If an astronaut spends a year on the International Space Station, by what amount does time slow down for him or her relative to someone who is on the ground?
As I understand it, the time dilatation factor is the square root of the following quantity: 1 - [(v squared) / (c squared)] , where v is the astronaut’s velocity and c is the speed of light. If I assume v can simply be stated in miles per second (not worrying about the vector), and further assume that the space station travels something on the order of one mile per second… then the resulting value for the time dilatation factor is so close to 1.0 that the difference is too small for my hand calculator to display. So I’m guessing that in the course of one year, time would slow down by less than a second. But would it be more like half a second, or more like a millionth of a second?
Using Windows XP built in calculator, and wikipedia’s value for the space station orbital velocity of 7694 m/s, I get a time dilation factor of 1.0000000003288. That’s 328 trillionths. Given 31.5 million seconds in a year, total dilation is about a hundredth of a second.
To answer the question you asked: the usual trick is to use the approximation sqrt(1 - x) = 1 - x/2, which is pretty close to being accurate if x is much smaller than 1. Since the ISS is orbiting at about 7.7 km/s, the time dilation factor is about 1 + 3.3 x 10[sup]-10[/sup]. This works out to about a 10 millisecond difference (i.e. one one-hundredth of a second) over the course of a year.
However, since the ISS is also higher in altitude, there are going to be gravitational effects as well; just as special relativity tells us that time passes differently according to moving observers, general relativity tells us that time passes differently for observers at different points in a gravitational field. The ISS is about 400 km higher in the gravitational field of the Earth; if I’ve done my math correctly, this causes the time dilation to be about 10% smaller than it would in the absence of gravity, or about 9 milliseconds per year instead.
The altitude isn’t the significant factor; it’s the fact that they’re in orbit. That is to say, you’re assuming that they’re in a field of about .88 g, but they’re actually in zero g. The net effect is that time is faster for them than for us, by an amount twice as large as the special relativistic effect muldoonthief and MikeS calculated.
Wait; in what way are they in zero g? Aren’t they constantly accelerating towards the earth? I mean, I assume you know what you’re talking about, but I was unaware that there was any sense in which objects in orbit could be said to be experiencing no gravity.
I’m going to stand by my calculation here. The OP asked for the time dilation relative to an observer at rest on the Earth’s surface, not relative to a distant observer (for whom there’s a nice relation between the potential difference and the orbital velocity.) Plus, the ISS is still moving relative to the distant observers (assuming that said observers are at rest relative to the Earth), so there’s still going to be a v[sup]2[/sup] term if you try to calculate the ISS’s proper time.
The basic idea behind general relativity is that “gravity” is what you experience when you’re not freely falling; in other words, observers in orbit do not experience gravity, and therefore so-called principle of equivalence says that there’s no way for you to distinguish (by yourself) between being in orbit and floating around in deep space. (Modulo a few caveats brought out towards the end of Cecil’s article on the nature of weight.) The “by yourself” part is important, though, because once you start making measurements that involve things far away from you (like, say, the Earth if you’re in orbit), you can pretty quickly figure out that you’re accelerating relative to them.
Right, and since we here on Earth are feeling gravity, we have to take that into account. If you were measuring relative to distant observers (presumably meaning far from any mass), then everyone’s in zero g, so there’s no gravitational effect and the special relativistic effect from velocity is the only one left.
I don’t understand what you are saying here. We have to pick a frame of reference, and the obvious one to me is relative to someone standing on the Earth. So the ISS is moving at a great speed relative to a guy standing on the equator. Therefore time will pass more slowly for the ISS.
However, the ISS is obviously accelerating due to the Earth’s gravitational pull. It is less acceleration than we experience standing here, so time will pass more quickly for the ISS.
For GPS, which is more or less the same case, the clocks run slower due to special relativity (velocity difference). This is about 7 microseconds per day.
The general relativity effect of the difference in the gravitational fields means the GPS clocks run faster - about 45 microseconds per day.
Thus, the difference between the clock on the GPS satellite and the clock on the equator is that the GPS clock is about 38 microseconds faster/day.
(Except that the real clocks on the GPS sats run at different speeds than atomic clocks on Earth, to help correct for relativity; but for this purpose I am assuming that the clocks on the sats are identical to the clocks on Earth)
On a bit of reflection, my post earlier wasn’t quite clear. I agree with Chronos as to the effect of general relativity (I think). However, I don’t agree that one should think of the satellite as being in zero-G.
A better way to think about it is to imagine universal time - no gravity, thus time runs at its fastest rate. Then you compare the satellite’s slowness due to the Earth’s gravity, and the observer’s slowness due to the (higher) gravity on the Earth’s surface.
So, the Earth’s gravity slows down the satellite (compared to Universal Time), but less so than it slows down our observer on the surface of the Earth. Thus, to our observer, the satellite appears to run faster – or it runs less slowly, depending on how you want to think about it.
Other than that, the effect of special relativity is small compared to that of general relativity, so the ISS gains time each day, just like GPS.
Advanced math: You use the Schwarzchild metric to find the dilation for both bodies and then compare those, giving you the difference between the two time systems.
No, the ISS is not accelerating, since there are no non-gravitational forces acting on it. We, however, having an unbalanced force applied to us by the ground, are accelerating.
Of course it is accelerating. The ISS is not moving in a straight line, therefore its velocity is changing, which is otherwise known as acceleration. How else is the ISS moving in a circle if it is not constantly changing its velocity?
This seems wrong to me. Suppose I fire a clock straight up into the air, and it follows a free-fall trajectory. Right before it reaches the apex of its arc at (say) 400 km above the Earth’s surface, I strap on my wristwatch and jump 5 cm into the air. My jump is timed so that according to distant observers (far away from from the Earth and its gravitational field), I reach the top of my jump at the same time as my ballistic clock reaches the top of its trajectory. At that instant, all three clocks — the ballistic clock, my wristwatch, and whatever the distant observers are using to tell time — are at rest relative to each other. However, they are not ticking off the seconds at the same rate; since my wristwatch is lower in the gravitational well of the Earth, proper time is elapsing more quickly relative to coordinate time (i.e. the time measured by the distant observers) for my wristwatch than the ballistic clock’s proper time is elapsing relative to coordinate time. By your logic, though, everyone’s in zero g and has no relative velocity, so there should be no difference between the three observers.
Or, to put it another way: The metric for a weakly curved spacetime is given by
where Phi is the gravitational potential (i.e. del[sup]2[/sup] Phi = rho) with the boundary condition This means that the proper time measured by an observer with coordinate velocity v is related to the elapsed coordinate time by
d tau = sqrt( 1 - 2 Phi(x) - v[sup]2[/sup]) dt
There will therefore be a difference in the rates of proper time elapsed for two observers at rest (relative to distant observers) at x[sub]1[/sub] and x[sub]2[/sub] given by
d tau[sub]1[/sub]/dt - d tau[sub]2[/sub]/dt = Phi(x[sub]2[/sub]) - Phi(x[sub]1[/sub])
even though they’re not moving relative to each other. Or have I gotten horribly confused somewhere?
There are no forces acting on the ISS, therefore it cannot be accelerating, therefore the path it is following, despite appearances, must be a straight line.
MikeS, give me a little more time to think about that one.
Struggling with this one a bit. To clarify, you’re talking about this in a relativistic frame? In other words, the ISS is following an inertial straight line path in curved spacetime, right? And looking at it this way, gravitational ‘force’ isn’t really a force? I think I’ve heard this line of thought before, but it’s been a while.
Likewise, is a skydiver not accelerating in this point of view? (neglecting air resistance, etc.)
Fine, but then you are just using misleading semantics. If the path of the satellite is a straight line on curved spacetime, then all we have to consider is that the ISS is higher up in a gravity field than our observer on Earth. It has less gravitational potential energy, therefore time moves faster on the ISS than on Earth, and light from the iss will be shifted accordingly.
Either way, I don’t understand what you are trying to claim. The ISS is affected both by special and general relativity, because the ISS is moving relative to Earth and because it is higher in the Earth’s gravity field, which is also exactly what MikeS is saying here. I think what you are doing is conflating the various frames of reference here in saying that the ISS has no forces acting on it vs gravity being a proper acceeleration from the viewpoint of an outside observer.
If we had a universe where there was only the Earth, the ISS, and an observer far enough away that he is not affected by the Earth’s gravity, he would not say that the ISS is in zero-g, nor would he say that the observer on the surface of the Earth is in zero-G. He would find that the ISS has a slightly slower clock than he does, since it is moving (assuming the observer and the earth are stationary relative to each other). He would find that the ISS’ clock was slower than accounted for only by special relativity, because it is in the Earth’s gravitational field. He would finally find that that the Earth clock was even slower because the clock on the Earth is deeper within a gravity field than the ISS.
Now imagine that the ISS starts its engine and comes to rest relative to the Earth and the observer. Its clock will STILL be faster than the Earth’s, and slower than the observer’s. It’s now undergoing an acceleration in our universal frame of reference, since it is straining against gravity. No special relativity applies, but general still does.
Of course, if you are actually saying something else, you should try to explain it in more than one sentence.
The height isn’t the only thing we have to consider: We also have to consider that the observer on Earth is in a non-inertial reference frame (relativistically speaking), while the observer on the ISS is in an inertial frame.
This is misleading: General Relativity contains Special Relativity in its entirety, so if you consider the GR effects and then also consider the SR effects, you’re double-counting the SR effects.
There is the problem. You are trying to measure the dilation from two different frames. Either we are using the frame of the observer on Earth or the observer on the ISS or a universal observer. You don’t need or want to mix and match. The GR and SRT dilations both need to be counted, because if you split the frames, then you postulate some sort of fictional gravitational field caused by the ISS’ motion – which is how GR solves the twins problem.
Looking at the GPS system, it has to be corrected for relativity or it will be inaccurate. The link above says by how much (7 microseconds/day slower + 45 microseconds faster = 38 microseconds/day fast). Here is a pdf with the math in it. These guys are rocket scientists, and GPS works, so the physics must be correct.
Fun fact: no clock in Earth orbit can beat slower than a clock on the Earth. The velocity at which SRT dilation equals the GR dilation is the escape velocity of the gravitational field.