special relativity - What's the difference between space and time?

I'm having a hard time understanding how changing space means changing time. In books I've read people are saying "space and time" or "space-time" but never explain what the difference is between the two concepts or how they are related.

How are the concepts of space, time, and space-time related?

Answer

Suppose you move a small distance $\vec{dr}$ = ($dx$, $dy$, $dz$) and you take a time $dt$ to do it. Pre-special relativity you could say three things. Firstly the distance moved is given by:

$$dr^2 = dx^2 + dy^2 + dz^2$$

(i.e. just Pythagorus' theorem) and secondly the time $dt$ was not related to the distance i.e. you could move at any velocity. Lastly the quantities $dr$ and $dt$ are invarients, that is all observers will agree they have the same value.

Special relativity differs by saying that $dr$ and $dt$ are no longer invarients if you take them separately. Instead the only invarient is the proper time, $d\tau$, defined by:

$$c^2d\tau^2 = c^2dt^2 - dx^2 - dy^2 - dz^2$$

In special relativity all observers will agree that $d\tau$ has the same value, but they will not agree on the values of $dt$, $dx$, $dy$ and $dz$.

This is why we have to talk about spacetime rather than space and time. The only way to construct laws that apply to everyone is to combine space and time into a single equation.

You say:

I'm having a hard time understanding how changing space means changing time

Well suppose we try to do this. Let's change space by moving a distance ($dx$, $dy$, $dz$) but not change time i.e. $dt$ = 0. If we use the equation above to calculate the proper time, $d\tau$, we get:

$$d\tau^2 = \frac{0 - dx^2 - dy^2 - dz^2}{c^2}$$

Do you see the problem? $d\tau^2$ is going to be negative so $d\tau$ is imaginary and has no physical meaning. That means we can't move in zero time. Well what is the smallest time $dt$ that we need to take to move ($dx$, $dy$, $dz$)? The smallest value of $dt$ that gives a non-negative value of $d\tau^2$ is when $d\tau^2$ = 0 so:

$$c^2d\tau^2 = 0 = c^2dt^2 - dx^2 - dy^2 - dz^2$$

or:

$$dt^2 = \frac{dx^2 + dy^2 + dz^2}{c^2}$$

If we've moved a distance $dr = \sqrt{dx^2 + dy^2 + dz^2}$ in a time $dt$, the we can find the velocity we've moved at the dividing $dr$ by $dt$, and if we do this we find:

$$v^2 = \frac{dr^2}{dt^2} = \frac{dx^2 + dy^2 + dz^2}{\frac{dx^2 + dy^2 + dz^2}{c^2}} = c^2$$

So we find that the maximum possible speed is $v = c$, or in other words we can't move faster than the speed of light. And all from that one equation combining the space and time co-ordinates into the proper time!