THE LIGHT BEAM RIDER

David Hilbert once seriously joked: Every school boy in Gottingen knows more hyperbolic geometry than Einstein, but it was Einstein who did it. There is something very uncommon about Einstein. He was not a child prodigy, but a lone thinker who did a grave mistake of bunking his Riemann geometry classes. He was called a ‘lazy dog’ by his teacher. Then, how was he different from other people? Had he obtained nice marks in class, had he been very regular and wasted time in solving problems out of books instead of reflecting, he would not have been Einstein. At the age of sixteen, he wondered how it would look to walk with light. Most of his words were words of pure logic. He was probably the man who least misunderstood the quantum mechanics, but he was never interested in developing its analytical part (like angular momentum laws, etc.). His first important paper showed that he was going to be a world leader of philosophy. Let us see how to work out relativity theory, viz. Lorentz transformation.

Light starts moving towards a frame with a velocity c. Suppose after time t, light reaches a particular position (x-ct) in L and (x-ct¢) in L¢. These points are related as:

                                x¢ - ct¢ = A(x – ct)   …… (1)

For light going away,

x¢ + ct¢ = B(x + ct)   …… (2)

 We have to write x¢ and t¢ in terms of x and t. But there are two unknowns A & B (which obviously must be) because we have not yet defined the time t and velocity v. Adding and subtracting, we get,

                                x¢ = px – qct                                                                                          

                                t¢ = pt – (q/c)x       

To measure the velocities of L¢, L will take a point in frame L¢ and measure the change in position (Δx) for this point of time Δt. But Δx¢ = 0 since point is still constant in L¢ frame. So v = Δx/Δt = qc/p.

So,                          x¢ = p(x – vt)           …… (3)

                               t’ = p(t – (v x) / c²)  …… (4)

Now, consider the measurement of a rod. Suppose there are two bulbs attached to their ends. A frame will observe the length of rod of in it both bulb glow simultaneously. If the rod is kept in L¢, Δx¢ = l (length of the rod). If L is able to see the rod, then Δt = 0 for L. So Δx = l¢ (length in the frame moving relative to rod).

So, l¢ = (1/p)l                                         ……(from (3))

If the rod is in L, then, Δx = l and Δx¢ = l¢. Now, Δt¢ = 0.

Using (2), Δt = (v/c²)Δx.

So, l¢ = p{1-(v²/c²)}l. So, comparing both l¢ and l, we get,

p = 1/{Ö(1-v²/c²)}.

Now, we have defined t (time in a particular frame) as a time which is used to measure the velocity of the frame or the length of the rod. Though the English word is not of importance, we can generalize the meaning of ‘time’. First, see that we get the Lorentz transformation:

                                 x¢ = (x-vt)/{Ö(1-v²/c²)}

                                t¢ = (t-vx/c²)/{Ö(1-v²/c²)}

Since p = 1/{Ö(1-v²/c²)}, q = v/{cÖ(1-v²/c²)}, we have,

                                A = Ö{(c+v)/(c-v)} and B = Ö{(c-v)/(c+v)}

Multiplying (1) and (2), we get,

                                x¢² - c²t¢² = x² - c²t²

Let R signify x² - c²t². So, R = R¢ = R² = ………. for many frames.

Another important relation is that of Doppler’s effect. Look at the relation:

x¢ - ct¢ = [Ö{(c+v)/(c-v)}] (x-ct) 

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