This document describes some of the ideas and consequences of Einstein's Theory of Special Relativity (or Special Theory of Relativity), which claims that the speed of light in a vacuum and all laws of physics (but not time and distance) are the same in all inertial frames of reference in the absence of gravity.

Special Relativity is a special case of General Relativity, where the latter adds gravity as a curvature of space and time (so frames in free fall are still considered inertial).

In this document, distance and time are measured in seconds (you might also say that distance is measured in light-seconds, because it is obtained by dividing by the speed of light c). Velocity is always relative to the velocity c, so it is unitless.

Unless otherwise stated, all statements about what observers "see", ignore the time it takes the light to travel from the observed event to the observer.

Let's assume we have two frames with the same orientation. Each has a single spatial dimension. We'll refer to one frame as the "lab frame" and the other as the "moving frame". The moving frame is moving with a constant velocity v relative to the lab frame, and v is considered positive when the movement is in the direction of plus infinity on the spatial axis of the lab frame. The time is zero in both frames, when their spatial zero locations coincide.

The Lorentz transformation gives the space-time coordinates
(x,t) of an event in the lab frame as a function of the coordinates (x_{1},t_{1})
of the same event in the moving frame:

where the Lorentz factor

is 1 for v = 0, and tends towards infinity as v tends towards ±1.

The inverse transformation is obtained by simply negating v:

Note that observers in both frames agree that the
coordinates (x,t) in the lab frame and (x_{1},t_{1}) in the
moving frame represent the same event.

Due to the so-called time dilation, a clock at rest in the
moving frame will appear to be slow to an observer at rest in the lab frame.
This is seen by setting x_{1} = 0 in the Lorentz transformation (to represent
the fixed clock location in the moving frame) and solving for t:

So, for the observer in the lab frame, 1s on the moving clock corresponds to γ seconds of his time. Hence, the clock appears to be slow.

A clock at rest in the lab frame will also appear to be slow
to an observer at rest in the moving frame. This can be seen by setting x = 0
and solving for t_{1}:

So, both observers see the other's clock as being slow.

The so-called Lorentz contraction is the spatial equivalent
of the time dilation described above. A rod (of length Δx_{1}) at
rest in the moving frame will appear to be shorter to an observer at rest in
the lab frame. This is seen by setting t = 0 in the Lorentz transformation (representing
the instant in lab time at which the locations of the two end points of the rod
are simultaneously determined) and solving for Δx:

The opposite is true as well: a rod at rest in the lab frame will appear to be shorter to an observer at rest in the moving frame.

Note that the concept of length only makes sense if the end
point locations are determined *simultaneously*, and two spatially
separate events, which are simultaneous in one frame, are not simultaneous in
the other. So, the Lorentz contraction is actually a direct consequence of the
time dilation.

If we add a second spatial dimension y, and still assume that the velocity v of the moving frame is in the direction of plus infinity on the x axis, then the Lorentz transformation becomes:

This is clearly true, because obviously there will be no effect on the spatial dimension perpendicular to the direction of movement. We'll call this 3-by-3 matrix L*.

Now let's consider the more general case where v is not
necessarily parallel to the x axis. We can handle this case by simply
transforming the spatial coordinates to a rotated frame of reference, where v *is*
parallel to the first spatial axis. Then we can perform the Lorentz transformation
above, and finally rotate the coordinates back again.

The following transformation will convert coordinates in the rotated frame to coordinates in the actual frame:

We'll call this 3-by-3 matrix (including the 1/v factor) R. Note that the columns of R are simply the base vectors of the rotated frame expressed in actual coordinates; the first base vector is the v vector normalized to a length of 1, the second base vector is the same, but rotated 90 degrees, and the third base vector representing the time axis is the same as the original time vector.

The inverse matrix R^{-1} will rotate the other way,
i.e. convert coordinates in the actual frame to coordinates in the rotated
frame. It is not difficult to see that a rotation in the opposite direction is
equivalent to negating v_{y} in the matrix R (negating the angle of
rotation from the x axis means negating the y coordinate), so we have:

where the 3-by-3 matrix (including the 1/v factor) is R^{-1}.

We now have the three matrices we need:

The total transformation matrix L is the matrix product of these:

So, the Lorentz transformation for arbitrary v≠0 in a 2-dimensional space is:

Note that this won't work for v=0, because our rotation
matrix R becomes undefined when v has no direction. However, we already know
the solution to this special case: it is simply the identity transformation
(x,y,t) = (x_{1},y_{1},t_{1}).

A space-time diagram shows the lab time t vs. the lab
location x. In this diagram we can also draw the t_{1} and x_{1}
axes of a frame moving with velocity v relative to the lab frame. The lines
representing the t_{1} and x_{1} axes can be found by setting x_{1}
= 0 and t_{1} = 0, respectively, in the Lorentz transformation. Thus,
we find that the lines x = vt and t = vx represent the t_{1} and x_{1}
axes, respectively (see example in Figure 1).

In a similar way, we can find the lines representing t_{1
}= ..., -2s, -1s, 1s, 2s, ..., and x_{1} = ..., -2s, -1s, 1s, 2s,
... It turns out that all the t_{1} = n lines are parallel, and all the
x_{1} = n lines are parallel. Here's a summary of the characteristics
of these lines:

t_{1} = n lines (where n is an integer number of
seconds):

- have slope v (in lab frame)

- intersect t axis at t = n/γ

- intersect t
_{1}axis at t = nγ and x = nγv

x_{1} = n lines (where n is an integer number of
seconds):

- have slope 1/v (in lab frame)

- intersect x axis at x = n/γ

- intersect x
_{1}axis at x = nγ and t = nγv

The area of one of the unit parallelograms in the (x_{1},t_{1})
coordinate system is always 1s^{2} (this area is equal to the
determinant of the x_{1} and t_{1} unit vectors, which
according to the information above are (γ, γv) and (γv, γ),
respectively).

From the above we can make the following observations about how the axes of the moving frame appear in the space-time diagram:

- For v = 0, the x
_{1},t_{1}axes are identical to the x,t axes. - As v increases from 0 to 1, the x
_{1}axis rotates up from the x axis towards a 45° angle, and the t_{1}axis rotates right from the t axis by the same angle. The units on both axes stretch out from 1 to infinity. - As v decreases from 0 to -1, the x
_{1}axis rotates down from the x axis towards a -45° angle, and the t_{1}axis rotates left from the t axis by the same angle. Again, the units on both axes stretch out from 1 to infinity.

Note that light (in a vacuum) always travels along a 45° line towards the upper right or upper left corner (corresponding to v=±1) in all frames.

A clock at rest at x_{1} = 0 in the moving frame
will be moving up the t_{1} axis in the space-time diagram as time
passes (i.e. the t_{1} axis is the clock's "world line" - the path it follows
through space and time).
An observer at rest at x = 0 in the lab frame will be moving up the t
axis as time passes (i.e. the t axis is the observer's world line).
From the observer's point of view, the clock and he will
be on the same horizontal line in the diagram at any given moment (because this
line represents the current lab time).

If both the observer and the clock start out at t = t_{1}
= 0, then they will be at the positions indicated in Figure 2 after 1s of the clock's time.
The clock will of course have moved one unit up the t_{1} axis
(because this is 1s of its time) corresponding to the lab coordinates
(γv,γ) determined previously
during the construction of the space-time diagram. The observer will be on the
same horizontal line, so his lab coordinates will be (0,γ). This shows us
that 1s of the clock's time corresponds to γ seconds of the observer's
time - in accordance with the time dilation formula derived in
section #!

Now, let's swap the observer
and the clock and see how this looks in the space-time diagram. The clock is
now at rest in the lab frame and moving up the t axis, while the observer is
moving up the t_{1} axis. From the observer's point of view, the clock
and he will be on the same line parallel to the x_{1} axis at any given
moment (because this line represents the current time in the moving frame).

Again, the observer and the
clock start out at t = t_{1} = 0, but this time they end up at the new
positions indicated in Figure 3 after 1s of the clock's time.
The clock will have moved one unit up the t axis, and the observer will have moved more than
one unit (γ seconds) up the t_{1} axis in order to remain at the same t_{1}
time as the clock (i.e. to remain on the same line parallel to the x_{1}
axis). So, again the observer sees the predicted time dilation.

Let's consider a rod at rest in the moving frame with one
end point at x_{a1} = 0 and the other at x_{b1} = 1s. As time
passes, the first end point will move up the t_{1} axis, and the second
end-point will move up a line parallel to the t_{1} axis. These two
lines (dashed red in Figure 4) are the world lines of the end points (i.e.
the "tracks" of the end-points as these move through space and time).

At any given moment, an observer in the moving frame will
measure the spatial distance between the two end points along a line parallel
to the x_{1} axis (e.g. the green rod in Figure 4), because such a line represents
simultaneity in the moving frame. Thus, this observer will see a rod length of
x_{b1}-x_{a1} = 1s.

An observer at rest in the lab frame, however, will measure
the spatial distance between the two end points along a line parallel to the x
axis (e.g. the green rod in Figure 5), because that is what *he* understands
to be a simultaneous measurement of the two end point locations. So, he will see
a shorter rod (of length 1s/γ, to be exact).

Note that in the situations shown in Figure 4 and Figure 5, the two observers
agree that they are determining the location of the leftmost end point of the rod
at the same moment (because this end-point happens to have the same coordinates
in both figures), although their
clocks do not show the same time at that moment. The observers also agree that
they are *not* determining the location of the rightmost end point at the
same moment. Each observer believes that he is determining the location of both
end points simultaneously, and that the other observer is *not* doing so,
so he does not find it surprising that the other observer comes up with a
different length.

The Lorentz transformation (see section #) tells us that the time in the moving frame at lab coordinates (x,t) is:

This means that an observer in the lab frame will see the
time in the moving frame at a fixed lab location (e.g. x = 0) as being *faster*
by a factor of γ than his own time.
In section # we determined that he will
see the time at a fixed location in the moving frame as being *slower*
by a factor of γ than his own time. In other words, if he observes
a row of clocks at rest and synchronized
in the moving frame, and shifts his attention from clock to clock as each of
them passes him, their time will appear to be faster than his
(e.g. the vertical instances of clocks B, C, and D in Figure 6 show times -2s, 0s, and 2s
when they pass lab location x=0 at lab times -1.73s, 0s, and 1.73s, respectively), but if he only
looks at a single clock all the time, then its time will be slower than his
(as illustrated in Figure 2 in section #).

The reason for this apparent paradox is that the moving clocks - due to their spatial separation - show different times in the lab frame, even though they are perfectly synchronized in the moving frame. If we fix the lab time t in the formula above, we see that at any given lab moment (t), the leading clock (with largest x if v>0) will show the earliest time, and each successive clock will show later and later times (e.g. the horizontal instances of clocks A, B, and C in Figure 6 show times -1s, -0.5s, and 0s, respectively, when they are observed simultaneously in the lab frame at lab time t=0s). The difference between the times shown on two clocks is equal to their lab distance multiplied by their velocity multiplied by γ.

The above observations help explain the well-known twin paradox, which asks what happens if one twin travels out into space and back again (with the dashed red world line in Figure 7), while the other twin remains at home (with the black t axis as world line). Since the twins are moving relative to each other, each twin should see the other one ageing more slowly than himself due to time dilation. So, when the traveling twin returns home, the two would supposedly disagree about which one is now the youngest. This is the paradox.

However, if we place synchronized clocks
throughout the lab frame and let the traveling twin shift his attention from
clock to clock as he passes them, then it is clear from our previous
observations that he will see the lab time on these clocks as being *faster*
than his own time, even though each individual clock (as well as the clock at
home) will be ticking slower than his. Since the traveling twin sees the lab
time at his own location to be consistently faster throughout his travel, he
will not be surprised to find that the other twin is older than himself when he
returns home. So, the twins agree that the one at home is the oldest.

But what if the traveling twin keeps his attention on the clock at home? Due to time dilation that clock will appear to be ticking slower during the outbound journey (where the blue coordinate system in Figure 7 represents the moving frame) - and also slower during the homebound journey (where the green coordinate system in Figure 7 represents the moving frame). So how does the twin at home end up being older? As previously shown, the clocks ahead of the traveling twin in his direction of movement will show later and later times, so when he reverses his direction of travel to move homeward, the clocks he passed on his outbound journey (including the one at home) will immediately jump from an earlier to a later time in order to fulfill the requirement that clocks ahead of him show later and later times (the clock at home jumps from the lower to the upper black circle in Figure 7). In other words, the twin at home will appear to age very fast during the short period of time while the traveling twin is reversing his direction of movement. Thus, there is no paradox.

Note that if the traveling twin were initially approaching the twin at home
(along the green t'_{1} axis in Figure 7), and then reversed his direction to move away
(along the blue t_{1} axis), he would see the twin at home jump *back* in time
(from the upper to the lower black circle).

As demonstrated earlier, the measured time and distance between two events depends on the
observer's velocity relative to the events. However, the so-called "space-time interval"
(Δs)^{2} between two events is independent of the observer (i.e. universally
invariant) and is determined as follows for 1 and 3 spatial dimensions, respectively:

There are 3 cases:

- (Δs)
^{2}>0: The space-time interval is said to be*timelike*because there is more time than space between the events. In this case, Δs is the so-called "proper time" between the events, and is the measured time between the events in a co-moving frame where the events occur at the same spatial location (i.e. the spatial terms are zero in the above equation). - (Δs)
^{2}<0: The space-time interval is said to be*spacelike*because there is more space than time between the events. In this case, no co-moving frame exists because it would have to move faster than light relative to other frames. Here, the quantity √-(Δs)^{2}(or imag(Δs)) is the so-called "proper distance" between the events, and is the measured distance between the events in a frame where the events occur simultaneously (i.e. Δt is zero in the above equation). - (Δs)
^{2}=0: The space-time interval is said to be*lightlike*because there is the same amount of time and space between the events, so a co-moving frame would have to move at the speed of light (unless the events are separated in neither time nor space in any frame). In such a frame, the measured time and distance between the events would be zero.

For one event to be the cause of another, the space-time interval between them must be timelike or lightlike (and the time difference must be non-negative).

An acceleration is defined as constant if co-moving observers see it as constant, i.e. if the instantaneous acceleration is the same in all inertial frames at the moment when the accelerating object is at rest in each of these frames.

The world line (in an inertial frame with a single spatial dimension) of an object with the
constant acceleration a_{0} is a hyperbola described by the following
equation:

where t_{0} is the time at which the object is at
rest in the inertial frame, and x_{c} is the x-coordinate where the hyperbola's
2 asymptotes intersect (see Figure 8). The distance from x_{c} to x(t_{0})
(i.e. the distance between the 2 black dots in Figure 8)
is 1/a_{0}. The left and right branches of the hyperbola correspond to a negative and
positive a_{0}, respectively.

Note that this constant acceleration hyperbola consists of all the space-time points whose
"proper distance" to (x_{c},t_{0}) is 1/a_{0} (see
section #).

As stated in section #, the Lorentz factor γ becomes infinite as the velocity v approaches 1 (the speed of light), so this is an upper limit on v, which cannot be exceeded. How does this affect space travel?

Well, if an observer in the lab frame determines the traveler's velocity v by measuring the traveled (lab frame) distance and dividing that by the (lab frame) duration of the travel, then it is true that this velocity v has an upper limit of 1. So, from the point of view of the observer in the lab frame, the traveler cannot travel faster than light.

Likewise, if the traveler himself determines his own velocity v by measuring the relative movement of some reference point at rest in the lab frame, and he makes sure that both distance and time are measured in his own moving frame, then this velocity will be the same as the one determined by the observer in the lab frame, and it will also have an upper limit of 1. So, if the traveler defines his velocity as the velocity at which objects at rest in the lab frame move past him, then also he will agree that he cannot travel faster than light.

However, one might argue that a
different definition of velocity would be more appropriate for the traveler. A
traveler going from location A to B (both locations being at rest in the lab
frame) will probably be more interested in knowing how long (measured in his
own time) it will take him to travel the (lab frame) distance from A to B. So
he might define his effective velocity as the traveled *lab frame*
distance divided by the duration of the travel measured in his *own* time.
Using the Lorentz transformation, we find that the relationship between this
effective velocity v_{eff} and the real velocity v is given by:

There is clearly no theoretical upper bound on v_{eff}, and it becomes greater than 1
("faster than light") when v is greater than 1/√2
(approx. 71% of the speed of light).
This means that a traveler can in fact (theoretically)
travel as far as he wants in as short time as he wants. To him it appears that the
Lorentz contraction is reducing the distance he has to travel, so he can do it
in a shorter time without exceeding the speed of light.

The disadvantage of traveling with a
high v_{eff} is that the lab time (at the traveler's moving location)
will speed up as demonstrated in section #,
so the interesting distant star that was the target of the travel, might be long gone
when the traveler gets there an hour later, and the folks at home will be long dead
when he returns after another hour of his own time.