A question about relativity and time dilation from a lay-person

Question

Observer A is in a spacecraft flying toward earth

Observer B is standing on earth

A is traveling at near light-speed and will reach the earth in 1 week.

When B looks at A's clock, he sees it moving more slowly than his own

However, from A's perspective, he is stationary and it is earth that is moving toward him. So when B looks at A's clock, he sees A's clock moving more slowly.

Based on the speed of A, when he is done with his trip, 100 years will pass on earth due to the time dilation

How can these things be true? If motion is relative, that means both A and B perceive themselves stationary and see the other moving at near light-speed and therefore, each sees his own clock moving more quickly than the other's clock.

But the reality of the situation dictates that B will arrive on earth 100 years in the future from when the trip started, from A's perspective. So A will be dead, and 100 years of history will have passed. As B flies toward earth for a week, why is he not seeing 100 years of earth pass in the span of a week? How can 100 years pass on earth, but the entire time B is moving, he sees the earth's clock moving more slowly than his own? So he would then see even less than a week of history on earth, but then when he stops, the earth skips forward 100 years? That doesnt seem right?

Since 100 years pass on earth while A is traveling, shouldnt he see that 100 year history unfold in the span of a week while looking out his window? If so, how can we say motion is truly relative? And if not, then at what point would observer A notice the earth was actually 100 years older?

Answer 1

It is better to think of a row of ships travelling to earth with the same velocity of A (Alice) and with synchronized clocks.

And a row of stationary clocks synchronized with the B (Bob) earth's clock.

When Alice passes by one of that stationary clocks, she sees $t = t_A$. One week later in her own time she passes by the earth, and sees $t = t_A + 100$ years at Bob's clock.

On the other hand, Bob sees t_B when one of the ships passes by the earth. When Alice passes by him after one week at his own time, he sees $t_B + 100$ years at the Alice's clock.

The point is that the clocks in Bob's frame are not synchronized with clocks in the Alice's frame. So, for Alice, the earth is already in the far future when she starts to count. And for Bob, Alice's ship is already in the far future when he starts to count.

Answer 2

When you say "he will reach the Earth in one week" you have to also specify whether this is as measured by A or as measured by B. They have different coordinate systems. They're both observing the same spacetime, but they divide it up differently -- for A (on the rocket) clocks on Earth are moving through space as they tick, so an Earth second occupies not only some time but also some space. The reverse is also true: for B (on Earth) an Earth clock tick is all time, with no space, but the rocket clock ticks are moving through space as well as time. Both agree that the speed of light is $c$, so they both agree on the length of the spacetime interval $\Delta s^2 = c^2 \Delta t^2 - \Delta x^2$ occupied by the clock ticks, but they will disagree on how much to assign to $\Delta x$ and how much to $\Delta t$.

Another thing to be careful of is to distinguish between what they literally see (the flashes of light from clocks, for example) and what they calculate based on knowing times, distances, and speeds. B on the rocket moving towards Earth will literally see Earth clocks moving fast. But based on the time for the light to reach him from Earth, and the speed the Earth is moving towards him, he will calculate that the Earth clock is actually ticking slower than his clock.

Answer 3

Alice is on earth. Bob travels from Far Away.

The story according to Alice:

At 2PM, Bob started his journey, with his clock correctly set to 2PM (as was mine). The journey took him four hours, but his clock ran at half-speed the entire time, so when he arrived here at 6PM, his clock had advanced only two hours and said 4PM.

The story according to Bob:

At 2PM by my clock, I started my journey, though Alice's clock at that moment was set for 5PM. The journey took two hours, but Alice's clock ran at half-speed the entire time, so when I arrived at 4PM (again, by my clock), her clock had advanced only one hour and said 6PM.

Note that Alice and Bob agree that when Bob arrives, Alice's clock says 4PM and Bob's says 6PM. They have to agree about these clock readings, because they are both in the same place and time as the clocks when these readings occur.

They do not agree on what Alice's clock said at the time Bob started his journey, and they also do not agree on whose clock ran at half-speed. They don't have to agree on these things, because they are not in the same place at the same time as these events.

Answer 4

Time dilation can be confusing until you view it in the right way, and then it becomes obvious.

Time dilation is caused because clocks in two moving frames are out of synch relative to each other- it is not because they tick at different rates.

Imagine you are walking down a corridor in which there are people holding clocks every ten metres or so, and each clock is set to run a second ahead of the pervious one. If you compare your watch with each of the clocks you pass, your watch will seem to be losing a second every time you pass another clock- you might think it is because your watch is running slow (ie, it is time dilated), but you would be wrong- your watch ticks at the same rate as each of the clocks, but it seems to lose a second simply because each successive clock you pass is set to run a second ahead of the previous one.

Now imagine that as you walk down the corridor, you are followed by a line of friends walking ten metres apart, and imagine each of your friends' watches is set a second ahead of the watch of next person in line in front of them. A person standing holding a clock will see you pass first, and then they will see each of your friends pass, and if they look at the watches of each of your friends, they will think their clock is losing another second each time one of your friends walks past them, ie they will think their clock is time dilated. But again they would be wrong- they seem to be losing time not because their clock is running slow but because each passing watch is set a second ahead of the next.

So in the thought experiment I have just described, the people holding the clocks think they are time dilated relative to the walkers with the watches, and the walkers with the watches think they are time dilated relative to the people holding the clocks, when in fact they are all ticking at the same rate. It is the fact that they are all out of synch that gives the impression of time dilation.

You need to apply that thinking to your example. Suppose the spaceship started its journey when the local time in the Earth's frame was the year 3000. Seven days later, according to the calendar on the spaceship, the Earth arrives, and people on the Earth say it is the year 3100. You need to ask yourself, what year was it, in the spaceship's frame, when and where the Earth started its journey? You will find that the Earth started its journey a hundred years earlier in the spaceship's frame.

Answer 5

The best way to see what's going on is with a special Minkowski Diagram called a Lodel Diagram.

Let's say the unprimed frame is on Earth, and the primed frame is the returning rocket ship. Since you chose:

$$\gamma= 5200$$

you get velocity

$$\beta=\frac 1 {\sqrt{1-\frac 1 {\gamma^2}}}=0.99999998$$

and rapidity

$$\omega = \tanh^{-1}\beta=9.24$$

We then make a third frame with half the rapidity:

$$\omega_0 = \frac 1 2 \omega = 4.62$$

so that:

$$\beta_0 = \tanh\omega_0 = 0.9998$$

and $$\gamma_0= \frac 1{\sqrt{1-\beta^2}}=51.0$$

The significance of the 3rd frame is that it sees the Earth and the rocket moving at the same speed, but in opposite directions.

You can plot all the axes of the 3 frames on a Minkowski diagram:

This is a special version that has the two frame moving equal and opposite directions.

This one is not to scale, as your case has:

$$\alpha = \tan{\beta}=44.995^{\circ}$$

The situation you described occurs below the origin, which is the event "rocket returns to earth". The red time axis is the world line of Earth, and the green one is the rocket ship approaching from the left.

What makes this useful, is that the time dilation in each frame is the same, and you can see the paradoxical result that each sees the other experience less time is caused by the relativity of simultaneity.

Lines of constant time are parallel to space-axes, so for instance, if you go to the 1st tic mark (from the bottom) and draw a line parallel to the $x'$ axis, you'll see it intercepts the $t$ axis closer to the origin. Likewise, drawing a line parallel to the $x$ axis from the $t'$ axis: it is complexly symmetric, each sees the other experiece less time until they are reunited.