The Lorentz transformations.

Special relativity (part three): an arithmetic example of the loss of simultaneous time between observers in relative motion comparable to light speed.

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Orbitsville.

Lorentz addition and subtraction of velocities.

Lorentz addition and subtraction of times.

Lorentz addition and subtraction of distances.

The Interval.

Kinematics and dynamics.


Orbitsville.

Suppose some advanced race created Orbitsville, an idea taken up by science fiction writers such as Bob Shaw. Freeman Dyson suggested that all but a tiny proportion of a suns energy is wasted in nurturing intelligent life on the odd planet orbiting it. A vastly more efficient way would be to engineer an egg-shaped surround at a distance of the earths elliptic orbit, about eight minutes light-speed distance from the sun. That way, life could exist on the entire inside surface of the egg (or rather an ellipsoid and not too different from a sphere). This concept is known as a Dyson sphere.

(Orbitsville would have to be manoevrable from collisions with stellar matter. It would be a mother ship to other space-ships.)

Imagine Orbitsville and a space-ship passing by it (or being passed by it, depending on your point of view) with a uniform velocity, at a significant fraction of light speed. We can use this scenario to illustrate Einstein demonstration that two observers in relative motion cannot agree on the time of an event: in the special theory of relativity, simultaneity is lost.

The sun is in the center of Orbitsville (we suppose, rather than at one of the two focuses of Orbitsvilles elliptic hull). The climate on its narrower part (the minor axis) will be warmer than on the length-wise direction (the major axis). From those living on the inside surface, observers are stationed at each end of the major axis. Suppose the engineers of Orbitsville have created a night-simulating mechanism, of eclipsing satellites or whatever, such that dawn breaks at exactly the same time for the observers at the far ends of Orbitsville. For instance, internal "astronomers" on the minor axis of Orbitsville might see a simultaneous swathe of light upon either end of the major axis.

Say, both ends of Orbitsville are about eighty-nine and one-third million miles from the sun. The speed of light is about 670 million miles per hour. Therefore, the internal astronomers will measure the sunlight taking {(89 1/3)/670} = 2/15 of an hour or eight minutes, to reach either end of Orbitsville.
(Light takes about eight minutes to reach the earth from the sun, and the distance between them, typically quoted, is about ninety-three million miles.)

Orbitsville might have a transparent band along the major axis of its shell. A space-ship, out-side, is moving in line with this at a fair fraction of light speed (say, one-half the speed of light). Relative to itself, it sees Orbitsville heading off at that 1/2 of light speed. But the space-ship doesnt agree with the internal astronomers that the shafts of sun-light, traveling from the middle of Orbitsville, arrive the same time at either end.
What is, from the space-ship, the forward-moving end of Orbitsville appears to be distancing itself from the suns rays. So, the space-ship measures a time longer than eight minutes for the sun-light to reach the end moving forward, relative to the space-ship.
Whereas, the aft end appears to be moving to meet the sun-light, so the space-ship measures it taking less than eight minutes. Orbitsville clocks show it took the same eight minutes for light to reach either end of Orbitsville. But the space-ship clock will show an inequality, dependent on how close its relative motion to light speed.

Lorentz addition and subtraction of velocities.

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Using the figures given, this inequality can be worked out by the arithmetic of the Lorentz transformations. These transformations also show that the space-ship will see Orbitsvilles sun-light has a constant speed, whether a light ray is moving from, or towards, the space-ship. This is an axiom of the special theory of relativity. We show this first.

Suppose the space ship passes the exact middle of Orbitsville, just as a night shield, covering its sun, opens before it. Light is made visible to either end of Orbitsville, at the same time. But, from the space-ship point of view, that does not mean both ends receive the light at the same time.

The common sense (of the Galilean transformations) would have us believe that if the relative motion of Orbitsville and space-ship is 1/2 light speed, then we have to add or subtract that speed, with respect to the light speed, depending on whether a light ray is moving with or against the direction of Orbitsville, with respect to the space-ship.

However, the Lorentz transformations, for the addition and subtraction of velocities, ensure that the space-ship sees light speed stay the same in both cases.

Take the case of the light ray moving against Orbitsvilles away-ward moving direction from the space-ship (parallel to it). This requires the Lorentz transformation for the addition of velocities. This is a bit more complicated than the Galilean transformation, that simply gives speed of light plus Orbitsvilles relative motion of one-half the speed of light.

[Post-script: Recalling my early efforts at understanding the Lorentz transformations, the notation for these formulas didn't follow the way, I expected. The variables, u, t, x refer to Orbitsville observation. In order to make the arithmetic work out, I found that I had to make the indexed variables, u', t', x' refer to (differing respective values of) the space-ships observations, of light rays moving relatively towards and away.]

Say velocity u' may refer to either of the differing speeds of two Orbitsville light rays, going both ways, seen from the space-ship. And velocity u is that light ray, which is no different, to those stationed in Orbitsville, from the ray moving to the other end of its shell. Those standing in Orbitsville have no relative motion to (add to or subtract from) the light rays going to both ends. Which end is forward or backward has no meaning to them. In this case (as well as the other case, we shall deal with next), the velocity, u, of the light ray, within Orbitsvilles stationary frame-work, can be set at a "c" for celerity (or constant): u = c.
Let v equal the relative motion, of Orbitsville to space-ship, given as one-half light speed or c/2.

The "aft" end of Orbitsville is only the view of the spaceship seeing what appears to be the back of Orbitsville moving away from it. Tho, front and back may have no meaning to the inhabitants of Orbitsville.
The speed of the sun-light ray, moving in the opposite (aft) direction, within Orbitsville, combines with the speed of Orbitsville moving away from the spaceship. This combined speed is given by the Lorentz transformation for addition of velocities:

u' = (u + v)/{1 + (uv/c²)}.

To simplify the working, light speed is often treated as unitary, or c = 1. This makes v = 1/2, and u = c, in this example. Therefore:

u' = (1 + 1/2)/{1+(1 x 1/2)/1} = 1.

Or, u' = c.

Therefore, the space-ship measures u', the speed of the light ray, moving against the away-moving Orbitsville, as just c, the constant speed of light, irrespective of an addition of relative motion.

The same applies for the light ray traveling to the opposite end of Orbitsville. (This is the "forward" or "front" end from the spaceship view-point.) This ray is traveling with the relative motion of Orbitsville from the spaceship. This means that it is playing catch-up to the forward end of Orbitsville, at one half light speed.

The common sense or Galilean transformation is that the combined velocity would be one minus one-half equals a velocity of half unitary light speed for the forward ray, relative to the space-ship.

But the Lorentz transformation, this time, for the subtraction of velocities again ensures the space-ships relative motion does not allow it to see light move at other than its constant speed.
In our example, the terms have the same meaning as before:

u' = (u-v)/{1-(uv/c²)}

u' = (c-v)/{1-(cv/c²)}

u' = (1 - 1/2)/{1-[(1 x 1/2)/1] }

u' = 1 = c.

Therefore, the Orbitsville sun-light moving relatively towards the space-ship is still measured at lights constant speed. This is the same velocity the space-ship measured for the Orbitsville sun-lights away-moving beam.


Lorentz addition and subtraction of times.

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The Lorentz transformations of time show why observers in relative motion, that is a fair fraction of light speed, cannot agree on the time they observed an event.
All standing in Orbitsville are agreed that both light rays, from centre to either end, took (time, t, of) 8 minutes, or 2/15 of an hour, to cross (distance, X, of) 89 1/3 million miles, at lights constant speed, c, which is about 670 million miles per hour.

But the space-ships clock will show different times, t', that it took the two light rays to reach the ends of Orbitsville. The away-moving light-ray will be timed as taking longer than eight minutes, as the Lorentz addition of times shows:

t' = {t+(Xv/c²)}/{1-(v²/c²)}^1/2.

Distance, X, equals velocity, u, times time, t. Also u = c. If c = 1, and v = 1/2, the square of v equals 1/4. So:

t' = 8(1 + 1/2)/(1 - 1/4)^1/2.

t' = 8(3)^1/2.

That is, the space-ship clock will record the away-moving Orbitsville light ray as taking about 13.9 minutes (to one decimal place) to catch up with the "forward-moving" end of Orbitsville, instead of the 8 minutes that Orbitsville clocks register.

Now we find out the time the space-ship clock tells us it takes the light ray, that is moving to meet the aft end of Orbitsville. From the space-ship view, the aft is moving, at half the speed of light, toward this ray, and so reducing the time they take to meet.
Using the Lorentz subtraction of times:

t' = {t-(Xv/c²)}/{1-(v²/c²)}^1/2.

Following a similar procedure to the previous example for Lorentz addition of times:

t' = 8 (1 - 1/2)/(1 - 1/4)^1/2.

t' = 8/(3)^1/2.

This time, t', of about 4.6 minutes (to one decimal place) is the space-ship clocking of the light ray, whose relative motion makes the apparent aft of Orbitsville appear to come to meet the ray.
This is not quite half the 8 minutes time, t, that Orbitsvilles clock shows for the same journey. With this other Orbitsville light ray, directed in the opposite direction, simultaneous time has been lost, again. Because of their high relative motion, the space-ship observers cannot agree with the Orbitsville observers when the light ray reaches from center to end of Orbitsville.


Lorentz addition and subtraction of distances.

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Besides the disagreement over times between the space-ship and Orbitsville observers, they also disagree over distances. Distance equals velocity times time. Knowledge of any two of these variables will tell us the third. The previous two sections have already found the respective velocities and times measured by the two lots of observers. So, the respective distances, measured by their respective "rods," follows from that information:

In Orbitsville, the distance, X, from centre to either end is eighty nine and one-third million miles. The space-ship sees one light ray appear to have to chase the end of Orbitsville moving away from the space-ship, thus elongating the distance, X', that the space-ship rod measures.

Or, X' = u'.t' = c.t'. This equals the light speed, of 670 million miles per hour, times (2/15) square root3 hours. (Eight minutes is 2/15 of an hour.) Therefore, X' works out at about 154.7 million miles (to one decimal place).

For the sake of completeness, we can repeat this result, by the Lorentz addition of distances:

X' = (X+vt)/{1-(v²/c²)}^1/2.

X' = {(89 1/3)+([670/2] x 2/15)}/{[(3)^1/2]/2}.

X' = {(402)/3}{2/[(3)^1/2]}

X' = (134 x 2)/3^1/2.

This cross-checked working for X' again equals about 154.7 million miles for the space-ships "expanded" rod reading.


Now we come to the case where the space-ship rod or spatial measure appears to "shrink" to less than the Orbitsville rod register of eighty-nine and one-third million miles. Again the simplest solution of, X', now as the shrunken distance, is u'.t', where u' still equals c, and where t' equals 8/3^1/2 (which is about 4.6) minutes, or 2/(15 x square root3) of an hour.
This puts X', as at about 51.6 million miles.

Had we not had the benefit of both velocity and time information, we might have had to find this result by starting with the Lorentz subtraction of distances:

X' = (X-vt)/{1 - (v²/c²)}^1/2.

Or,

X' = t(u-v)/{1 - (v²/c²)}^1/2.

Where u = c = 670 million miles per hour.

X' = (2/15)(670/2)2/3^1/2.

X' = 268/{3 x square root(3^1/2)}.

X' ~ 51.6.

This repeats the result we got directly because we happened to have the information for X' = u'.t'.

In the above examples, we illustrated the special relativity principle of the constant speed of light, despite an observer adding or subtracting the other observers relative motion to it. In general, observers respectively measured velocities, u' and u, need not be of light beams, whose constant speed they are both bound to equal, but can measure things at slower velocities, so that, like distance and time, velocities are not generally agreed by observers in relative motion, significantly approaching the velocity of light.


The Interval.

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The value of the Interval, for our example, also illustrates its nature as a particular case. The Lorentz transformations, as exemplified above, show that observers in high speed relative motion usually measure different distances and times for a given event.
The Interval recovers a common measurement between observers of an event. But it is a combined space-time measurement, upon which they depend for agreement as to the "where-when" something happened.

The Interval measures a four-dimensional space-time common to observers. Our simplified example only dealt in one dimension of space, X, and the one time dimension, t. From the above sections, the Orbitsville co-ordinates are X and t, and the space-ship co-ordinates are X' and t'.

As described on the previous web page, the square of the Interval, I, equals:

I² = X² - c².t² = X'² - c².t'².

Our example is of the Orbitsville observers and space-ship observers of light beams. Special relativity assumes both sets of observers must measure the same velocity for light. That is both u = c and u' = c.

In this particular case, since X = ut and X' = u't', the Interval for both observers is zero. The general point is that the Interval is always the same value for both observers.


Kinematics and dynamics.

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My explanations of special relativity were in terms of "kinematics" or motion, not "dynamics," which further involves the concept of mass or energy.

Fortunately, the conversion of the Lorentz transformations, from their kinematic form to their dynamic form, is fairly straight-forward. The Lorentz transformations correspond different observers differing distance and time measurements of an event. Call these observers differing distances, x and x', and differing distances , t and t', respectively. Consequently, the differing velocities of a body, they observe, will be, say, u = x/t and u' = x'/t'.

If the mass of the body is also to be considered, then an observer, who is at rest relative to the motion of the body, will measure its "rest mass."
In classical mechanics, there only is a rest mass: the mass is deemed constant. In relativistic mechanics, the mass of a body is known to noticably increase as its motion noticably approaches the speed of light. (The relevant Lorentz transfomation shows that a body would have to be of infinite mass to reach the speed of light. And this is why light speed is natures speed limit.)

Essentially, the Lorentz transformations for distance and for time are both multiplied by the same quantity, the mass divided by the time, to convert them into Lorentz transformations for momentum and energy, respectively.

Momentum, p, is mass times velocity. In the system of mechanics, distance has three dimensions, the x, y and z directions, to refer to all spatial motions. Therefore, velocity, as distance over time, also has direction with reference to the three co-ordinates. For simplicity, we have considered observers measuring events, which differ only in the x-direction, according to their respective measurements. That is to say where distances y = y' and z = z' between respective observers. But their respective measurements do not make x equal to x'. So, velocities in the x-direction, u and u' also need not be equal.

Consequently, the Lorentz transformation for momentum is in terms of differently circumstanced observers differing measures of both mass and velocity, or: p = mu and p' = m'u'.

The differing times, observers measure, are proportionate to the differing masses they measure: t'/t = m'/m. Note, also, that E = mc² and E' = m'c², where E and E' are the respective energies measured by the observers, according to the famous equation, not here demonstrated.

Then, multiply the Lorentz transformation for distance, x, by m/t, and x' by m'/t', for the Lorentz transformation between observers differing measures of momentum (in the x-direction):

mx/t = mv = p = m'(x' + vt')/ t'square root(1 - v²/c²)^1/2.
= (p' + vE'/c² )/square root(1 - v²/c²)^1/2.

The same procedure, of multiplying by mass over time, is followed for converting the Lorentz transformation of time into the Lorentz transformation of energy:

tE/c²t = ( t'E' + vx'E'/c² )/ c²t'square root(1 - v²/c²)^1/2.

This simplifies to:

E = (E' + vu'm')/(1 - v²/c²)^1/2

= (E' + vp')/square root(1 - v²/c²)^1/2.

The Lorentz transformation of velocity follows from dividing the Lorentz transformation for distance by that for time:

u = (u'+v)/(1+vu'/c²).

In dynamics, compare the result of dividing momentum by energy. The denominators of both transformations are the Lorentz contraction factor, which cancels:

p/E = (p' + vE'/c²)/(E' + vp').

Or: u.m/mc² = m'(u' + v)/m'c²(1 + u'v/c²).

This reduces to the Lorentz transformation for (addition of) velocities.

Furthermore, the Minkowski Interval, as well as the Lorentz transformations, convert from kinematics to dynamics, in a straight-forward manner. The Minkowski Interval is a four-dimensional space-time geometry. It is like Euclid geometry in three dimensions. But time is treated like a fourth dimension of space. The only formal mathematical difference between the three space dimensions and the one time dimension is that the time is multiplied by the square root of minus one.

This factor gives the time, t, the opposite sign to the three space dimensions x, y and z, when the Interval extends Euclid geometry, of Pythagoras theorem in three dimensions, into a 4-D space-time version of the theorem.

The Lorentz transformations are the means by which an observer changes their measurements into those of differently situated observers of the same event. The point of the Interval is that it combines their respective measurements into a "space-time" measurement, of a given event, that is one and the same for all observers, in uniform relative motion.

As for the Lorentz transformations, the Interval, I, is simply given here for only one dimension of space and the one dimension of time. As discussed on previous pages, this is:

I² = t²(c²-u²) = t'²(c²-u'²).

Or:

t²c²-x² = t'²c²-x'².

The space-time Interval changes to its momentum-energy version, as with the Lorentz transformations, thru replacing time by mass:

(mc)²t²(c²-u²)/t²c² = (m'c)²t'²(c²-u'²)/t'²c².

Or: (E/c)² - p² = (E'/c)² - p'².




Richard Lung.
2001; revised & corrected June 2015 & january 2019.

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