Imagine for a second that you are an old timey scientist and you don't know a thing about the theory of relativity (note that the purple words in the post are very relevant to relativity and you should make sure you understand them). The recent Michelson-Morley null result has demonstrated that the historically hypothesized luminiferous aether does not exist, and they have measured the speed of light as a constant, \(c=299792458\) m/s. What can we find out about relativity ourselves?
You're living on Planet 1, our destination planet, when you observe two spaceships moving with equal speed, \(v = 0.597c\), with respect to the ground. They each fire a laser beam at the other, creating a chain of events that we will call event A, B, C, and D:
| Event A | Left spaceship fires laser from position \(x=x'=0\) at time \(t=t'=0\) (this event, which is the same in all frames of reference, is what is called the event at the origin). |
|---|---|
| Event B | Right spaceship fires laser |
| Event C | Leftmost explosion |
| Event D | Rightmost explosion |
These events can be considered points in space and time. The reason why we have one marked coordinate, and one unmarked, is to differentiate between two different frames of reference:
- Rest frame: This is where the relevalt events are happening in the same place, and often uses marked coordinates \((t',x')\).
- Laboratory: This is where we observe events in another place, and often uses unmarked coordinates \((t,x)\).
One thing we know for certain, is that the speed of light is an invariant size, meaning that it has the same value in all frames of reference!
You've reports from an observer (code name M) right in the middle of the two spaceships that event A and B happen simultaneously and event C and D happen simultaneously.
The question here is, do you agree, based on what you've seen?
Before you answer, consider why observer M observes events A and B simultaneously. She is in the same frame of reference as the spaceships, always an equal distance from both (meaning she is moving with the same velocity as the spacecrafts, they seemingly stand still to her).
If this is hard to imagine, here's a more relatable example. Imagine two cars, both driving
past a tree towards the right. The tree is standing still (in the earth's frame of reference), and measures the red car has a velocity of 100 km/t, while the blue car has a velocity of 80 km/t. However, if you're sitting in the blue car, you won't measure the same. In your frame of reference, you're standing still and the tree is moving relative to you at -80 km/t, while the red car is moving 20km/t.
So, as long as she is moving with the same velocity as the spaceships, she will remain in the middle of them. We know that the speed of light is constant in all frames of reference, and that they have to travel the same distance to get to the middle. Then it makes sense that they reach the middle point at the same time! The laser beams have to travel the same distance from the spaceships to the middle in their own frame of reference, and therefore the observations of observer M are true.
You may suspect that what you see from the planet isn't the same as what observer M sees. We have defined event A as happening simultaneously in both frames of reference, this is an event at the origin, defined to be the same in all frames of reference. The beams will still cross at the middle position, as the observer M cannot both see that the beams cross simultaneously and that they don't. If we think of the two beams being at the point of observer M as two events, they will both happen at the same time in the same place, making them a singular event. Therefore every observer must agree on this, and it must be true in all frames of reference.
Now, as you're in the planet's frame of reference, the spaceships will appear to be moving while you are standing still. The right spaceship is moving away from the beam fired from the left spaceship, meaning the left beam will have to travel further than the right beam. They still need to cross the middle point at the same time, and so you will observe that the beam from the right spaceship seems faster than the beam from the left spaceship. This also means that the right beam will be fired later than the left beam, and as it's still moving faster after both have crossed the middle, it will reach the left spaceship first (i hope the colors helped you keep track of things here).
In your frame of reference, event A happens first, then event B, then the beams cross at the origin at the same time, then event C happens, and finally event D.
So the events are totally different depending on the frame of reference? Let's explore these further, and see if we can figure out why this is.
We begin by finding expressions for the positions as functions of time t, velocity v, the time tA and the distance between the spaceships L. We've only used basic physics, keeping in mind that distance is given as \(x = x_0 + v\Delta t\):
| The leftmost spaceship | \(x_l = v\Delta t = v(t-t_A)\) |
|---|---|
| The observer in the middle | \(x_M = L/2 + x_l = L/2 + v(t-t_A)\) |
| The beam from the leftmost spaceship | \(x_{lb} = c\Delta t = c(t-t_A)\) |
These are simply based on the equations of motion, where the observer in the middle is a length \(x_0 = L/2\) from the leftmost spaceship. We know that when the beams cross at the position of the middle observer \(t_M\), the position of the middle observer, \(x_M\), equals the position of the beam emtted from the leftmost spaceship, \(x_{lb}\). We can set these up in order to find the time \(t_A\) of the first event:
So the time of event A can be written as \(t_A = t_M - \frac{L/2}{1-v}\).
Notice how we dropped the variable for the speed of light, \(c\)? We are now using something called natural units / geometrized units, which are units based on universial physical constants. By giving distance and time the same unit, we set the speed of light in vacuum, \(c\), as well as certain other constants, equal to unity (one). The conversions made in order for this to make sense are made with respect to the velocity of light, as this is constant for all observers, \(t=\frac{x}{c}\). Therefore every velocity is given as a percentage of the velocity of light and thus lack dimension.
We would also like to know how much time passed between event A and C, and so we find an expression for the beam from the rightmost spaceship as a function of time t expressed in terms of the time tM, L and v:
| The rightmost spaceship | \(x_r = v\Delta t = v(t-t_B)\) |
|---|---|
| The observer in the middle | \(x_M = -L/2 + x_l = -L/2 + v(t-t_A)\) |
| The beam from the rightmost spaceship | \(x_{rb} = -c\Delta t + L/2= c(t-t_M)+L/2\) |
and use this in order to find an expression for the distance to the middle point, as we know that at the time of event C, \(t=t_C\), the beam from the rightmost spaceship, \(x_{rb}\), hits the leftmost spaceship, \(x_l=x_{rb}\):
So the time of event C can be written as \(t_C = t_M + \frac{L/2}{1+v}\).
Now we have some idea of when the events happen (in the frame of the spaceships), meaning we can find the timeframe from event A to event C \(\Delta t' = t_M' - t_A' =\frac{L}{1-v^2}\) (simply linsert the expressions we found above, and do the math). In the frame of the planet the timeframe from event A to event C from the planet can be written as \(\Delta t = \frac{L}{c}= L \).
Okay, it can be quick to get lost inbetween these functions, but what we've really done here is to stand in the spaceship frame of reference, and find expressions that let us track the movement of the spaceships and lasers as functions of time. From these expressions, we could find the time of the first event, and of the first spaceship being hit! And from there, we could figure out how much time passed in the spaceship frame of reference. Using basic equations of motion, we can compare this timeframe to the timeframe in earths frame of reference, and we see that they are different!
As we do not yet know about special relativity, we may wonder why. Have we made some mistake? Do the laws of physics change depending on our frame of reference? Or is the implication that time is relative, that it depends on your frame of reference (as velocity does)?
We can find a relation between the two timeframes,
\(\frac{\Delta t}{\Delta t'} = \frac{L}{L/(1-v^2)}=\frac{1}{1-v^2} \leftrightarrow \Delta t = \frac{\Delta t'}{1-v^2}\)
and get a familiar expression (to those familiar with special relativity, at least).
Einstein found an expression for the relationship between the proper time (the time we observe), and the time observed from another system:
\(\gamma=\frac{1}{\sqrt{1-v^2/c^2}}\)
We can use this to transfor between frames of reference, both from marked to unmarked:
\(t=v\gamma x'+\gamma t'\)
\(x=\gamma x' + v\gamma t'\)
and unmarked to marked:
\(t'=-v\gamma x + \gamma t\)
\(x'=\gamma x - v\gamma t\)
where the time we measure on the moving object from the rest frame, the proper time \(\Delta \tau = \Delta t'\), is used in order to find the time and position.
The Lorentz factor is similar, but not the same as what we found. Why is that? Well, in our equations we've only accounted for time dialation, but if you're familiar with special relativity you know that time isn't the only thing that we need to consider- length is also relative, and we've yet to think about length contraction.
Not to worry, though, we're not done with relativity just yet.