# Does special relativity imply that I can reach a star 100 light years away in less than 100 years?

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.

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.

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.