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Special Relativity 1: Lorentz Transformation


  1. Lorentz Transformation 🢀
  2. Spacetime


Our common sense physics that is based on the Newton's laws is only an approximation of the actual reality that is described by Einstein's relativity theory. This approximation is excellent for moderate speeds and fails only for speeds approaching the speed of light $$ c = 299'792'458 \; m/s . $$ This value is exact because the new definition of 1 meter is the distance travelled by light in 1/299'792'458 seconds. A sufficient approximation of $c$ is $c \approx 3 \cdot 10^8 m/s$.

The theory of special relativity rests on two experimental facts that are adopted as postulates:

  1. The principle of relativity: the laws of physics are the same in all inertial frames. The observer in an inertial frame cannot distinguish whether she is "at rest" or "moving".
  2. The speed of light in vacuum is the same in all inertial frames.

The principle of relativity was introduced already by Galileo Galilei (1564 - 1642) and is the basic of the classic physics developed by Isaac Newton (1643 - 1727). This principle is in full accordance with our everyday experiences. The constance of light speed postulated by Albert Einstein (1879 - 1955) contradicts our common sense concerning relative velocities. If I move with a velocity $v$ in the same direction as the light I should measure the light velocity as $c - v$ with respect to me. Einstein claims that I measure the light velocity $c$ independent of my velocity $v$.

Relativity of simultaneity

Consider a train moving with a velocity $v$ with respect to the platform and passing it at time 0. At this time a flash is fired in the middle of the carriage. The following picture shows the situation on the train.

On the train v t' B t' F L/2 L/2
Fig. 1: Light hits both walls simultaneously

The observer on the train does not perceive any motion. The light from the flash expands with velocity $c$ and hits the front wall F and the back wall B at the same time (measured by the observer on the train). $$ t'_F = t'_B = \frac{L}{2c}. $$

The observer on the platform has a different view:

View from the platform v t B t F L/2 v t B v t F
Fig. 2: Light hits the walls at different times

For him the light travels a longer distance to the front wall than to the back wall because the train moves to the right. Since the light speed is constant the light reaches the respective walls at different times. The events that are simultaneous for the train passenger are asynchronous for the ground-based observer. The distances travelled by the light to the respective walls are: $$\begin{align} c t_F & = L/2 + v t_F \\ c t_B & = L/2 - v t_B \end{align}$$ From these two equations we obtain easily the difference between the two hitting times: $$ \Delta t = t_F - t_B = \frac{L v}{c^2 - v^2}. $$

The described situation is not just a one-way perspective of the ground-based observed looking at the train. The train passenger would observe the same asynchronicity when looking at the platform where the same experiment would be performed as follows: The ground-based observer sets up two walls with the distance $L$ between them. At time $t = 0$, when the train passes, a flash is triggered in the middle of the walls. The light reaches the respective walls at the same time for the ground-based observer but at different times for the train passenger. The train passenger is at rest with respect to the train and the platform with the erected walls is moving with the velocity $-v$ with respect to him. He observers that the light reaches one wall before it reaches the other wall.

The described asynchronicity is very small for every day situations and hence is not observed. An example is: $L = 25 m, v = 300 km/h \Rightarrow \Delta t = 2.3 \cdot 10^{-14} s$.

There is no time difference in the Newtonian world, of cource. Here, the ground-based observer reasons as follows. The "relative" velocity of the light with respect to the train is $c$. The train moves with the velocity $v$ to the right. So, the light speed with respect to the ground-based is $c + v$ forward and $c - v$ backward. The corresponding equations are: $$\begin{align} (c + v) t_F & = L/2 + v t_F \\ (c - v) t_B & = L/2 - v t_B \end{align}$$ and their solution is $$ t_F = t_B = \frac{L}{2c}. $$ which equals the observation of the train passenger.

Time dilation

The described asynchronicity of events makes it obvious that the time depends on the motion. We will measure the time using a light clock consisting of two mirrors A and B and a light pulse ("light ball") moving vertically between them. The time period or the "tick" of this clock is defined by the time it takes the light to travel from A to B and back to A.

Bob with his clock v B A L = c Δ𝜏/2
Fig. 3: Clock at rest measures the proper time

Bob on a train moving with velocity $v$ has installed this clock in his car. Bob is at rest with respect to the car and the clock. This clock at rest measures the "proper time". The duration of one tick is: $$ \Delta \tau = \frac{2 L}{c}. \tag{1} $$ where $L$ is the distance between A and B.

Alice standing on the platform sees a different path of the light ball because the train moves:

Alice's clock v L A B A d d v Δt
Fig. 4: Moving clock (blue) runs slower than stationary clock (red)

The path of the light is longer and so the time of a tick (A-B-A) is also longer. The distance $d$ is according to the Pythagorean theorem: $$ d^2 = (v \Delta t / 2) ^2 + L^2 $$ Since $d$ is also the distance travelled by light in the time $\Delta t / 2$ we get $$ \Delta t = \frac{2 L}{\sqrt{c^2 - v^2}} = \frac{2 L}{c} \gamma = \gamma \Delta \tau \tag{2} $$ where $$ \gamma = \frac{1}{\sqrt{1 - v^2 / c^2}} \ge 1 \tag{3} $$ is the so called Lorentz factor. One tick on the moving clock takes $\gamma$-times longer than one tick on the stationary clock. Since the time itself is measured by number of ticks, the moving clock runs $\gamma$-times slower than the stationary clock.

For Bob the situation is analoguos. He observes Alice's clock moving with respect to him and getting late by the factor $\gamma$. This seemingly paradoxical situation can be illustrated by our perception in everyday life: When two persons look at each other from a distance both think that the other person is smaller than herself.

The described property is valid not only for the special light clock but for any clock. Let us take any CLOCK and synchronize it with the light clock at rest. Now attach the two clocks together and move them with velocity $v$. The CLOCK will remain synchronized with the light clock. If it did not, we could detect the motion of an inertially moved system and this would contradict the Postulate of inertial system. We conclude that time runs slower in a moving system than in a system at rest. Equivalently, the clock at rest shows the longest possible time. Therefore the phenomenon is called time dilation (time lengthening). As already described in the previous section we can always switch the roles of the system at rest and the moving system: An observer sitting on the moving clock would claim that his clock runs faster than the laboratory clock that is moving with respect to him.

The observer that is in the frame of the clock sees the light beam moving between fixed points A-B-A and measures the shortest time interval between the events A-A because the light travels the shortest possible distance $2L$. All observers looking at a moving clock observe the light beam moving between moving points A-B-A and measure longer times because the light travels a didtance longer than $2L$. The shortest possible time duration between two events is the time duration measured by an observer who sees those two events occur in the same place. This time duration is called the proper time.

Length contraction

Alice is standing on the ground and observing Bob who is racing with velocity $v$ on his skateboard towards the finish line that is $L_0$ meters away. He will reach the finish at the time $\Delta t = L_0 / v$ (measured on Alice's clock). $L_0$ is the distance to to the target seen by Alice. This distance is called the proper length because the target is at rest with respect to Alice.

v L0 = v Δt Bob Alice's view H

Fig. 5: Proper length L0

Bob has a different perspective. He is at rest with respect to his skateboard and observing Alice who is moving away from him at velocity $v$. He also sees the finish line at distance $L$ that is speeding towards him with velocity $v$. He will meet the finish line at the time $\Delta \tau = L / v$ measured on his clock. Bob and Alice agree, of course, on the relative velocity $v$.

v v L = v Δ𝜏 L0 = v Δt Bob's view Alice H
Fig. 6: Length contraction

We have $$ \frac{L_0}{v} = \Delta t = \gamma \Delta \tau = \gamma \frac{L}{v} $$ and hence $$ L = L_0 / \gamma \tag{4} $$ The length in a moving system is shorter than the proper length in the rest system. This effect is also known as Lorentz contraction. The lengths are contracted only in the direction of the movement. Orthogonal lengths like $H$ are not contracted. They appear constant for both the stationary and moving observer.

Example: Muons on Earth

The relativistic effects of time dilation and length contraction have been confirmed by many real experiments. A famous one is the observation of muons on Earth as performed in 1941. Cosmic rays entering the Earth's atmosphere create short-lived particles called muons in the height of about 10 km. The half-life of a muon at rest is $\Delta \tau = 2.2 \mu s$. Its velocity is about $v = 0.999 c$. In classical mechanics, a muon would travel the distance $$ s_{classic} = v \Delta \tau = 0.999 \times 3 \times 10^8 \times 2.2 \times 10^{-6} = 659 m $$ and could never reach the earth surface. Since muons move so fast, relativistic effects must be considered. The Lorentz factor is $$ \gamma = 22.37 $$ The lifetime of a muon observed from Earth is $$ \Delta t = \gamma \Delta \tau = 22.37 * 2.2 = 49.2 \mu s $$ and the travelled distance is $$ s = v \Delta t = 14.7 km \gt 10 km $$ so the muons reach the Earth surface.

We could also argue from the perspective of the muon. In its frame the lifetime is $\Delta \tau = 2.2 \mu s$. The muon is at rest and Earth is moving toward it at the velocity $v = 0.999 c$. The distance that the ground travels before the muon decays is $v \Delta \tau = 659 m$. The distance 10 km to the Earth surface is contracted in muon's frame to $$ L = 10 / \gamma = 447 m $$ so the ground will reach a living muon.

Lorentz transformations

Consider two inertial systems $S$ and $S'$ moving with relative speed $v$ along the parallel axes $x$ and $x'$. The Lorentz transformation relates the coordinates $(x, t)$ in $S$ to $(x', t')$ in $S'$. Since both $S$ and $S$ are inertial frames, the map $(x, t) \rightarrow (x', t')$ must map straight lines to straight lines; such maps are, by defnition, linear. Moreover, an observer sitting at the origin, $x' = 0$, of $S'$ moves along the trajectory $x = v t$ in S, and hence the points $x = v t$ must map to $x' = 0$: $$ x' = \gamma (x - v t) \tag{5} $$ When we look at things from the perspective of $S'$, relative to which the frame $S$ moves backwards with velocity $-v$, we obtain: $$ x = \gamma (x' + v t') \tag{6} $$ In S, a light ray has trajectory $$ x = ct $$ In $S'$, the same light ray has trajectory $$ x' = ct' $$ because we require that the ligt speed $c$ is constant. Substituting these trajectories into (4) and (5), we get two equations relating $t$ and $t'$, $$\begin{align} c t' & = \gamma (c - v) t \\ c t & = \gamma (c + v) t' \end{align}$$ When we eliminate $t$ and $t'$ we obtain $$ \gamma = \sqrt{\frac{1}{1 - v^2 / c^2}} \tag{7} $$ and see that $\gamma$ is the Lorentz factor that we have already established in (3) using a "Gedankenexperiment". When we substitute the expression (5) for $x'$ in (6) and rearrange, we get $$ t' = \gamma \left( t - \frac{v}{c^2} x \right) \tag{8} $$ It is common to write the Lorentz transformation in the following equivalent form: $$\begin{align} c t' & = \gamma \left( c t - \frac{v}{c} x \right) \\ \tag{9} x' & = \gamma \left( - \frac{v}{c} c t + x \right) \end{align}$$ Now, both equations have the same unit of "length" (meters).

The inverse transformation is obtained by solving for $ct', x'$ or more easily by subtituting $-v$ for $v$: $$\begin{align} c t & = \gamma \left( c t' + \frac{v}{c} x' \right) \\ \tag{10} x & = \gamma \left( \frac{v}{c} c t' + x' \right) \end{align}$$ We will consider only relative movement along the x-axis such that $y' = y, z' = z$.

The transformation (9) was already developed by the Dutch physicist H. A. Lorentz (1853 - 1928) around the year 1900 but under wrong assumptions about the existence of an "ether". Albert Einstein (1879 - 1955) derived the transformation in 1905 based on his two postulates of relativity. He showed that the transformation does not depend on any kind of ether but is the property of the space-time.

Addition of Velocities

Consider a particle moving 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 trajectory of the particle in $S'$ is $$ x' = u' t' $$ The velocity in $S$ is (using the inverse Lorentz transformation): $$ u = \frac{x}{t} = \frac{\gamma (x' + v t')}{\gamma (t' + v x' / c^2)} $$ Sutbstituting $x' = u' t'$ we obtain: $$ u = \frac{u' + v}{1 + u' v / c^2} \tag{12} $$ Note that when $u' = c$, this gives us $u = c$ as expected.

The formula for the addition of velocities can also be obtained by the composition of two Lorentz transformations. The Lorentz transformation can be written in matrix form as: $$ \begin{pmatrix} \tag{13} c t'\\ x' \end{pmatrix} = \begin{pmatrix} \gamma & - \gamma v/c \\ - \gamma v/c & \gamma \end{pmatrix} \begin{pmatrix} c t\\ x \end{pmatrix} = \Lambda[v] \begin{pmatrix} c t\\ x \end{pmatrix} $$ A particle moving with velocity $v_1$ with respect to frame $S$ can be thaught of as a particle at rest at the origin of frame $S_1$ that is moving with velocity $v_1$ with respect to frame $S$. Analoguosly, a particle moving with velocity $v_2$ with respect to frame $S_1$ can be thaught of as a particle at rest at the origin of frame $S_2$ that is moving with $v_2$ with respect to $S_1$. Multiplication of the two respective Lorentz matrices yields the formula for the movement of $S_2$ with respect to $S$: $$ \Lambda[v_1] \Lambda[v_2] = \Lambda \left[ \frac{v_1 + v_2}{1 + v_1 v_2 / c^2} \right] \tag{14} $$


Read further in Special Relativity 2: Spacetime