5c, how long would I travel for? Still a year? $\endgroup$ Commented Oct 4, 2019 at 7:09
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    $\begingroup$ An observer at your starting point would observe you to have travelled for 2 years. However, due to time dilation, the trip would take you only 1.73 years, subjectively. Since you both agree on the distance travelled, you will assume you were traveling at ~0.58c. On the other hand, if you measure yourself to be moving at 0.5c, you will not be surprised to find that you have taken exactly 2 years to arrive, but the observer will see you to have taken longer than 2 years, and thus traveling a bit slower than 0.5c. $\endgroup$
    – chepner
    Commented Oct 4, 2019 at 13:44
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    $\begingroup$ This seems to me like a circular argument: "A light-year is constant because a light-(1/299729458 seconds) is constant" $\endgroup$
    – JiK
    Commented Oct 4, 2019 at 14:18
  • $\begingroup$ @JiK: you misread it, the speed of light is the constant, therefore distances based on it cannot change $\endgroup$ Commented Oct 4, 2019 at 14:32
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    $\begingroup$ @AdrianHoward The asker already knows that the speed of light is constant. He is confused about time dilation, and this answer does nothing to explain it. $\endgroup$
    – JiK
    Commented Oct 4, 2019 at 14:33

The definition of a meter is not constant. Presume that one day we scientists determine a way to more accurately measure the speed of light in a vacuum. The constant (c) would not change in this scenario but rather the definition of a meter.

  • $\begingroup$ Speed of light is the definition of the metre. The metre, symbol m, is the SI unit of length. It is defined by taking the fixed numerical value of the speed of light in vacuum c to be 299 792 458 when expressed in the unit m s–1, where the second is defined in terms of the caesium frequency Delta nu Cs. bipm.org/metrology/length/units.html $\endgroup$
    – D Duck
    Commented Oct 6, 2019 at 10:05
  • $\begingroup$ @DDuck: The second is still "only" accurate to about 2 nanoseconds per day. Given better technology, we might change the accuracy of our current measurement of time, and therefore of the meter. The difference here is between the abstract "meter" that's explicitly defined in terms of c, and the practical meter that's measured as best as possible relative to c. tycho.usno.navy.mil/cesium.html $\endgroup$
    – MichaelS
    Commented Oct 6, 2019 at 23:07
  • $\begingroup$ Although I'm not sure this is really related to the question being asked. $\endgroup$
    – MichaelS
    Commented Oct 6, 2019 at 23:07

To give an example, take our universe, but with no spacetime curvature or expansion, to keep things simple. Then to observers on Earth, the distance to the star Alpha Centauri is around 4 light years. However to an alien zooming past with a Lorentz factor of $\gamma \approx$ 600 000, 4 light years stretches from Earth to the Andromeda galaxy!

Light itself (technically, I mean a null curve) does not have any distance or time associated with it. But if one chooses an "observer" -- that is, a velocity -- then that observer can measure space and time intervals for the light. However this is dependent on the observer's motion.


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