At any given moment you are traveling through time. Only forwards though, and since you're stuck on earth you can only affect the speed at which you travel minimally. You can also travel in space, which you do any time you're walking, or driving etc.
Your movement through both space and time can be described as the spacetime distance:
\(\Delta s^2 = \Delta t^2 - (\Delta x^2+\Delta y^2 + \Delta z^2)\)
This is an invariant size \(\Delta s = \Delta s'\), equal to the proper time, and represents the line-element between two 4-vectors. 4-vectors are vectors in spacetime, of which the first component is the timecomponent \(t\), and the last three are the space components, \(x,y,z\) (usually). All components of a 4-vector must me physically measurable sizes and you must be able to transform them between frames of reference using the Lorentz-transformation.
It's useful to represent your movement through both space and time, and we do this using a spacetime diagram. Let's try to draw up a few.
The gif shows three spaceships traveling with different velocities with respect to a space station. We can draw up a spacetime diagram in order to show how each of them move through time and space:
This first diagram shows how each spaceship moves in the space station frame of reference. The space station is standing still in its own frame of reference and is therefore only moving through time, straight up. Both ship 1 and 2 have constant velocity, and so they move, and the faster they are, the closer their line is to the \(t=x/c\) line. At this line the velocity would be at the speed of light, and as they have mass they cannot pass this line (they would seemingly move backwards in time if they did!).
What about ship 3? This one represents the ship that at first goes slower than the rest, then accelerates untill it's going as fast as Red, and at that point deaccelerates again.
The space station frame of reference is pretty intuitive, once you get the hang of spacetime diagrams. But what about the spaceships? Here, you have to imagine that the ship you're on (the frame of reference you are in) is standing still, while the other ships and the space station are moving. In the case of ship 1, the other ships would pull ahead, while the space station would move backwards at (negative) ship 1's velocity.
We have basically just rotated the coordinate system counter clockwise! The same logic applies to ship 2, because it too is travelling at constant velocity!
What about the accelerating ship 3? In its own frame of reference it will be standing still, while everything around it moves, accellerating and deaccelerating. You will see soon that acceleration messes up Lorentz transformations. Here you can see this visually, as we no longer simply get a rotated coordinate system, but rather different lines.
For ship 3, every other line would have to be curved, to show the acceleration happening, and it would curve as a mirror to the ship 3 line in the other time diagrams.
Very cool. But we cant just show you spacetime diagrams and other pretty things without also discussing some events! From now on we will go back to, and remain in, the space station frame:
| Event 1 | When all the spaceships are aligned at \(x=0\) and \(t=0\) |
|---|---|
| Event 2 | When spaceship 3 catches up with ship 2 and they are at the same position |
Look at the spacetime diagrams to see if you can tell how Event 1 (at the origin) takes place at the same time and place for all the frames of reference! See how they are all at the same spot there? Anyway.
Measured on the clock in the space station it takes 10 ms between the two events, on the clock in the frame of ship 2, it takes 8 ms. We assume that the clocks make a tick every milisecond, with the first tick at event 1 and the last tick at event 2.
We can deduce from this that the faster the ship, the fewer ticks on the clock. Spaceship 3 moves slowly at first, and then has to accelerate a lot in order to catch up with ship 2. In the space it takes for it to catch up, it must be moving faster than ship 2, as it covers the same distance in less time! So if we add a clock to ship 3, we expect it to tick just a tinybit slower than the space station at first, before ticking even slower than ship 2 as it catches up!
Our argumentation here bases itself on time dilation, which we became familiar with in earlier posts. However, what we actually observe here is that the spaceships follow the principle of maximal aging! This principle states that an object always will follow the path which gives the maximal distance in spacetime.
Let's look at the formula for distance in spacetime (for one dimension, as the ships are only traveling in the x-direction): \(\Delta s^2 = \Delta t^2 - \Delta x^2\)
If we are standing still we have that \(\Delta x^2 = 0\) and so we are left with \(\Delta s^2 = \Delta t^2\). If we now divide the expression by \(\Delta t^2 \) on both sides, we get:
\(\frac{\Delta s^2}{\Delta t^2}=c\)
Which implies that so long as we are standing still, we are still moving at the speed of light through spacetime! We move through spacetime no matter what distance we have, meaning that when you move through space (thats just regular movement, like walking), you'll move even slower through time, as the sum of \(\Delta t^2 \) and \(\Delta x^2 \) stay equal to \(\Delta s^2\). And now you know the real reason why we experience time dilation! The faster the spaceship, the slower the time (and the longer space between each tick).
More relativity to come, you know, relatively soon.