Traveling between two planets at rest to one another [duplicate]

If I travel at relativistic speed from planet A to planet B which are at rest relative to one another, I will be younger than people on A or B when I arrive. However how does this mesh with the fact that the change in proper time should be symmetrical, i.e. I should observe events on A as well as B as moving at a slower rate while they observe events for me to be moving at a slower rate, so when I arrive at B why would I be younger? I understand this is similar to the twin paradox and other questions I have asked but I still don't understand how you can resolve the discrepancy since you remain in one inertial frame for the entire journey. Is it because I have to de-accelerate and thus change reference frames to arrive at B, and if so would the effect be the same if I never accelerated or de-accelerated from A to B I just merely flew past them with some velocity set for me at lets say the big bang?

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marked as duplicate by Brandon Enright, Colin McFaul, Qmechanic♦Jul 6 '14 at 23:29

possible duplicate of How can time dilation be symmetrical? – ACuriousMind Jul 6 '14 at 18:38
I asked that question, the reason I'm asking this one is that I did not find the answer on the last one satisfactory so figured I'd ask a better version of the question. Linking it as a duplicate does not help me much. – Krel Jul 6 '14 at 18:40
And, as to all your other questions, the answer is: Proper time is a Lorentz invariant. I really don't know where you get the weird idea from that it should be "symmetrical". – ACuriousMind Jul 6 '14 at 18:40
Because it has been taught that if you are flying in a ship at relativistic speeds and pass another ship you will see time running slower for them while they will see time running slower for you. Thus it seems symmetrical. – Krel Jul 6 '14 at 18:43
@bright magus: SR speaks clearly about Lorentz invariance. The whole talk of frames and who sees what is an unfortunate pedagogical failure, just as dmckee says. Proper time is Lorentz invariant no matter whether anything is accelerating or not. If you do not understand the power of Lorentz invariant quantities, you have not understood SR. dmckee is absolutely right. – ACuriousMind Jul 6 '14 at 21:27

so when I arrive at B why would I be younger?

I believe I addressed this in another question of yours.

Once again, assume that when you pass planet A, your clock and planet A's clock both read $t = t_A =0$.

Now, according to the inhabitants of planet A, planet B's clock is synchronized with their clock.

However, in your inertial frame of reference, planet B's clock is ahead of planet A's clock.

Assume for simplicity that planet B is 1 light-second from planet A, in the rest frame of the planets, and that your are travelling at $0.5c$ towards planet B.

Then, when you pass planet A, you observe planet A's clock to read $t_A = 0$ and you observe planet B's clock to read $t_B = 0.5s$

When you fly past planet B, you observe your clock to read $t = 1.732s$ and you observe planet B's clock to read $t_B = 2s$.

So, in fact, you observe that planet B's clock runs slow; your elapsed time is $\tau = 1.732s$ while planet B's elapsed time is $\Delta t_B = 2s - 0.5s = 1.5s$

Moreover, the inhabitants of planet B observe your clock to run slow. They observe that you passed planet A when their clock read $t_B =t_A = 0$ so, according their clock, you took $2s$ to make the trip while your clock only showed $1.732s$.

Thus, the time dilation is symmetrical - you observe planet B's clock to run slow and they observe your clock to run slow.

Note that this is not a contradiction and is made possible by the fact that the two planetary clocks are not synchronized in your frame of reference.

These are the calculations for the above numbers...

When your clock reads $t=0$, planet B's clock reads

$$t_B = \frac{0.5c \cdot 1ls}{c^2} = 0.5s$$

Since you cover 1 light-second at a speed of $0.5c$ in the rest frame of the planets, your elapsed time is

$$\Delta t = 2s \cdot\sqrt{1 - 0.5^2} = 1.732s = \tau$$

Since, according to you, planet B's clock is moving, you should calculate that

$$\Delta t_B = 1.732s \cdot \sqrt{1 - 0.5^2} = 1.5s$$

which agrees with what you observe

$$\Delta t_B = 2s - 0.5s = 1.5s$$

Still what would happen if the traveler decided to drastically decelerate as they passed B? How would their clock go from reading 2 seconds to 1.5 seconds for me?

As long you stay inertial, the time dilation is symmetric.

However, if you suddenly decelerated to zero speed (relative to the planets) upon arriving at planet B, you would now find that your clock runs at the same rate as the planetary clocks, which you now observe to be synchronized, and that you are behind them by $2s - 1.732s = .268s$.

Since you know that your clock read $t=0$ when $t_A=0$ you know you aged less than the inhabitants on planet A. Essentially, you would 'see' that the inhabitants of planet A 'jumped' in age by 0.5s during the deceleration

Just before the deceleration, you would observe planet A's clock to read $t_A = 1.5s$.

Just after the deceleration, you would observe planet A's clock to read $t_A = 2s$.

Since you are co-located with planet B just before and after the deceleration, you would not observe planet B's clock to change.

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Ah right simultaneity is broken between A & B if your in a different inertial frame. I have continued to make that mistake in the scenarios I've asked about. Still what would happen if the traveler decided to drastically decelerate as they passed B? How would their clock go from reading 2 seconds to 1.5 seconds for me? Also if B was to drastically accelerate to match the travelers velocity as they passed, I'm assuming the inhabitants of planet B would then be the younger ones? – Krel Jul 6 '14 at 22:05
What I've been trying to ask is how the separate inertial frames come to agree on their previous observations when they align themselves in the same frame. If I've been orbiting a planet at .5 c for some time and I would observe everyone on the planet as experiencing less time until I decelerate, at which point I'd realize they all experienced more time. That process is confusing to me. Likewise if someone from the planet were to match my speed while I orbit, suddenly they become the younger ones and I the older. Again confusing. What am I missing here? – Krel Jul 6 '14 at 22:11
Alfred Centauri: "as long you stay inertial, the time dilation is symmetric. However, if you suddenly decelerated to zero speed..." - this means that the source of time dilatation here is claimed to be deceleration only, and not the difference in uniform speed before. Otherwise, the total difference after landing on the planet would also depend on the time of travel at uniform speed. But then time dilatation due to uniform speed would not be symmetric. – bright magus Jul 6 '14 at 22:31
@krel, see updates to my answer – Alfred Centauri Jul 6 '14 at 22:31
@krel, and have you drawn the spacetime diagrams of your various scenarios yet? You really should because, once you start thinking in terms of spacetime diagrams and events, you can 'picture' the solution in your mind's eye. – Alfred Centauri Jul 6 '14 at 23:09

While travelling in an inertial reference frame, you perceive the time of objects moving relative to you to go slower than your time. For such situations, you may apply the naive notion of time dilation. As soon as you accelerate anywhere, you should forget about time dilation as a way to gain what you will read on any clock. Time dilation is not the only concept in SR. The right concept for elapsed times is:

Regardless of acceleration, for any path $\gamma$ in spacetime travelled, the elapsed time on a clock on the end of that path is the proper time $\tau = \int_\gamma \sqrt{\mathrm{d}x^\mu\mathrm{d}x_\mu}$.

In order for you to meaningfully say that you are "younger" or "older" than anyone else, you must both be in the same inertial frame.

If anyone is travelling from anywhere to anywhere else, they must always accelerate or decelerate in order to compare their ages to the people living at the end of such paths.

Therefore, time dilation will not yield any meaningful result about who is "older" or "younger" since it is only formulated for inertial frames.

There is no paradox because calculating the proper time for each travelled path through spacetime will yield unambiguous results about which clock reads what since proper time is a Lorentz invariant.

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Many comments on this answer (by myself, and others) have apparently been removed recently. I hereby restate my objection. ACuriousMind: "for any path $\gamma$ in spacetime travelled, the elapsed time on a clock on the end of that path is the proper time $\tau = \int_{\gamma} \sqrt{dx^\mu dx_\mu}$. [...] calculating proper time for each travelled path [...] will yield unambiguous results about which clock reads what" -- What do you suppose do "readings" of a clock have to do with "proper time" values for any or all particular path segments of a given clock at all; or even unambiguously? – user12262 Jul 9 '14 at 16:21
@user12262: Why proper time is the reading of a clock at the end of a spacetime path is discussed here. This is unambiguous, as by Lorentz invariance all observes must agree on the proper time a path in spacetime has. – ACuriousMind Jul 9 '14 at 16:38
ACuriousMind: "Why proper time is the reading of a clock at the end of a spacetime path is discussed here [ Why do clocks measure arc-length? ]" -- There's no mentioning on that page of "clock reading", "a clock reads", or variations thereof. So ... what (which terminology) on that page do you call "reading of a clock", please? – user12262 Jul 9 '14 at 16:54
@user12262: The reading of a clock is, quite naturally in my opinion, the $I_{ba} = \int_{\lambda_a}^{\lambda_b} \mathrm{d}\lambda \dot{t(\lambda)}$ looked at in joshphysics' question, defined as simply the time $t$ that passes in the frame in which the clock is stationary at all times. It is then shown that this can be expressed as the invariant proper time. – ACuriousMind Jul 9 '14 at 17:05
ACuriousMind: "The reading of a clock is, quite naturally in my opinion, [...] simply the time $t$ that passes in the frame in which the clock is stationary at all times." -- Is the phrase "the time $t$ that passes" just another way of saying "the elapsed time on a clock", or "the proper time" (as in your answer above), or "the duration of that clock"? If so, then why the separate symbol: "$t$" instead "$\tau$"? (And for what it's worth, in my humble opinion any "reading of a clock" is quite simply some real or integer number read off an en.wikipedia.org/wiki/Clock#Indicator ) – user12262 Jul 9 '14 at 17:45

If I travel at relativistic speed

... let's say at constant speed $\beta ~ c$ ...

from planet A to planet B which are at rest relative to one another

then

(1) A and B succeed in determining which indication of A had been simultaneous to which indication of B (and vice versa); and

(2) A's duration from indicating your departure, until the (A's) indication simultaneous to B's indication of your arrival is equal to
B's duration from the (B's) indication simultaneous to A's indication of your departure until indicating your arrival; and

(3) your duration from indicating A's departure until indicating B's arrival is
$\sqrt{ 1 - \beta^2 }$ of
the duration(s) described above in (2).

I will be younger than people on A or B when I arrive.

That's not at all guaranteed, but it depends on

• whether you had been as young as the people of A at your departure,

• whether the people of B, at their indication simultaneous to A's indication of your departure, had been as young as you (and the people of A) at your departure,

• whether the people of B and the people of A had been aging equally (as determined by comparing their simultaneous indications), and

• whether the people of A or B had been aging equally as you did; in proportion to the duration ratio described above in (3).

In other words:
The duration ratio of (3) can be derived completely independent of any comparisons of "youthfulness of appearance".

However how does this mesh with the fact that the change in proper time should be symmetrical [...]

For a symmetric setup we should consider someone, say Q, who is and remains at rest to you, such that A travelled at speed $\beta~c$ from you to Q.
Then, symmetric to (3) above:
A's duration from indicating your departure until indicating Q's arrival is
$\sqrt{ 1 - \beta^2 }$ of
your duration from indicating A's departure until the (your) indication simultaneous to Q's indication of A's arrival.

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You have got a synchronization problem because you are describing 2 events (departure planet $A$ and arrival planet $B$), but there are 4 events in spacetime to be taken into account (state of planet $B$ at the departure and state of planet $A$ of the arrival are missing).

The contradiction is relatively simple to track, only applying the time dilation formula $$T=\gamma\tau$$ and the proper time formula $$\tau = T/\gamma$$:

Lets say the travel from $A$ to $B$ $(v=0,6, \gamma=1,25)$ takes $10$ years, so the proper time of the spaceship is $\tau_1 = 10$ years. Now you will see that we will get two different values for the proper time $\tau_2$ of the planets:

1.The observed time of the travel, observed by the planets' reference frame $T_1 = 12,5$ years. Thus the proper time of the planets is $\tau_2=10$ years.

2.The observed time of the relative movement of the planets, observed by the spaceship $T_2 = \tau_1=10$ years.

Applying the proper time formula, the spaceship will calculate for the planets a proper time of $\tau_2=8$ years.

The difference between these two different values for $\tau_2$ is due to the fact that one starting time and one arrival time have not been defined for each planet. Thus, the planets consider a proper time of $10$ years, while the spaceship considers only a proper time for the planets of $8$ years.

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It gets worse, because "the state of planet B at the departure" is not uniquely defined as departure occurs at A and they are spatially separated. Likewise for "state of planet A at arrival" with arrival occurring at B. – dmckee Jul 6 '14 at 20:31