We left off (Part 1) with the last two equations related to the Lorentz transformation, e.g.

^{2}/(1 - v

^{2}/c

^{2})

^{½}

and

t' = t - xv/c

^{2}/(1 - v

^{2}/c

^{2})

^{½}

Referring back to Fig. 2 in Part (1), suppose there’s a clock located in system S’ and an observer in system S sees this clock moving with velocity v. At any time t, the position of the clock with reference to the S –observer’s system is given by:

T = T’/ (1 - v

^{2}/c

^{2})

^{½}

Since T > T’ then it seems to Observer S his clock is running slower than it does to S’. This applies not only to the clock but to all physical processes that depend on time, e.g. the vibrations of electrons in atoms, rates of chemical reactions, biological processes (heart beat, respiration) etc. In effect it appears to the observer in S that his counterpart in S’ is living at a slower rate than he is. However, to the observer in S’ it is the observer in S who seems to be living at a slower rate.

This is a paradoxical result but one which can’t be escaped if we carry special relativity to its logical conclusion as Einstein did. So long as the two systems have a constant relative motion, we cannot say that one is moving and the other standing still, or that one’s clock is moving slowly and the other quickly.

A number of subtle consequences arise out of this. For one thing, the notion of

*simultaneous events*, i.e. for observers in two different reference frames, is no longer tenable. Einstein, in fact, seems to have been the first human being to recognize that simultaneity between two events is a provincial illusion. In his landmark (1905) paper on special relativity (“

*’) he remarked:*

__Does the Inertia of a Body Depend on Its Energy Content?__*“We have to take into account that all our judgments in which time plays a part are always judgments of simultaneous events. If, for instance, I say ‘That train arrives here at 7 o’clock’ I mean something like this: ‘The pointing of the small hand of my watch to seven o’clock and the arrival of the train are simultaneous events.”*

Einstein agreed that observing such simultaneity was a reasonably accurate way for a person holding a watch to tell the time of an event happening

**next to the watch**, but insisted that on principle the method couldn’t be relied upon for timing an event far away from the watch, especially by someone moving in relation to the other things involved.

Two bolts of lightning then strike the
extremities of the train and do so simultaneously according to Bob’s watch.
Bill, however, is situated on a stationary platform as shown – defined in
system S- at the exact midpoint between the strikes and swears that he recorded
the rear flash a sizeable fraction of a second earlier than the forward flash.

Who is correct? A non-relativistic
thinker would undoubtedly incline towards the view of Bill, the stationary
observer. The argument here is that Bob’s simultaneity is only apparent, not
actual since the train would have carried him closer to the forward flash by
the time he received their simultaneous light from points that were, by then,
no longer equidistant from him.

For a relativist, however, it is just
as true to say the train was standing still and the ground sped by it, so the
ground observer (Bill) could have been carried beyond the equidistant point as
easily as Bob.

In Einstein’s cosmic viewpoint, the
lightning flashes (and their midpoint) are free to be considered an integral
part of the train OR Earth. OR any other frame of reference. Each and every
observer may select his own viewpoint. Any viewpoint is true and right, none is
wrong. This example embodies the essence of relativity, that is, that time is
truly relative and not absolute as Newton had believed.

As an illustration of a type of
experiment that attempts to reckon simultaneity, consider Fig. 2 below.

*Fig. 2: Graphic depiction of simultaneity*By Pythagoras’ theorem (square of the hypoteneuse equals the sum of the squares of the other 2 sides, then:

(c (Dt)/2)

^{2}= (v (D t)/2)

^{2}+ d

^{2}

This implies:

D t = 2d/(c

^{2}– v

^{2})

^{½}= 2d/ c[(1 - v

^{2}/c

^{2})

^{½}]

^{ }

D t’ = 2d/ c

which implies:

D t = D t’/ (1 - v

^{2}/c

^{2})

^{½}

We call D t the “proper time” or that time interval between two events as measured by an observer who sees the events at the same place. It’s always the time measured by a single clock at rest in the frame.

*Example Problem*:

The period T of a pendulum is measured to be T= 3.0 s in the inertial frame of reference of the pendulum. What is the period of the pendulum when measured by an observer moving at a speed of 0.95c with respect to the pendulum?

__Solution:__*We note that the time measurement taken in the inertial frame is the proper time, and this is t' = 3.0 s. Then we need to obtain t, for which:*

t = t’/ [1 - v

^{2}/c

^{2}]

^{½}

Where v = 0.95c, so:

t = 3.0 s/ [1 - (0.95c)

^{2}/c

^{2}]

^{½}= (3.0s) 1/ [0.0975]

^{½}

and t = (3.0 s)(3.2) = 9.6s

*Other Problems*:1) With what speed would a clock have to be moving to run at a rate that is one half the rate of a clock at rest?

2) An atomic clock is placed on a Jumbo Jet. The clock measures a time interval of 3600 s when the jet is moving at v = 300 m/s. What corresponding time would an identical clock left on the ground measure? (Hint: whenever v << c (e.g. v/c << 1), note that we have 1 + v

^{2}/2c

^{2 }and not [1 - v

^{2}/c

^{2}]

^{1/2})

3) A muon formed high in the Earth's atmosphere travels at v = 0.99c for a distance of 4.6 km before it decays into an electron, a neutrino and an anti-neutrino.

a) How long does the muon survive as measured in its rest frame?

b) How far does the muon travel as measured in its frame?

4) The average lifetime of a pi meson in its own frame of reference is 2.6 x 10

^{-8}s. If the meson moves with v = 0.95c, what is its mean lifetime as measured by an observer on Earth?

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