Special relativity

Special Relativity

Newtonian mechanics provides an excellent description of nature but is not universally valid, especially at extreme conditions (very small, very heavy, or very fast).
Newtonian mechanics encapsulates our common sense view of the world, which doesn't apply when looking away from our everyday world.
Special relativity, developed by Einstein, replaces Newtonian mechanics for particles traveling very fast, comparable to the speed of light in a vacuum.
The speed of light is $c = 299792458 \text{ ms}^{-1}$ , an exact value that defines the meter as the distance traveled by light in $1/299792458$ seconds.
For this course, we approximate the speed of light as $c \approx 3 \times 10^8 \text{ ms}^{-1}$ .
Nothing can travel faster than $c$ .
The theory of special relativity is based on two postulates:
- Postulate 1: The principle of relativity: the laws of physics are the same in all inertial frames.
- Postulate 2: The speed of light in a vacuum is the same in all inertial frames.
The second postulate seems nonsensical from a common-sense view, which is encapsulated in Galilean transformations.
Galilean transformations:
- Relate Cartesian coordinates of two inertial frames, S and S', moving relative to each other with velocity $v = (v, 0, 0)$ .
- $x' = x - vt, y' = y, z' = z, t' = t$
- If light travels in the x direction in frame S with speed c, then $x/t = c$ .
- In frame S', the trajectory of the light ray is $x'/t' = c - v$ , implying the speed of light should be $c - v$ , which contradicts the second postulate.
To obey both postulates, the assumption of absolute time ( $t' = t$ ) must be abandoned, and time must tick at different rates for observers in different inertial frames.

Lorentz Transformations

Consider two inertial frames, S and S', moving with relative speed v, with Cartesian coordinates (x, t) for S and (x', t') for S'.
The most general relationship between these coordinates is $x' = f(x, t), t' = g(x, t)$ for some functions f and g.
The law of inertia implies the trajectory of a particle is a straight line in the (x, t) plane, which means the map $(x, t) \rightarrow (x', t')$ must be linear.
Therefore, the functions f and g must be of the form $x' = \alpha1 x + \alpha2 t, t' = \alpha3 x + \alpha4 t$ , where $\alpha_i$ are functions of v.
Since S' is traveling at speed v relative to S, an observer at the origin $x' = 0$ of S' moves along the trajectory $x = vt$ in S.
With the assumption that the origin $x' = 0$ coincides with $x = 0$ when $t = 0$ , this restricts the coefficients to $x' = \gamma (x - vt)$ , where $\gamma$ is a function of v: $\gamma = \gamma(v)$ .
$\gamma(v)$ must be an even function, so $\gamma(v) = \gamma(-v)$ , due to rotational invariance or by considering inertial frames S˜ and S˜' where the x-coordinate is measured in the opposite direction.
From the perspective of S', frame S moves backward with velocity -v, leading to $x = \gamma (x' + vt')$ , where $\gamma = \gamma(-v) = \gamma(v)$ .

Derivation of the Lorentz Transformations

Insisting on absolute time ( $t = t'$ ) forces $\gamma = 1$ , leading back to Galilean transformations, which are incompatible with the second postulate of special relativity.
The speed of light is equal to c in both S and S'.
In S, a light ray has trajectory $x = ct$ , while in S', it has trajectory $x' = ct'$ .
Substituting these trajectories into $x' = \gamma (x - vt)$ and $x = \gamma (x' + vt')$ , we have two equations relating t and t':
- $ct' = \gamma (c - v)t$
- $ct = \gamma (c + v)t'$
These equations are compatible only if $\gamma$ is given by $\gamma = \sqrt{\frac{1}{1 - v^2/c^2}}$ .
For v << c, we have $\gamma \approx 1$ , and the transformation approximates the Galilean transformation. As $v \rightarrow c$ , $\gamma \rightarrow \infty$ .
$\gamma$ becomes imaginary for v > c, making no sense for inertial frames with relative speed v > c.
The temporal transformation law is $t' = \gamma(t - \frac{v}{c^2}x)$ , derived by substituting the expression for x' into $x = \gamma (x' + vt')$ and rearranging.
When v << c, we recover the trivial Galilean transformation law $t' \approx t$ .
The spatial and temporal transformation laws are the Lorentz transformations.

Lorentz Transformations in Three Spatial Dimensions

Transformations of coordinates y and z perpendicular to the relative motion are trivial ( $y' = y, z' = z$ ).
The final form of the Lorentz transformations, also known as Lorentz boosts, is:
- $x' = \gamma (x - \frac{v}{c}ct)$
- $y' = y$
- $z' = z$
- $ct' = \gamma (ct - \frac{v}{c}x)$
For v/c << 1, the Lorentz boosts reduce to Galilean boosts.
The transformations must take the same form if we invert them to express x and t in terms of x' and t', except with v replaced by -v.
For light traveling in the x direction, $x = ct$ in frame S; in S', it follows $x' = ct'$ , ensuring the speed of light is the same in all inertial frames.
For an object traveling in the y direction at the speed of light ( $y = ct$ ) in S, its trajectory in S' is $y' = ct'/\gamma$ and $x' = \gamma vt'$ , with speed $v'^2 = v^2 + c^2(1 - v^2/c^2) = c^2$ .

Spacetime Diagrams

Spacetime diagrams illustrate the physics of relativity, with one direction of space (x) along the horizontal axis and time (ct) on the vertical axis.
Space and time in special relativity are called Minkowski space.
Each point P represents an event with coordinates (ct, x).
A particle moving in spacetime traces out a worldline.
A light ray in the x direction moves at 45 degrees.
No object can move faster than light, so worldlines of particles must always move upwards at an angle steeper than 45 degrees.

Inertial Frames

The horizontal and vertical axes in the spacetime diagram are coordinates of the inertial frame S.
Axes corresponding to an inertial frame S' moving with relative velocity $v = (v, 0, 0)$ can also be drawn.
The t' axis sits at $x' = 0$ and is given by $x = vt$ .
The x' axis is determined by $t' = 0$ , given by the equation $ct = \frac{v}{c}x$ .
These axes are symmetric about the light ray, reflecting that the speed of light is equal to c in both frames.

History of Light Speed

The first evidence that light does not travel instantaneously was presented by Ole Rømer in 1676.
Rømer noticed that the periods of the orbits of Io, the innermost moon of Jupiter, are not constant.
When the Earth is moving towards Jupiter, the orbits are a few minutes shorter; when the Earth moves away, the orbits are longer by the same amount.
Rømer correctly deduced this was due to the finite speed of light and gave a rough estimate for the value of c.
By the mid-1800s, the speed of light had been determined fairly accurately using experiments involving rotating mirrors.
Maxwell showed that light could be understood as oscillations of electric and magnetic fields.
Maxwell related the speed of light to two constants, $\epsilon0$ and $\mu0$ , the permittivity and permeability of free space, $c = \sqrt{\frac{1}{\epsilon0 \mu0}}$ .
Newtonian physics tells us that speeds are relative.
It was thought that Maxwell’s equations were only valid in a preferred reference frame.
This material was dubbed the luminiferous ether and it was thought that Maxwell’s equations must only be valid in the frame at rest with respect to this ether.

Michelson-Morley Experiment

In 1881, Michelson and Morley performed an experiment to detect the relative motion of the Earth through the ether.
At some moment, the Earth is moving in the x-direction relative to the ether with some speed v.
The Newtonian addition of velocities tells us that light propagating in the x-direction should have speed $c + v$ going one way and $c - v$ going the other.
The total time to travel backwards and forwards along a length L should therefore be $T_x = \frac{L}{c + v} + \frac{L}{c - v} = \frac{2cL}{c^2 - v^2}$ .
Light making the same journey in the y-direction will have to travel (by Pythagoras) a total distance of $\sqrt{L^2 + v^2(T_y/2)^2}$ on each leg of the journey.
It makes this journey at speed $c$ , meaning that we can equate $\frac{cTy}{2} = \sqrt{L^2 + v^2(Ty/2)^2}$ , so $T_y = \frac{2L}{\sqrt{c^2 - v^2}}$ .
The goal of the Michelson-Morley experiment was to measure the time difference between $Ty$ and $Tx$ using interference patterns of light ray making the two journeys.
The experiment didn’t work: there seemed to be no difference in the time taken to travel in the x direction and y direction.
Towards the end of the 1800s, the null result of the Michelson-Morley experiment had become one of the major problems in theoretical physics.
Several explanations were proposed, including the idea that the ether was somehow dragged along with the Earth.
The Dutch physicist, Hendrik Lorentz, went some way to finding the correct solution.
Lorentz had noticed that Maxwell’s equations had the peculiar symmetry that we now call the Lorentz transformations.
He argued that if a reason could be found that would allow distances between matter to change as $x' = \gamma (x - vt)$ then lengths would be squeezed in the direction parallel to the ether, explaining why no difference is seen between $Tx$ and $Ty$ .

Einstein's Relativity

Lorentz set to work trying to provide a mechanical explanation for this transformation law.
Although Lorentz had put in place much of the mathematics, the real insight came from Einstein in 1905.
He understood that there is no mechanical mechanism underlying the Lorentz transformations.
Nor is there an ether.
Instead, the Lorentz transformations are a property of space and time themselves.
With Einstein’s new take on the principle of relativity, all problems with Maxwell’s equation evaporate.
There is no preferred inertial frame.
Instead, Maxwell’s equations work equally well in all inertial frames.
However, they are not invariant under the older transformations of Galilean relativity; instead they are the first law of physics to be invariant under the correct transformations of Einstein/Lorentz relativity.
From this perspective, we could dispense with the second postulate of relativity all together.
We need only insist that the laws of physics – which include Maxwell’s equations – hold in all inertial frames.
Since Maxwell’s equations predict $c = \sqrt{\frac{1}{\epsilon0 \mu0}}$ , this implies the statement that the speed of light is the same in all inertial frames.

Relativistic Physics

In this section we will explore some of the more interesting and surprising consequences of the Lorentz transformations.

Simultaneity

In Newtonian physics, two events, $P1$ and $P2$ , happen at the same time if $t1 = t2$ .
In the relativistic world, things are not so easy.
An observer in inertial frame S decides that two events, $P1$ and $P2$ , occur simultaneously if $t1 = t2$ .
Lines of simultaneity for this observer are drawn on the spacetime diagram on the left of Figure 48.
For an observer in the inertial frame S', simultaneity of events occurs for equal $t'$ .
Using the Lorentz transformation, lines of constant $t'$ become lines described by the equation $t - vx/c^2 = \text{constant}$ .
These lines are drawn on the spacetime diagram on the right of Figure 48.
Two events simultaneous in one inertial frame are not simultaneous in another.
An observer in S thinks that events $P1$ and $P2$ happen at the same time. All other observers disagree.
The fact that all observers cannot agree on what events are simultaneous is a direct consequence of the fact that all observers do agree on the speed of light.

Lights on Trains

Consider a train moving at constant speed, with a lightbulb hanging from the middle of one of the carriages.
A passenger on the train turns on the bulb and, because the bulb is equidistant from both the front and back wall of the carriage, observes that the light hits both walls at the same time.
A person standing on the platform as the train passes through disagrees.
The light from the bulb travels at equal speed $\pm c$ to the left and right, but the back of the train is rushing towards the point in space where the light first emerged from.
The person on the platform will see the light hit the back of the train first.

Causality

Different observers disagree on the temporal ordering of two events.
Causality issues arise because there are only some events which observers can disagree about.
Because Lorentz boosts are only possible for v<c, the lines of simultaneity cannot be steeper than 45 degrees.

Light Cones

Take a point P and draw the 45 degree light rays that emerge from P.
This is called the light cone.
The light cone is really two cones, touching at the point P.
They are known as the future light cone and past light cone.
For events inside the light cone of P, all observers will agree that Q occurred after P.
For events outside the light cone, some observers will see R as happening after P; some before.
The events which all observers agree can be causally influenced by P are those inside the future light cone.
Similarly, the events which can plausibly influence P are those inside the past light cone.
Causality is preserved, only if nothing can propagate outside the light cone.
That's the same thing as travelling faster than the speed of light.
If we ever see particles that travel faster than the speed of light, we are in trouble.
We could use them to transmit information faster than light.
But another observer would view this as transmitting information backwards in time.

Rigid Objects

There are no perfectly rigid objects.
Suppose that you push on one end of a rod.
The other end cannot move immediately since that would allow us to communicate faster than the speed of light.
Of course, for real rods, the other end does not move instantaneously.
Instead, pushing on one end of the rod initiates a sound wave which propagates through the rod, telling the other parts to move.
The statement that there is no rigid object is simply the statement that this sound wave must travel slower than the speed of light.
When talking about waves as opposed to point particles, there is a slight subtlety in exactly what must travel slower than light.
There are at least two velocities associated to a wave:
- the group velocity is (usually) the speed at which information can be communicated. This is less than c.
- In contrast, the phase velocity is the speed at which the peaks of the wave travel. This can be greater than c, but transmits no information.

Time Dilation

Consider a clock sitting stationary in the frame S' ticks at intervals of $T_0$ .
The tick events in frame S' occur at $(ct'1, 0)$ then $(ct'1 + cT_0, 0)$ and so on.
Inverting the Lorentz transformations (7.6) gives $t = \gamma (t' + vx'/c^2)$ .
The clock sits at $x'=0$ , so the interval between ticks is $T = \gamma T_0$ .
The gap between ticks is longer in the stationary frame.
A moving clock runs more slowly.
The same argument holds for any process, be it clocks, elementary particles or human hearts.
The correct interpretation is that time itself runs more slowly in moving frames.

Lights and Trains

If the train has height h, a passenger on the train will measure time $t_0 = h/c$ for the light to travel from the light bulb to the middle of the floor (i.e. the point directly below the light bulb).
The train has moved forward at speed v. To hit the same point on the floor, the light has to travel a distance $\sqrt{h^2 + (vt)^2}$ .
The time taken is therefore $t = \frac{\sqrt{h^2 + (vt)^2}}{c} = \frac{h}{c} \sqrt{\frac{1}{1 - v^2/c^2}} = \gamma t_0$ .

Experimental Consequences of Time Dilation

This phenomenon is tested accurately in particle accelerators where elementary particles reach speeds close to c.
The protons spinning around the LHC have $\gamma \approx 3500$ .
The previous collider in CERN, called LEP, accelerated electrons and positrons to $\gamma \approx 2 \times 10^5$ .
The effect of time dilation is vivid on unstable particles which live much longer in the lab frame than in their own rest frame.

Early Demonstration of Time Dilation

An early demonstration was seen in muons in 1941.
These are heavier, unstable, versions of the electron.
They decay into an electron, together with a couple of neutrinos, with a half-life of $\tau \approx 2 \times 10^{-6} s$ .
Muons are created when cosmic rays hit the atmosphere, and subsequently rain down on Earth.
Yet to make it down to sea level, it takes about $t = 7 \times 10^{-6} s$ , somewhat longer than their lifetime.
The reason that they do not is because the muons are travelling at a speed $v \approx 0.99c$ , giving $\gamma \approx 10$ .
From the muon’s perspective, the journey only takes $t_0 = t/\gamma \approx 7 \times 10^{-7} s$ , somewhat less than their lifetime.
Elementary particles are structureless.
The reason that they live longer can’t be explained because of some mechanical device which slows down: it is time itself which is running slower.

Hafele and Keating Experiment

A direct test of time dilation was performed in 1971 by Hafele and Keating.
They flew two atomic clocks around the world on commercial airliners; two more were left at home.
When they were subsequently brought together, their times differed by about $10^{-7} s$ .
There are actually two contributions to this effect:
- the time dilation of special relativity that we’ve seen above,
- a related effect in general relativity due to the gravity of the Earth.

Twin Paradox

Two twins, Luke and Leia, decide to spend some time apart.
Leia stays at home while Luke jumps in a spaceship and heads at some speed v to the planet Tatooine.
With sadness, Leia watches Luke leave but is relieved to see — only a time T later from her perspective — him safely reach the planet.
However, upon arrival, Luke finds that he doesn’t like Tatooine so much. It is a dusty, violent place with little to do.
So he turns around and heads back to Leia at the same speed v as before.
When he returns, he finds that Leia has aged by $T_{Leia} = 2T$ .
And yet, fresh faced Luke has only aged by $T_{Luke} = 2T/\gamma$ .
We see, that after the journey, Luke is younger than Leia.
In fact, for large enough values of $\gamma$ , Luke could return to find Leia long dead.
This is nothing more than the usual time dilation story.
The resolution to this “paradox” is that there is no symmetry between Luke’s journey and Leia’s.
Leia remained in an inertial frame for all time.
Luke, however, does not. When he reaches Tatooine, he has to turn around and this event means that he has to accelerate.
This is what breaks the symmetry.

Length Contraction

Moving clocks run slow, moving rods are shortened.
Consider a rod of length $L_0$ sitting stationary in the frame S'.
When we say that a rod has length $L0$ , it means that the distance between the two end points at equal times is $L0$ .
Drawing the axes for the frame S', the situation looks like the picture on the left.
The two, simultaneous, end points in S' are $P1$ and $P2$ .
Their coordinates in S' are $(ct', x') = (0, 0)$ and $(0, L_0)$ respectively.
Now let's look at this in frame S. This is drawn in right-hand picture.
Clearly $P_1$ sits at $(ct, x) = (0, 0)$ .
Meanwhile, the Lorentz transformation gives us the coordinate for $P2$ $x = \gamma L0$ and $t = \gamma vL_0/c^2$ .
To measure the rod in frame S, we want both ends to be at the same time.
And the points $P1$ and $P2$ are not simultaneous in S.
We can follow the point $P2$ backwards along the trajectory of the end point to $Q2$ , which sits at $x = \gamma L_0 - vt$ .
We want $Q2$ to be simultaneous with $P1$ in frame S. This means we must move back a time $t = \gamma vL0/c^2$ , giving $x = \gamma L0 - \frac{v^2 \gamma L0}{c^2} = \frac{L0}{\gamma}$ .
The length $L$ measured in frame S is $L = \frac{L_0}{\gamma}$ .
It is shorter than the length of the rod in its rest frame by a factor of $\gamma$ .
This phenomenon is known as Lorentz contraction.

A question

Ladders and Barns

Take a ladder of length 2L and try to put it in a barn of length L. If you run fast enough, can you squeeze it?
There are two arguments, each giving the opposite conclusion
- From the perspective of the barn, the ladder contracts to a length $2L /\gamma$ . This shows that it can happily fit inside as long as you run fast enough, with \gamma >2
- From the perspective of the ladder, the barn has contracted to a length L / . This means there's no way you're going to get the ladder inside the barn. Running faster will only make things worse
As usual, to reconcile these two points of view we need to think more carefully about the question we're asking. What does it mean to fit a ladder inside a barn?
Any observer will agree that we've achieved this if the back end gets in the door before the front end hits the far wall. But we know that simultaneity of events is not fixed, so the word 'before' in this definition suggests that it may be something different observers will disagree on.

Addition of Velocities

A particle moves with constant velocity $u'$ in frame S', which, in turn, moves with constant velocity $v$ with respect to frame S. What is the velocity $u$ of the particle as seen in S?
The Newtonian answer is just $u = u' + v$ . But we know that this can’t be correct because it doesn’t give the right answer when $u' = c$ .
The worldline of the particle in S' is $x' = u't'$
So the velocity of the particle in frame S is given by $u = \frac{x}{t} = \frac{\gamma(x' + vt')}{\gamma(t' + vx'/c^2)}$ , which follows from the Lorentz transformations (7.6).
Substituting $x' = u't'$ into the expression above gives us the result:
$u = \frac{u' + v}{1 + u'v/c^2}$
When $u' = c$ , this gives us $u = c$ as expected.
|u'| < c and |v| < c then c<-u<c.
If a particle is travelling slower than the speed of light in one inertial frame, it will also be travelling slower than light in all others.

The Geometry of Spacetime

Time is relative, length is relative, simultaneity is relative. Is nothing sacred anymore? Well, the answer is yes: there is one measurement that all observers will agree on.
The invariant interval

The Invariant Interval

Let's start by considering a spacetime with just a single spatial coordinate, x. In frame S, two events P1 and P2 have coordinates $(ct1, x1)$ and $(ct2, x2)$ .
The events are separated by $\delta t = t1 - t2$ in time and $\delta x = x1 - x2$ in space.
We define the invariant interval $\delta s^2$ as a measure of the distance between these two points:
$\delta s^2 = c^2\delta t^2 - \delta x^2$
The advantage of the invariant interval is that it is something all observers agree upon. In frame S0, we have
\delta s^2 = \gamma^2 (c \delta t' + \frac{v \delta x'}{c})^2 - \gamma^2 ( \delta x' + v \delta t')^2}
The cross terms cancel out. after the factorization. This includes all three spatial dimensions, the definition of the invariant interval is.
$\delta s^2 = c^2 \delta t^2 - \delta x^2 - \delta y^2 - \delta z^2$
The spacetime of special relativity is topologically $R^4$ .
When endowed with the measure of distance (7.11), this spacetime is referred to as Minkowski space.
Although topologically equivalent to Euclidean space, distances are measured differently.
To stress the difference between the time and spatial directions, Minkowski space is sometimes said to have dimension $d = 1 + 3$ (for once, it’s important that you don’t do this sum!).

Distance between two infinitesimally close points

In later courses — in particular General Relativity — you will see the invariant interval written as the distance between two infinitesimally close points. In practice that just means we replace all the (something)s with d(something)s.
$ds^2 = c^2dt^2 - dx^2 - dy^2 - dz^2$

Observer Independent Characterisation

The invariant interval provides an observer-independent characterisation of the distance between any two events.
Two events separation of which is: \delta s^2 > 0 are said to be timelike separated. They are closer together in space than they are in time. Pictorially, such events sit within each others light cone.
events with separation \delta s^2 < 0 are said to be spacelike separated. They sit outside each others light cone. Two observers can disagree about the temporal ordering of spacelike separated events. Note that since \delta s^2 < 0 for spacelike separated events, if you insist on talking about s itself then it must be purely imaginary. However, usually it will be perfectly fine if we just talk about $\delta s^2$ .
Finally, two events with $\delta s^2 = 0$ are said to be lightlike separated.
It means that they can be connected by a light ray.

Analogy

Understand the meaning of the invariant interval with an analogy in Euclidean space.
Consider three-dimensional Euclidean space with coordinates $x = (x, y, z)$ .
An observer measures the position of a stationary object—let’s say a helicopter—and announces the x, y, and z coordinates.
A second observer shares the same origin but has rotated his axes to use coordinates $x' = (x', y', z')$ , where $x' = Rx$ for some rotation matrix R.
The helicopter is seen by the second observer, who declares that it sits at coordinates $x' , y'$ and $z'$ .
There’s no reason the coordinates of the two observers should agree with each other.
One quantity should be invariant: the distance from the origin (shared by both observers) to the helicopter.
$s^2_{Euclidean} = x^2 + y^2 + z^2 = x'^2 + y'^2 + z'^2$
This is true if the rotation matrix obeys $R^TR = 1$ .
The Lorentz boosts are a rotation between space and time.
The individual spatial and temporal coordinates differ for the two observers, but there remains an invariant distance.

The Lorentz Group

We have defined the interval (7.11) as the measure of distance which is invariant under Lorentz transformations. However, it is actually better to look at things the other way: the invariant interval is the primary object. This is a property of spacetime which defines the Lorentz transformations. Let’s see how the argument runs this way around.
If we sit at the origin in a fixed frame S, the coordinates of an event can be written as a four vector X. We won’t denote that this is a vector by bold font or squiggly underlines (which we’re really saving for three-dimensional spatial vectors). We’re getting sophisticated now and just the capital letter will have to suffice. However, we will sometimes use index notation, in which the components of the 4-vector are.
$X^\mu = (ct, x, y, z)$ , $\mu = 0, 1, 2, 3$
Note that we write the indices running from mu = 0 to mu = 3 rather than starting at 1. The zeroth component of the vector is time (multiplied by c).
The invariant distance between the origin and the point P can be written as an inner product, defined as
$X · X ≡ X^T \eta X = X^\mu \eta_{\mu\nu}X^\nu$
The matrix $\eta$ is given by
$\eta = \begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & -1 & 0 & 0 \ 0 & 0 & -1 & 0 \ 0 & 0 & 0 & -1 \end{pmatrix}$
This matrix is called the Minkowski metric. With this expression for the Minkowski metric, the inner product becomes
$X · X = c^2 t^2 - x^2 - y^2 - z^2$
which is indeed the invariant distance (7.11) between the origin and the point X as promised.

Lorentz Transformation

Let $\Lambda$ be a 4x4 matrix that rotates the coordinates in frame S to coordinates in frame S', such that the four vector becomes $X' = \Lambda X$ .
This can also be written index notation as $X'^\mu = \Lambda^{\mu}_{\nu}X^\nu$ .
Our definition (7.13), we see that this is true only if $\Lambda$ obeys the matrix equation
$\Lambda^T \eta \Lambda = \eta$
The matrix $\Lambda$ has four rows and four columns, so 16 components.
The equation only provides 10 constraints on the coefficients of $\Lambda$ .
There are therefore 16 − 10 = 6 independent solutions.

Solutions to equation 7.14 split into two classes: Rotations and Boosts

Rotations

\Lambda = \begin{pmatrix} 1& 0 & 0 & 0\0& R\ \end{pmatrix}
where R is a 3 × 3 matrix. These transformations change space, but leave time intact. The condition (

Boosts

boost along the x axis is given by
$\Lambda = \begin{pmatrix} \gamma & -\gamma v/c & 0 & 0 \ -\gamma v/c & \gamma & 0 & 0 \ 0 & 0 & 1 & 0\ 0 & 0 & 0 & 1\ \end{pmatrix}$

he set of all matrices obeying (7.14) form the Lorentz group, denoted $O(1, 3)$ .
Taking the determinant of