Science and education

The theory of relativity made simple? OK......Albert Einstein is driving a Volkwagen Beetle and Niels Bohr is driving a Ferrari Testarossa. The impartial arbitrator is Werner Heisenberg, who is sitting by the road at a position he calls x=0.
Niels Bohr is much faster and at the time t=0 (as clocked by Heisenberg) he passes Einstein at position x=0. Wolfgang Pauli, who is moonlighting as a traffic cop, sees them coming and prior to t=0 he switches on his radio to call for backup. A spherical electromagnetic wave emanates from his position and reaches the speeding pair at t=0 (at x=0). It is obvious that Heisenberg sees the wave coming at the speed of light, but what about the two drivers? They are moving towards the wavefront so common sense dictates that they see the wave coming at a higher speed than Heisenberg and that Bohr sees it coming at a higher speed than Einstein. Interestingly enough, relativity begins where common sense ends: all three observers see the wave coming at the same speed.

This highly dramatic result is called Einstein's Second Postulate. Actually this is inherent to Maxwell's equations, which, when solved for the electromagnetic wave equation, produce a constant, of which we mathematically know that it's the speed at which the electromagnetic wave moves. Electromagnetism has it's apparent paradoxes equivalent to our above-mentioned Bohr-Einstein automobile experiment. This made electromagnetist Anton Lorentz (before Einstein) come up with his very own Lorentz transformations. This is how they go:

Deriving the Lorentz transformations

Here we will employ only one spatial dimension x (the distance the drivers cover on the road) and for simplicity we will drop Einstein. Let's say Bohr is driving his Ferrari at 100 km/h and Heisenberg is stationary (with respect to the ground), by the roadside, so the relative velocity v between Bohr and Heisenberg is 100 km/h.

Our next important point is that in special relativity everybody moves at constant velocity, that is, not accelerated. This is because accelerated motion is a whole different ballgame and belongs to general relativity, so forget about acceleration for now. It is assumed that you are familiar with the Galilean transformation equations:

x' = x - vt (1a)

x = x' + bt (1b)

Here primed coordinates are Bohr's, unprimed coordinates belong to Heisenberg and v is their relative velocity (100 km/h ~= 28 m/s ). Heisenberg and Bohr set their watches so that at t=t'=0 Bohr just passes Heisenberg (x=x'=0). Then at t=1 (in seconds) Bohr is at x=28 (m) according to Heisenberg.

However, if a spherical electromagnetic wave emanated from Bohr (who is moving at a 100 km/h with respect to Heisenberg), then the Galilean picture would require that the center of the spherical light wave moved at a velocity of 100 km/h with respect to Heisenberg. This conflicts with the constancy of the speed of light so we must find an alternative for the Galilean transformation equations, which are incorrect in the relativistic picture, as we have just demonstrated. We are looking for equations that are linear:

x = ax'+ bt' (2a)

x' = ax - bt (2b)

In relativity linearity is mathematically expressed by the constancy of a and b and that the equations above have no powers of x, x', t or t' (that is, there are no x², (t')³, etc). Remember that a parabola, like t(x) = x² is not a straight line while equations (2a) and (2b) are straight lines. The physical meaning of this is that in a spacetime diagram straight lines correspond to uniform (not accelerated) motion, while curved lines correspond to accelerated motion. This is show in figure 1.

figure 1

Now if the relationship between Heisenberg and Bohr's time and space coordinates wasn't linear (as expressed by eq 2a and 2b), then uniform motion according to Heisenberg wouldn't be uniform according to Bohr and different observers must always agree whether or not an object is accelerating. If they didn't, then according to one observer there would be a force acting on an object (after all, f = ma) and according to another observer there would be no such force, which is impossible. Hence we assume equations 2a and 2b to be relativistically correct.

If we choose to investigate the motion of the origins of Bohr and Heisenberg's coordinate systems, then we may put x = 0 in (2a), (this is Heisenberg's origin) and x' = 0 in (2b), (Bohr's origin). We then get:

0 = ax'+ bt' 0 = ax - bt	<=>	ax' = -bt' ax = bt	<=>	x'/t'= - b/a x/t = b/a
\|x'/t'\| = \|x/t\|	<=>	dx/dt = dx'/dt'	, because we have rectilinear motion..
in other words: the derivative along a straight line is just x/t, for every value of x and t.
= v	= \|± b/a\|	= b/a.
v being the relative velocity between our two observers, who we called Bohr and Heisenberg.

So now we know a little more about the constants a and b of equations 2, a result we will use later.
We know that the speed of light c is constant and therefore it's path through spacetime will be a straight line. Hence,

c = dx/dt = x/t

<=>

x = ct and x' = ct' ,

inserting this into equations 2a and 2b gives:

ct = act' + bt'
ct' = act - bt

<=>

ct = (ac + b)t'
ct' = (ac - b)t

From the first of these four equations, we get:

t = t'/c(ac + b), inserting this into the last of the four equations above gives:

ct'= (ac - b)t'/c(ac + b)

<=>

c² = (ac - b)(ac + b)

c² = a²c² + abc - abc - b² b = av	<=>	c² = a²c² - a²v²	c² = a²(c² - v²)	a² = c²/(c² - v²)
a² = 1/(1 - v²/c²)	=>	a = 1/(1 - v²/c²)½	and:	b = v(1 - v²/c²)½

Then equation 2a becomes:

x = ax' + bt' b = av	<=>	x = ax' + avt' = a(x' + vt') = (x' + vt')/(1 - v²/c²)½
hence:	x = g(x' + vt')	where g(v) = (1 - v²/c²)-½
and analogously:	x' = g(x - vt)
some reshuffling also gives:	t = g(t' + vx'/c²)
and:	t' = g(t - vx/c²)

Science and education

Deriving the Lorentz transformations

Science and education resources

education science education 2 essays and papers education 3