Special Relativity

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In another article we discussed the Galilean Transformation. In early physics before the 19th centruy physicists believed that there is some special frame of reference in which all the planets move, including the earth. Relative to this special frame of reference they suspected some medium called the aether, which has the property that it is at rest in this reference frame. If such an aether exists - so was the hypothesis - light moves faster when traveling with the direction of the eather than against the direction of the eather. Just like you swim slower against current, than with it. Thus they expected the light to obey the Galilean Transformation.

MichelsonMorley.png

If the system at rest according to the aether has the referece frame $x,y,z$ and we denote the frame of the earth's orbit with $x',y',z'$, we call the velocity with which $x',y',z'$ is moving with respect to $x,y,z$ $\mathbf{u}$. Say the earth moves at time $t1$ in the positive $z$ direction with velocity $v$ and at time $t2$ in the negative $z$ direction. How fast is the speed of light measured in each direction? Following the rules of the Galilean Transformation and assuming light travels in some medium - the aether - the speed of light would be $c+u+v$ in the $z$ direction and $c-u-v$ in the negative $z$ direction. However this is not what Michelson and Morley discovered. They found that the speed of light is the same in every frame of reference. This raises the importance of the speed of light to that of a universal constant. The findings of Michelson and Morley where also the starting point for Albert Einstains theory of Special Relativity.

  • To illustrate the meaning of this result, imagine you are flying along with a beam of light, with velocity $v$ in some reference frame $A$. If the Galilean Transformation would be valid, the beam of light would be at rest, provided your reference frame moves with the same speed, that of light. In this case you could see the light standing still. However, this is not true. We know now, that the light can not stand still it needs to have the speed of light in every refrence frame, including the frame which is moving according to $A$.
  • Another interesting Gedankenexperiment is the following. Imagine you travel with speed $v$ in some reference frame $A$. If you send out a light beam from your moving reference frame, let's call it $B$, than the speed of light according to the Galilean Transformation would in the reference frame $A$ be $v+c$. But we know that the speed of light is as fast as you can get and so the Galilean Transformation must be invalid, since their predictions contradict our experience.

We will now develop the theoretical framework that is able to explain the described phenomena above and do so by roughly following Albert Einsteins argumentation in his famous work "Zur Elektrodynamik bewegter Körper". But let's first introduce the simplest ideas of what a Minkowsky Diagram is. To examine an event in two different frames of reference, we want to know how one reference frame $A$ is described in another $B$. Let the velocity with which $A$ moves according to $B$ be denoted $v$. If we just consider one coordinate axis (horizontal) and a time axis (vertical), an event can be described by a point in this graph. Moreover, the system $B$ moving with $v$ in the $x$-direction, is in the reference frame $A$ characterized by a line $x=v \cdot t$

Minkowski.png

Notice, while an event occurs at $x=v \cdot t$ in System $A$, the same event in the moving system $B$ is described by the coordinate $x'=0$. If the relative velocity of the reference frames would be $c$ this would be very unconvenient to draw in standard units. The speed of light is so large that the line $x=ct$ would be practically horizontal. Thus we choose lightseconds as a measure for distance so that $c$ is one lightsecond per second. In this case $x=ct$ follows the median at a $45°$ angle. A light ray with velocity $-c$ would be represented by the line $x=ct$ mirrored at the time axis.

Eventslike.png

From this we can immediately tell, that a moving frame can only reach points above the graph $x=ct$ of light, as time goes forward, such events are called time-like (they satisfy $t^2-x^2/c^2>0$), since you can reach them as time moves forward. Events below the graph $x=ct$ of light are only reachable if you start from a point $x \not =0$, so you have to move in space to reach such a point, thus this events are called space-like (they satisfy $x^2-t^2c^2>0$). Not suprisingly the points on the line $x=ct$ are called light-like; they behave like light.

Now, why does the Galilean Transformation fail to explain the phenomena oulined above? It turns out that the assumption, that time is the same in each reference frame does not longer hold. Time depends on the reference frame. Two events that happen simultaniously in reference frame $A$, do not happen simultaniously in another reference frame. Let us study this fact in more detail.

Simultaneity

Two systems $S$ and $S'$ more with velocity $v$ against each other. Each system consists of two lasers positioned at $A$ and $B$ or likewise $A'$ and $B'$ seperated by a distance $L$. Inbetween this lasers is an Observer $O$ and $O'$ respectivley. To synchronise clocks located at $A$ and $B$ a light beam is sent out to the observer, traveling the same distance $\overline{AO}=\overline{BO}$. We define clocks as synchronous if they measure the same time as their lighbeam reaches the obeserver $O$ in this set-up.

Simultaneity.png

If the clocks are synchronised in the system $S$, both the light beam from $A$ and $B$ would reach $O$ simultaniously. But what does the observer $O'$ see? We assume that clocks at $A'$ and $B'$ are synchronised according to $O'$. During the time the light needs to travel the distance $\bar{AO}$, say $\Delta t=t_1-t_0$, the moving system $S'$ has moved by $\Delta x$ to the right.

Simultaneity2.png

But this means, a light beam coming from $B$ reaches the observer $O'$ sooner than the light beam from $A$. Thus $O'$ would say that $B$ emitted the lightbeam first, before $A$. In the frame of reference of system $S'$ the clocks at $A$ and $B$ would therefore be asynchronous.

Which events in $S$ are synchronous according to the frame of reference of system $S'$? To answer this qustion we draw a Minkowski diagram of the situation above. For simplicity we choose $\overline{AO}=\overline{BO}=1$. The points $A'$ $O'$ and $B'$ of system $S'$ are represented by $x=vt$, $x=vt+1$ and $x=vt+2$ in system $S$.

Simderiv.png

The important question is, at which time $t_b$ does the signal from $B$ have to be emitted, so that $O'$ would say $A$ and $B$ are synchronous? A light beam sent from $A$ travels from the origin to $a$ with speed $c$. A light ray sent at the right time $t_b$ travels from $b$ to $a$ with speed $-c$. Now, let us find the coordinates of the points $a$ and $b$. We do so by simply calculating the intersection of $x=ct$ and $x=vt+1$. For the point $a$ we find the coordinates

\begin{align} ct&=vt+1\\ t(c-v)&=1\\ t&=\frac{1}{c-v} \end{align} and \begin{align} x&=ct\\ x&=\frac{c}{c-v}\\ x&=\frac{1}{1-v/c}\\ \end{align} All light rays with $-c$ satisfy the equation $x=-ct+const.$ or $x+ct=const.$ Thus we can easily find $b$ by plugging in the coordinates of $a$ in this equation. \begin{align} x+ct&=const.\\ &=\frac{1}{1-v/c}+c\frac{1}{c-v}\\ &=\frac{2}{1-v/c} \end{align} The intersection of this line with $x=vt+2$ will give us the coordinates of point $b$. Subtracting the two equations \begin{align} x+ct&=\frac{2}{1-v/c}\\ x-vt&=2 \end{align} from another we have \begin{align} t(c+v)&= \frac{2}{1-v/c}-2\\ t&=\frac{2}{(c+v)(1-v/c)}-\frac{2}{c+v}\\ &=\frac{2v/c}{(c+v)(1-v/c)}\\ &=\frac{2v/c}{c-v^2/c}\\ &=\frac{2v/c^2}{1-v^2/c^2}\\ \end{align} we find $x$ by the equation $x=\frac{2}{1-v/c}-ct$ and plugging in for $t$ what we just found \begin{align} x&=\frac{2}{1-v/c}-c\frac{2v/c^2}{1-v^2/c^2}\\ &=\frac{2(1+v/c)}{(1-v/c)(1+v/c)}-c\frac{2v/c^2}{1-v^2/c^2}\\ &=\frac{2+2v/c)}{(1-v^2/c^2)}-\frac{2v/c}{1-v^2/c^2}\\ &=\frac{2}{(1-v^2/c^2)} \end{align} We could have arrived here in a less tedious way, by setting $c=1$ and restoring the units afterwards (watch the video lecture for details). However, the coordinates for the point $b$ are. \begin{align} x&=\frac{2}{(1-v^2/c^2)}\\ t&=\frac{2v/c^2}{1-v^2/c^2}\\ \end{align}

All points laying on the line with slope $t/x=v$ correspond to some time for which an event in $S$ is simultanious in $S'$, thus this line corresponds to the time $t'=0$ in system $S'$.

Simderiv2.png

Notice the symmetry in $x$ and $t$ of the equations $x=vt$ and $t=vx$. Graphically this means (provided we choose the same scaling for both axis), if you have one equation (or graph) you find the other by interchanging $x$ and $t$, which says reflect one graph (e.g. $x=vt$) about the median to get the other graph (e.g. $t=vx$). This symmetry leads to the same angle $\alpha$ between the time and coordinate axis respectively.

Relsymmetry.png

Finally we would like to know how to transform between the two frames of reference $S$ and $S'$.

The Relativity Principle

In the moving System $S'$ the coordinates are given by $t'=0=t-vx$ and $x'=0=x-vt$ relative to system $S$, but this could be incorrect by some factor depending on the speed $v$, so the transformation is \begin{align} x'=(x-vt)f(v)\\ t'=(t-vx)f(v)\\ \end{align} The relativity principle says, that there is no special coordinate system. In other words we could imagine $S$ at rest and $S'$ moving, but it is equally true to say $S'$ is at rest and $S$ moves. All we can tell is, that $S$ and $S'$ move relative to each other. Thus from the perspective of $S$, $S'$ moves in the other direction and the equations \begin{align} x=(x'+vt')f(v)\\ t=(t'+vx')f(v)\\ \end{align} are valid. Using both pairs of equations we can find the factor $f(v)$ \begin{align} x'=((x'+vt')f(v)-v(t'+vx')f(v))f(v)\\ x'=x'(1+v^2)f^2(v) \end{align} We find the factor \begin{align} f(v)=\frac{1}{\sqrt{1+v^2}} \end{align} to make the factor dimensionless we deived the velocity by $c^2$ \begin{align} f(v)=\frac{1}{\sqrt{1+v^2/c^2}} \end{align} We find the

Lorentz Transformation \begin{align} x'=\frac{x-vt}{\sqrt{1+v^2/c^2}}\\ t'=\frac{t-vx/c^2}{\sqrt{1+v^2/c^2}}\\ \end{align}

with the dimensions of $vx$ fixed to be time. Let us no examine an event, in two different frames of reference by using a Minkowski diagram.

Contractiondilation.png

In $S$ the event occurs as a distance, say $L$ given by the intersection of the vertical dotted (black) line and the $x$-axis. The same event in $S'$ has the coordiante $L'$ given by the intersection of the vertical dotted (black) line and the (blue) $x'$-axis. $L'>L$ so for a moving observer in system $S'$ the length $L$ is contracted. The length contraction is given by \begin{align} L=\frac{L'}{\sqrt{1+v^2/c^2}}\\ \end{align} For an observer in reference frame $S$ time in System $S'$ appears to go slower. Follow the dotted (blue) line from the event to the time axis. The time on the $t'$ axis - let's call it $T'$ - is larger than in the time $T$ measured on the $t$-axis, given by the intersection of the dotted (blue) line with the $t$-axis. The phenomenon that for a moving observer time in a system at rest appears to go slower is called time dilation. Time is dilated by the same factor $\gamma =\sqrt{1+v^2/c^2}$, as $L'$ is contracted \begin{align} t'=t\sqrt{1+v^2/c^2}\\ \end{align}




Video Lectures:

  • Leonard Susskind - Special Relativity | Lecture 1 [1]
  • Some examples that help to form an intuition for relativistic phenomena [2][3][4]

Further Reading:

  • Zur Elektrodynamik Bewegter Körper [5]