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## Classical space and time

Physics is build on the idea that the universe is composed of things, matter, that moves through something, space. This point of view of our universe follows directly from our commonsense observations of moving objects. If we observe two things, matter and space, it is reasonable to define two independent quantities for each of these observations. We call the quantities we observe dimensions. To measure dimensions, we need a measurement procedure and an arbitrarily choosen unit of measurement. We denote the dimension for matter with $M$ and the dimension of space with $L$. We measure matter with mass. We use a standard body of 1 kg as our unit of mass. To measure space we must assume a type of geometry. For now we assume an Euclidean geometry. In an Euclidean geometry we measure the distance or lenght between points in space, $\Delta s$, with the Pythagorean theorem. We choose a ruler with a standard length of 1 m as our unit of lenght.

### Measurement of space

We can go in space from our current position upwards or downwards, to the north or to the south and to the east or to the west. We can choose each of these directions independent of the other. Our experience suggests that space is three dimensional $L^3$. With Pythagoras we find for the distance between two points: $(\Delta s)^2=(\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2$

The first thing we must arrange to study motion is to assign coordinates to each point in space. We choose three straight lines, one for each dimension of space, that make a right angle with each other. We call them coordinate axes. The point of intersection we call the origin and attach it to a rigid body. The rigid body should have a constant velocity. We mark positions on these axes with a meter stick. The coordinates of any point in space are defined by dropping lines from the point perpendicular to each of the coordinate axes. Physicists call this system a inertial reference frame and it defines the point of view of an inertial observer in the universe.

### Relativity principle

We assume that any inertial observer in the universe will observe the same laws of nature. This principle that the laws of nature are the same in each inertial reference frame is called the relativity principle. It makes sense, why would nature prefer one system over the other.

The trajectory of a stone dropped from a train to the ground is a straight line for observers in the train, but a parabolic curve for observers on the ground. These two types of curves represent however the same physical phenomenon. To describe a law of nature for a falling stone, it should be based on something else than the curve traced in space. A law should describe a quantity of a phenomenon that remains invariant in any inertial reference frame.

### Time

As position is not frame independent we must look at other quantities derived from position. To do that we first introduce a new quantity, time, denoted by $T$. We use the concept of time to turn the change in position into a change in position per unit of time $v =\frac{dx}{dt}=\frac{\Delta s}{dt}$ this derived quantity, $LT^{-1}$ we call velocity. We can continue and derive the change of velocity per unit of time $a= \frac{d^2 x}{dt^2}$ This newly derived quantity,$LT^{-2}$, we call acceleration.

To measure time we use a clock. A clock is based on any regular phenomenon. An operational definition of time depends on the existence of uniform cyclic motion. Motion along a straight line can not be used, as we can not count it. Uniform rotational motion is a good candidate. For example we can define a solar day as the time between the instant when the sun is directly overhead and the next instant it is directly overhead. With this definition of time we have a clock that counts days. We can compare any other motion now with this clock and measure velocity and acceleration.

In the case of the solar clock the unit of time is called a day. The speed of a body is nothing else than the distance it travels per unit clock tick. However the motion on which a clock is based, is physically not different from any other type of motion. In one clock tick the clock mechanism travels a certain distance. If the earth travels 40.000km in one cycle, solar day, then one could say 1day=40.000km. A speed of 40km/day is nothing else than: $40\mbox{km/day}=\frac{40\mbox{km}}{40.000\mbox{km}}=1/1000$ This 1/1000 is a dimensionless ratio. However we never use this ratio. We use the number of clock ticks to convert a distance to a speed or a speed to a distance.

We choose one uniform cyclic motion as our standard reference for time, and call one cycle of this motion our unit of time: 1 second, $1s$. Let us further assume that all clocks used by inertial observes tick at the same speed and are synchronized, then we have a universal time. We can use our clocks to measure not only speed and velocity, but also to uniquely order events. An event $E=(x,y,z,t)$ defines a particular event at a particular position relative to an inertial reference frame but at a global time. The stone hits the bottom of the train at the same time for the observer at the train as for the observer on the embankment. There are no disputes about elapsed time. This is our commonsense experience of time and is the basis for classical physics.

### Law of inertia

Having a clock, which ticks in seconds $s$, we can now further investigate the motion of objects by their velocity. First we analyze the motion of objects along a straight line. If we mark the position of an object each second, we can then measure the distances between these marks. We discover that either the distances between the marks are the same, or they vary. The first type of motion we call, uniform motion, and the other non-uniform motion. So all objects are either in the state of uniform or non-uniform motion. The state of uniform motion along a straight line, is what we call the natural state of an object. In fact we can not tell whether the object is in motion or the observer measuring the distances. So a state at rest or in uniform linear motion are equivalent.

The fact that uniform motion is the natural state of an object is a physical law. This is called the law of inertia. It implies that an object at rest with its environment or in a constant motion along a straight line will remain in this state unless based on another physical law a change occurs. Two objects in uniform motion can either at rest relative to each other or in constant motion. There is no such thing as absolute rest. The only motion one can detect from an inertial reference frame is relative motion.

### Galilean transformations

Let $S$ and $S'$ be two intertial reference frames, let $S'$ move relative to $S$ with velocity $u$, and at $t'=t=0$ both frames are in the same position, then the coordinates of any measurement in $S$ can be simply transformed to coordinates in $S'$ by the Galilean transformation: \begin{align*} x' &= x - u_x t \\ y' &= y - u_y t \\ z' &= z - u_z t \\ t'&=t \end{align*}

Now assume we have an object in uniform motion described in $S$ by $r(t)=v(t)t+r(0)$: \begin{align*} x(t) &=v_x t + x(0) \\ y(t) &=v_ y t +y(0) \\ z(t) &= v_z t + z(0) \\ \end{align*} If we observe this same motion from $S'$ we get: \begin{align*} x'(t) &= (v_x - u_x) t + x(0) \\ y'(t) &= (v_y - u_y) t + y(0)\\ z'(t) &= (v_z - u_z) t + z(0)\\ \end{align*} So also in $S'$ this motion is uniform.

We conclude that if we observe a body in motion the measured speed of an object depends on the speed of the inertial reference frame of the observer relative to the inertial reference frame which carries the body in motion. By the Galilean transformation laws we notice that this difference in speed is a constant, and is just the difference in speed between the two reference frames.

This outcome is in line with our daily experience. If a man in a train walks with a constant speed, $v$, in the direction of the moving train from the rear to the front, the observers in the train will measure a speed $v$ whereas observers of the passing train on the embankment will measure a speed of $v+w$ where $w$ is the speed of the train relative to embankment. Observers on the moon measure even another speed $v+w+x$ where $x$ is the speed of the earth relative to the moon.

When an object is in non-uniform motion, which physicists call acceleration, the velocity per unit of time changes. This change can be constant or variable.

### Newtons laws

Acceleration is a quantity that remains invariant between inertial reference frames, as can be easily shown: \begin{align*} a'(t)&=\frac{dv'(t)}{dt} \\ &=\frac{dr'(t)/dt}{dt} \\ &=\frac{d(r(t)-ut)/dt}{dt} \\ &=\frac{v(t)-u}{dt} \\ &=a(t) \end{align*}

When we apply the knowlegde we gained sofar to our example of the falling stone, then both observers, on the ground and in the train, will notice by measuring the distances from their perspective, a constant change in velocity of $9.8m/s^2$ in the vertical direction towards the ground or in mathematical symbols: $9.8=\frac{d^2 x}{dt^2}$ This happens to be true for all falling bodies and can be seen as a specific form of a law of nature of falling bodies near the surface of the earth.

Once we have a law, like $a=\frac{d^2 x}{dt^2}$ where $a$ is the acceleration we can calculate the trajectory of the object if we know its initial position in any reference frame at rest.

Physicists tried to explain the cause of $a$ and this led to Newtons discovery of the second law of motion (the first law is the law of inertia and due to Galilea): $a=\frac{F}{m}$ The accelaration of a body is proportional to a force (the cause) and inversely proportional to its mass (resistance). The force $F$ is a mathematically derived quantity with dimension: $MLT^{-2}$ with unit $1N=1kg.m.s^{-2}$.

Assume an isolated system $S$ of bodies which are in rest relative to each other. Suppose an observer from another inertial reference frame $S'$ studies the bodies in $S$. He writes down $p=mv=C$ and calls this quantity the momentum of the system $S$. Based on Newtons law he knows that once this quantity changes a force must have been applied to the system $S$. Suppose one body $m_x$ within $S$ suddenly accelerates for $1s$. Then there must have been a force $F_x$ applied from within the system to $m_x$ for $1s$ as we assume that there are no external forces. The observer definitely notices $a_x$ for the body $m_x$. If he would notice nothing else than he must conclude that $p=m*v=C+m_x(v_x-v)$. But this is against the law of inertia, so we derive a new law: $m_x(v_x-v)=-m_{\bar{x}}(v_{\bar{x}}-v)$ or equivently: $F_x=-F_{\bar{x}}$ This is Newtons third law of motion. It is implied by the first law. The first law implies the second and the third.

## Relativistic space and time

### Speed of light is a universal constant

At the begin of the twentieth century Einstein dramatically changed our view on space ant time. It all started with Maxwell's Electromagnetic theory. One outcome of this theory was that light is an electromagnetic wave, which spread from its source spherical in all directions, the points on the wavefront move along straight lines with a constant speed. The speed of light in a vacuum: $c \approx 3 \times 10^8 m/s$

### Space and Time depends on the observer

The discovery of the constancy of speed for all observers had a number of important consequences.

In Newtonian mechanics we were used to the idea that a body takes on the speed of its container. A man sitting in a train moves at the speed $v$ of the train relative to an observer on the embankment. Also sound and water waves take on the speed of their medium for inertial observers in motion relative to the medium. This behavior implied the Galilean transformation rule where we can add the relative speeds. Now with the constancy of the speed of light relative to each inertial observer this principle breaks down. The speed of a light flash in the direction of the moving train, measured both by the man in the train and by the observer on the embankment gives the same value $c$. This is remarkably. Suppose the distance for the light to travel within the train is $x$, then as the train moves, the front of the train is moving away from the flash of light. For an observer on the embankment the distance the flash travels before it reaches the front is definitely greater than $x$ like $x^\prime$. But an observer on the moon, who would also notice the rotation of the earth, would notice an even larger distance if the train moved in the direction of the rotating earth.

In Newtonian physics one should have to conclude that the speed of light is $c+v$ for the observer outside to justify $c$ for the observer inside or $c-v$ inside to justify $c$ outside. However by theory and experimental fact both observers measure $c$.

It is impossible to understand this phenomenon with our common sense, as we have grown up with the idea that the distance between two points can not be $x$ and $x^\prime$ depending on the point of view you take. Furthermore we believe that the light flash must travel a definite distance and not arbitrarily many. But it is not only space that is different. As speed is the ratio between space and time also the time depends on the point of view if the speed of light remains constant.

So space and time are relative concepts. Space and time are not existing as definite quantities which are absolute, that is the same for any observer. This was exactly the insight of Einstein. One meter on my ruler is not necessarily the same as on yours. If our meters are different than why could not our clocks tick different ? It was shown that time and space are intertwined. This special fabric of the geometry, where units of meters and seconds change with relative motion, had a huge advantage, all known laws of physics look exactly the same in each inertial reference frame. There is no priviliged frame of reference. The only consequence is that we need to transform measurements between different inertial reference frames in a different way than we were used to with the Galilean transformation.

### Simultaneity depends on the observer

The fact that space and time are relative to an observer also has a consequences on the simultaneity of events. If a light detector in the middle of a train detects two light flashes, one emmitted from the rear and the other from the front, an observer in the train will conclude that these light flashes were emitted simultanously in time. This directly follows from the constancy of the speed of light and the same spatial distance travelled in the train for the two light flashes. However for an observer outside the train the light flash emitted from the rear of the train must have travelled a longer distance than the light flash emitted from the front of the train. So for this observer the light flash at the rear must have been emitted before the light flash at the front. So both observers do not agree on the simultaneity of the emmission of the two light flashes. Which is the same thing as saying that the clock at the rear is set ahead of the clock at the front for the observer outside the train. Moving clocks which are spatial separated are not synchonized for the observers.

Relativity of simultaneity is different than the appearance simultaneity due to the fact that two signals of the same event are observed at different times due to different speed of transmisson or that two signals of different events are detected simultaneous due to difference in space. Appearance simultaneity can always be corrected by a correction factor for the transmission speed.

### Invariant Spacetime Interval

We can turn these consequences also in an algebraic form. We have found by experimental fact for all observers: $\Delta s = c\Delta t \\$ we know from Pythagoras: $\Delta s^2 = \Delta x^2 + \Delta y^2 + \Delta z^2$ So we can derive the following relation: $$c^2\Delta t^2 - \Delta x^2 - \Delta y^2 - \Delta z^2=0$$ This equation relates the time for light to travel with the space it travels during that time. If we measure a light flash than we are sure that for all observers this equation is true. But it turns out that for any measurement of an event measured in an inertial reference frame $S$ and some other $S'$ with coordinates $(\Delta t,\Delta x)$ and $(\Delta T,\Delta X)$ respectively we have: $c^2\Delta t^2-\Delta x^2=c^2\Delta T^2-\Delta X^2$ We call the following equation the spacetime metric and $s$ the spacetime interval between two points in this space, which are called events: $\Delta s^2=c^2\Delta t^2-\Delta x^2$ Each observer will always find the same $\Delta s$ for the same event based on measurements of the coordinates in his reference frame. We call this feature the invariance of the spacetime interval. This equation defines the geometry of our universe, and is quite different from the Euclidean geometry, which only related to a distance in space: $\Delta s^2=\Delta x^2+\Delta y^2+\Delta z^2$. Now we define our universe as consisting of events with a spacetime distance between them which is calculated by the above metric.

We now define how we measure a kind of proper time and proper distance in this geometry. We have three situations. First two events E and F in our spacetime can be events with a light flash, in that case we have $\Delta s=0$, and the events are lightlike separated. A line connecting lightlike events lies on a light cone. A second possibility is that the two events are marked by the time recording of a clock at rest in an inertial reference frame. The time between the two recordings is then called the proper time and the events are called timelike separated events and we have $\Delta s=ct$. The line connecting these events is a non-horizontal line within the light cone. The remaining possibility is that the events are separated by space only, in this case a rigid rod can have its ends simultaneously at E and F. (Simultaneously in the sense that light flashes emitted at E and F reach the center of the rod simultaneously, or equivalently, that E and F are simultaneous in the rest frame of the rod.) These events are spacelike separated and we define the proper distance as $\|\Delta s\|$. The line connecting these events is a horizontal line within the light cone.

### Lorentz Transformation of coordinates

This new geometry of our flat spacetime also replaces the Galilean transformation of coordinates between different inertial reference frames by another type of transformation. This transformation has still to be a linear transformation as the separation in space and time between events in each reference frame must be proportional to a constant factor. Only then it does not matter which event has been taken as the reference for our measurements. In fact this proportionaly scale is also a precondition for the second law of Newton to hold.

Let $S$ and $S'$ be two inertial reference frames with coordinates $(t,x)$ and $(t',x')$ respectively and $S'$ has a relative speed $v$ with respect to $S$ and the reference event coincides $(t=t'=0,x=x'=0)$. We tactically assume that $y$ and $z$ coordinates coincide between these two reference frames. The transformation, which is called the Lorentz transformation, must therefore have the following form: \begin{align*} t&=Bx'+Dt' \\ x&=Gx'+Ht' \\ \end{align*} Furthemore these equations must satisfy the invariance of the spacetime interval. We proceed the derivation in two steps. First for the particular case $x'=0$ and then for the general case of any pair $(x',t')$.

Suppose $x'=0$, we have a timelike separation of events in the $S'$ frame. The space location of an event in the $S'$ frame has $x=vt$ space location in the $S$ frame. Now we align the space timeintervals for both $S'$ and $S$ \begin{align*} c^2(t')^2&=c^2t^2-x^2=c^2t^2-v^2t^2=t^2\left(c^2-v^2\right) \\ (t')^2&=t^2\left(1-v^2/c^2\right) \\ t&=\frac{t'}{\sqrt{1-v^2/c^2}}\\ t&=t'\gamma,\;\; \gamma \text{ is called the boost factor } \end{align*} Substituting this result in $x=vt$ gives: $x=v\gamma t$

This gives us the Lorentz equations for the specific case $x'=0$: \begin{align} x&=v\gamma t \\ t&=t'\gamma \\ \gamma&=\frac{1}{\sqrt{1-v^2/c^2}} \end{align}

Now we generalize this result voor any pair of coordinates in the $S'$ frame. For the same event (measured from a common reference event) we have following transformation equations: \begin{align*} t&=Bx'+Dt' \\ x&=Gx'+Ht' \\ \end{align*} Which must satisfy: $c^2t^2-x^2=c^2t'^2-x'^2$ We substitute first $D=\gamma$ and $H=v\gamma$ based on the particular solution we found when $x'=0$. Then we subsitute the two equations for $t$ and $x$ into the metric equation and get following result: $c^2\left(Bx'+\gamma t'\right)^2-\left(Gx'+v\gamma t'\right)^2=c^2t'^2-x'^2$ We multiply out the squares and group terms by $t'^2$, $x'^2$ and $x't'$ after we have the following equation: $\gamma^2\left(c^2-v^2\right)t'^2+2\gamma\left(Bc^2-vG\right)x't'-(G^2-c^2B^2)x'^2=t'^2-x'^2$ Now we must have: $2\gamma\left(Bc^2-vG\right)=0$ So, $B=\frac{vG}{c^2}$ We also must have: $G^2-c^2B^2=1$ We substitute the previous found value for $B$ into this equation: \begin{align*} G^2-\frac{c^2v^2G^2}{c^4}=1 \\ G^2(1-\dfrac{v^2}{c^2})=1 \\ G=\frac{1}{\sqrt{1-v^2/c^2}} \\ G=\gamma \end{align*} Then we find for $B$: $B=\frac{v\gamma}{c^2}$ Finally we substitute the values for $B,D,H,G$ into the transformation equations to get the general Lorentz transformation: \begin{align} t&=\dfrac{t'+vx'/c^2}{\sqrt{1-v^2/c^2}} \\ x&=\dfrac{x'+vt'}{\sqrt{1-v^2/c^2}} \end{align}

The inverse transformation equations follow simply from symmetry. From the $S$ frame the speed of $S'$ is $v$, but from the $S'$ the speed of $S$ is $-v$. The only thing we need to do is switch the $t,x$ with the $t',x'$ and change $v$ to $-v$: \begin{align} t'&=\dfrac{t-vx/c^2}{\sqrt{1-v^2/c^2}} \\ x'&=\dfrac{x-vt}{\sqrt{1-v^2/c^2}} \end{align}

As the speed of light is a constant we could put $c=1$ that implies that $1s=3\times 10^8 m$, so whenever we measure a time in seconds we multiply it with this factor to get an amount in meters. Then we can get rid of the $c$ factor in the metric and Lorentz transformations. We could also in any physical equation eliminate $m/s$ with the factor $1/(3\times 10^8)$. As an example take $1J$ which is a unit for energy: $1J=kg\frac{m^2}{s^2}=kg\frac{m^2}{s^2}\left(\frac{1}{3\times 10^8}\frac{s}{m}\right)^2=\frac{1}{9\times 10^{16}}kg$

### Lorentz contraction

Take again the inertial reference systems $S$, $S'$ as defined above. Let Joe be at rest in $S$ and Moe in $S'$. When Moe measures the distance of a point in $S'$ from the origin, he lays down a meter stick $x'$ times. So he will report a distance $x'$. However from the viewpoint of Joe in the $S$ system this distance will be: $x=x'\sqrt{1-v^2/c^2}+vt$ Joe will say that the meter stick of Moe is foreshortened by the factor $1/\gamma=\sqrt{1-v^2/c^2}$. This is called lenght contraction. Although the meter stick in $S'$ contracts relative to $S$, the distance $x$ measured by Joe is always larger than $x'$ as one also need to add the distance $S'$ moved: $vt$.

The same reasoning we can apply for the time measurements. If Moe measures $t'$ then Joe will say that its measurement is: $t=t'\sqrt{1-v^2/c^2}+vx/c^2$ Joe will say that clocks of Moe tick slower than his clocks by the same factor $1/\gamma=\sqrt{1-v^2/c^2}$ as we have seen for the length contraction. This is called time dilation. But that is not all, the time also is different due to the location at which Joe see the event. But this implies that when Moe measures in his frame two events at different locations $x'_1, x'_2$ simultaneous at $t'$ then they are not simultaneous for Moe. Moe will measure a time difference: $t_2-t_1=v(x_1-x_2)/c^2$ This circumstance is called failure of simultaneity at a distance

### Examples

Now we have mastered the principles of Special Relativity. We turn now to applying these principles to concrete examples. As the speed of light is a huge number when measured in meters per second, we use another unit for time, the nanosecond ($1ns=10^{-9}s$) and define $1 ft = 0.3m$, then we have $1c = 1ft/ns$.

#### Moving frames

We setup a laboratory $S$ with a scientist Joe and a rocket $S'$ with pilot Moe flying through the laboratory from $x_0$ to $x_1$ with constant speed $v=3/5c$. We define two events, $E1=$rocket at $x_0$ in laboratory and $E2=$rocket at $x_1$ in laboratory. As seen from Moe's perspective he is at rest and the laboratory moves along him. So he will define $E1=x_0$ of laboratory passes $x'_0$ and $E2=x_1$ of laboratory passes $x'_0$. We choose $x_0=x'_0=0$ and also $t_0=t'_0=0$. This implies for both Moe and Joe $E1(t,x)=E(t',x')=(0,0)$.

The time between $E1$ and $E2$ as measured by Moe is $\Delta t'=120ns$. So he has following coordinates: $E2(t',x')=(120,0)$. With the laboratory passing him with a constant speed of $v=3/5c$ he calculates the length between $x_0$ and $x_1$ as $\Delta x'=120ns*3/5c=72ft$. The measurements of Joe of these same events are quite different. We can only find out when applying the Lorentz transformations. We have two choices, either we assume the reference frame of Joe is the rest frame, or we assume that it is moving. We assume first Joe is at rest. We convert the coordinates of $E2$ of Moe from the moving reference frame to the rest frame of Joe. As $x'_1=0$ for Moe's measurements we only need to apply the boost factor $\gamma$ to the measurements of Moe: $t_1=t'_1*\gamma=120ns*5/4=150ns$ and for the space coordinate: $x_1=vt'*\gamma=72ft*5/4=90ft.$ So Joe records $E2(t,x)=(150,90)$ from his reference frame for $E2$. Let us check the invariance of the spacetime metric: $120^2 (\text{ Moe })=150^2-90^2 \text{ (Joe) }=120^2$

Now let us assume that Joe is moving, this is the view that Moe will take. Now we take the inverse Lorentz transformations and put in $t'=120ns$ and solve for $t$, this gives $t=120ns*1/\gamma=120*4/5=96ns$.

Now suppose that in $S$ Joe had placed two synchronized clocks $C_0$ and $C_1$ at $x_0$ and $x_1$. At $E1$ these clocks both read $t=0$, and at $E2$ $150ns$ has passed away in the reference frame of Joe so they both read $150ns$. From the perspective of Moe the clock $C1$ also reads $t'=0$ at the $E1$.

But what reads the clock $C_2$ at $E1$ from the perspective of Moe ? We have $t'=0$ and the speed of Joe relative to Moe is $-3/5c$. We also know already that the distance of the clocks as measured by Joe is $90ft$. We can plug this information in the Lorentz equation for $t'$ and the solve $t$. This gives: $t=\frac{-vx}{c^2}=\frac{3/5c90ft}{c^2}=54ns$ So from the perspective of Moe the clock $C_2$ in the reference frame starts at $54ns$ and the time separation between $E1$ and $E2$ is from the same perspective $96ns$ to the clock must read $150ns$. There is no ambiguity in the reading the same clock when Moe and the clock are in the same physical location. There is only a disagreement on the simultaneity of events between mving frames. At $E1$ and $E2$ both Moe and Joe read the same time on the clocks which are at rest in Joe's frame. But at $E1$ they disagree about the clock time of $C_2$ and at $E2$ they disagree about the clock time of $C_1$.

## References

• [1]    Richard P. Feynman,The Feynman Lectures on Physics,vol 1,Addison Wesley,1963.