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I am just starting familiarizing myself with special relativity, and want to understand if the following is a correct thought process:

Consider a star that is 100 light years away from Earth, from my perspective, where I am a stationary observer on earth at some moment in time.

Consider another observer that is passing me heading towards that star with a speed of 0.8c.

I see it as both myself and that other observer to be at position x=0 at time t=0, from both of our perspectives, and the star to be at position x=100 at t=0 from my perspective.

However, it appears to me that at time t=0 from perspective of the moving observer the star is at position 60.

The way I arrive to this conclusion is the following: say at time t=80 from my perspective there's an explosion on the star. The coordinates of the event for me are t=80, x=100. Applying the Lorenz transformation I get the coordinates of the event for the moving observer as t'=0, x'=60, thus I conclude that at time t'=0 (when the moving observer passes me on Earth) for them the star is 60 light years away.

So, from perspective of the moving observer it will take them 75 years to reach the star after the moment they passed me by.

The next step in my thinking is the following: say at time t=0 I start accelerating towards the star with acceleration of 1g (such that from my perspective Earth moves 9.8 meters / s faster away from me every second). In approximately 300 days I will be moving with the speed of 0.8c relative to Earth, and thus be in the frame of reference of the moving observer who recently passed me. Given I was moving away from Earth, the Earth in my frame of reference is behind me, thus the star must be closer than 60 light years, and it should take me less than 75 years to reach it. Correspondingly, the total time to reach the star will be bounded above a 75 + 2 years to accelerate and decelerate. Thus, I reached a star that was 100 light years away from me in less than 100 years.

If my understanding is correct, does it mean that if we find a way to reach speeds close to the speed of light, interstellar travel won't require multiple generations to reach distant stars?

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    $\begingroup$ That's exactly correct. Although if you ever come back to Earth you'll find that a couple hundred years passed while you were gone (see: the twin paradox). $\endgroup$ Feb 5 '21 at 20:33
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    $\begingroup$ This is how muons from the upper atmosphere can reach the surface of the earth. $\endgroup$
    – user137289
    Feb 5 '21 at 21:36
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    $\begingroup$ In a nutshell: Yes for the traveler, no for those watching from earth. $\endgroup$
    – R. Rankin
    Feb 6 '21 at 4:09
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    $\begingroup$ In its own frame, a photon can reach it in zero time, which is still 100 years in our reference frame. $\endgroup$
    – Neil_UK
    Feb 6 '21 at 6:38
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    $\begingroup$ I just want to point out that "the rest frame of the photon" is not a valid inertial frame, so it is difficult to properly formulate statements like "photons don't see time passing". $\endgroup$
    – gabe
    Feb 6 '21 at 21:26
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You are right that the traveler need not age much when voyaging from one star to another many lightyears away. If the distance plotted from Earth is $d$ in the measurements appropriate to the rest from of Earth or Sun, then the proper time accumulated by the traveler is $$ \tau = \frac{1}{\gamma} \left(\frac{d}{v}\right) $$ if they journey mostly at speed $v$, relative to Earth, where $\gamma = 1/\sqrt{1 - v^2/c^2}$. Thus if $v$ is large enough then their proper time $\tau$ (i.e. the amount they age) will increase by less than $d/c$.

Here's an example. Say $d$ is $100$ lightyears and we manage to get to the amazing speed of $v = 0.99c$. Then $\gamma = 7.09$ so the proper time taken to get to the distant star is $100 / (0.99 \times 7.90) = 14.25$ years.

However, by undertaking such a journey the traveler cannot avoid parting company with their friends on Earth and when they come back their friends will have aged more. In the case of a long journey and many lightyears, their friends will be be long dead on their return. So interstellar voyaging still has this time cost associated with it.

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    $\begingroup$ If I remember correctly, the speed at which you travel $n$ light-years in $n$ years of proper time is $v = c/\sqrt{2}$. $\endgroup$ Feb 6 '21 at 3:13
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    $\begingroup$ @MichaelSeifert See last sentence of hft's answer. It's n light years in n years of ship time. (Whether that is "proper time" depends on definition). $\endgroup$ Feb 6 '21 at 12:40
  • $\begingroup$ Can your friend on Earth avoid aging so much by orbiting a singularity? $\endgroup$
    – Michael
    Feb 6 '21 at 20:42
  • $\begingroup$ @Michael The answer is yes: the proper-time period of a circular orbit around a nonrotating black hole can be arbitrarily close to zero by making the orbit's radius arbitrarily close to $3R/2$, where $R$ is the Schwarzschild radius. (Circular orbits with smaller radii don't exist: you can travel in a circle at any radius $> R$, but you won't be in free-fall unless the radius is $> 3R/2$.) In case you're interested, I posted a related question a while ago. $\endgroup$ Feb 7 '21 at 4:00
  • $\begingroup$ @ChiralAnomaly yes I like that example of black hole orbit (or even an orbit around a neutron star); it was invoked in at least one science fiction story by Ursula K. Le Guin. I have not checked the other considerations such as what forces are needed to arrive and get away, or the dangers of hitting other orbiting material. $\endgroup$ Feb 8 '21 at 11:59
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Does special relativity imply that I can reach a star 100 light years away faster than in 100 years?

The time that you experience in your spaceship on your way to the star could be more than or less than (or equal to) 100 years depending on how fast your spaceship travels.

The time that an earthbound observer sees you take to reach the star is necessarily more than 100 years.

The relevant calculations are presented below along with explanatory examples. (Note, in my calculation I have used the typographic simplification of assuming the units are chosen such that $c=1$ during the calculation. This means I don't write the factors of $c$ explicitly and I just re-insert the necessary factors of $c$ at the end by dimensional considerations. It's actually quite a useful practice when doing these quick back-of-the-envelope calculations).

I will call the change in time observed from the Earth frame $\Delta t$, the change in distance in the earth frame $d$, the change in time in the ship frame $ \tau$, and the change in distance in the ship frame $0$.

The Lorenz Transformation from the Earth frame to the spaceship frame is: $$ \begin{bmatrix} \tau \\ 0 \end{bmatrix} = \gamma \begin{bmatrix} 1 & -v \\ -v & 1 \end{bmatrix} \cdot \begin{bmatrix} \Delta t \\ d \end{bmatrix}\;. $$

This matrix equation gives us two individual equations: $$ \tau = \gamma (\Delta t - vd)\;, $$ and $$ 0 = -v\Delta t + d\;. $$

The second equation gives us the obvious relationship between the time change observed on earth and the velocity of the ship: $$ \Delta t = \frac{d}{v} = \frac{d}{c}\frac{c}{v} = (100 years)\frac{c}{v}\;, $$ which is clearly greater than 100 years since v is less than c (as it is for any massive object).

The first equation can similarly be simplified to read: $$ \tau = (100 years)\sqrt{\frac{c^2}{v^2} - 1}\;. $$

So, for example, if your ship travels at $v=0.01c$ it will feel like it takes you about 9999.5 years to get to the star.

As another example, if your ship travels at $v=0.99c$ it will feel like it takes you about 14 years to get to the star.

As a final example, if your ship travels at $v=\frac{c}{\sqrt{2}}$ it will feel like it takes you 100 years to get to the star.

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    $\begingroup$ I like the examples that you've picked, especially the last one. +1 $\endgroup$ Feb 7 '21 at 12:43
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Don't worry about Lorentz transformations, just realise that the proper time interval $\Delta \tau$ for something to travel between two events is the same judged in all inertial frames (an invariant).

$$(\Delta \tau)^2 = (\Delta t)^ - (\Delta x)^2 = (\Delta t')^2 - (\Delta x')^2\ ,$$ where $x, t$ refer to spacetime coordinates in the Earth's frame of reference, with time in years and distances in light years, and where $x', t'$ can be coordinates in the frame of a travelling astronaut.

Thus for travel to a star 100 light years away, $\Delta x=100$, while $\Delta x'=0$. If the spaceship sets of at a speed $v = dx/dt$, as measured on Earth, then $\Delta t = \Delta x/v$ and the proper time measured on an astronauts watch is $$\Delta \tau = \Delta x \left(\frac{1}{v^2} - 1 \right)^{1/2}\ ,$$ where $v$ is expressed as a fraction of light speed.

If you want to get there in a time less than $\Delta x$ years, then the bracket just needs to be less than 1. $$ \frac{1}{v^2}-1 < 1$$ $$ v^2 > \frac{1}{2}$$ So the premise of your question is satisfied if $v > 1/\sqrt{2}$ (as a fraction of$c$) and the proper time interval can become arbitrarily small as $v \rightarrow 1$.

Meanwhile, back on Earth, $$\Delta t = \sqrt{(\Delta \tau)^2 + (\Delta x)^2}\,$$ which is always $>\Delta x$.

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Yes, this is predicted by special relativity. Due to time dilatation, in your frame of reference, time will proceed slower than in the reference frame on earth. The closer your speed is to light speed, the larger the effect. As seen from earth, your trip will last more than a hundred years. In your own reference frame, the trip can approach instantaneous duration.

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Time dilation is a tricky subject, But mostly vast differentials in time perception require increasing high acceleration towards light speed. enter image description here

Over all, the time frame is negligible. Even if velocities of 1/10th lightspeed were achievable by technical means, the time differentiation to an observer would be 0.005 seconds. It's exponential, speed increases perceived time index. At 1/4 (25%) light speed, observed time 0.03 seconds Put it this way 1% lightspeed, a days worth of travel adds 4.32 additional seconds to the observer' 25% lightspeed, a day of travel to astronaut, is 47 minutes longer to observer .5c o r 50% lightspeed, a day of travel for astronaut vs 3.7 hours added 0.75c a day of travel; = 36.2 hours (day and a half) 0.9c. Observer 55 hours (over 2 days) 0.95c. Observer 76.8 hours (Over 3 days) 0.99c. observer will have spent a week 0.999c. observer (22 days) .9999c. observer (2.5 months)

Traveling 100 lightyears at 0.25c would take the astronaut 400 years but would be 413 years to the observer. At 0.9c, 111 years to reach 100 lightyears, observer would wait 254 years.

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