Wednesday, October 17, 2018

Special Relativity Revisited (3)

We now extend the special relativity dynamic to place two observers ("Bill" and "Bob") in the roles of two astronauts in spaceships referred to systems S and S'. Initially each is moving at a constant speed relative to the other. Subsequently, Bob is accelerated to some higher speed so that he is displaced. To return to Bill, Bob will have to slow down at some point (decelerate) then speed up again to overtake Bill, and finally change his speed once more to fall in step with Bill.
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During this entire time interval, Bill hasn't experienced any acceleration, but has maintained his constant normal speed. We now ask what effect the acceleration would have had on Bob? It turns out that time has passed more slowly for him and so he's aged less than Bill. If twins to begin with, Bill would now be older than Bob.

    Superficially, at least, one may be tempted to dispose of this improbable conclusion using a simple symmetry argument, i.e. basing our coordinate system on Bill's reference frame. This is wrong because there is no fundamental difference between the twins: one of them (Bob) has undergone acceleration (experienced a force) which the other (Bill) has not.

    While all positions and speeds are equivalent in relativity theory, accelerations do have physical consequences, so the relationship between the two twins is not symmetric in this instance and there can be no symmetry argument precluding aging of the un-accelerated twin.

    This result of special relativity (which some diehards still can't accept) is the basis for lots of sci-fi stories that depict space explorers roaming the galaxy at great speeds and returning to Earth after having aged only a few years while millennia have passed on the home planet in the meantime. This effect is called time dilation and there is an experimental basis for it.

    It has been found that mesons moving very rapidly have a longer average lifetime than mesons not in motion. In fact, this very difference was the basis for two problems given previously. To recap just one of the problems (#4, SR Revisited Pt. 2) ):

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?

    We found the proper time t' = 2.6 x 10-8 s for the meson's own frame

and for the mean meson lifetime for an Earth observer:

t = t’/ [1 - v2/c2] ½

t = (2.6 x 10-8 s)/ [1 - (0.95c)2/c2] ½

t = (2.6 x 10-8 s)/ [0.0975] ½ = (2.6 x 10-8 s)/ 0.312

t = 8.3 x 10-8 s

    In other words, t > 3t' or, more than three times as much time has elapsed for the Earth observer in his stationary frame. This result can be checked for the muon moving at relativistic speeds in problem (3).

Time dilation has been found to occur when comparing the clocks in a Jumbo Jet and those on the ground. This was the basis for problem (2) in the last set which found that the clock time aboard a Jumbo Jet traveling at 300 m/s indicated an hour clocked on Earth as 3600 +  1.8 x 10-9 s, or slightly longer than one hour (by about two billionths of a second). Such a tiny difference can be determined by a high precision atomic clock.

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An additional example is provided by the well-known Doppler effect (diagram above)  by which any form of electromagnetic radiation changes its wavelength and frequency as the source moves toward or away from an observer. The Doppler formula for z, the red shift is:  z = (Dl/ l) = v/c

tells us exactly how this change should vary with velocity. However, when a source is moving very close to light speed we will find the change doesn't quite agree with what is observed, so the formula must be modified for relativistic effects - just as ordinary motions must be modified whenever v ~ c. Thus, the correct relativistic effect (say for changing frequency) is given by:

f o/f = [[1 - v2/c2] ½/ (1 + v/c) =

[(1 - v/c) ½]/(1 + v/c) ½]

In terms of z, the red shift

1 + z = [1 + v/c] ½]/[1 - v/c] ½

Note that z tends to infinity when v approaches c.

In any problems where z becomes comparable to 1, one must apply the relativistic formula. Failure to do so results in an incorrect interpretation that the redshifts indicate recessional velocities greater than light speed.

Problems for the Ambitious:

1) Assume two astronauts are traveling at v = 0.95c on a journey to the system of Alpha Centauri. We on Earth would say that it takes 4.2 / 0.95c = 4.4 years to reach the system 4.2 light years distant. But the astronauts dispute this.

(a) How much time passes on the astronauts' clocks?

(b) What is the distance to Alpha Centauri as measured by the astronauts? (Hint: this is an exact analog of the muon path length problem (#3) from the previous problem set)

2) According to Hubble's law, the distant galaxies are receding from us at speeds proportional to their distances, d, e.g. v = Hd. (Where H = 2.26 x 10-18 s-1, currently).

a) How far away would a galaxy be in light years whose velocity relative to the Earth is c?

b) Would it be observable from Earth? (Take 9.5 x 1015 m = 1 LY)

3) A galaxy in Hydra emits light with a red shift corresponding to a recessional velocity of 6 x 104 km/s.

a) What is its distance according to Hubble's law?

b) What is the value of z?

c) Assume this galaxy passed Earth T years ago and has moved with constant velocity ever since, what is the value of T?

4) Some observations reported on the quasar 3C-9 suggest that when it emitted the light that just reached Earth it was receding at a velocity of 0.8c. One of the lines identified in its spectrum has a wavelength of 1200 Å (angstroms) when emitted from a stationary source.

a) At what wavelength must this spectral line have appeared in the observed spectrum of the quasar?

b) What is its red shift, z?

c) Find its corrected velocity, v.


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