Will you see distance contraction inside the space ship at near light speed?

Question

We know that distance to destination contracts for a passenger in a spaceship traveling at near the speed of light. If a spaceship is traveling at 0.865C to the outside observer, then the distance to the destination will contract to half in the passenger's frame of reference. I have a few questions about how this is observed by the passenger.

1. If the passenger had a ranging device for looking at the star they're traveling to, would the ranging device report half the distance? Or would that depend on the type of ranging device?
2. Will the front of the spaceship appear to be half as close to the passenger? Will the people and objects next to them look half as thick?

Answer 1

The answer to 2 is no; earlier answers have covered that. The answer to 1 is complicated. It does depend on how you measure the distance.

It's important to understand that you don't see distant objects directly. What you see is light at your location in spacetime. The light that is available to be seen only depends on your spacetime location, and not your speed. Your speed does affect what you see (Doppler shift and aberration), but you can work out what someone at your location with a different velocity would see, without having to change your velocity.

If you're at rest relative to the star, you can determine your distance to it by measuring its angular size or its relative magnitude, if you know its absolute size or magnitude. You can find your speed relative to the star by measuring the Doppler shift. Therefore, you can measure the distance to the star by these methods regardless of your speed (at least in principle). That distance doesn't contract when you change your speed, because it doesn't depend on your speed at all, only your position. It's the distance in the rest frame of the star, or more abstractly it's the spacetime version of the distance from a point to a line, where the point is you when you make the measurement, and the line is the star's worldline. Cosmological distances and speeds are measured more or less in this way; they aren't relative to the rest frame of the solar system (which isn't even well defined at a cosmological scale).

Another way of measuring the distance to an object is by bouncing a light beam off of it. The time delay of the signal times $c/2$ is the distance to the object in your rest frame at the time halfway between the emission and detection times. It's a bit difficult to bounce a light beam off of a star, but never mind that. The other problem with this method is that it's very slow. It takes 2 years to measure a distance of 1 light year, and by the time the answer comes back, you'll have covered most of that distance already. So you can't just measure the distance while at rest and get $L$ and then quickly accelerate to $0.865c$ and measure it again and get $L/2$. Even if you accelerate instantly and then instantly start the measurement, the distance you get back will be much less than $L/2$.

Answer 2

I'll start with 2.

No, for anyone in the spaceship, the spaceship is at rest and the outside moves near the speed of light. Thus everything inside the spaceship and the ship itself will appear perfectly normal.

Answer 3

The pilot will not see that the ship or the people in it are shortened, because they are not moving relative to the pilot. A ranging device moving with the pilot would show that the distance to the star is halved.

Incidentally, you might find it useful to consider how length contraction comes about. If an object is moving (as the star is moving towards the spaceship), if you ask 'How far away is it?' what you really mean is 'How far away is it now?' as its position is constantly changing on account of its motion. 'Now' on the spaceship is different from 'now' in the stationary frame, which is why there is a disagreement about how far away the star is.