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.
Answer
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:
$$ t' = \gamma \left( t - \frac{vx}{c^2} \right) $$
$$ x' = \gamma \left( x - vt \right) $$
where
$$ \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} $$
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 $t/\gamma$.
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