$\>\!$hysics   

February 19, 2001

Special Relativity,
Lorentz Transformations, and Minkowski Diagrams

Chapter 38 of your text (RH&W) is ordered so that it will be confusing as little as possible, or at least this is the intent of the authors. Nonetheless, I find the order of presentation somewhat illogical from a deductive point of view. Therefore, here I try (only) to present the material in a more logical order, emphasizing which formulae and concepts are the most important.
As you know, we are ALSO using the additional text by French (new this Spring, 2003, semester. It has an approach more like this writing. It is lengthy, and is therefore better read FIRST, with this as a backup for reference.

I also append the comment that the paperback book ``Spacetime Physics,'' by Edwin F. Taylor and John Archibald Wheeler, W. H. Freeman & Co. (2nd Edition, 1992) is an extremely good source book, at the current level, although probably even more verbose than French.

Based on both experimental observations and philosophical notions about the foundations of physical science, Einstein introduced the two

Postulates of Special Relativity

1. The laws of physics have the same form in all inertial reference frames.

2. The speed of light in free space has the same value, c, in all inertial reference frames.

To discuss the consequences of these postulates, we need to have available to us a minimum of 2 inertial reference frames, i.e., two observers who make measurements of whatever physical quantities they desire. We will refer to these two reference frames as S and S', and presume that the observer in S , Barney, makes measurements to confirm that the reference frame S' is moving with velocity $\bf\vec v$ . Likewise the observer in S', Sheila, makes measurements to confirm that the reference frame S is moving with velocity $\bf\vec v^{{{\lower1.1pt\hbox{{\ponts /}}}\!\!}}$. It is of course true that $\bf\vec v^{{{\lower1.1pt\hbox{{\ponts /}}}\!\!}}$= -$\bf\vec v$ ; however, I will always use ${{\raise0.8pt\hbox{{\ponts /}}}\!\!}$ to indicate measurements made by the observer in S'. Therefore, we may now discuss the relationship of measurements of position and time as made by the two observers. At least the simplest way to understand these measurements seems to be to suppose that they are made on some object that both observe. It is customary to refer to events when discussing these measurements.

For example, suppose that a car is going down a particular street. At a certain moment, the car is passing a streetsign; that passing is called event 1. Some time later the car passes a pedestrian; that passing we can refer to as event 2. One can see that each event has associated with it a set of coordinates, i.e., a value for $\{x,y,z\}$, as measured relative to some choice of origin, and also some time, t, at which the event occurred. Therefore, we may associate with each event 4 ``coordinates''; event 1 will have coordinates $\{x_1, y_1, z_1, t_1\}$, while event 2 will have coordinates $\{x_2, y_2, z_2, t_2\}$. The usual approach to measurements made by an inertial frame of reference is that the observer in that frame keeps the origin next to her; put differently, if an observer, such as S', is moving, as observed by some other observer, then the origin for S' is also moving. Therefore, it should be clear that the values of these coordinates will be different. For instance, suppose that Sheila has a small pet dragon, sitting next to her, and that S says that Sheila is moving at 10 miles per hour, in the $\hat x$ direction. As our first event, we suppose that both Barney and Sheila are at the same location at some particular time, and that they synchronize their watches then, and both also agree to watch the dragon for a while. Therefore for that event, both observers agree, giving us coordinates for the dragon at that time as $x = 0 = x^{{{\lower1.1pt\hbox{{\ponts /}}}\!\!}}, t = 0 = t^{{{\lower1.1pt\hbox{{\ponts /}}}\!\!}}$. Then after 1 hour S would say that the coordinates of the dragon are x = 10  miles, t = 1  hour, while Sheila, in S', would say that the dragon's coordinates are $x^{{{\lower1.1pt\hbox{{\ponts /}}}\!\!}}= 0, t^{{{\lower1.1pt\hbox{{\ponts /}}}\!\!}}= 1$  hour. (We have ignored the y- and z-coordinates since the motion is in the x-direction!)

The Lorentz transformation equations are the relativistic equations that relate measurements of location and time as made by observers in two different inertial reference frames, as promulgated by Einstein so as to be consistent with the two postulates noted above. We write them so that they relate measurements made by S and S', as above, but restrict ourselves to the case when the coordinate axes have been chosen so that the relative velocity of the two frames is in the x-direction. We may therefore write down the formulae which allow us to determine the values of measurements made in S', once the measurements have been made in S, which are Eqs.(38-20) in your text:

$\parbox{2in}{ \begin{displaymath}\left\{ \begin{array}{rcl} x^{{{\lo... ...wer1.1pt\hbox{{\ponts /}}}\!\!}}& =& z \end{array}\right\}\end{displaymath} }$$\parbox{2.8in}{,\hspace{0.4in}$\hbox{where~~~} \gamma \equiv {1\over \sqrt{1-(v/c)^2}}\ge 1\;.$ }$ $\parbox{1cm}{\begin{eqnarray} \end{eqnarray}}$

It is useful to notice that we could, instead, have asked what were the values of measurements in S , relative to those made in S'. These are easily obtained from Eq.(1) by remembering that the velocity of S , as measured by S', is just -$\bf\vec v$ :

$\parbox{2in}{ \begin{displaymath}\left\{ \begin{array}{rcl} x & = &\... ... z^{{{\lower1.1pt\hbox{{\ponts /}}}\!\!}} \end{array}\right\}\end{displaymath}}$$\parbox{2.8in}{,\hspace{0.4in}$\hbox{where~~~} \gamma = \gamma^{{\raise1.6pt\hbox{{\ponts /}}}\!\!}\equiv {1\over \sqrt{1-(v/c)^2}}\ge 1\;.$ }$ $\parbox{1cm}{\begin{eqnarray} \end{eqnarray}}$

While these equations do not explicitly all appear in your text, the important parts of them, i.e., those relating $\{x,t\}$ and $\{x^{{{\lower1.1pt\hbox{{\ponts /}}}\!\!}},t^{{{\lower1.1pt\hbox{{\ponts /}}}\!\!}}\}$, are given in Table 38-2, on page 933. From quite a different point of view, these are in fact the equations one would obtain if Eqs.(1) were solved for the other set of variables. NOTE:   Most measurements we consider will in fact be measurements of differences in coordinates, i.e., coordinates describing pairs of events. Two very important examples, which we will soon discuss in more detail, are

• a length is the difference of two measurements of location, made at the same time,
• interesting time intervals are, again, the difference between two successive measurements of the time, at the same location.

However, since one could always have arranged to set up one's origin and to re-set one's clock so that the first measurement would have occurred at zero-values of all coordinates, differences in coordinates also satisfy exactly the same transformation equations as we have written above; therefore, we now bring those two sets together, the transformation equations and their inverses, noting that the relevant content is in Table 38-2, on p. 933 of your text:

 $\parbox{1cm}{\begin{eqnarray} \end{eqnarray}}$

I.  Length Measurement and Length Contraction:The measurement of the length of some object by two observers in relative motion leads to the famous Lorentz contraction formula, comparing these two measurements. It is important to note that the English-language meaning of the word length is a measurement of the location of the two ends of ``something'' at exactly the same time. For simplicity, let us agree to measure the length of a desk in Sheila's office, at rest in her reference frame, S'. She locates one corner of the desk at her origin, and then measures the next corner to have $x^{{{\lower1.1pt\hbox{{\ponts /}}}\!\!}}= L^{{\raise1.6pt\hbox{{\ponts /}}}\!\!}$. Both measurements are made at the same time, say $t^{{{\lower1.1pt\hbox{{\ponts /}}}\!\!}}= 0$. Of course, only the difference between the values of $x^{{{\lower1.1pt\hbox{{\ponts /}}}\!\!}}$ at the two corners was important. She resolved this by locating one corner at the origin. On the other hand, she could just as easily have simply measured some value for $x_1^{{{\lower1.1pt\hbox{{\ponts /}}}\!\!}}$ and another for $x_2^{{{\lower1.1pt\hbox{{\ponts /}}}\!\!}}$ and then calculated $x_2^{{{\lower1.1pt\hbox{{\ponts /}}}\!\!}}- x_1^{{{\lower1.1pt\hbox{{\ponts /}}}\!\!}}\equiv \Delta x^{{{\lower1.1pt\hbox{{\ponts /}}}\!\!}}$. We may describe all this by saying that the observer in S' measures $\Delta x^{{{\lower1.1pt\hbox{{\ponts /}}}\!\!}}= L^{{{\lower1.1pt\hbox{{\ponts /}}}\!\!}}$ and $\Delta t^{{{\lower1.1pt\hbox{{\ponts /}}}\!\!}}= 0$! Since the desk is at rest in this frame, this value is also often called L₀, the rest length, or the proper length, of the desk. Now we want to know about Barney's measurement of the length of the desk, in S , we may use the Lorentz transformation equations for differences of coordinates, i.e., Eqs.(3), to calculate this. Firstly, we say that

$$L_0 = L^{{\raise1.6pt\hbox{{\ponts /}}}\!\!}\equiv \Delta x^{... ....6pt\hbox{{\ponts /}}}\!\!}= \gamma(\Delta x + v\Delta t)\quad,$$

but if we want the $\Delta x$ in the equation to be the length, L, in Barney's frame, then we must require that the corresponding $\Delta t = 0$, which leaves us with the following result, which is Eq.(38-13) or Eq.(38-16), on p. 930 and p. 931 of your text, although it should truly be compared with Eq.(38-25), on p. 934, since there it is obtained in the same way I did.

$$\begin{array}{lrcl} &L_0 & = &\gamma L \\ [4pt] \hbox{or, bet... ...}& L & = & {L_0\over\gamma} = \sqrt{1-(v/c)^2}\,L_0 \end{array}$$ (4)

II.  Time Dilation:We now consider the relation between measurements of time interval (or, if you prefer, `age') as made by the two observers. The first thing we note about measurements of age is that they must be applied to the same object. Therefore, let us consider the simplest possible example: Sheila, in S', watches her clock for some time period, $T^{{{\lower1.1pt\hbox{{\ponts /}}}\!\!}}\equiv \Delta t^{{{\lower1.1pt\hbox{{\ponts /}}}\!\!}}$. Since these are measurements of her clock, in her own system, she also measures $\Delta x^{{{\lower1.1pt\hbox{{\ponts /}}}\!\!}} = 0$. In addition, since that clock is at rest in that frame, we will refer to it as the proper time for that pair of events, $\Delta T_0$. We now want to know what time period has passed as measured by Barney, in S ! Referring again to the Lorentz transformation equations, Eqs.(3), we may write that

$$T \equiv \Delta t = \gamma (\Delta t^{{{\lower1.1pt\hbox{{\po... ...}}+ v\Delta x^{{{\lower1.1pt\hbox{{\ponts /}}}\!\!}}/c^2)\quad.$$

However, the two observations Sheila made of the clock had $\Delta x^{{{\lower1.1pt\hbox{{\ponts /}}}\!\!}} = 0$, the requirement for an `age' or time interval of a single object at rest in that frame. Therefore, we have the following result, which is Eq.(38-7), (38-9), or (38-23) in your text:

$$\begin{array}{lrcl} &T \equiv \Delta t = \gamma \Delta t^{{{\... ...ta T & = &\gamma \Delta T_0 = T_0/\sqrt{1-(v/c)^2} \end{array}$$ (5)

III.  Velocity Addition: For a more complicated example, we suppose that both of our standard observers are watching an object that is moving. They might, for instance, be watching a butterfly. I would like to be able to say that S measures the speed of the butterfly to be one quantity, and S' another quantity. However, those two quantities are related by $\bf\vec v$ ; therefore, in order to continue further, we need a transformation law for velocity. The first thing I will do is to note that if I refer to velocity by the symbol $\vec v$, I will be in trouble, since I have already taken over the use of this symbol in order to characterize the relative velocity of our two frames. Therefore, during this entire set of notes, I will use the symbol $\vec u$ to denote the velocity of some object, as measured by the observer in S ; at the same time, I will use the symbol $\vec u^{{\lower1.1pt\hbox{{\ponts /}}}\!\!}$to denote the velocity of that same object, as measured by the observer in S'. Then, we may write that the vector $\vec u$ has, in general, 3 components, given as follows:

$$u_x = \lim_{t\rightarrow 0}{\Delta x\over \Delta t}\;,\quad u... ...;,\quad u_z = \lim_{t\rightarrow 0}{\Delta z\over \Delta t}\;.$$ (6)

Likewise, we have similar expressions for $\vec u^{{\lower1.1pt\hbox{{\ponts /}}}\!\!}$, where all objects of course have primes, including $t^{{\raise0.8pt\hbox{{\ponts /}}}\!\!}$. Remembering that for our approach, the relative velocity of the 2 frames is in the $\hat x$-direction and that the $\hat x^{{\lower1.1pt\hbox{{\ponts /}}}\!\!}$-direction is parallel to the $\hat x$-direction, it is then straightforward to use the Lorentz transformation equations to perform these divisions. However, this will treat the component of $\vec u$ in the $\hat x$-direction in a different way than the other two directions. Therefore, it is also useful to divide $\vec u$ into those parts which are parallel to $\bf\vec v$ , which we denote by $u_{\scriptscriptstyle \parallel}$, and that part which is perpendicular to $\bf\vec v$ , which we denote by $u_\perp$.

$${\vec u}_{\scriptscriptstyle \parallel}\equiv u_x = {\Delta x... ... + \vec v\cdot\vec u^{{\lower1.1pt\hbox{{\ponts /}}}\!\!}/c^2}$$ (7)

$${\vec u}_\perp = \left\{\begin{array}{rcl} u_y = & {\Delta y... ... + \vec v\cdot\vec u^{{\lower1.1pt\hbox{{\ponts /}}}\!\!}/c^2}$$ (8)

Note that the denominator in these expressions comes into them because of the fact that time differences are not measured to have the same numerical values in the two different inertial reference frames! Also note that it is exactly the denominator in the expressions that causes the addition of velocities to never give an answer greater than c!

The inverse of these expressions, i.e., formulae that determine the values of the components of $\vec u^{{\lower1.1pt\hbox{{\ponts /}}}\!\!}$ when one is given the values of $\vec u$ are just determined by changing primes and ``not-primes'' everywhere, and remembering that $\bf\vec v^{{{\lower1.1pt\hbox{{\ponts /}}}\!\!}}$= - $\bf\vec v$ . The parallel part of these equations is then in Eq.(38-28) in your text.

IV. World Velocity and Momentum-Energy:Since the velocity, $\vec u$, has reasonably complicated transformation expressions, people have wanted a quantity which would transform in a simpler way. This desire has led to the introduction of the ``world velocity,'' which transforms in ``the same way'' as the coordinates themselves. To see this fact, we first continue determining the transformation of quantities related to velocity by asking how the quantity $\gamma_u \equiv \left\{1-{u^2\over c^2}\right\}^{-1/2}$ transforms from one reference frame to another. In other words, how is $\gamma_u$ related to $\gamma_{_{u^{\smash{{\scriptscriptstyle {{{\lower1.1pt\hbox{{\ponts /}}}\!\!}}}}}}}$? Beginning with Eqs.(7) and Eqs.(8), we can first calculate $\vec u^2$ in terms of $(\vec u\smash{{}^{{{{\lower1.1pt\hbox{{\ponts /}}}\!\!}}}})^2$ and then go on to determine the transformation of $\gamma$:

$$\begin{eqnarray*}{\vec u}^2 = \vec u_{\scriptscriptstyle \parallel}^2 + \vec u_... ...\perp^{{{\lower1.1pt\hbox{{\ponts /}}}\!\!}})^2(v^2/c^2)\} \quad.\end{eqnarray*}$$

which allows us to show that

$$\begin{eqnarray*}\sqrt{1-{u^2\over c^2}} & = \{1+{\vec v\cdot\vec u^{\smash{{\sc... ...\left\{1-{u^2\over c^2}\right\}\left\{1 - {v^2\over c^2}\right\} \end{eqnarray*}$$

or, simply substituting in definitions of $\gamma$, we have the following nice result:

$$\gamma_u = \gamma_v\gamma_{_{u^{\smash{{\scriptscriptstyle {{... ...{{{\lower1.1pt\hbox{{\ponts /}}}\!\!}}\over c^2}\biggr\}\quad.$$ (9)

Since the factor in the numerator, for the transformation of $\gamma_u$, is the same as the factor in the denominator for the transformation equations for the various components of $\vec u$, it follows that the product of $\gamma_u$ and $\vec u$ will have very much nicer transformation properties. This is more easily written down if we define the world velocity as

$$\vec w \quad\equiv \quad\gamma_u\vec u \quad = \quad {\vec u\over \sqrt{1-u^2/c^2}}\quad,\eqno(10)$$

from whence we can calculate that

$$\vec w_{\scriptscriptstyle \parallel}= \gamma_v(\vec w_{\scri... ... \vec w_\perp = w_\perp^{{{\lower1.1pt\hbox{{\ponts /}}}\!\!}}$$ (11)

These look suspiciously similar to the behavior of the parallel and perpendicular components of location, except for the rather troublesome `extra term' in the transformation equation for $\vec w_{\scriptscriptstyle \parallel}$. The usual approach to dealing with the transformation equation for x, which can be said to have an `extra term' in it, which is proportional to the time, t, is to create the notion of

$$\hbox{a 4-dimensional vector,~~}\qquad \widetilde x\,\equiv (x,y,z,ct) \equiv (\vec r , t)\;.$$ (12)

Then the various components of the 4-vector $\widetilde x$ simply transform among themselves as one considers observations made in different reference frames. The next obvious thing to do is to create a new `4-vector' out of $\vec w$ by appending to it a fourth component to take care of that ``troublesome extra term'' in the equation above:

$$\hbox{We define }\ \quad \vec w \equiv \gamma_u\,{\vec u}\;,\... ... c\quad\hbox{and}\quad \widetilde w \equiv (\vec w , w^4)\quad,$$ (13)

so that we may re-write the transformation equations for $\vec w$ and, as well, append the transformation equation for $\gamma_u$ into the set:

$\parbox{3in}{\begin{eqnarray*} w_{\scriptscriptstyle \parallel}& = ... ...scriptstyle \parallel}^{{{\lower1.1pt\hbox{{\ponts /}}}\!\!}}) \end{eqnarray*}}$ $\parbox{1cm}{\begin{eqnarray} \end{eqnarray}}$

V.  Momentum and Energy

Having generalized the notions of location (and time) and also velocity to be consistent with the postulates of special relativity, one might now decide to go ahead and try to also generalize the notions of momentum and energy. The most important properties of these quantities is that they are conserved during interactions of particles: More precisely, the 3-dimensional vector, linear momentum, is conserved when there are no exterior forces, and energy is conserved when the collision is totally elastic. Therefore, it is very important to preserve these properties when new definitions are made, if they are needed. As it turns out, the discussion above, of world velocity is extremely useful in this regard. We follow your text, in its Eqs.(38-29) and (38-45), and simply define the 3-dimensional linear momentum, $\vec p$, and the scalar energy, E, as follows:

$$\vec p \equiv m\vec w = \gamma_u m\vec u\quad,\quad E \equiv mc w^4 = \gamma_u mc^2\;,$$ (15)

or, in the notation of 4-vectors:

$$\widetilde p \equiv m\widetilde w = m\pmatrix{\vec p\cr E/c\cr} = \pmatrix{\gamma_u m\vec u\cr \gamma_u mc}\;.$$ (16)

Then, via Eqs.(13), we have the desired simple form of the Lorentz transformation equations and also the property that $\widetilde p$ is conserved in collisions or interactions without external forces:

$\parbox{3in}{\begin{eqnarray*} p_{\scriptscriptstyle \parallel}& = &... ... \vec v\cdot\vec p^{\,{{{\lower1.1pt\hbox{{\ponts /}}}\!\!}}}) \end{eqnarray*}}$  $\parbox{1cm}{\begin{eqnarray} \end{eqnarray}}$

VI.  Kinetic Energy, Mass Energy, and WorkThe relativistic energy, E, defined as the 4-th component of the momentum-energy 4-vector, $\widetilde p$, includes the usual notion of kinetic energy, generalizing it to a form that is in agreement with experiment for velocities close to the velocity of light, and also contains an additional term, the mass energy, which is a constant unless the mass changes. One approach to this is to look at its behavior for velocities that are relatively small when compared to c:

$\parbox{5in}{\begin{eqnarray*} E = \gamma_u mc^2 & = {mc^2\over \sq... ...2} m\,u^2 + {\textstyle {3\over 8}} m\,u^4/c^2 + \ldots \quad. \end{eqnarray*}}$  $\parbox{1cm}{\begin{eqnarray} \end{eqnarray}}$

The second term in the expansion above is the `old,' classical term for the kinetic energy of a particle of mass m and speed u. This relativistic version tells us that there are additional terms which must be added when the speed is sufficiently large, and also implies that there is a rest-mass energy which a particle has even if it is not moving. That this energy is real is approved, experimentally, by the decay of particles into other, lighter particles, when mass is NOT conserved, but relativistic energy is in fact conserved.

The second interesting notion that can be obtained from this form of the equation is to consider the work required to change the speed of a particle. Letting KE(u) be the amount of work required to boost a particle from rest up to some speed, u; this amount of work we will refer to as the kinetic energy of the particle:

$$KE(u) = m c^2\{\gamma_u - 1\}\;.$$ (19)

We may then ask about the amount of work, $\Delta W (u)$, required to boost the speed yet further, from u to $u+\Delta u$:

$\parbox{5in}{\begin{eqnarray*} \Delta W (u) & = mc^2\{\gamma_{u+\Del... ...\\ & = \gamma^3 m\,u\,\Delta u + \ \hbox{O}(\Delta u)^2\quad. \end{eqnarray*}}$  $\parbox{1cm}{\begin{eqnarray} \end{eqnarray}}$

Since $\gamma \longrightarrow \infty$ as $u \longrightarrow c$, we see that the amount of work required to boost the speed of a particle of mass m increases to infinity as the speed approaches c. From this point of view, the reason a particle can never travel at speed c is that, beginning from rest, it would be necessary to perform an infinite amount of work on it before that would happen. It is worthwhile doing some simple calculations concerning this fact, perhaps. Suppose we have a particle of mass m = .5 MeV/c², which is approximately the mass of the electron in these useful units. (Recall that MeV means million electron Volts.) When u=0, its value of E is .511 MeV. We now do work on it to increase its speed to u = 0.9c, corresponding to $\gamma_{.9c} = 2.294$. By the previous equation this requires an amount of work, $W = m\,c^2\{\gamma_u -1\} = .511(2.294-1) = .661$ MeV, so that its total energy, E = .661 + .511 = 1.172 MeV. Now, we increase its speed to u = 0.99c, a net change of only $\Delta u = .09c$, corresponding to $\gamma_{.99c} = 7.089$. This requires an amount of work $W = m\,c^2\{\gamma_{.99c} - \gamma_{.90c}\} = .511\,(7.089 - 2.294) = 2.245$ Mev, so that it took approximately twice as much additional energy, i.e. work, to create a $\Delta u = .09c$, on top of u=.9c, as it did to get to .9c in the first place. Now we go ahead and consider going to u=.999c, for a $\Delta u = .009c$. For u=.999c, we have $\gamma_{.999c} = 22.37$. Therefore, the extra work required to get to u=.999c, starting from u=.99c, is $W= .511\,(22.37-7.089) = 7.81$ MeV. For the last try, let us go to u = .99999c, corresponding to $\gamma_{.99999c} = 223.61$, and an additional work of $W = .511\,(223.61-22.37) = 102.8$ MeV. You can see that as we get closer and closer to the speed of c, it begins to cost more and more energy, with the result that as one goes ever closer, it costs ever more work. Just to finish the discussion, we now try to go to $u = (1-10^{-12})\,c$, corresponding to $\gamma = 707,115$, which then costs an additional work of 361,283 MeV!