Space travel and time dilation

[Stas1234]

Hi everyone, help me figure this out please.

There are 2 stars (A and B) 1000 light years apart.

We know that time slows down as speed increases. (Or does it? I'm not a physicist)

There is a clock on the ship, we start it at the moment of takeoff from star A.

At what speed should the spaceship move (or how should it move at all) from star A to star B so that the clock on the ship shows 10 years at the moment of arrival at star B? The pilot on the ship must age by ten years.

 

[PeroK]

Hi everyone, help me figure this out please.

There are 2 stars (A and B) 1000 light years apart.

We know that time slows down as speed increases. (Or does it? I'm not a physicist)

There is a clock on the ship, we start it at the moment of takeoff from star A.

At what speed should the spaceship move (or how should it move at all) from star A to star B so that the clock on the ship shows 10 years at the moment of arrival at star B? The pilot on the ship must age by ten years.

Is this homework?

 

[Stas1234]

Is this homework?

No, this is my argument with a friend, we are not physicists but we like to watch videos on YouTube about physics and astronomy. From the videos it is not clear at all how time dilation works

 

[berkeman]

Thread closed for Moderation...

 

[berkeman]

Thread is reopened provisionally in the non-schoolwork forums for now.

we are not physicists but we like to watch videos on YouTube about physics and astronomy.

Depending on which videos, they are probably not the best resource to be using. Please read through this introductory article at Wikipedia, and ask specific questions about that article if you are having trouble understanding something. Thanks.

https://en.wikipedia.org/wiki/Time_dilation

 

[berkeman]

Also, here is a Relativity Calculator recommended by @DaveC426913

Just point them here: https://www.1728.org/reltivty.htm

 

[DaveC426913]

Just point them here: https://www.1728.org/reltivty.htm

It's pretty simple - maybe a little too simple - but useful, once you grok it.

You want your pilot to make a trip of 1000 light years in 10 years subjective time.
That's a dilation factor of 100:1.
So input "100" and then hit "Factor of Change" and it will tell you the fraction of c you need to travel at (it is going to be a decimal point, followed by many nines).

(The scenario gets interesting when you ask yourself what the pilot sees out his window on that 1000 light year journey. hint: He sees a large flat pancake, much closer than expected.)

 

[Ibix]

"Time slows down as speed increases" is a massive oversimplification and will lead you into contradictions if you use it as a basis for thinking about relativity. However, in this case it works well enough to do calculations with as long as we use the Earth's rest frame.

The time dilation factor is $1/\sqrt{1-v^2/c^2}$. If you want that to be 100 then you need to be travelling at about 99.995% of the speed of light. That calculation ignores acceleration time, so you'd actually have to go faster - possibly much faster depending on how long is the acceleration phase.

 

[jbriggs444]

It's pretty simple - maybe a little too simple - but useful, once you grok it.

You want your pilot to make a trip of 1000 light years in 10 years subjective time.

This is a good (very good) approximation, though not quite exact.

If the trip is made at nearly light speed then it will take almost exactly 1000 years as reckoned in the shared rest frame of the two stars. We want 10 years of subjective time, so we need approximately a hundred to one time dilation ratio.

But that means that we were not really travelling at light speed. So it will really take just a bit more than 1000 years as reckoned in the rest frame of the stars. So we need a time dilation ratio just a bit more than one hundred to one.

Which means that we pretty much have to write down an equation. Let us use $T_\text{subjective}$ to denote the time in years experienced by the pilot and $T_\text{rest}$ to denote the time in years as reckoned in the stars' rest frame. $v$ will be the speed of the craft as a fraction of the speed of light and $d$ is the distance to be covered in light years.$$10 = T_\text{subjective} = \frac{T_\text{rest}}{\gamma} = T_\text{rest} \times \sqrt{1-v^2} = \frac{d}{v} \times \sqrt{1-v^2} = \frac{1000}{v} \times \sqrt{1-v^2}$$ or $$1 = \frac{100}{v} \sqrt{1-v^2}$$We need to solve that for $v$. It looks rather nasty. But let us try. We can square both sides:$$1 = \frac{10000}{v^2}(1-v^2)$$ and multiply by $v^2$$$v^2 = 10000 - 10000v^2$$ $$10001 v^2 = 10000$$ $$v^2 = \frac{10000}{10001}$$ $$v = \sqrt{\frac{10000}{10001}} \approx 0.99995000375$$

This is negligibly different from the naive estimate given by both @DaveC426913 and @Ibix

 

[PeroK]

In general, we have:
$$\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$$Which can be rewritten as:
$$\frac{v}{c} = \sqrt{1 - \frac {1}{\gamma^2}}$$If $\gamma$ is large, we can use the binomial expansion to get:$$\frac v c \approx 1 - \frac 1 {2\gamma^2}$$In this case, $\gamma = 100$, so:
$$\frac v c \approx 1 - 0.5 \times 10^{-4} = 0.99995$$

 

[SiennaTheGr8]

This "ultrarelativistic" approximation can even be done without a binomial expansion (in case OP or other lurkers aren't familiar with that technique). Using $\beta = v/c$:
$$
\begin{align*}
\gamma^{-2} &= 1 - \beta^2 \\
& = 1 - \left( 1 - \left( 1 - \beta \right) \right)^{2} \\
& = 1 - \left(1 - 2 \left( 1 - \beta \right) + \left( 1 - \beta \right)^2 \right) \\
&= 2 \left( 1 - \beta \right) - \left( 1 - \beta \right)^2.
\end{align*}
$$
Now since $(1 - \beta)$ is very small, we can neglect its square:
$$
\gamma^{-2} \approx 2 \left( 1 - \beta \right),
$$
and:
$$
\beta \approx 1 - \dfrac{1}{2 \gamma^2}.
$$
If the thing is very fast, your calculator might just round this to $1$. In that case, write it like $1 - \beta \approx 1/(2 \gamma^2)$ so you have leading zeros.

 

[robphy]

Obligatory rapidity comment:
Since \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} is \cosh\theta = \frac{1}{\sqrt{1 - \tanh^2\theta}}.,
then $$(v/c)=\tanh\theta=\tanh( \rm arcosh \gamma).$$
https://www.wolframalpha.com/input?i=tanh(arccosh(100)) = 0.99994999874

 

[SiennaTheGr8]

And the ultrarelativistic approximation in terms of rapidity:
$$1 - \beta \approx \dfrac{1}{2} \textrm{sech}^2 \, \theta$$
or just as good:
$$1 - \beta \approx \textrm{sech} \, (2 \theta)$$

 

[Pencilvester]

That calculation ignores acceleration time

I’m always surprised at the time difference, (and I know this isn’t exactly the same scenario that OP described), but if the spaceship is providing just 1g of acceleration in the direction of travel for the first half of the journey and in the opposite direction for the second half (starting and ending at rest wrt Earth), then the passengers on board will age less than 14 years for a 1000 LY trip. In this scenario, the passengers have to spend 3.8-ish more years aboard the ship than they would in the OP’s scenario, but they’ll be able to board and de-board the spaceship without dying (necessarily).

 

[Trysse]

In the scenario of inertial motion, there’s a straightforward formula for the required speed $\beta$:
$$\beta=\frac{d}{\sqrt{\Delta \tau^2+d^2}}$$
This follows directly from the invariant interval. The proper time $\Delta \tau$ is related to the coordinate time and spatial distance by:
$$\Delta \tau=\sqrt{\Delta t^2-d^2}
\Rightarrow
\Delta t=\sqrt{\Delta \tau^2+d^2}$$
Substituting into $\beta=\frac{d}{\Delta t}$ gives the result.

The nice thing about this approach is that it allows for a geometric interpretation/solution. The speed is the ratio between $AB$ and $AC$. This diagram also nicely shows the relation between $d$, $t$ and $\tau$.

 

[Dale]

My approach would also be with the spacetime interval. $c^2 \Delta \tau^2=c^2 \Delta t^2-\Delta x^2$.

For this problem we set $\tau=10$, $x=1000$, $c=1$, and $t=1000/v$. Then solve for $v$.