# Rapid (ac/de)celeration in relativity does what to inertial clocks?

EDIT (to clarify my question): I think some of the answers here are accounting for light travel time and telling me what I'd actually see on Earth's clock, so I've edited my first paragraph to clarify. I don't think this clarification changes the meaning of this question, but it might.

Summary: if I accelerate to $0.8 c$ in 1 second, how much time passes for observers in my starting inertial reference frame?

This seems like a simple question that has probably been answered, but I couldn't find a simple answer for what appears to be a simple question:

• I start 8 light years from Earth, at rest with respect to Earth, and observe Earth's time is t=0. Of course, technically, I'm seeing Earth the way it was 8 years ago (t=-8), but I know I'm 8 light years away from Earth, so automatically add 8 years to the time I see.

I make this assumption throughout the question. In other words, when I say "Earth's clock time", I mean: "the time I'm seeing on Earth's clock right now plus my distance from Earth in light travel time".

I believe this is the norm in relativity questions, but could be wrong about that.

• Keeping an eye on Earth's clock, I accelerate to $0.8 c$ in 1 second. Because I'm accelerating, I know Earth's clock will go faster than mine. The question is: how much faster, and where will it end up after I've finished my one second of acceleration to $0.8 c$?

• At $0.8 c$ the distance to Earth is now 4.8 light years (minus the little bit I traveled during acceleration). Earth's clock now runs slower than mine by time dilation. So, when 6 of my years have passed, fewer than 6 years have passed on Earth's clock.

• As I get close to Earth, I "decelerate" to the Earth's reference frame so that I will be at rest when I actually arrive at Earth. Of course, deceleration is just acceleration in a different direction, so, once again, Earth's clocks run faster than mine.

• And, once again, the question is: in that 1 second of deceleration, how much time elapsed on Earth's clocks?

• In the 6 years I was traveling at $0.8 c$, Earth's clocks ticked off only 3.6 years by time dilation.

• By the time I arrive at Earth, Earth's clocks must have ticked off 10 years, since they say me traveling at 0.8c for (most of the) 8 light years.

• The only way I can reconcile these numbers (10 years minus 3.6 years, or 6.4 years) is that my 1 second of acceleration and deceleration each took 3.2 Earth years (about 10^8 seconds).

• This seems high, and I can't get the numbers/formulas to yield this, but...

• On the other hand, it seems somewhat reasonable that the amount of time that passes depends only on my final velocity ($0.8 c$) and not how fast I reached that velocity.

Note that I don't think there's a simultaneity issue here, since I start and end in Earth's reference frame.

• If you accelerate under a short time, the (de)acceleration, will not significatly affect the travel time. The main difference in the clocks will happen since you are moving in the first place. – Mikael Fremling Mar 21 '16 at 14:15
• @MikaelFremling I agree that's true if you're the one accelerating. I'm not as sure it's true if you're "standing still" (in an inertial frame) and watching someone else accelerate. – barrycarter Mar 21 '16 at 14:24
• Of course there is a simultaneity issue. You ask "what will earth's clock show at the moment I stop accelerating?", but different observers will disagree about that. The main moral of relativity is that questions like the one you're asking make no sense in the first place. – WillO Mar 21 '16 at 15:11
• @barrycarter Of course clock on earth that you actually see will tick forward inevitably, but the clock on earth that you percieve as simultaneous to you will move backwards if you accelerate away from the earth. – Mikael Fremling Mar 21 '16 at 17:32
• @barrycarter No, i'm saying the oposite. You will never be able to actually see into the earths past. Instead what happens is that your perception of what happens simoultaneously will steach into earths past. (I'll post a propper answer with space time diagrams) – Mikael Fremling Mar 22 '16 at 9:26

That's a harder question than you think, because to answer it requires the calculation of a geodesic in the Rindler coordinates.

If you are accelerating at some constant acceleration $a$ then the metric in your (accelerating) coordinates is given by:

$$ds^2 = -\left(1 + \frac{ax}{c^2}\right)^2c^2dt^2 + dx^2$$

You're accelerating towards the Earth, so let's put the Earth at positive $x$, which means the acceleration $a$ is also positive. In these coordinates the Earth now accelerates towards you, tracing out a world line as it does so, and you need to solve the geodesic equation, put in the initial conditions then calculate the length of the Earth's world line corresponding to 1 second that you accelerate for.

I have to confess I don't know how to do this calculation and in fact I don't even know if the geodesic has a closed form equation. I've been Googling for the solution to this for some time with no luck. If anyone knows how to do this calculation I'd be interested to hear how it's done.

But we can get a rough idea as follows. Suppose you accelerate for a short enough time that the Earth doesn't move significantly, then the path length of the Earth's world line is just its elapsed time. And we can calculate this very easily just by setting $dx = 0$ and rewriting the metric as:

$$d\tau^2 = \left(1 + \frac{ax}{c^2}\right)^2dt^2$$

which immediately gives us:

$$\frac{d\tau}{dt} = 1 + \frac{ax}{c^2}$$

In this equation $x$ is the distance to the Earth (8 light years) and $a$ is your proper acceleration (0.8$c$/sec). Putting these values into your equation I get:

$$\frac{d\tau}{dt} \approx 2 \times 10^8$$

So during the $1$ second you accelerate about $2 \times 10^8$ seconds passes on Earth.

• This seems odd. If the same traveler acclerates from 0 to $.8c$ in an instant, then 6.4 earth years pass in that instant (in the sense that if the traveler sees an earth-clock reading $0$ as simultaneous with his departure immediately before departing, he will see an earth-clock reading $6.4$-years as simultaneous with his departure immediately after departing). You've spread the acceleration over a second, and reduced that 6.4 years to about 4 hours. Can this be right? Or are you answering a subtly different question? – WillO Mar 21 '16 at 21:29
• Okay, I looked at this again. You appear to be using the Rindler coordinates to compute the length of a segment of the earth's worldline --- which is to say that you are calculating (I think) a proper time as measured by observers on earth. But the OP (in comments) said that he is interested in clock time on earth as it appears to the traveler, not time as measured on earth. The answer to that question is clearly (approximately 6.4 years). You have answered a different (but considerably more interesting!) question. Do I have this right? – WillO Mar 22 '16 at 1:42
• @WillO: there is always confusion about the difference between observing and seeing, where the latter is normally taken to mean what appears on the retina i.e. including travel time of the light. I confess that I'm now not sure what is intended here. As you say, I've simply compared proper time to the acceleratee's coordinate time. I'll have to find the time to calculate the whole trajectory properly as it is an interesting problem. – John Rennie Mar 22 '16 at 6:18
• @WillO, John Rennie: About calculating things from an accelerated observer's view, you may find interesting Sec.3 of arxiv.org/abs/physics/0405038. – udrv Mar 22 '16 at 6:46

Here is a spacetime diagram showing what happens and how one resolves the question of what is simultaneous and not.

You need to know two things about space-time diagrams. Diagonal distances are computes using a Pythagoras theorem but with a minus sign. Thus, the time it takes for a traveller to cover the distance $X$ in time $T$ (ass seen on earth) is $$t^2 = T^2 - X^2$$.

See the diagram bellow where these different times are marked.

Let's do the math: You are $X=8$ light years away from earth. You have synctronized your clocks to that both you and the earth have you start at $T=0$. However the clock on earth that you actually see is showing $T=-X=-8$ yrs since that is the time it took the light to reach you. (The light cone is drawn in dotted orange).

Then you accelerate (insanely fast) so that you reach $v=0.8c$ in a second. You will still read $T=-X=-8$ yrs on the earth clock (it's the same light arriving).

Now however, your view of what is simultaneous on earth has changed. This is because your line of simultaneity has changed. In this diagram above (and bellow) I have also drawn the lines of simultaneity as dot-dashed pink lines. (The light cone is drawn in dotted orange). The rule in space-time diagrams is that the line of simultaneity makes the same angle with a ray of light as the world line of a traveller (the purple lines).

Since you are travelling at $v=0.8$ ($c=0$) your line of simultaneity in the diagram has a slope that is $1/v$. This line intersects the world-line of earth at $T=X*v=8*0.8=6.4$ yrs. Thus you conclude that you are now simultaneous with earth 6.4 yrs into the future. But you are still seeing the light form 8 yrs in the past.

Now you can see that when you are approaching earth, you consider earth future to be simultaneous with your time. However the clock on earth is slower than yours and you will see how you steadily catch up with earth’s future and by the time you have arrived you are at the same time.

Iv'e illustrated this below, the iv'e removed the light cone for clarity.

This is how you feel it on the ship though. Your are $X=8$ yes away and the trip will (from the earths perspective) take $T=X/v=10$ years. For you the trip will take $$t^2 = T^2 - X^2 = 10^2 - 8^2 = 100 - 64 = 36$$ so $t=6$ yrs. Thus as you approach earth you are going towards the light emmited from earth and the clock you are wathich will gor from $T=-8$ yrs when you start to $T=10$ yrs when you finish. The observed clocks on earth will thus look like they are ticking at a rate of $(10+8)/6=3$ times as fast as yours.

Finally in the last figure bellow, iv'e drawn what happens when one travels away form the earth.

In this case you you will be simultaneous with earth’s past (purple lines). But the information the reaches you from earth will still be in chronological order (orange light lines).

This even though relativity changes you perspective of what is constitutes simultaneity you can never peer into the future by going fast (in any direction).

Hope this helps a bit.

• Thanks. I was actually hoping for a more mathematical answer. I don't have a problem with diagrams, but I believe the underlying math is more important. Could you tell me, in numbers, what time I see on Earth's clock 1) after I'm dong accelerating, 2) as I approach Earth, and 3) after I'm done decelerating? These are the numbers I can't reconcile. – barrycarter Mar 22 '16 at 12:06
• I also like the math, but when it commies to relativity I, I find space-time diagrams very useful (they are equivalent to the math anyway) since you can easier pinpoint where people become confused. – Mikael Fremling Mar 22 '16 at 13:22
• I find diagrams are more confusing, but I do like your edited answer. You're saying what I actually see is continuous, but the times I assign it are not. You're also saying that the "Doppler shift" exactly cancels out the Lorentz transform. I want to look into this a little more, but I think this is the best answer I've seen so far. – barrycarter Mar 25 '16 at 16:42

I'm going to address the latter part of your question, namely how to reconcile the $10$ years that passed on the Earth's clock with the $3.6$ years that the traveller might have naively expected. I'm not going to focus on the details of the accelerating phase, and I'll assume that the change in velocity from $0$ to $0.8c$ happens instantaneously.

The solution to the paradox, as is so often the case in relativity, has to do with simultaneity. In particular, the observer on Earth and the traveller will disagree on whether or not their clocks started at zero at the same time.

The situation is sketched in the Minkowski diagram below. The $(x,t)$ coordinates represent the Earth frame, and $(x',t')$ the traveller's frame when he's moving with constant velocity $v=0.8c$ towards the Earth. The Lorentz factor is $$\gamma = \frac{1}{\sqrt{1-v^2/c^2}} = \frac{5}{3}.$$

Suppose that the observer has a clock $\text{A}$ on which he reads a time $t_\text{A}$, and the traveller has a clock $\text{B}$ displaying a time $t'_\text{B}$. Let's also assume that $\text{A}$ and $\text{B}$ are synchronized before the traveller starts accelerating, such that $t_\text{A}=0$ and $t'_\text{B}=0$ at the same time, in the observer's frame. At that moment, the traveller accelerates to a velocity $v=0.8c$ towards the observer on Earth, and after this rapid acceleration he is at rest in the $(x',t')$ inertial frame. Also, his initial distance is $d=8$ ly, in the observer's frame.

The relation between both frames is given by the Lorentz transformations: \begin{align} x' &= \gamma\left[(x-d) + vt \vphantom{1^1_1}\right]\tag{1},\\ t' &= \gamma\left[t + (x-d)v/c^2\vphantom{1^1_1}\right]\tag{2}. \end{align} (notice the $+$ signs because the traveller moves in the negative direction, and the extra $d$'s because of the offset between the origins of the frames). The traveller is at rest in his own frame, i.e. $x'=0$, and from eq. (1) we get his path in the observer's frame: $$x = d - vt.$$ The observer (at $x=0$) reads on his clock $\text{A}$ that the traveller reaches him when $$t_\text{A} = d/v = 10\;\text{y},$$ and from eq. (2) we find the corresponding travel time for the traveller: $$t'_\text{B} = \gamma\left[t_\text{A} -dv/c^2\vphantom{1^1_1}\right] = \gamma\left[t_\text{A} -t_\text{A}v^2/c^2\vphantom{1^1_1}\right] = t_\text{A}/\gamma = 6\;\text{y},$$ as expected. So, from the observer's point of view, clocks $\text{A}$ and $\text{B}$ were initially synchronized, but went out of sync due to the traveller's time dilation.

What happens from the point of view of the traveller? First of all, he finds that the distance to the observer has changed: he measures the distance $d'$ along the $x'$-axis, which means setting $t'=0$ and $x=0$. From eq. (2) we have $$t = dv/c^2\tag{3},$$ and plugging this into eq. (1) we get $$d' = -x' = -\gamma\left[-d + dv^2/c^2 \vphantom{1^1_1}\right] = d/\gamma= 4.8\;\text{ly},$$ consistent with $t'_\text{B} = d'/v$. But what does he conclude about clock $\text{A}$? Due to his acceleration, he changed from the $(x,t)$ inertial frame to the $(x',t')$ inertial frame. With it, his notion of simultaneity has also changed: in the $(x',t')$ frame, clocks $\text{A}$ and $\text{B}$ did not start at the same time.

Indeed, let's give the observer on Earth a second clock $\text{C}$, which is set to zero at the same time as clock $\text{B}$ according to the traveller in the $(x',t')$ frame. In other words, $t_\text{C}=0$ at $t'=0$ and $x=0$. And we already calculated in eq. (3) what the corresponding time is on clock $\text{A}$: $$t_\text{A} = dv/c^2 = 6.4\;\text{y,}\quad\text{for }t_\text{C}=0\;\text{y},\tag{4}$$ in other words $$t_\text{C} = t_\text{A} - dv/c^2.$$ And now the paradox is resolved, because when the traveller arrives on Earth ($x=0$), we find from eq. (2): $$t'_\text{B} = \gamma\left[t_\text{A} - dv/c^2\vphantom{1^1_1}\right] = \gamma\, t_\text{C},$$ or $$t_\text{C} = t'_\text{B}/\gamma = 3.6\;\text{y},$$ which is consistent with $$t_\text{C} = t_\text{A} - dv/c^2 = 10 - 6.4 = 3.6\;\text{y}.$$

UPDATE

What does the traveller actually see as he's receiving signals from clock $\text{A}$? Will he see a sudden jump? No, after his acceleration he'll see that clock $\text{A}$ is ticking 3 times faster than his own clock, despite the effect of time dilation. The reason for this is the Doppler effect: it takes less and less time for light signals to travel from the observer to him, as he approaches Earth. I've added a few of those light rays in the figure.

Suppose we have a signal that is sent from clock $A$ at time $t_\text{A,s}$. This signal will follow a path $$x = c(t - t_\text{A,s}).$$ The traveller moves on the path $x=d-vt$, so he'll receive the signal when $c(t - t_\text{A,s}) = d-vt$, or $$t = \frac{d+ct_\text{A,s}}{c+v}.$$ This corresponds with a time $t'_\text{B,r}=t/\gamma$ on the traveller's clock (the subscript $\text{r}$ stands for 'received'), which means $$t'_\text{B,r} = \sqrt{\frac{1-v/c}{1+v/c}}\left(d/c +t_\text{A,s}\right) = \frac{1}{3}(8 + t_\text{A,s}). \tag{5}$$ Thus he sees that clock $\text{A}$ is ticking 3 times as fast as his own. He'll receive the signal $t_\text{A,s}=-8$ y at $t'_\text{B,r}=0$ y; he'll receive the signal $t_\text{A,s}=-6$ y at $t'_\text{B,r}=2/3$ y, and so forth, until he sees $t_\text{A,s}=10$ y when $t'_\text{B,r}=6$ y as he arrives on Earth.

• If I understand correctly: in the one second it takes the traveler to accelerate, the Earth time he computes jumps from 0 to 6.4 years, and the Earth distance he computes jumps from 8 light years to 4.8 years. Is this correct? If so, accounting for light travel time, he sees Earth at -8 years before the acceleration, and Earth at 6.4-4.8 or +1.6 years afterwards. In other words, his view of Earth jumps 9.6 years in one second? – barrycarter Mar 29 '16 at 15:43
• @barrycarter No, he doesn't see any jumps. What he will see is that the observer's clock is now ticking 3 times as fast as his own, due to a Doppler effect. I've updated my answer with this info. – Pulsar Mar 29 '16 at 18:28

This is not a new answer, but it's a little too long for a comment and might help avoid some momentary confusion for anyone new who stumbles on this.

The OP has a traveler 8 light years from earth who accelerates from earth's rest frame to $v=.8$ (in the direction of the earth) in one second (as measured by the traveler), and asks how much time passes on earth during that acceleration.

There are two very different ways to interpret this question.

Question A is: How much time does the traveler say passes on earth clocks during the acceleration?

Question B is: How much time does an earthbound observer say passes on earth clocks during the acceleration?

If the acceleration is instantaneous, it's very easy to see that the answers to these questions are "6.4 years" for Question A and "zero" for Question B.

Because the actual question makes the acceleration not instantaneous but rather 1 second long according to the traveler, the exact answers are approximations to 6.4 years and to zero.

To work out the correction for Question A is both easy and boring. To work out the correction for Question B is both interesting and difficult. John Rennie's answer addresses only Question B, where he finds that the correction is approximately 14,000 seconds.

• I think Question A is the only possible interpretation of my question, but OK. Your answer at least makes the numbers add up properly, but what about my 1 second deceleration when I reach Earth? If 6.4 years pass during my 1 second acceleration, shouldn't 6.4 years also pass during my 1 second deceleration? If each of these were 3.2 years, the numbers would still add up. There is no need to call my question "boring". – barrycarter Mar 23 '16 at 14:26
• @barrycarter : 6.4 years don't pass on Earth while you accelerate to 0.8c in a second. You know this, because you know you could accelerate to 0.8c and decelerate to zero repeatedly all the way to Earth. – John Duffield Mar 26 '16 at 15:42
• @JohnDuffield If I did that, lots of time would pass on Earth, 100s to 1000s of years depending on how often I accelerated/decelerated. Remember, acceleration means you age less than all inertial frames. And deceleration is just acceleration in another direction. – barrycarter Mar 26 '16 at 15:46
• @barrycarter : it wouldn't. Your average speed would be 0.4c, it would take you 20 of your years to get to Earth, during which time circa 24 Earth years would have elapsed. – John Duffield Mar 26 '16 at 15:50
• @JohnDuffield Please join me in the h bar, this discussion works better in chat. – barrycarter Mar 26 '16 at 15:51

Rapid acceleration/deceleration in relativity does what to inertial clocks?

Nothing much. Think about the parallel-mirror light clock used in the Wikipedia simple inference of time dilation due to relative velocity. Note that it's due to relative velocity. You need some acceleration to get that relative velocity, but it's the relative velocity that matters.

Summary: if I accelerate to 0.8c in 1 second, how much time passes for observers in my starting inertial reference frame?

One second! LOL! But let's just say your average speed is 0.4c. Using the Lorentz factor $$\Delta t' = \frac{\Delta t}{\sqrt{1-\frac{v^2}{c^2}}}$$ gives a time dilation factor of 0.9165. If we say you're moving and they aren't to avoid getting bogged down with the "Twins paradox", one second of your time is circa 1.1 seconds of their time. Note though that it isn't actually correct to use your average speed. The time dilation increases square-wise, not linearly. See Wikipedia for some more complex expressions.

I start 8 light years from Earth, at rest with respect to Earth, and observe Earth's time is t=0. Of course, technically, it will take me 8 years to know this (speed of light travel time), but I believe special relativity is OK with my saying "it's currently t=0 on Earth", since I'm in the same frame as Earth.

No problem.

Keeping an eye on Earth's clock, I accelerate to 0.8c in 1 second. Because I'm accelerating, I know Earth's clock will go faster than mine. The question is: how much faster, and where will it end up after I've finished my one second of acceleration to 0.8c

You should forget about the acceleration. We're talking about a fraction of a second. That's nothing compared to your ten-year trip. Work out the time dilation factor, and it's $\sqrt{1-0.8^2}$ or 0.6. Your trip will take six years ship time.

At 0.8c the distance to Earth is now 4.8 light years (minus the little bit I traveled during acceleration). Earth's clock now runs slower than mine by time dilation. So, when 6 of my years have passed, fewer than 6 years have passed on Earth's clock.

No. The distance to Earth is 8 light years. You know full well that this distance doesn't change just because you're going fast. You know about length contraction and that if you tried to measure the distance you'd get 4.8 light years. But you know that it's your measurements that's changed, that's all. And you should know that you're time dilated, and that the Earth's clock will clock up ten years whilst you're travelling.

As I get close to Earth, I "decelerate" to the Earth's reference frame so that I will be at rest when I actually arrive at Earth. Of course, deceleration is just acceleration in a different direction, so, once again, Earth's clocks run faster than mine.

Forget about the acceleration and deceleration. The Earth's clock end up four years different to your own.

And, once again, the question is: in that 1 second of deceleration, how much time elapsed on Earth's clocks?

We're talking about a tenth of a second. This is nothing compared to the four years. Forget it.

What vexes me about this problem: In the 6 years I was traveling at 0.8c 0.8c, Earth's clocks ticked off only 3.6 years by time dilation.

You're getting sucked into the Twins Paradox here. Steer clear. Just focus on the light paths in parallel-mirror light-clocks, like this /\/\/\/\ and this: ||.

By the time I arrive at Earth, Earth's clocks must have ticked off 10 years, since they say me traveling at 0.8c for (most of the) 8 light years.

Yep.

The only way I can reconcile these numbers (10 years minus 3.6 years, or 6.4 years) is that my 1 second of acceleration and deceleration each took 3.2 Earth years (about 10^8 seconds).

Nope.

This seems high, and I can't get the numbers/formulas to yield this, but on the other hand, it seems somewhat reasonable that the amount of time that passes depends only on my final velocity (0.8c) and not how fast I reached that velocity.

Like I said, you need acceleration for the relative velocity, but it's the latter that makes the clock readings different. Think up a scenario with the same acceleration and double the distance. Now there's an eight-year difference between your clocks and the Earth clock. Give or take a few tenths of a second or two.