# A little confused about how Relativity works?

1. Aug 15, 2010

### coldbreeze10

hello ppl
Imagine that you are on a spaceship flying very close to the speed of light and we know that time slows down significantly at this speed relative to that on earth. now say every day for the persons on the spaceship is a whole year for a person on earth. So for a person on the spaceship the distance travelled is the 'speed(Km/Hr)' of the craft multiplied by the '24' Hr in a day.. while for a person on earth the distance covered by the space ship would be the speed(Km/hr) of the spaceship multiplied by 24Hr multiplied by 365 days .(of course in an year on earth but it's still a day on the space ship.) So it gives two completely different answers for the distance travelled. How is that?? may be it's a dumb question but i am unable to figure it out!!

2. Aug 15, 2010

### AJ Bentley

Oh, it's far worse than that!
You see, the situation is absolutely symmetric.
You see me moving slowly. I see you moving slowly. I see your clock has slowed to 10 ticks in a day, You see my clock has slowed to 10 ticks in a day.

Under these circumstances, the question you are asking has no meaning - you can't talk about the distance travelled by the spaceship because the two observers can't agree on what time it is. They can agree on the speed, it's something they both can measure as one passes the other. And they agree about the speed of light. But that's about it.

The important point to grasp is that the idea of two events happening at the same time in different places has absolutely no meaning.
It's a really strange idea - we're used to thinking of simultaneous events - what's happening right now in China for example - but it's a completely wrong concept.

3. Aug 15, 2010

### Janus

Staff Emeritus
Its accounted for by length contraction. Let's say that you put a buoy 1.000004 ly from Earth (as measured from The Earth), and your ship is traveling just fast enough to get there in 365 days as measured from the Earth.

Sine the Earth and buoy are moving in respect to the ship as seen from the ship, they both, and the distance between them undergo length contraction as measured from the trip. In other words, the distance between them has contracted to just a tad over 1 light day. Thus the ship measures that it has traveled one light day in distance in one year, and the Earth measures tha traveled 1 light year in distance in one day, but they both agree that the ship reached the buoy.

4. Aug 15, 2010

### AJ Bentley

The easiest way to get a proper understanding of this one is to learn about 'Lorentz invariants'.
These are things that don't change when you move from one reference frame to another.
One of the most fundamental is the space-time interval between two events.

It's analogous to the distance between two points in ordinary geometry - no matter how you rotate and translate your piece of paper, the distance between two points remains the same. In Cartesian coordinates it would be the familiar root-of-x-squared-plus-y-squared.

In relativistic space-time, the equivalent is very similar but rather strange, it's root-of-x-squared-minus-t-squared.
(Sounds crazy but it's true!) This doesn't change - it's the same for all inertial observers, no matter what their velocity.

Now lets see how this applies:-
I'm going to set c=1 by using years and light years - makes it easier.

Let's say the moving observer is heading for a beacon at such a speed that he's going to reach it in one day.
What does the moving observer see? Well, first - he isn't moving.
He can see the beacon - and he can see the earth, but both are moving at pretty close to the speed of light. As the earth whips past him he sees that the beacon is about 1 light day away, sure enough, roughly 24 hours later, it too passes by close to the speed of light. Wow!.

Now, what is the space-time interval between these two events? Well, there was no distance involved - no x to square. Why not? - because both the earth and the beacon went past him at roughly the same place - nearly hit him in fact. The only thing to deal with is the root-of-nothing-plus-t-squared - which was 1/365th of a year so the space-time interval he computes is minus 1/365.

What about the earth observer? For him, the beacon isn't moving - it's space distance is fixed - about a light year away.
But what we have to compute is the space-time interval between the two events. The first event was the ship passing earth and the second was it reaching the beacon - just a tiny bit over a year later.
So whats the interval - it's (the square root of) 1^2 minus 1-and-a-bit^2 - very nearly zero - in fact it comes out to be minus 1/365 - same as for the moving observer.

Invariants make Relativity easy.