# Lorentz Transformations

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18.5Lorentz Transformations

In this Physics tutorial, you will learn:

• What are the Lorentz spacetime transformations?
• How do they relate to spacetime Galilean transformations?
• How to obtain the Lorentz transformations for velocity?
• In what condition(s) do the Lorentz transformations converge with Galilean transformations?

## Introduction

Earlier, we have explained that a physical event (otherwise known as elementary process) has an absolute nature in itself; it is an objective fact for any observer. However, a given event looks different in different inertial frames of reference. In fact, it was so even in Newtonian inertial systems (Galilean transformations) except for time, which was absolute. We henceforth will use the new concepts of relative space and time to obtain new equations that replace the Galilean transformation which already are outdated as they belong to classical relativity.

## The Spacetime Lorentz Transformations

Now, let's find in a simple way to obtain the equations which show the transformation of spacetime coordinates of physical events observed in different inertial frames of reference. The need for such a new transformation is evident, especially now that we already are aware for the ultimate speed of light, which must be the same in all systems. Another thing that requires a new approach on relative transformations of spacetime coordinates is the time dilation and length contraction - two elements not prevised in Galilean transformations, in which the time is absolute. Obviously, Galilei could not imagine these things, because they are observed only when objects are moving very fast, at speeds comparable to the speed of light; for V << c such effects are almost zero. Therefore, we must find a set of transformations, which include the Galilean transformations but as a special case, where the time be the same in all inertial systems only in such a case but not in general.

Just like in the other situations discussed so far, we consider a system S' moving due right at velocity V relative to another system S assumed at rest. Obviously, the system S moves at velocity -V relative to the system S' (i.e. the two systems move at the same speed but in opposite direction relative to each other). For simplicity, we will consider any movement only due X and X', despite the two inertial systems S' and S contain all three spatial dimensions, X', Y', Z' and X, Y, Z respectively.

Let's suppose an event appears "somewhere" in S at coordinates x, y, z at a given instant t. The same event appears in S' at coordinates x', y', z' at the instant t'. We choose as time origin the instant in which there is a convergence of origins of the two systems S and S'. In this instant, t0 = t0' = 0. Then, the systems separate from each other at velocity V (S' slides away from S in the X-direction at velocity V). Moreover, since now we are dealing with very fast movements, t ≠ t' despite the time origin is the same for both systems. Let's suppose the event P (for example, the generation of an elementary particle, for which we will discuss in Section 21) does appear in the instant t for S and t' for S' at the corresponding point P shown in the figure. When both systems were at the same position, the particle did not exist yet. Obviously, when the particle is generated, the system S' has moved due right of S by V · t while the system S has not moved. It is evident that the x-coordinate of P at the moment of its generation, is

x(P) = V ∙ t + x'(P) ∙ √1 - V2/c2

This is because the origin O' of the system S' is displaced by V · t from the origin O of the system S, while the length has been contracted by the known factor

x = x' ∙ √1 - V2/c2

Rearranging the previous formula, we obtain for the coordinate of the point (event) P in S':

x' (P) = x(P) - V ∙ t/1 - V2/c2

Even if we assume S' at rest and take the system S as moving at -V relative to S', nothing changes except the sign. In this case, we obtain

x' (P) = -V ∙ t' + x(P) ∙ √1 - V2/c2

Combining the last two equations, we obtain

x(P)-V ∙ t/1 - V2/c2 = -V ∙ t' + x(P) ∙ √1 - V2/c2
V ∙ t' = x(P) ∙ √1 - V2/c2 - x(P)-V ∙ t/1 - V2/c2
= x(P) ∙ 1 - V2/c2 - x(P) + V ∙ t/1 - V2/c2
= V ∙ t-x(P) ∙ V2/c2/1 - V2/c2

Thus,

t' = V ∙ t-x(P) ∙ V2/c2/V ∙ √1 - V2/c2
= V ∙ (t - x(P) ∙ V/c2 )/V ∙ √1 - V2/c2

Hence, we obtain for the time t' of the event measured in S':

t' = t - x(P) ∙ V/c2/1 - V2/c2

Taking y = y' and z = z' as explained earlier, we obtain the Lorentz Transformation of spacetime coordinates:

x' = x - V ∙ t/1 - V2/c2
y' = y
z' = z
t' = t - V/c2 ∙ x/1 - V2/c2

These simple but very famous equations are named after Hendrik Lorentz - the famous Dutch scientist, who in 1902 was the first who published them. The only drawback is that Lorentz did not consider the time t' as a real time, still relying on the false concept of cosmic ether.

### Example 1

An airplane makes a 6000 km flight at 400 m/s (linear motion at constant speed) when viewed from Earth.

1. How much does the flight last for an observer at rest on the Earth surface?
2. How much does the flight last for the passengers?
3. What conclusion do you draw from the two results obtained at (a) and (b)?

### Solution 1

Clues:

x = d = 6000 km = 6 000 000 m = 6 × 106 m
V = 400 m/s = 4 × 102 m/s
(c = 3 × 108 m/s)
t = ?
t' = ?

1. Since the airplane travels a 6000 km distance at 400 m/s, the flight lasts for
t = d/V
= 6000 km/400 m/s
= 6 000 000 m/400 m/s
= 15 000 s
= 1.5 × 104 s
2. For the passengers, the flight lasts for
t' = t - x ∙ V/c2/1 - V2/c2
where the distance d given in the clues is denoted by x in the formula. Thus,
t' = (1.5 × 104 s) - (6 × 106 m) ∙ (4 × 102 m/s)/(3 × 108 m/s)2/1 - (4 × 102 m/s)2/(3 × 108 m/s)2
= (15 000 - 0.0000000278) s
3. As you see, the time difference is very small. Therefore, for speeds much smaller than light speed, it is not suitable to use the relativistic approach, as the results are more or less the same.

## Galilean Transformations as Limit of Lorentz Transformations

It is easy to see that for V << c, the Lorentz transformations point towards Galilean transformations. This is because V is assumed as very close to zero. Thus, we have

x' = x - V ∙ t/1 - V2/c2 = x - 0 ∙ t/1 - 02/c2 = x/√1 = x
y' = y
z' = z
t' = t - x ∙ 0/c2/1 - 02/c2 = t - 0/√1 = t

These equations are the same as those we obtained when studying the Galilean transformation for spacetime coordinates. In other words, the modern relativistic effects are evident only when we have systems moving relative to each other at velocities comparable to that of light, which is finite, albeit very high. If the speed of light were infinite, we would still be in the field of classical physics for any value of light speed.

Now, let's consider once again the clocks synchronization in inertial systems. If c pointed to infinity, then we wouldn't need to do any operation for clocks synchronization. In other words, we would have an absolute time and as a result, there was no need to do the clocks synchronization individually. If we wished, we would take two clocks, turn them on simultaneously in the two systems S and S' by starting from zero and in this way, they would always show the same time for the same event.

However, the Lorentz transformations do not allow this. Any clocks synchronization in S does not necessarily bring a clocks synchronization in S'. If we place standard clocks in S and S', they will not be synchronized. Let's explain this point through the example shown in the figure below. We have the two clocks O (in S) and O' (in S') synchronized and placed at the same position. Likewise, the clock M (of S) and M' (of S') are initially at the same position as well. However, the time of M' is not the same as that of M despite the fact that t(O) = t(M) = t(O') = 0. We can find this time by applying the Lorentz Transformations:

t' (M' ) = t(M) - V/c2 ∙ x(M)/1 - V2/c2
= 0 - V/c2 ∙ d/1 - V2/c2
≠ 0

It is obvious that the clock M' does not show the same time as O' because they are not synchronized. If the speed of light pointed towards infinity, or if V << c, then we would have a synchronization (perfect for c → ∞ and very close to perfect for V << c).

Giving that the Galilean transformations represent a limit of Lorentz transformations, we can always use the Lorentz transformation but for practical purposes, we often use the Galilean transformations for normal velocities. This approach is similar to situations involving gravitation, when it is not necessary to use the general formula of gravitation that involves the masses of objects and the distance between them [F = (G · M · m) / r2], but we often use the simplified formula F = m · g instead.

## Quick Recap of Dilation and Contraction

Let's consider a bar at rest in S' which lies in the X'-axis. The coordinate of the left end of the bar is x'1 and that of the right end is x'2. The bar's length when measured in S' is

L' = x'2 - x'1

We want to calculate the length of the same bar when measured in the system S considered at rest. Let's suppose that we decide to measure the coordinate x1 of the left end of the bar at the instant t1 and in another instant t2 we measure the coordinate of the right end of the bar. Obviously, we obtain different results for

L = x2 - x1

because the bar meanwhile is moving. Hence, it is more logical to make the measurement of the above coordinates in S in a fixed instant t, in order to avoid issues caused by the bar's displacement. The following equations derive from Lorentz transformations:

x1' = x1 - V ∙ t/1 - V2/c2 and x2' = x2 - V ∙ t/1 - V2/c2

Subtracting the first coordinate in S' from the second one, we obtain

x2' - x1' = x2 - V ∙ t/1 - V2/c2-x1 - V ∙ t/1 - V2/c2
= x2 - V ∙ t-x1-V ∙ t/1 - V2/c2
= x2 - x1/1 - V2/c2

or

L' = L/1 - V2/c2

Rearranging the last formula, we obtain

L = L' ∙ √1 - V2/c2

This is the well-known formula of length contraction in relativistic events.

Yet, let's suppose that two events occur successively at a fixed point of X' with coordinate x'0. The instants of events' occurrence are t'1 and t'2 respectively. Applying the inverse Lorentz transformation, (i.e. exchanging the places and signs), we obtain

t1 = t1' + V/c2 ∙ x0'/1 - V2/c2 and t2 = t2' + V/c2 ∙ x0'/1 - V2/c2

Again, subtracting the first equation from the second we obtain

t2 - t1 = t2' + V/c2 ∙ x0'/1 - V2/c2 - t1' + V/c2 ∙ x0'/1 - V2/c2
= t2' + V/c2 ∙ x0' - t1'-V/c2 ∙ x0'/1 - V2/c2
= t2' - t1'/1 - V2/c2

Thus,

∆t = ∆t'/1 - V2/c2

This too is the known equation of time dilation we have discussed in previous articles.

### Example 2

A 20 m long rocket moves vertically upwards at 30 km/s. What is its length when measured from Earth surface?

### Solution 2

Clues:

L' = 20 m
V = 30 km/s = 3 × 104 m/s
c = 3 × 108 m/s
L = ?

Using the equation

L = L' ∙ √1 - V2/c2

derived from Lorentz transformations, we obtain after substitutions

L = (20 m) ∙ √1 - 3 × 104 m/s2/(3 × 108 m/s)2
= (20 m) ∙ √1 - 10-8
= (20 m) ∙ √108 - 1/108
= (20 m) ∙ 9999.99995/10000
= (20 m) ∙ 0.999999995
= 19.9999999 m

Thus, the contraction of length is only 0.0000001 m or 0.1 μm - practically undetectable.

## Lorentz Transformation of Velocity

Lorentz transformations of spacetime coordinates allow us find the relativistic formulae of velocity transformations for a particle moving and observed in two inertial systems S and S'. The particle's motion can be random; we will discuss only for the instantaneous velocity of particle in the two systems, as defined in Kinematics section.

Let's suppose a particle is at the point x, y, z of the system S in the instant t (this can be thought as an event). The same event is characterized by the coordinates x', y', z' of the system S'. After a short time interval Δt has elapsed in S, the particle (event) has the new coordinates x + Δx, y + Δy, z + Δz in the time t + Δt. This can be thought as the second event, which has the new spacetime coordinates x' + Δx', y' + Δy', z' + Δz', t' + Δt' in S'. If the event occurs only according X or X', based on the Lorentz transformation of spacetime coordinates we have:

x' = x - V ∙ t/1 - V2/c2
y' = y
z' = z
t' = t - x ∙ V/c2/1 - V2/c2

and

x' + ∆x' = (x + ∆x) - V ∙ (t + ∆t)/1 - V2/c2
y' + ∆y' = y + ∆y
z' + ∆z' = z + ∆z
t' + ∆t' = (t + ∆t)-(x + ∆x) ∙ V/c2/1 - V2/c2

Subtracting the first set of equations from the second, we have

(x' + ∆x' ) - x' = (x + ∆x) - V ∙ (t + ∆t)/1 - V2/c2 - x - V ∙ t/1 - V2/c2
(y' + ∆y' ) - y' = (y + ∆y) - y
(z' + ∆z' ) - z' = (z + ∆z) - z
(t' + ∆t') - t' = (t + ∆t) - (x + ∆x) ∙ V/c2/1 - V2/c2 - t - x ∙ V/c2/1 - V2/c2

After doing the operations, we obtain

∆x' = ∆x - V ∙ ∆t/1 - V2/c2
∆y' = ∆y
∆y' = ∆y
∆t' = ∆t - V/c2 ∙ ∆x/1 - V2/c2

Dividing each of the first three equations by the fourth one, we obtain:

∆x'/∆t' = ∆x - V ∙ ∆t/∆t - V/c2 ∙ ∆x
∆y'/∆t' = ∆y/∆t - V/c2 ∙ ∆x ∙ √1 - V2/c2
∆z'/∆t' = ∆z/∆t - V/c2 ∙ ∆x ∙ √1 - V2/c2

Dividing the numerator and denominator in the right side of each equation by Δt, we obtain

∆x'/∆t' = ∆x/∆t - V ∙ ∆t/∆t/∆t/∆t - V/c2∆x/∆t
∆y'/∆t' = ∆y/∆t ∙ √1 - V2/c2/∆t/∆t - V/c2∆x/∆t
∆z'/∆t' = ∆z/∆t ∙ √1 - V2/c2/∆t/∆t - V/c2∆x/∆t

or

∆x'/∆t' = ∆x/∆t - V/1 - V/c2∆x/∆t
∆y'/∆t' = ∆y/∆t ∙ √1 - V2/c2/1 - V/c2∆x/∆t
∆z'/∆t' = ∆z/∆t ∙ √1 - V2/c2/1 - V/c2∆x/∆t

Assuming the time intervals in the two systems as very small (Δt' → 0 and Δt → 0), i.e. taking the above rates as limits, we find the components of particle's velocity in the two inertial systems S' and S:

vx' = vx - V/1 - V ∙ vx/c2
vy' = vy ∙ √1 - V2/c2/1 - V ∙ vx/c2
vz' = vz ∙ √1 - V2/c2/1 - V ∙ vx/c2

If we consider a one dimensional light ray (for example a light ray emitted by a laser) moving only in the X (X') direction, we have vx = c, vy = 0 and vz = 0 (in S). Thus, we find for the velocity in S' (giving that V << c):

vx' = c - V/1 - V ∙ c/c2 = c - V/1 - V/c = c/1 = c
vy' = 0 ∙ √1 - V2/c2/1 - V ∙ vx/c2 = 0
vz' = 0 ∙ √1 - V2/c2/1 - V ∙ vx/c2 = 0

Therefore, the observer in S' measures the same velocity c in the positive direction of X' while the velocity of light in the other directions is zero (as expected).

The Lorentz transformations for velocity converge with the classical formulae for V << c or for c → ∞. This is obvious given their structure.

### Example 3

A spaceship leaves the Earth at v1 = 0.9c and at a certain point, it emits a spatial module at v2 = 0.7c relative to itself. What is the velocity of module when viewed from Earth?

### Solution 3

Clues:

v1 = V = 0.9c
v2 = v'x = 0.7c
vx = ?

Using the equation for Lorentz transformation of velocity

vx' = vx - V/1 - V ∙ vx/c2

we obtain after rearranging and substituting the given values:

vx' - V ∙ vx/c2 ∙ vx' = vx -V
vx + V ∙ vx'/c2 ∙ vx = vx' + V
vx ∙ (1 + V ∙ vx'/c2 ) = vx' + V
vx = vx' + V/1 + V ∙ vx'/c2
= 0.7c + 0.6c/1 + 0.6c ∙ 0.7c/c2
= 1.3c/1 + 0.42
= 1.3c/1.42
= 0.915c

This result is reasonable as it is smaller than c. If we used the classical approach, the calculations would give v = 0.9c + 0.7c = 1.6c - a wrong result which contradicts the findings of the Special Theory of Relativity discovered by Einstein.

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