MATHEMATIZING THE LORENTZ TRANSFORMATION

If you have read Mister Anthony's Lecture on Euclidean Algebra and have gained some sense of how algebra generalizes and abstracts arithmetic, you may now want to see a little of how physicists express the laws of Relativity in algebraic form. First, I will repeat the derivation of the description of time dilation, using algebra to augment the logical deduction, then I will show you what the equations of the Lorentz Transformation look like and offer some examples of how we may use them to work out descriptions of new effects.

We actually start out with a bit of plane geometry, because we want to set up a triangle from which we can draw our initial formulae. Thanks to the Ancient Greeks, Euclid and Pythagoras in particular, we know how to set up and use deductions involving triangles. And although the Greeks did not use algebra, we can easily convert the Greek style description of a triangle into the corresponding algebraic description.

In our deduction of the basic description of time dilation we imagined standing next to a track and photographing a Feynman clock on a passing train. In the resulting picture the pulses in the clock traced a sawtooth pattern (or a series of vees) and we took one of the lines in that pattern to be the hypotenuse of a right triangle in order that we might work out the relation between time counted on the moving clock and time counted on our stationary clock. We drew the right triangle by drawing a vertical straight line from the bottom point of one vee and by drawing a horizontal straight line that just touched the upper points of the vee and then we listed what we knew about the three lines that comprised our right triangle:

Line A; the length of the vertical line, from the bottom of the vee to the horizontal line is simply the length of the train clock's optical axis, the distance between its laser and mirror. By Lorentz Rule 1 it is also equal to the length of our clock's optical axis, because the train clock's optical axis is oriented perpendicular to the direction in which the train is moving;

Line B; that part of the horizontal line that extends from the vertical line to one of the vee's upper points is simply the distance that the train moves in the time it takes the train clock's pulse to go from one end of the clock to the other; that is, the distance that the train moves in one half of its clock's cycle. To calculate that length we must multiply the speed of the train by the time elapsed on our clock during that half cycle of the train's clock;

Line C; the hypotenuse is the line traced by the train clock's pulse as it moves from one end of its clock to the other. By Einstein's second postulate, it must move at the same speed of light (100 miles per hour in our fantasy world) in both our frame and the train's frame. Thus the length of the line is equal to the product of multiplying the speed of light by the time elapsed on our clock, the same time that we used to calculate the length of Line B.

Now how shall we convert those descriptions into their algebraic equivalents? We apply the technique that Rene Descartes devised when he invented analytical geometry: we set up a coordinate grid based upon mutually perpendicular axes to give us a frame of reference in which we establish our triangle. In this case we only need a two-dimensional grid, so we establish our x-axis parallel to the railroad track in our imaginary experiment and then make our y-axis run vertically. I'll further specify that upper-case letters will represent measurements made in our inertial frame (the one fixed to the ground alongside the track) and that lower-case letters will represent measurements made aboard the train. We thus have the algebraic description of the three lines that comprise our triangle as:

Line A; we represent the length of the train clock's optical axis with the letter y and the time required for a pulse to go from the top of the clock to the bottom of the clock with the letter t, so we have

y = ct.

(Eq'n 1)

By Lorentz Rule 1 we also have

y = Y.

(Eq'n 2)

Line B; the distance that the train moves in our inertial frame in the time T that the pulse takes to go from the top of the train clock to the bottom of the train clock. We represent the train's speed with the letter V, so we have

X = VT.

(Eq'n 3)

Line C; the hypotenuse of our triangle is the distance that we represent with the letter H, the distance that the train clock's pulse travels in our inertial frame. We calculate that distance with the simple product

H = cT.

(Eq'n 4)

But we can also calculate the length of Line C by using the Pythagorean theorem; that is, by extracting the square root of the sum of the squares of Lines A and B. For convenience I will delay extracting the square root and simply write

H2 = X2 + Y2.

(Eq'n 5)

To deduce the time dilation effect I will replace H, X, and Y in Equation 5 by substituting the appropriate formulae from the preceding equations; that is, I will replace H with cT from Equation 4, I will replace X with VT from Equation 3, and I will replace Y with y from Equation 2 and then replace that with ct from Equation 1. When I have made the substitutions, Equation 5 becomes

(cT)2 = (VT)2 + (ct)2.

(Eq'n 6)

I want to solve that equation for T because I want to calculate how much time elapses on my clock for a given time t elapsed on the train clock. That means that I must get T on one side of the equation, so I'll start by subtracting the square of VT from both sides of the equation. That subtraction gives me

(cT)2 - (VT)2 = (ct)2.

(Eq'n 7)

Then I'll divide the whole equation by the square of cee and remove the parentheses by carrying out the indicated squarings;. That procedure transforms Equation 7 into

(Eq'n 8)

Because the square of T appears in both terms on the left side of the equation, I can factor that side of the equation; that is, I can rewrite it as an explicit multiplication of the square of T by the remaining part of the formula. Thus Equation 8 becomes

(Eq'n 9)

Finally I can extract the square root of that equation and divide the result by the coefficient that multiplies T. I will thus have my solution;

(Eq'n 10)

And, again, because I am a lazy fellow who does not like to keep writing the same formula over and over again, I will represent the Lorentz factor by L and rewrite Equation 10 as

T = Lt.

(Eq'n 11)

By comparing Equations 10 and 11 you can see that the Lorentz factor must come from the formula

(Eq'n 12)

Compare that with the recipe for calculating the Lorentz factor that I gave you as Lorentz Rule 0 in the essay The Lorentz Transformation: take the velocity between the two inertial frames, square it, divide the square by the square of the speed of light, subtract the resulting fraction from the number one, extract the square root of the result, and divide that square root into the number one.

That shows you a fair example of how physicists mathematize a problem. Consider what we did in the laboratories of our minds. We imagined a scenario in which we included a feature that embodies something that we know in a relatively familiar form. In this case we imagined a Feynman clock set up next to a railroad track in such a way that its optical axis is oriented perpendicular to the direction in which any train on the track will move, thereby implicitly preparing to exploit Lorentz Rule 1. Then we imagined a feature intended to reveal something that we did not know; specifically, we imagined a Feynman clock on a passing train with the intent of discerning whether it would count time any differently from the manner in which our trackside clock did. From that scenario we abstracted a geometric figure, a right triangle, whose parts we could represent with simple algebraic formulae and we combined those formulae into an equation by invoking the theorem of Pythagoras. Finally we solved that equation for the time measured on the train's clock and found that it is, as we suspected, different from the time elapsed on our trackside clock. Equation 10 then gives us the description that relates the two different measurements of time.

The other Lorentz Rules can be mathematized in a similar fashion and their algebraic descriptions then combined into the four equations of the Lorentz Transformation. Those four equations are:

X = L(x + Vt)

(Eq'n 13)

Y = y

(Eq'n 14)

Z = z

(Eq'n 15)

T = L(t + xV/c2)

(Eq'n 16)

Compare those equations with the corresponding recipes that I laid out in the essay The Lorentz Transformation;

LORENTZ TRANSFORMATION EQUATION 1: To calculate my measurement in the x-direction of the distance between two events take your measurement in the x-direction of the distance between the events, add the product of the velocity between our frames and the time that you measure between the events (by subtracting the time of the following event from the time of the leading event as measured on clocks synchronized in your frame), and multiply the sum by the Lorentz factor between our frames. This recipe combines Lorentz Rules 2 and 4 with Galilean Rule 1 from the essay Events In Inertial Frames.

LORENTZ TRANSFORMATION EQUATION 2: My measurement in the y-direction of the distance between the two events is equal to your measurement in the y-direction of the distance between the two events. This is simply Lorentz Rule 1.

LORENTZ TRANSFORMATION EQUATION 3: My measurement in the z-direction of the distance between the two events is equal to your measurement in the z-direction of the distance between the two events. Again, this is simply Lorentz Rule 1.

LORENTZ TRANSFORMATION EQUATION 4: To calculate my measurement of the time interval between the events take your measurement in the x-direction of the distance between the events, multiply it by the speed between our frames, divide that product by the square of the speed of light, then add to that number the time that you measure between the events (again, by subtracting the time of the following event from the time of the leading event as measured on clocks synchronized in your frame), and multiply the result by the Lorentz factor. This recipe combines Lorentz Rules 2 and 3.

That may seem rather simplistic, but we use those equations for much more than the translation of measurements of distance and duration between inertial frames. Consider the frequency of light from a source that's moving away from us. We represent the time between successive crests of the electromagnetic waves that comprise the light rays by lower-case tee in the source's inertial frame and by upper-case tee in our frame. To convert the former into the latter we would use Equation 16 above, but what we want is the inverse of the time interval, the frequency, so we must invert the equation. Before we take that step we must convert one factor in the formula on the right side of the equation. The distance represented by the lower-case eks is the distance between two clocks whose readings differ by one period of the wave's oscillations and is thus, in the source's frame, the distance that one part of the wave travels in one cycle, so we have

x = ct.

(Eq'n 17)

We now have Equation 16 in the form

T = L(t + Vct/c2)

T = Lt(1 + V/c).

(Eq'n 18)

Did you notice how I factored the lower-case tee out of the parentheses and used the cee in the numerator of the second term on the right to cancel one of the cees in the denominator? Those moves have put the equation into the form I want in order to invert it. Now I will represent the frequency of the light in question by the letter eff, so I have now

(Eq'n 19)

Now I can perform one of the neatest tricks in the algebraic playbook. I factor both the numerator and the denominator in that equation's right side; that is, I split each formula into the two formulae that yield it when we multiply them together. Thus I transform Equation 19 into

(Eq'n 20)

Now I just cancel out the identical factors in the numerator and in the denominator of that formula and I get

(Eq'n 20)

That's actually one form of the equation describing the relativistic Doppler shift of the frequency of electromagnetic radiation. Now look at what that equation becomes when I multiply the numerator and the denominator on the right side by the numerator;

(Eq'n 21)

That equation just describes the negative Doppler effect as I described it in the essay E=mc2. Compare that equation with Equation 18 and you should be able to see that if Equation 18 had had a minus sign where it has a plus sign, the resulting Equation 21 would have a plus sign where it has a minus sign in its numerator and, thus, that it would then describe the positive Doppler effect. You may recall that I applied both the positive and negative Doppler shifts to the same frequency of light coming from opposing spotlights to deduce Einstein's mass-energy equation, E = mc2.

Finally, for this appendix, I want to show you one way of explaining why no object can ever go faster than light. Let's imagine that you are moving away from me in the x-direction at a speed that we represent with upper-case vee and that you have sent a small rocket in the same direction at a speed that we represent with a lower-case vee. To assign a number to that latter speed you calculate the ratio of the distance that the rocket crosses (represented by lower-case eks) to the time that elapses (represented by lower-case tee) as it crosses that distance; that is,

v = x/t.

(Eq'n 22)

In my inertial frame I will represent the speed of the rocket with an upper-case double-you (because we have run out of vees) and I calculate it in the same way from the distance and duration that I measure for the rocket as it runs its defined course (between two points, one on the orbit of Mars and one on the orbit of Jupiter, let's say, that are the same for both of us); thus, I have

W = X/T.

(Eq'n 23)

Now I want to relate my W to your v.

I'll start by using Equation 13 to replace X with the equivalent formula containing only your measurements and using Equation 16 to make the same replacement for T:

(Eq'n 24)

The Lorentz factors in the numerator and the denominator canceled each other out, leaving me with the formula on the second line. If I multiply Equation 22 by lower-case tee, I can replace the lower-case ekses in Equation 24 with their equivalent (that is, x = vt) and obtain

(Eq'n 25)

In the second step in creating Equation 25 I have factored the lower-case tee out of the formulae in the numerator and the denominator. What that means is that I have seen that the numerator is the sum of two velocities multiplied by the same time interval, so I have replaced the formula with one that instructs me to add the velocities together first and then multiply by the time interval. Then I did the same trick with the formula in the denominator. But now the lower-case tees are in the formula in the same way in which the Lorentz factors were in Equation 24, so they will also cancel each other out of the formula. Thus we end up with

(Eq'n 26)

Now let's suppose that lower-case vee represents the speed that the second stage of a rocket can achieve relative to the inertial frame from which it begins its acceleration and that upper-case vee represents the speed that the first stage of the rocket can give the second stage. Let's also suppose that both speeds are equal to three-quarters of the speed of light; that is, v = V = 0.75c. A naive calculation of the final speed, based on our everyday experience, would lead us to add those numbers directly and to declare that the second stage will achieve a speed relative to its launch point of one-and-a-half times the speed of light. But Equation 26, the correct relativistic calculation tells us something different. It gives us

(Eq'n 27)

The final speed of the second stage is still less than the speed of light. It doesn't matter what velocities we use in Equation 26 (as long as they are all less than the speed of light, which is what we would have to begin with in any case) or how many times we repeat the procedure of Equation 27 (that is, no matter how many stages we give our rocket), the result will always be less than the speed of light. Equation 26, then, is one mathematical statement telling us that nothing can fly faster than light, at least by addition of velocities (which is what straightforward acceleration is). If we want to fly faster than light, as the characters in "Star Trek" do, we will have to find a way to fly outside our space and time, but that's a subject for a different book.

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