*The Lorentz Contraction:*

Having
dealt with time dilation in the context of special relativity, we now consider what
happens to length. We return to the diagram of systems S and S' moving relative
to each other ('

*Special Relativity**Revisited 2*, Oct. 11, Fig. 1) and consider a meter stick of length L (= 1 m) pointed in the +x direction and moving in that direction with velocity v.We can also think of it as being at rest in the S' system with one end at x' = 0 and another end at x' = L', initially.

Its length in the system S' is therefore L'. At time t = 0 we take a photograph of the meter stick with a camera located in the S system. We inquire: What are the positions of the two ends of the meter stick in the S system? We can answer this by using the first of the Lorentz-Einstein transformations:

x = x' + vt'/(1 - v

^{2}/c

^{2})

^{½}

and

x' = x - vt/ (1 - - v

^{2}/c

^{2})

^{½}

We opt to use the lower form rather than the upper, having already specified an instant at t = 0 and hence looking for simplification. Then:

x' = x / (1 - v

^{2}/c

^{2})

^{½}

Now, the end of the meter stick that is at x' = 0 in S' is found to be at x = 0 in S. The

*other*end in S is seen to be at:

x = L' (1 - v

^{2}/c

^{2})

^{½}

This informs us the length of the moving meter stick in coordinate system S so we can say that the length of the moving meter stick as seen by the camera is:

L = L' (1 - v

^{2}/c

^{2})

^{½}

As can readily be seen, this implies

*length contraction*. For example, say the rod is moving at v = 0.6c, then the length L is (given L' = 1m):

L = (1 m) (1 - (0.6c)

^{2}/c

^{2})

^{½}= 1m [(1 - 0.36c

^{2}/c

^{2})

^{½}

L = (1m) (0.64)

^{½}= 1m (0.8) = 0.8 m (or 80 cm)

(Remember that by symmetry arguments of relativity, the observer in the other system will argue that L' = L (1 -v

^{2}/c

^{2})

^{½}so that

*from his viewpoint*a meter stick will be similarly foreshortened.)

Readers may recall that something similar occurred in the time transformation, i.e. each of two relatively moving clocks ran slower with respect to the other than to itself.

This brings us back to the solution of the Michelson-Morley experiment discussed in an earlier chapter. It is the curious, symmetric relativity in time and length (in direction of motion) which solves the paradox of the "missing ether wind" quite simply and logically. Moreover it is solved more fundamentally and satisfactorily than either Fitzgerald or Lorentz could manage.

Let's derive the

*new law for addition of velocities.*Assume an object in the S' system starts at the point x' = 0 at time t' = 0. It moves with constant velocity u' (relative to S') and in the time t' it travels a distance x'.

By definition, u' = x'/ t'

We ask: 'How fast does this object travel according to the observer at rest in System S?'

This observer will see a velocity given by the formula:

u = x/t = (x' + vt')'/ t' + x'v/c

^{2}or

u=
(t' + v)/(1 + x'v/t' c

u = (u' + v)/ 1 + u'v/ c

This is

u = (c + c)/ 1 + c

u = 2c/ 1 + 1 = 2c/2 = c

If u' and v

Say two objects are moving at 3c/4 towards each other, then what is their relative velocity as recorded in a system S observing their approach?

We have: u' = 3c/4 and v = 3c/4

then:

u = (u' + v)/ 1 + u'v/c

where: (3c/4 + 3c/4) = 3c/2

and:

u' v = (3c/4) (3c/4) = 9c

Then:

u = (3c/2)/ [1 + 9c

u = (3c/2)/ (1 + 9/16) = (3c/2)/ (25/16)

u = (3c/2) (16/25) = 24c/25

^{2})u = (u' + v)/ 1 + u'v/ c

^{2}This is

*the relativistic formula for addition of velocities*. If u' and v = c then the formula yields:u = (c + c)/ 1 + c

^{2}/c^{2}u = 2c/ 1 + 1 = 2c/2 = c

If u' and v

*are less than c*then u must always be less than c.*Example Problem*:Say two objects are moving at 3c/4 towards each other, then what is their relative velocity as recorded in a system S observing their approach?

We have: u' = 3c/4 and v = 3c/4

then:

u = (u' + v)/ 1 + u'v/c

^{2}= (3c/4 + 3c/4)/1 + u'v/c^{2}where: (3c/4 + 3c/4) = 3c/2

and:

u' v = (3c/4) (3c/4) = 9c

^{2}/16Then:

u = (3c/2)/ [1 + 9c

^{2}/16/c^{2}]u = (3c/2)/ (1 + 9/16) = (3c/2)/ (25/16)

u = (3c/2) (16/25) = 24c/25

**Problems for ambitious readers**:

1)
Assume that the lifetime of quasar 3c-9 is 1 million years as measured in

2) Suppose that you happen to be moving at a velocity of 3c/4 past a remote observer who picks up a stopwatch and then sets it down. Using a high power telescope you observe he held the watch for 9 seconds. How long would HE think that he held it?

3) An astronaut orbits the Earth at a distance of 7 x 10

4) Consider 3 galaxies: A, B and C. An observer in A measures the velocities of B and C and finds they are moving in opposite directions - each with a speed of 0.7c relative to him., i.e.

*its own rest frame*. Over what total time span (in Earth-measured time) would its radiation be received at the Earth? (Assume 3c-9's velocity relative to the Earth remains constant)2) Suppose that you happen to be moving at a velocity of 3c/4 past a remote observer who picks up a stopwatch and then sets it down. Using a high power telescope you observe he held the watch for 9 seconds. How long would HE think that he held it?

3) An astronaut orbits the Earth at a distance of 7 x 10

^{6}m from its center for a week. How much younger than his twin on Earth is he when he lands? Assume standard orbital speed of 18,000 mph and neglect the rotation of the Earth.4) Consider 3 galaxies: A, B and C. An observer in A measures the velocities of B and C and finds they are moving in opposite directions - each with a speed of 0.7c relative to him., i.e.

*(0.7c) <-----------(B)-----(A)-----(C)--------->(0.7c)*

What is the speed of A observed by someone in B?

What is the speed of C observed by someone in B?

The observer in A thinks that the two other galaxies are receding from him at a rate 1.4c. Show him how this is wrong, by providing the correct result.

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