On most other pages in this section on the
Special Theory of Relativity, we have been considering
just two
observers, the famous A and B. 

Here we will introduce a third observer... yes, C. 



The aim here is to answer this question: 

If we know the relative
velocity of A and B and A has also measured the velocity of C
relative to him/herself, how can we calculate the velocity of C
relative to B? 

In the diagram above (which is drawn from
A's point of view) we are suggesting that A has measured the
velocity of B relative to him/herself and found it to be u. 

A has also measured the relative velocity of
C and found it to be v. 



The Galilean Relativity answer
(the "common sense" answer) to this
question would be simple, let's put some numbers in: 

if v = 200ms^{1} and u = 150ms^{1}
then the velocity of C relative to B, v' would be... give yourself a
gold star if you got 50ms^{1}. 

In other words, we used: v' = v  u 



We will use the "flashing lights" method
introduced here to show that, unfortunately, things are not so
simple. 



In order to measure the velocities u and v
we imagine that A has sent out flashes of light towards B and C but
that B will also send out flashes to C. 

Remember that to use this method of
measuring velocities we defined the constant, k such that 



We assume the following
procedure: 

A sends flashes of light to B and B sends
flashes of light to C. 

B sends a flash to C at
the same instant as he/she receives flashes from A. 

It is clear that C will receive both sets of
flashes simultaneously. 



This means that the two values of k
(corresponding to light going from A to B then from B to C)
combined, are equivalent to the value of k for light going directly
from A to C. 



Let 

T_{A} = time period of pulses
transmitted by A 

T_{B} = time period pulses
received and transmitted by B and 

T_{C} = time period of pulses
received by C 



We can therefore write 



from which it should be clear that the way
we combine the "k" values is 



and, as we will be interested in what
happens to the pulses of light sent by B to C, we rearrange this to
give 



therefore, the velocity of C relative to B
(v') is given by 



In general we have shown
that, for a relative velocity, v 


which means that 



so, in this case, we have 



which can be written as 



and, as any 10 year old could tell you*,
this expands to 



which rather charmingly
reduces to 



finally giving us the
following useful equation for the relativistic combination of
velocities 





Note that, for uv << c^{2} (which is
frequently the case), this equation reduces to Galilean case, v' = v
 u 



A couple of numerical examples. 

Consider the following situation: 



Let u = 0.5c and v = 0.9c 

The equation above gives 

v’ = 1.4c/(1 + 0.45c^{2}/c^{2})
= 0.965c 

whereas, the Galilean result would be 1.4c 

However, if u = 100ms^{1
}and v = 300ms^{1}
the results are effectively the same, 400ms^{1} 



Now imagine that the third "body" is
actually a pulse of light as shown below. 



If the velocity of B relative to A (u) is
again 0.5c then the equation gives 

v’ = 1.5c/(1 + 0.5c^{2}/c^{2})
= c 



This example simply illustrates that the
equation used to calculate v’ agrees with Einstein’s initial
postulate that the velocity of light is a constant... which isn't
really surprising, as all the equations of special relativity are
derived based on this assumption! 





*go out and find me a 10 year old, I can't make head or tail of
it... extra points if you can say which movie this phrase is taken
from... 
