Selected Questions (And Answers) From All Experts Astronomy Forum (Big Bang, proper time, quantum cosmology etc.)

The following were culled from cosmology questions asked on the All Experts Astronomy forum over 2010--2012:


1-What is the evidence for the Big Bang and how was it discovered?

The evidence for the Big Bang is the 2.7 Kelvin degree, cosmic background radiation. The key aspect that makes it evidence is that it has the same intensity in all directions- wherever one observes in the sky. No ordinary radiative or local explosive source (e.g. supernova) would produce such a characteristic. The only feasible explanation to account for it would be a vast “explosion” occurring in the distant past and affecting the entire universe. Its entire volume.

The isotropic (all directions) radiation was discovered during a sky survey at the Bell Laboratories in 1965, by Arno Penzias and Robert Wilson. They had pointed their radio telescope in different directions in the sky and received the same “buzz” no matter where they observed. After rigorous data reduction (removal of data detritus and effects of bat guano), plus later confirmation of the “universal buzz” - it was later identified by Robert Dicke as the cosmic microwave background radiation (CMB), the radio remnant of the Big Bang, they realized they’d discovered the relic radiation left over from the Big Bang.

In 1978 the pair received the Nobel Prize in Physics for their discovery

2- If there was a Big Bang can we say the universe had a beginning, or not? What came before the Big Bang?

Asking ‘what came before the Big Bang’ is like asking ‘what’s north of the north pole?’ It’s the kind of question that at first glance seems logical, but really isn’t. The key thing here is that the time continuum we use to measure instants or epochs starts at the Big Bang, using the Hubble constant. Going to epochs before the Big Bang isn’t feasible because there is no temporal template to go by.

Though we can use the current Hubble constant to measure the proper time since the Big Bang as 13.7 billion years we can’t go any further than that. Or, into any “negative times” – as defined “before the Big Bang”. The reason is that so far there is no physically operational meaning to these.

3- You mentioned “proper time” – what’s that?

Proper time   τ   is defined in terms of the relativistic factor  (from special relativity):

g  =   [(1 - v  ²/c²)  ^-1/2]  

where v is a specific velocity- say of an observer’s frame, c is the speed of light.

Then the proper time is:   dτ   = dt/ g

Reversing the above, we can say that when the (relativistic) clock advances by increment dτ   then  the clock time that elapses on Earth is given by:

dt  =    g   dτ  

Why use proper time? Basically, because normal time measures are unreliable and subjective owing to relativity. For example, even saying or asserting the energy is greater or less at one time (t2) than time (t1) is impossible since energy (like mass, time, length) varies depending on the observer’s frame.

For example, consider the energy of an object moving at velocity v in the simple diagram below:


But, to the observer O moving in a reference frame with the same velocity, the object appears at rest so its kinetic energy = 0. In more technical consideration, on the basis of the Lorentz transformation (google!) the energy E’ in a new frame will depend on the energy E in an older frame according to:

E’ =  g _o (E –  v _o  p _x )

Where  v _o is the velocity and  p _x   the momentum in direction (x).

The factor  g _o   =   1  / [(1 - v _o ²/c²)^½]  


The point is that the energy in the new frame E'  depends on the momentum in the old frame, p _x . If this momentum is not conserved, then energy won’t be conserved in the new frame.

Note here that the relation of the new p' _x momentum  to the old,  p _x is:

p' _x    =  g _o (p _x   –  v _o  E/   c² )

Bottom line, unless one employs corrections using proper time to make energy estimates or statements, he will end up talking nonsense.

For example, let’s say you are looking at an object receding at a velocity of 0.5 c, and assert based on careful ordinary time measurements that it emerged 1 billion years after the Big Bang. Or – 12.7 billion year removed from your observation. But how does this compare to the proper time?

We have: dτ  = dt/ g

Where dt = 12.7 billion yrs.

g  = [1 – (0.5c/c) ²]  
 
 = 1/  Ö[1 – (0.25)] = 1/  Ö[0.75] = 1.155

Then the proper time interval dτ   is: (12.7 x x 10 ⁹  yr)/1.155 = 1.1 x x 10¹⁰  yr

Or 11 billion years not 12.7 billion.

This obviously also means that one cannot say or assert that from the Big Bang there has been “an infinite” amount of time, or instants. No, there hasn’t, and one can do the computation for the actual time interval using the proper time format.


4- What is meant by Quantum Cosmology and how does it work? Can it explain the Big Bang?

Quantum cosmology means that we use terms, conditions and parameters that are peculiar to the quantum level – for example quantum variables – in examining how they might work in explaining aspects of cosmology and yes, origins.

Quantum cosmology also is not for the faint of heart, or those who haven’t done a lot of physics – including taking quantum physics courses. In other words, one can’t simply approach its basis or claims using common sense or even ordinary logic. I believe it was Richard Feynman in one of the prefaces to his lectures who wrote: “Don’t even try to go down the rabbit hole of understanding quantum mechanics, or you will be lost forever.”

What he meant by that is that we don’t approach quantum formulations on the basis of trying to understand them, like one might wish to understand how an electronic amplifier works. We approach them from the mathematics we use, and let those math results lead the way.

At least one theory of the origin of the cosmos uses quantum cosmology in what we call “quantum bootstrapping”. This means, formulating a quantum fluctuation which could have incepted the cosmos via the Big Bang.

The starting point for these sorts of theories actually resides in a real world phenomenon. Particle production is one- wherein particles are created then quickly destroyed with energy release.

The typical pi meson (call it p ) lasts 10 ^-16 sec then vanishes yielding two gamma ray photons in its wake,viz.

 p    -> g   +   g

Thus, rest energy is real energy and is capable of doing work. In the case of the pion above, the total mass 2.4 x 10^-28 kg, is converted to electromagnetic energy.

The amount of the energy can be estimated using the Heisenberg Uncertainty Principle in the energy-time format:

D E D t   <  ħ  (where   ħ   is the  modified Planck constant: 

 h/ 2p =   1.05 x 10^-34 J-s

Thus,  if the time uncertainty for disappearance of the pi meson is:  D E = 10^-16 s

Then the energy available that comes off is:

   D E =    ħ  / D t   = 1.054 x 10 ^-18  J = 6.5 eV

In the same manner, a number of theoreticians (e.g. T. Padmanabhan) have speculated the universe could have emerged like the pi meson using the same basis for energy arising out of “nothing”.

Padmanabhan’s full paper: Universe Before Planck Time – A Quantum Gravity Model, can be found in Physical Review D, Vol. 28, No. 4, p. 756.

Solutions To Special Relativity Problems

Alas, there were no submissions of solutions to the special relativity problems - for award of a  $50 book prize.  Anyway, the solutions are given below:

1) Given x' =  1/a (x - vt) and t' = 1/a (t - vx/c²),

Then: x' = x/a - vt/a and t' = t/a - vx/ac²

and: x' + vt/a = x/a and t' + vx/ ac²  = t/a

so:

x = a(x' + vt/a) and t' = a(t' + vx/ ac² )

finally: x = a(x' + vt) and t = a(t' + vx/c²)


2) We have: x' = 60m, t' = 8 x 10^-8 s and y' = y, z' = z

v = 0.6c = 1.8 x 10⁸ m/s 

Then:

x = [60m + (1.8 x 10⁸ m/s)( 8 x 10^-8 s)]/ (0.64)^½

x = [60m + 14.4m]/ 0.8 = 74.4m/0.8 = 93m

and t =

[(8 x 10^-8 s)+ (1.8 x 10⁸ ms^-1)(60m)/(3 x 10⁸ ms^-1)²/0.8

t = 2.5 x 10^-7 s/ 0.8 = 2.33 x 10^-7 s


The space time coordinates are: (93 m, 2.33 x 10^-7 s)


3) The problem requires no relative motion defined specifically in the x-direction so the equations:

t = t' + x'v/c²/(1 - v²/c²)^½

and

t' = t - xv/c²/(1 - v²/c²)^½

are immediately simplified by the terms in x becoming zero, so:

t = t'/(1 - v²/c²)^½

and

t' = t /(1 - v²/c²)^½

Here: t = time passage on Earth clock

and t' = time passage on astronaut's clock

For t = 1 Earth year = 365 ¼ days:

t' = (365 ¼ days)/ [1 - (0.9c)²/c²]^½

= (365 ¼ days)/(1 - 0.81)^½

t' = (365 ¼ days)/0.436 = 837.7 days

    This is the time elapsed on the astronaut's clock when the Earth has made one revolution equal to 365 ¼ days. In other words, each of his days is roughly equal to 2.29 Earth days. Hence, his clock is obviously running slower than the Earth clock.

    A more intuitive way to look at the result would be in terms of the time transformation: 

t = t'/ (1 - v²/c²)^½  

-  or asking how much time elapses on an Earth clock for each year elapsed on the astronaut's? The result will be found to be 2.29 years, or in other words his Earth counterparts are aging 2.29x faster.

Selected Questions- Answers From All Experts Astronomy Forum (Proper Time)

Question:  While reading a book on special relativity I came across the term proper time. Can you explain what it is, how used?

Answer:

Proper time is a time defined in terms of the relativistic interval (from special relativity):

g   =        [1 – (v/c) ²] ^-1/2

where v is a specific velocity- say of an observer’s frame, c is the speed of light.

Then the proper time is: dt(p) = dt/ g

Why use proper time? Basically, because normal time measures are unreliable and subjective owing to relativity. For example, even saying or asserting the energy is greater or less at one time (t2) than time (t1) is impossible since energy (like mass, time, length) varies depending on the observer’s frame.

For example, consider the energy of an object (denoted by the Greek letter sigma) moving at velocity v in the simple diagram below:

But, to the observer O moving in a reference frame with the same velocity, the object appears at rest so its kinetic energy   KE= 0. In more technical consideration, on the basis of the Lorentz transformation (google!) the energy E’ in a new frame will depend on the energy E in an older frame according to:

E’ =    g _o  (E –   v _o   p _x )

Where   v _o   is the velocity and p _x  the momentum in direction (x).

The factor  g _o   =      m _o v/  (1 - v²/ c² ) ^-1/2

and   m _o   is the rest mass.

The point is that the energy in the new frame E depends on the momentum in the old frame, p _x. If this momentum is not conserved, then energy won’t be conserved in the new frame.

Bottom line, unless one employs corrections using proper time to make energy estimates or statements, he will end up talking or writing nonsense.

For example, let’s say you are looking at an object receding at a velocity of 0.5 c, and assert based on careful ordinary time measurements that it emerged 1 billion years after the Big Bang. Or  12.7 billion year removed from your observation. How does this compare to the proper time?

We have: dt(p) = dt/ g

Using a finite difference equation:

 D t(p) =  D t / g  

  D t= 12.7 billion yrs.  where:

g  = [1 – (0.5c/c)² ]  ^-1/2   = [1 – (0.25)]  ^-1/2   = [0.75]  ^-1/2  = 1.155

Then the proper time interval (finite difference result)  is  D t (p) = 

  (12.7 x 10 ⁹ )/1.155 = 1.1 x 10 ¹⁰       yrs.

Or 11 billion years not 12.7 billion.

This obviously also means that one cannot say or assert that from the Big Bang there has been “an infinite” amount of time, or instants. No, there hasn’t. One needs to do the computation for the actual time interval using the proper time format.

Introduction to Special Relativity (3)

Let’s now consider the implications of the Lorentz-Einstein transformation and refer again to the original diagram (Fig. 1). Recall:

(11)

x = x' + vt'/(1 - v^2/c^2)^½

and

x' = x - vt/ (1 - v^2/c^2)^½


(12) y = y' and y' = y

(13) z = z' and z' = z

(14) t = t' + x'v/c^2/ (1 - v^2/c^2)^½

and

t' = t - xv/c^2/(1 - v^2/c^2)^½

Now, supposed, there’s a clock located in system S’ and an observer in system S sees this clock moving with velocity v. At any time t, the position of the clock with reference to the S –observer’s system is given by x = vt. If the length of time between two ticks is of this clock is T’ in the S’ system, the transformation to the S system (assuming x’ = 0) makes the time interval T appear to be:

T = T’/ (1 - v^2/c^2)^½

Since T > T’ then it seems to Observer S his clock is running slower than it does to S’. This applies not only to the clock but to all physical processes that depend on time, e.g. the vibrations of electrons in atoms, rates of chemical reactions, biological processes (heart beat, respiration) etc. In effect it appears to the observer in S that his counterpart in S’ is living at a slower rate than he is. However, to the observer in S’ it is the observer in S who seems to be living at a slower rate.

This is a paradoxical result but one which can’t be escaped if we carry special relativity to its logical conclusion as Einstein did. So long as the two systems have a constant relative motion, we cannot say that one is moving and the other standing still, or that one’s clock is moving slowly and the other quickly.

A number of subtle consequences arise out of this. For one thing, the notion of simultaneous events, i.e. for observers in two different reference frames, is no longer tenable. Einstein, in fact, seems to have been the first human being to recognize that simultaneity between two events is a provincial illusion. In his landmark (1905) paper on special relativity (“Does the Inertia of a Body Depend on Its Energy Content?’) he remarked:

“We have to take into account that all our judgments in which time plays a part are always judgments of simultaneous events. If, for instance, I say ‘That train arrives here at 7 o’clock’ I mean something like this: ‘The pointing of the small hand of my watch to seven o’clock and the arrival of the train are simultaneous events.”

Einstein agreed that observing such simultaneity was a reasonably accurate way for a person holding a watch to tell the time of an event happening next to the watch, but insisted that on principle the method couldn’t be relied upon for timing an event far away from the watch, especially by someone moving in relation to the other things involved.

As an illustrative example, consider Fig.2 which features an enormously long train (defined as system S’) with an observer ‘Bob’ at its very center. (E.g. if the total length of the train is L, he is at L/2). Two bolts of lightning then strike the extremities of the train, and do so simultaneously according to Bob’s watch. Bill, however, is situated on a stationary platform as shown – defined in system S- at the exact midpoint between the strikes and swears that he recorded the rear flash a sizeable fraction of a second earlier than the forward flash.

Who is correct? A non-relativistic thinker would undoubtedly incline towards the view of Bill, the stationary observer. The argument here is that Bob’s simultaneity is only apparent, not actual since the train would have carried him closer to the forward flash by the time he received their simultaneous light from points that were, by then, no longer equidistant from him.

For a relativist, however, it is just as true to say the train was standing still and the ground sped by it, so the ground observer (Bill) could have been carried beyond the the equidistant point as easily as Bob.

In Einstein’s cosmic viewpoint, the lightning flashes (and their midpoint) are free to be considered an integral part of the train OR Earth, OR any other frame of reference. Each and every observer may select his own viewpoint. Any viewpoint is true and right, none is wrong. This example embodies the essence of relativity, that is, that time is truly relative and not absolute as Newton had believed.

As an illustration of a type of experiment that attempts to reckon simultaneity, consider Fig. 3. Here a light pulse is directed at a moving mirror M traveling with velocity v to the right. By the time the pulse reaches the mirror it has moved a distance v (delta t)/2 horizontally.

According to an outside observer, if light is to hit the mirror it must depart at an angle to the vertical as shown.

By Pythagoras’ theorem (square of the hypoteneuse equals the sum of the squares of the other 2 sides):

(c (delta t)/2)^2 = (v (delta t)/2)^2 + d^2

This implies:

delta t = 2d/(c^2 – v^2)^ ½ = 2d/ c[1 - v^2/c^2]^ ½


But, for the stationary observer:

delta t’ = 2d/ c

which implies:

delta t = delta t’/ [1 - v^2/c^2]^ ½

We call delta t’ the “proper time” or that time interval between two events as measured by an observer who sees the events at the same place. It’s always the time measured by a single clock at rest in the frame.


Example Problem:

The period T of a pendulum is measured to be T= 3.0 s in the inertial frame of reference of the pendulum. What is the period of the pendulum when measured by an observer moving at a sped of 0.95c with respect to the pendulum?

We note that the time measurement taken in the inertial frame is the proper time, and this is t' = 3.0 s.

Then we need to obtain t, for which:

t = t’/ [1 - v^2/c^2]^ ½


where v = 0.95c, so:

t = 3.0 s/ [1 - (0.95c)^2/c^2]^ ½ = (3.0s) 1/ [0.0975]^ ½

and t = (3.0 s)(3.2) = 9.6s


Other Problems for the Bold and Adventurous:

1) With what speed would a clock have to be moving to run at a rate that is one half the rate of a clock at rest?

2) An atomic clock is placed on a Jumbo Jet. The clock measures a time interval of 3600 s when the jet is oving at v = 300 m/s. What corresponding time would an identical clock left on the ground measure? (Hint: whenever v << c (e.g. v/c << 1), we have 1 + v^2/2c^2 and not [1 - v^2/c^2]^1/2)

3) A muon formed high in the Earth's atmosphere travels at v = 0.99c for a distance of 4.6 km before it decays into an electron, a neutrino and an anti-neutrino.

a) How long does the muon survive as measured in its rest frame?

b) How far does the muon travel as measured in its frame?

4) The average lifetime of a pi meson in its own frame of reference is 2.6 x 10^-8 s. If the meson moves with v = 0.95c, what is its mean lifetime as measured by an observer on Earth?

An Introduction to Special Relativity (2)

In a last ditch efort to satisfactorily explain the riddle of the null result (from the Michelson-Morley experiment), a fantastic idea was put forward by George F. Fitzgerald in 1890. Using the analogy of a rubber ball which is deformed upon striking a wall, Fitzgerald conceived that the ether would distort matter.

This distortion would take the form of a contraction of length in the direction of the motion through the ether. Such a contraction would explain the null result of the Michelson -Morley experiment. That is, the arm L, of the apparatus, moving against the ether would be shortened by "ether pressure" just enough to compensate for the slowing down of light by the ether wind.

A similar but more mathematical theory was worked out by the physicist Hendrik A. Lorentz, who expressed the length contraction in the direction of motion by:

L' = L[1 - v^2/c^2]^½

This famous equation became known as the "Lorentz-Fitzgerald contraction".

Understandably, the hypothesis of Lorentz and Fitzgerald gradually gained general acceptance, except for a Swiss Patent clerk, who refused to be deluded by such a contrived idea (based as it was on absolute motion). The clerk's name was Albert Einstein, born in Ulm, Bavaria, in 1879.

Einstein had graduated as a physics major, but he managed to produce such a poor impression as a teacher that he was dismissed from three teaching jobs in succession. Having been reduced to a 'hand-to-mouth' existence, he was lucky to find a job processing patent applications in Bern, Switzerland. Fortunately, he also has a leisurely schedule that permitted him to while away a good many hours contemplating space, time and energy. (A good thing his time was nearly a century ago, otherwise - today- he'd be accused of "gold bricking"!)

After much thought, Einstein was forced to conclude that motion is never observable as motion with respect to spact and that there is no basis for the introduction of "absolute motion". In Einstein's mind, the only kind of motion was relative rest perceived from different viewpoints (e.g. "reference frames"). Einstein called this insight the Principle of Relativity. According to Einstein's Principle of Relativity: "All the laws of physics are the same in all inertial reference frames."

It's instructive at this point to commence a quantitative approach to see exactly how Einstein reasoned. This necessitates we first get accustomed to the idea of a coordinate system. Basically, the purpose of any coordinate system is to enable us to identify a particular point in space. The standard procedure (for 3-dimensional space) is to take three mutually perpendicular axes with coordinates in x, y and z, representing distanced to where the axes meet at the origin, or 0. In special relativity such coordinate systems are typically designated: S, S', S" etc. or alternatively, S1, S2, S3.

In Fig. 1 we have two coordinate systems, S and S', with S having coordinate x, y and S' having coordinates x', y'. (We confine the systems to 2 dimensions here for initial simplicity.) Thus, S' has its origin at 0' and S at 0. In the figure, S' is shown moving with constant velocity v in the x - x' direction (which obviously coincide for S and S').

Keep in mind though we arbitrarily assigned S' as moving this depends on which coordinate system is taken to be at rest. An observer attached to system S may very well consider himself moving and S' stationary. This would resemble the well-known example of a passenger in a stationary train observing a train parallel to his through his window and deducing he is moving, though it is actually the train parallel to his. The key point is that the systems S and S' have a constant relative velocity. Such coordinate systems occupy a special place in relativity and are called "inertial reference frames" or "inertial coordinate systems". Their primary feature is that they lack and acceleration of one to the other.

Now, think about this carefully: if the origins O and O' coincide at time t = 0, and we are observers in S', then we will see 0' move along OX with velocity v. Then the two sets of coordinates, representing the same point (in S and S') are related by:

x' = x - vt and y' = y

If we chose we could append a 3rd axes (z) i.e. coming out of the page, and have also:

z' = z

The preceding equations describe the "Galilean transformation". From these it is very easy to obtain velocities, and we need only differentiate with respect to t:

E.g.

dx'/dt = dx/dt = v

dy'/dt = dy/dt

dz'/dt = dz/dt

(Note here that never once did we write t' = t since the notion of an absolute, universal time was the very cornerstone of Newtonian theory!)

We now start with the preceding transformations and see how Einstein reasoned. Reference may be made to Fig. 2 which displays a three-dimensional perspective of the relative motion and hence is a bit more complicated. Our aim in using it is to find an alternative to the Galilean transformations - which obviously can't be correct for all situations.

For example, according to the Galilean transformations, a pulse o flight sent out from S' would move with the velocity c, the speed of light, as measured from S', but with the velocity (c + v) measured from S. But this directly contradicts the demopnstrated fact (Michelson-Morley) that the speed of light is always constant when computed from one reference frame relative to another.

Einstein thus began afresh by using not only the Principle of Relativity (just stated, i.e. the laws of physics are the same in all inertial reference frames)) but also:

The speed of light is always found to have the same value no matter what the motion for the source or the observer.

From these two postulates, Einstein deduced a number of surprising results which would have been totally unacceptable to a more conservative mind.

Start then with the two systems depicted in Fig. 2 which are coincident in space at at the instant a flash bulb (say) is set off when the origins coincide. The observer S sees S' moving in the x-direction with the velocity v and observer S' sees S moving with the same velocity in the opposite (-x) direction.

Both observers see the flash of light travel away from the origin with the same velocity c. The distance the light travels in the S system during time t, is ct. This must be true in any direction. If the light spreads out equally in all directions, then by the end of time t it has extended to fill a sphere whose radius is r, and we can write:

(1) r^2 = x^2 + y^2 + z^2 = c^2 t^2

The exact same equation holds in the S' system since the flash was set off when O coincided with O. Thus, while the 2 systems are separating from each other, the observer in S' also sees the light fill a spherical shell in his own system, so writes (for r'):

(2) r'^2 = x'^2 + y'^2 + z'^2 = c^2 t'^2

Note the velocity of light is the only thing that's the same in both systems. The two preceding equations thus describe a point lying on the spherical shell. Even though they describe the same point in space, the observer in S sees the point at position (x, y,z, t) abd the observer S' at position (x', y', z', t').


Subtracting equation (2) from equation (1) and transposing terms:

x^2 + y^2 + z^2 - c^2 t^2 = x'^2 + y'^2 + z'^2 - c^2 t'^2


We now look for a transformation similar to the Galilean transformation, but which will allow c to be the same in both S and S'. Since the y and z coordinates of the position are not affected by the motion in the x-direction we can say y' = y and z' = z. For the x-coordinate, we try a transformation of the form: x = a(x' + vt') and x = a(x - vt), where a is an invariant determined by the two fundamental postulates (i.e. the same quantity is used in going from x to x' as from x' to x).

Further, we expect a to depend on the velocity v in such a way that it becomes equal to 1 when v becomes very small compared with the speed of light. When this happens, the x and x' transformations become the same as the ordinary Galilean transformations.

So we begin by using x = a(x' + vt') to solve for t' and obtain:

(3) t' = 1/v (x/a - x')

For x' above, we now insert the value for x' (e.g. x' = a(x- vt)):

(4) t' = 1/v(x/a - ax - avt) = at - x^2(a^2 -1)/ va

Similarly, we find for t:

(5) t = -at' + x'(a^2 - 1)/ va


If we now substitute x' = a(x - vt) and equation (4) into the right hand side of equation (2), we obtain:

(6) x^2 + y^2 + z^2 -c^2t^2 = a^2(x - vt)^2 + y^2 + z^2

= c^2[at - x/v (a^2 - 1)^2/a]

Now, re-arrange terms and cancel the z and y terms, which are the same on both sides of the equation, to get:

(7) x^2 - c^t^2 = [a^2 - (a^2 - 1)c^2/ a^2v^2]x^2

+ 2[(a^2 - 1)c^2/v^2 - a^2]xvt - (c^2 - v^2)a^2t^2

If the preceding is to hold true for any value of x and t, each term on the left side must equal each term on the right. Since there are no terms with the combination xt on the left, the xt term on the right must be zero. This means:

(8) (a^2 - 1)c^2/v^2 - a^2 = 0

Solving for a:

(9) a^2 = 1/ (1 - v^2/c^2) and a = 1/ (1 - v^2/c^2)^½

Finally:

(10) (a^2 - 1)/a = v^2/c^2/ (1 - v^2/c^2)^½

Substituting the preceding into our x, x' transformation equations and equation (3), we arrive at the following transformations to replace the Galilean:

(11)

x = x' + vt'/(1 - v^2/c^2)^½

and

x' = x - vt/ (1 - v^2/c^2)^½


(12) y = y' and y' = y

(13) z = z' and z' = z

(14) t = t' + x'v/c^2/ (1 - v^2/c^2)^½

and

t' = t - xv/c^2/(1 - v^2/c^2)^½


Equations (11)- (14) are known as the Lorentz transformation or the Lorentz-Einstein transformation.

Note the important feature is that the time must be given as well as the position, because the respective clocks in S and S' will cease to read identical times after they have parted from one another. This is the significance of (14) in the above set. Also, the fact that time is given the same importance as space (i.e. as another dimension) shows there's nothing special or mystical about "the fourth dimension".

Problems:

1- Given that x' = 1/a (x - vt) and t' = 1/a (t - vx/c^2), derive similar equations for x and t in terms of x' and t'.
(Recall: 1/a = (1 - v^2/c^2)^½)

2- An event in space-time occurs at x' = 60 m, t = 8 x 10^-8 s, in a frame S' (y' = 0, z' = 0). The frame S' has a velocity of 0.6c along the x-direction with respect to a frame S. The origins O and O' coincide at time t = t' = 0. Find the space-time coordinates of the event in S.

3- Suppose an astronaut is traveling at 0.9c in a space ship with respect to the Earth. How long a time interval will his clock indicate when the Earth has revolved once around the Sun? (Take the duration of one standard revolution of Earth around the Sun to be 365 ¼ days.)