Question On Lorentz Transformation...

[afchris]

I'm currently reading Einstein's Theory of Special and General Relativity and am kind of stumped on his deviation of the Lorentz Transformation.

There's just a couple of questions that I have.

Why does he introduce a variable (lambda and mu) for the equivalence of the two reference reference for the positive and negative axis.
(x'-ct') = lambda(x-ct)
(x'+ct') = mu(x+ct)

Also, I don't quite grasp how he adds those two equations, and then substitutes:
a=(lambda+mu)/2, b=(lambda-mu)/2

and how does adding these two equations produce a system of equations:
x' = ax-bct
ct' = act -bx

Thank you very much for all who are happy to enlighten me!

---
afchris

[rpenner]

In appendix A, Einstein is talking about two observers watching the same light and used that light flash as the zero of their x and t (or x' and t') coordinate systems. This is diagrammed in Figure 2 of Chapter XI.

So obviously what is desired is a relationship between x (the distance observer 1 sees the light move before hitting a target), t (the elapsed time observer 1 sees the light move before hitting that target), x' (observer 2, same light, same target), and t' (observer 2 again). (To emphasize these are distances and elapsed time and not arbitrary x coordinates or time coordinates, many books will write Δx and Δt for distance and elapsed time, respectively. But Einstein was writing before computer typography.)
So we get x - ct =0 and x' - ct' = 0. The two agree on the speed of light. But this holds no matter how far between source of light and target.

Now we could say x - ct = x' - ct' and that would be true, but we have an extra freedom here (because for any number r, 0 = 0 r), and Einstein uses that freedom to explore what happens if start with λ(x - ct) = x' - ct'. What Einstein is saying is really λx - λct = x' - ct' -- that some mysterious linear factor relates x, and x' and t and t'. We know that λ(x - ct) + ξ = x' - ct' can only mean ξ = 0. And if we try to make the relationship non-linear for x or t, we can't make f(x) - c f(t) = 0 for all possible targets unless f(x)/f(t) = c. Everything screams a linear relationship.

Then μ(x + ct) = x' + ct' follows from the same setup but with a different direction of light.

Now we have λ(x - ct) = x' - ct' and μ(x + ct) = x' + ct' and from linear algebra we know we can add (or subtract) left and right sides of equations and we will have valid equations. (What needs to be remembered is that λ and μ are mere numbers. We don't know what they are.)

Adding: λ(x - ct) + μ(x + ct) = ( x' - ct' ) + (x' + ct') ⇒ (λ + μ)x - (λ - μ)ct = 2x'
Subtracting: λ(x - ct) - μ(x + ct) = ( x' - ct' ) - (x' + ct') ⇒ (λ - μ)x - (λ + μ)ct = -2ct'

Dividing everything by 2 is useful, but too boring to bother with.

Then we change variables again, with a = (λ + μ)/2 and b = (λ - μ)/2 and we have
ax - bct = x'
bx - act = -ct'

(Since λ and μ are mere numbers, a and b are mere numbers also. But soon we will be able to relate them in terms of the relative velocity of the inertial observers.)

Finally we multiply the second by -1 to get:
x' = ax - bct
ct' = act - bx

So what we have done is gone from an assumption that any two observers in inertial motion will agree that the speed of light is the same, to say that there is a linear relationship between the distances and elapsed time one measures and the distances and elapsed time the other measures. Further, not just any linear relationship will work -- it has to look like:

x' = a x - b ct
ct' = -b x + a ct

In matrix form, the diagonal is one value and the off-diagonal elements are another value. Einstein then used the definition of inertial motion to write b = a (v/c) so that there is only one unknown, a. Obviously a has to be 1 when v = 0. But when v is not 0, then we get departure from Newtonian assumptions.