Time dilation and length contraction for a rocket

[PainterGuy]

Homework Statement

I was trying to do a simple problem related to special theory of relativity and need to clarify few points. It's not a homework.

Relevant Equations

I'm including all the equations in the post. Thanks.

Hi,

It's not a homework but still thought to post it here as advised in the past.

A rocket is going to leave Earth's surface and it is decided that a data pulse encoding emission time of pulse will be sent every second from Earth station to the rocket, and the rocket would do the same.

The rocket's speed is 0.5c where 'c' is speed of light, i.e., 300000 km/s. Assuming the rocket accelerates to 0.5c instantaneously. (Not sure if it's legitimate assumption. I think while acceleration the gravitational time dilation will be quite dominant and should also be factored in for more accuracy but ignoring it in favor of simplicity.)

t₀ is proper time in Earth's frame of reference. From Earth's reference, the first pulse will get to the rocket when time on rocket is 2 seconds after departure, as the calculation below shows, and by that time the rocket would have traveled distance of c(sec) according to Earth station. I hope I have it correct this far.

$$
\begin{aligned}
&\mathrm{c}\left(\mathrm{t}_{0}-1\right)=0.5 \mathrm{ct}_{0} \\
&\mathrm{ct}_{0}-\mathrm{c}=0.5 \mathrm{ct}_{0} \\
&\mathrm{t}_{0}-1=0.5 \mathrm{t}_{0} \\
&\mathrm{t}_{0}-0.5 \mathrm{t}_{0}=1 \\
&\mathrm{t}_{0}=\frac{1}{1-0.5} \\
&\mathrm{t}_{0}=2
\end{aligned}
$$

The rocket's crew would think that their clock is running fine but the clock on rocket is time dilated as shown below.

$$
t^{\prime}=\frac{t_{0}}{\sqrt[2]{1-\frac{v^{2}}{c^{2}}}}
$$
taking $t_{0}=1$ and $\mathrm{c}=300000 \mathrm{~km} / \mathrm{s}$, and $\mathrm{v}=0.5 \mathrm{c}$
$$
\begin{aligned}
t^{\prime} &=\frac{1}{\sqrt[2]{1-\frac{(0.5 c)^{2}}{c^{2}}}} \\
t^{\prime} &=1.1547
\end{aligned}
$$

The result above means that for every 1.1547 seconds on earth, the rocket's clock will read 1 second. Please correct me if I'm wrong.

But by the time t₀=2 the rocket has traveled distance of 346410 km as result of length contraction as is shown below.

\begin{aligned}
L^{\prime} &=\frac{L_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}} \\
L^{\prime} &=\frac{300000}{\sqrt{2 / 1-\frac{(0.5 c)^{2}}{c^{2}}}} \\
L^{\prime} &=346410 \mathrm{~km}
\end{aligned}

Do I have it correct this far? I think it's better to confirm this before I proceed with the main queries. Thanks for the help, in advance!

 

[Halc]

From Earth's reference, the first pulse will get to the rocket when time on rocket is 2 seconds after departure, as the calculation below shows, and by that time the rocket would have traveled distance of c(sec) according to Earth station. I hope I have it correct this far.

The first pulse will will get to the rocket when time in Earth's frame is 2 seconds after departure, but that's not the time on the rocket, which is going to read 1.73 seconds which corresponds in the ship frame to 1.15 seconds for the Earth receding at 0.5c to put out the pulse and .58 sec for the pulse to travel the distance back to the stationary ship.

 

[PainterGuy]

@Halc , thank you for the help and keeping it simple!

If the first pulse gets to the rocket when time in Earth's frame is 2 seconds then how much distance the rocket would have traveled by that 2 seconds time in Earth's frame? 3,00,000 km?

 

[PeroK]

@Halc , thank you for the help and keeping it simple!

If the first pulse gets to the rocket when time in Earth's frame is 2 seconds then how much distance the rocket would have traveled by that 2 seconds time in Earth's frame? 3,00,000 km?

Yes, in the Earth's frame the rocket moves at $v = 0.5c$. So, you can do the usual kinematics to get the position of the rocket in terms of coordinate time of the Earth's frame: $x_R = x_0 + vt = 0.5c \ t$.

Meanwhile the position of a light pulse send from Earth in the direction of the rocket after a time $T$, say is: $x_L = c(t - T)$.

If you put those two equations together, you can calculate where and when the light pulse reaches the rocket in the Earth's frame. I.e. calculate the $x$ and $t$ coordinates in terms of $T$.

 

[Halc]

I was keeping it simple because I had to leave in a moment. I can comment on a bit more.

Yes, as PeroK shows, simple kinematics (as you have done in your OP) will show when and where the pulse will overtake the ship. The mistake I pointed out was where you said this was time in ship frame instead of time as measured in Earth frame.

OK, so both rocket and Earth are putting out regular pulses every second. I presume the rocket pulses are emitted every 1 second in rocket frame. This makes the situation entirely symmetrical.

Assuming the rocket accelerates to 0.5c instantaneously. (Not sure if it's legitimate assumption. I think while acceleration the gravitational time dilation will be quite dominant and should also be factored in for more accuracy but ignoring it in favor of simplicity.)

The rocket doesn't need to have accelerated at all. It might be already moving past Earth and everybody zeros their clocks when they're momentarily in each other's presence. Gravity doesn't play a significant role in this exercise. At half light speed, the momentary glitch of Earth's gravity isn't going to make any difference to the ship. I'm presuming Earth is massless and its clock stays in sync with any other clock that is stationary in that frame. I.E. we're doing a special relativity exercise here.

As for acceleration, relativity doesn't have any problem with arbitrarily short acceleration times for point masses, but extended objects take time to accelerate to a given speed, and the time taken depends on where it is measured. Earth might say one thing, and clocks at opposite ends of the extended accelerating object will each read different values. So we're also assuming the ship, like Earth, is effectively a point object.

t₀ is proper time in Earth's frame of reference.

t₀ seems to be the coordinate time for the first Earth pulse to catch up to the rocket, in Earth frame.

From Earth's reference, the first pulse will get to the rocket when time on rocket is 2 seconds after departure, as the calculation below shows, and by that time the rocket would have traveled distance of c(sec) according to Earth station. I hope I have it correct this far.

Yes.

The rocket's crew would think that their clock is running fine but the clock on rocket is time dilated as shown below.

The rocket crew, if using the inertial frame of the rocket, will compute that their own stationary clock runs normally and that the Earth clock is dilated since Earth is moving away at 0.5c in that frame.

The rocket's crew would think that their clock is running fine but the clock on rocket is time dilated
...

$$
\begin{aligned}
t^{\prime} &=\frac{1}{\sqrt[2]{1-\frac{(0.5 c)^{2}}{c^{2}}}} \\
t^{\prime} &=1.1547
\end{aligned}
$$
The result above means that for every 1.1547 seconds on earth, the rocket's clock will read 1 second. Please correct me if I'm wrong.

Only according to Earth frame. Relative to the rocket frame, which you specified, it is the rocket clock that advances 1.1547 seconds for every second measured on Earth. Nobody considers their own clock to be dilated since it by definition measures its own proper time.

But by the time t₀=2 the rocket has traveled distance of 346410 km as result of length contraction as is shown below.

No. It is moving at 0.5c in Earth frame, so in 2 seconds it has moved ~300000 km. In the rocket frame, it is stationary and hasn't gone anywhere after any amount of time.

Interested to see where the main queries are going, but there are definitely corrections to make before then.

 

[PainterGuy]

OK, so both rocket and Earth are putting out regular pulses every second. I presume the rocket pulses are emitted every 1 second in rocket frame. This makes the situation entirely symmetrical.

Yes, they both emit pulses every 1 second in their respective frame of reference.

The rocket doesn't need to have accelerated at all. It might be already moving past Earth and everybody zeros their clocks when they're momentarily in each other's presence.

So, when the rocket passes by Earth, the crew sees the Earth moving at speed 0.5c. On the other hand, Earth crew things the ship is moving at 0.5c

Let's focus on the first pulse which will get to the rocket when time in Earth's frame is 2 seconds after departure, but the rocket's clock reads 1.73 seconds. What would rocket's crew think of the situation? They know that Earth was moving at speed of 0.5c and it was decided that a pulse will be emitted after 1 second by each party. The rocket's crew would know that they were supposed to receive the pulse at 2 seconds after departure but they had received it earlier. How would they reach the conclusion that Earth's clock is running slow?

A side question. Can't the rocket/spaceship's crew look at the stars outside and noticing the changing star background find that it is really them who are moving?

Thanks for the help, in advance!

 

[Halc]

Let's focus on the first pulse which will get to the rocket when time in Earth's frame is 2 seconds after departure, but the rocket's clock reads 1.73 seconds. What would rocket's crew think of the situation? They know that Earth was moving at speed of 0.5c and it was decided that a pulse will be emitted after 1 second by each party.

Yes. 1 second on each of their respective clocks, not as measured by anyone frame.

The rocket's crew would know that they were supposed to receive the pulse at 2 seconds after departure but they had received it earlier. How would they reach the conclusion that Earth's clock is running slow?

Had there been no time dilation, the pulse would have been sent by Earth after only 1 second after it had traveled only half a light second away, so it would take an additional half second for that signal to get back to the stationary ship. They would have expected it after 1.5 seconds, not 1.73 seconds. It would only take 2 seconds for the ship pulse to reach Earth.
But that would be a asymmetry, violating Galilean relativity.

A side question. Can't the rocket/spaceship's crew look at the stars outside and noticing the changing star background find that it is really them who are moving?

Maybe it's all the stars moving. Yea, at 0.5c, they'd definitely notice the stars aren't more or less stationary on average.

 

[PeroK]

A side question. Can't the rocket/spaceship's crew look at the stars outside and noticing the changing star background find that it is really them who are moving?

Thanks for the help, in advance!

The question is not whether there is a natural rest frame for the stars. There is, approximately. The question is whether that frame (or any frame) is special regarding the laws of physics. For example, if you move relative to the distant stars, do you not observe the normal laws of electromagnetism? Does the measured speed of light remain invariant?

The answer to these questions is yes. Motion relative to the distant stars does not mean you need to adjust the laws of physics.

In the same way that the surface of the Earth is a natural frame for us, but not special regarding the laws of physics.

 

[PainterGuy]

Had there been no time dilation, the pulse would have been sent by Earth after only 1 second after it had traveled only half a light second away, so it would take an additional half second for that signal to get back to the stationary ship. They would have expected it after 1.5 seconds, not 1.73 seconds. To ship crew members, it'd mean that Earth's clock is running slow.

Thank you!

So, the rocket/ship crew were expected to get the pulse by 1.5 seconds and not 1.73 seconds.

The time dilation formula gives:
$$
\frac{1}{\sqrt[2]{1-\frac{(0.5 c)^{2}}{c^{2}}}}=1.1547
$$
But if the ship crew do simply analysis, they will get this answer:
$$
\frac{1.73}{1.5}=1.1533
$$

Is ship's crew supposed to know the actual time dilation formula to reach correct time dilation factor?

 

[PeroK]

Thank you!

So, the rocket/ship crew were expected to get the pulse by 1.5 seconds and not 1.73 seconds.

The time dilation formula gives:
$$
\frac{1}{\sqrt[2]{1-\frac{(0.5 c)^{2}}{c^{2}}}}=1.1547
$$
But if the ship crew do simply analysis, they will get this answer:
$$
\frac{1.73}{1.5}=1.1533
$$

Is ship's crew supposed to know the actual time dilation formula to reach correct time dilation factor?

It's not clear what that second calculation is supposed to be. The Earth's clocks are dilated as measured in the rocket frame.

You seem to have the idea that calculations should be done using classical physics to get what is "expected". But, then, what happens is a surprise!

Nothing is a surprise if the correct, relativistic laws of physics are used!

 

[PainterGuy]

Thank you!

So, it means that to get to the correct calculations they already need to have Lorentz equations at hand? If the ship crew knows that speed of light is a constant, how would they get to correct conclusion about time dilation?

I think we should do it in detail after a few days once I get more time. Thanks for the help and time!

 

[Halc]

If the ship crew knows that speed of light is a constant, how would they get to correct conclusion about time dilation?

The Lorentz equations, length contraction, time dilation, and relativity of simultaneity all follow directly from the two premises of special relativity (Galilean relativity and constant light speed relative to any inertial frame). If the crew wishes to derive it all from the premises, they can do that. Or they can just utilize what Einstein derived and save a lot of work.

 

[PeroK]

So, it means that to get to the correct calculations they already need to have Lorentz equations at hand?

You need to use the correct laws of physics, of course!

If the ship crew knows that speed of light is a constant, how would they get to correct conclusion about time dilation?

Either from their physics books. Or, more likely, it would be programmed into the ship's computer!