Suppose you're travelling at 0.9c toward the sun, and you tag past the Earth and start a clock. Would Lorentz contraction/time dilation cause you to get to the sun faster than (about) 9 minutes (measured in your own reference frame)?

If so, does this allow us to create a measure of "effective Newtonian speed" by dividing the distance you would have travelled without Lorentz contraction by the time it took you to reach the Sun from the earth? If not, why, and is there any thought experiment that demonstrates this?

(I know that it would take 9 minutes from the point of view of somebody actually at the Sun.)

A little research on Wikipedia about "faster-than-light travel" reveals what what I'm talking about is known as four velocity.


3 Answers 3


From the comments to user16307's answer I'm guessing you're fairly new to special relativity. Until you get familiar with the subject it's very dangerous to throw around concepts like time dilation and length contraction because you can easily fall into traps like the pole in a barn paradox. The only safe way to work out what happens is to use the Lorentz transformations to compare what happens in your frame with what happens in some other frame.

Let's define our and the spaceships frames to coincide at time zero. Then as we watch the spaceship fly towards the Sun we can define two spacetime points $(t, x)$: the spaceship passes the Earth at $(0, 0)$ and it passes the Sun at $(d/v, d)$. I written the distance to the Sun as $d$ to keep things general, and the time taken to reach the Sun is just the distance $d$ divided by the speed $v$ i.e. $t = d/v$.

Now, the Lorentz transformations allow us to transform these two events into the spaceship's frame, and from this we can work out how much time has elapsed when the spaceship passes the Sun, and also how far it's travelled. The Lorentz transformations are:

$$ t' = \gamma \left( t - \frac{vx}{c^2} \right) $$

$$ x' = \gamma \left( x - vt \right) $$


$$ \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} $$

We've defined our frames so that the point $(0, 0)$ is the same in both frames, so we just have to work out where the point $(d/v, d)$ is in the spaceship's frame. let's do the time $t'$ first. We need to substitute $d/v$ for $t$ and $d$ for $x$, so:

$$ \begin{split} t' &= \gamma \left( \frac{d}{v} - \frac{vd}{c^2} \right) \\ &= \gamma \frac{d}{v}\left( 1 - \frac{v^2}{c^2} \right) \\ &= \frac{d}{v} \frac{1}{\gamma} \\ &= \frac{t}{\gamma} \end{split} $$

It might be a bit hard to see what I did in the third step. If you look above to see how $\gamma$ is defined then you'll see that $\gamma (1 - v^2/c^2)$ is just equal to $1/\gamma$, and that's how we get from the second to the third step. The last step just notes that $d/v = t$.

Now let's calculate $x'$:

$$ \begin{split} x' &= \gamma \left( d - v \frac{d}{v} \right) \\ &= 0 \end{split} $$

What! How can the spaceship pass the Sun at a distance $x' = 0$? Well of course it does. By definition, in the spaceship's rest frame it's at rest and it's the Sun moving not the spaceship. If the spaceship is stationary it's position cannot change from the initial value of 0, so the Sun has to pass it at $x' = 0$.

So in the spaceship's frame the Earth passes it at $(0, 0)$ and the Sun passes it at $(t/\gamma, 0)$. In the spaceship's frame the Sun takes less time to reach the spaceship (by a factor of $1/\gamma$) than the ship takes to reach the Sun in our frame. Both frames have to agree about their relative velocity, so that means in the spaceship's frame the Sun is traveling towards the ship at speed $v$ and if it takes less time the distance travelled by the Sun must be less than $d$. In fact the distance the Sun travels is just velocity times time, and we already worked out that time = $t/\gamma$ = $(d/v)/\gamma$, so the distance travelled by the Sun is just $d/\gamma$.

So in the Earth frame we see the spaceship travel at speed $v$ for a distance $d$ in a time $t$ to reach the Sun from the Earth. In the spaceship's frame the crew see the Sun travel at speed $v$ for a distance $d/\gamma$ in a time $t/\gamma$.

  • $\begingroup$ I am indeed fairly new to special relativity. By "effective Newtonian speed" I was talking about the more naive conception of velocity as "distance travelled while 'accounting for' distance contraction". $\endgroup$
    – Joe Z.
    Feb 13, 2013 at 18:32
  • 2
    $\begingroup$ You need to be awfully careful with naive concepts as they are rarely helpful! $\endgroup$ Feb 13, 2013 at 18:52
  • 2
    $\begingroup$ Yes, if you travel at near enough the speed of light the elapsed time for the traveller can be made arbitrarily small. No end of science fiction books have made use of this plot device! $\endgroup$ Feb 14, 2013 at 7:01
  • 1
    $\begingroup$ @JoeZeng: read The Forever War by Joe Haldeman for a passable illustration of the results of time dilation. $\endgroup$ Feb 14, 2013 at 15:48
  • 3
    $\begingroup$ @JoeZeng: Yes, you theoretically can get anywhere in the galaxy, and even further, in your own lifetime. But not in mine! (Unless I go with you.) It's a bit hard to "just account for those". Long distance communication becomes a radically different business when the person speaking to you dies of old age before you can get a word in edgewise... That is why sci-fi generally needs the faster-than-light gimmick. $\endgroup$ Feb 14, 2013 at 17:54

Yes, you would measure less than 4 minutes, as others have said. And I'm not really sure what all their disclaimers are about. I imagine that if we ever develop relativistic interstellar travel, it would be common to talk about an "effective speed" which is distance, in some galactic frame of reference, divided by time, as measured by the traveller.

There is no theoretical upper bound to this effective speed.

In fact, even with just 1 g of sustained acceleration, you can reach quite high speeds in a year, and get anywhere in the galaxy, or beyond, in your own lifetime. It's "just" (!) a technical problem -- carrying a whole lot of fuel to sustain the acceleration, and dealing with the dangers of ultra high speed travel, like radiation and running into stray bits of dust. (See the 1 g rocket for some numbers and further explanation.)

Of course on the return trip, you would find that ages and ages had passed at home. But you knew that.

  • 1
    $\begingroup$ "But of course on the return trip, you would find that ages and ages had passed at home." That's fine. That's relativity in effect. $\endgroup$
    – Joe Z.
    Feb 14, 2013 at 18:20

(1) If a person travels towards the Sun at relativistic speed, they will see the distance to the Sun length-contracted, so it will take less time (in their frame) to get there than it does in the frame of an observer on Earth. Whether they could get there in 9 minutes or not you'd have to stick the numbers in: take the Earth-Sun distance as $l$, then the person in the spaceship would see that distance as $L = l/\gamma$, where \begin{equation} \gamma = \frac{1}{\sqrt{1-(\frac{v}{c})^2}} \end{equation} is the Lorentz factor. Then use $t=L/v$. Use $v=0.9 c$ as the speed that the person in spaceship will see the Sun coming towards them. (I'm curious to know why you specify 9 minutes).

(2) The speed that an observer travels in their own reference frame is always zero as far as I am aware. In this case, no, you wouldn't be able to derive some kind of speed. A possible thought experiment: suppose you are the only entity in the Universe. How can you tell if you are moving, nevermind how fast? I don't think you can.

Disclaimer: special relativity is a tricky concept and to apply common sense rules/ Newtonian physics is not a good idea. Plenty of practice examples are a good way I found when studying it.

  • $\begingroup$ I specified 9 minutes because that's (roughly) the distance from the Earth to the Sun divided by 0.9c. $\endgroup$
    – Joe Z.
    Feb 13, 2013 at 16:06
  • $\begingroup$ I don't mean the speed that an observer travels in their own reference frame. I might need to make some calculations based on the answer you just gave me to tell you what I really meant. $\endgroup$
    – Joe Z.
    Feb 13, 2013 at 16:09
  • $\begingroup$ Are you learning from a book? $\endgroup$
    – user12345
    Feb 13, 2013 at 16:10
  • $\begingroup$ No, not yet. In fact I haven't yet taken a physics class in university. $\endgroup$
    – Joe Z.
    Feb 13, 2013 at 16:11
  • $\begingroup$ Okay, so what I meant was, will $t$ be $l/\gamma v$ instead of just $l/v$, due to length contraction (about 4 minutes in the example I gave)? $\endgroup$
    – Joe Z.
    Feb 13, 2013 at 16:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.