by Steve Bryson
We now have our spacetime diagram, a way of visualizing spacetime. We know how to express our both own coordinates on it and the coordinates of someone moving relative to us. In a spacetime diagram, we have our own coordinates represented by two perpendicular lines:
In our spacetime diagram, the coordinate axes of someone moving relative to us will look like two lines tilted towards each other, and the faster the motion the more tilted towards each other they are. Call the moving coordinates x' and ct':
Remember that the reason for this funny tilt is so that the speed of light will be the same for all observers. Now we want to ask how to compare actual measurements in the two coordinate systems. In other words, we are asking the question "given an event with coordinates ct and x in my coordinates, what will it's coordinates ct' and x' be for someone moving relative to me?" We can visually find the ct' and x' coordinates on our spacetime diagram by using the following simple rules:
To find the ct' coordinate of an event, draw a line parallel to the x' axis through the event. Where this line meets the ct' axis is the ct' coordinate of the event.
To find the x' coordinate of an event, draw a line parallel to the ct' axis through the event. Where this line meets the x' axis is the x' coordinate of the event.
So here are two diagrams showing how to find the ct' and x' coordinates of an event:
Here we can see that if an event takes place at, say, ct = 1 then in the moving coordinate system, the event will have ct' less than 1. This means that time seems to me to be passing slower for the moving coordinate system. Similarly, if the event had x coordinate for me, then it x' coordinate will be less than one. This means that my moving friend would think that the event is not as far from the center of the diagram as I do.
Now remember that these diagrams cannot be used to get the actual numerical values of the ct' and x' coordinates. To do this we must use actual formulas. I will give you these formulas, but in this class I cannot show you why they are true. That is the material for a technical handout, though it will be the easiest thing in the handout. Here are the formulas, where v is the velocity of the moving coordinates relative to mine, and c is, as usual, the speed of light:
These formulas are true in all units of measurements, miles, seconds, etc. If you have a calculator with a square root function, you may wish to plug in some sample velocities and see what happens. You will discover that if v is much smaller than the speed of light, then v2/c2 is very small and so 1 minus this is still very close to 1. As the square root of 1 is 1, then the factors effecting the coordinates in the above formulas are very close to 1 and so the coordinates are not very different. In fact, for speeds that we are used to, the difference is entirely undetectable. This is why we do not have an intuitive feel for relativity.
Let's now see what this all means. We need to look at a case where you will go on a trip and travel at some speed close to the speed of light and go a very great distance. In this way, the difference in coordinates between us will be large enough to appreciate.
Say that you needed a real change of scene and decided to fly to the star Alpha Centauri, this being the nearest star. Even though this is the nearest star it is still quite far away, at a distance of 4.3 light years (=25,277,980,000,000 miles), and so you choose to travel at 90% of the speed of light, or at 167,654 miles per second (603,554,400 miles per hour). Let's say that you accelerated at this speed in an instant of time, not worrying about how you did this or whether or not you could survive this acceleration. Then this is what the situation would look like on my spacetime diagram:
We see that while you arrive at the star after 4.78 years have passed for me, the time that passed for you will be much shorter! (In evaluating the above diagram, there is a subtlety in that here we are noticing what you would say the time on Earth is on your arrival.) In fact, the time that passes for you is only about 2.1 years. This is because at this speed the conversion factor in the above formulas is only about .436, and .436 times 4.78 years is about 2.1 years. Thus you would say that you arrive at Alpha Centauri after only 2.1 years of travel. If you immediately took off again and flew back to Earth at the same speed, only 4.2 years would have passed for you while for me 9.56 years would have passed. Thus I would have aged 5.36 years more than you since we last met! This is a very physical result of living in spacetime.
What if I managed to take a picture of your spaceship as you flew by at 90% of the speed of light? Let's say your ship is a mile long, so that if you orient your coordinate system so that the tail of the rocket is at your origin then the nose will have x' coordinate = 1 mile. For me, the x coordinate of the nose of your ship will be at 1 mile times .436 or at .436 miles. Your ship would appear shorter to me by more than half!
In fact, you would see both the Earth and Alpha Centauri moving at 90% of the speed of light, and so the distance between them would seem to be 4.3 light years times .436 = 1.87 light years. This is exactly the distance you think you would travel at 90% of the speed of light in 2.1 years, and so everything looks perfectly natural to you. This is the situation from your spacetime diagram:
So to me, you traveled 4.3 light years in 4.78 years traveling at 90% of the speed of light, but you only aged 2.1 years. To you, you traveled 1.87 light years at 90% of the speed of light and so you aged 2.1 years, exactly how long such a voyage should have taken. The unusual thing for you is that the distance you had to travel was much shorter than you thought it should be.
But wait! If you are moving relative to me then I am moving relative to you!
Why don't the effects exactly cancel out?
If you flew to Alpha Centauri as in the previous section and flew back again, then you would age 4.2 years and I would age 9.56 years. Yet we have been saying all along that motion is relative--who is to say whether it was you moving or if it was I? This is known as the twin paradox, as it is commonly stated with regard to comparing the ages of twins.
In fact, it is not a paradox at all, for it involves a confusion about the statement "motion is relative". While it is true that steady motion is relative to the person observing the motion, acceleration is not. When you accelerate, you cannot close your eyes and pretend that nothing is happening. You will feel the acceleration, and you can do experiments, such as watching a hanging string, which measure your acceleration and do not refer to other observers. So if you take off in your rocketship, you must accelerate (in our example instantaneously) and this makes your situation different from mine in an absolute way. The exact difference is in the rotating of your coordinate axes by the act of acceleration. When you turn around to come back, you undergo another acceleration which causes your axes to 'swing around' and in so doing you do not experience as much time as I do. This can be understood by carefully staring at the following spacetime diagram:
If, on the other hand, you just happened to be going at 90% of the speed of light relative to me and did not speed up or slow down, then our situations would be reflexive and the effects would exactly cancel. I would think that time was passing slower for you and your objects were flattened, and you would think it was I who was flattened and my time that was running slow. We could not, however, ever compare our observations without agreeing on a coordinate system, and we cannot do this without one of us accelerating. That would destroy the symmetry of the situation.
The twin paradox is an excellent object lesson that one must be very careful about how one interprets things in spacetime.
More on the spacetime distance
Remember the spacetime distance, labeled s from the last session? It represents the distance of an event from the origin, and in terms of coordinates it is given by the formula
s2 = c2t2 - x2
This distance has a special property, that it is the same in all coordinate systems. It has tremendous significance for what happens in spacetime, and in fact you can start with this formula for the distance and derive all of relativity theory (this is the way that I actually approach the subject). We will only mention a few aspects of this distance here.
First, what is the spacetime distance from my origin to a point on my world line? Any point on my world line will have a time coordinate t given by the time that passes for me, and an x coordinate with the value zero (it is where I am). So the spacetime distance is given by
s2 = c2t2 - 0
= c2t2
so that s = ct. This means that for me, the spacetime distance is simply the time that passes for me. That is to say that the spacetime distance between two events on my world line is given by the time that I would say passed between the two events. This is true only for events on my world line.
We can use this to quickly analyze the situation from above where you flew to Alpha Centauri at 90% of the speed of light. Here we are interested in the spacetime distance between two events, your launch and your arrival at the star. The first event is at the origin of both of our coordinates, so they have coordinates t = t' = 0 and x = x' = 0. The second event is given in my coordinates by t = 4.78 years and x = 4.3 light years. We want to find the coordinates of the second event (your arrival) in your coordinate system. We automatically know the x' coordinate-- it is zero (you arrived there). Thus all we have to do is plug my coordinates into the spacetime distance to get (because we are measuring distance in light years the speed of light c = 1):
s2 = t2 - x2
= (4.78)2 - (4.3)2
= 4.36
= (2.1)2
= t'2
So we see that you will arrive at the star after 2.1 years had passed for you. Using this method we have seen that time passes slowly for you without actually using the coordinate transformation formulas at the beginning of this handout.