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Relativistic Combination of Velocities
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  
TA = time period of pulses transmitted by A  
TB = time period pulses received and transmitted by B and  
TC = 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 << c2 (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  
ví = 1.4c/(1 + 0.45c2/c2) = 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  
ví = 1.5c/(1 + 0.5c2/c2) = c   
   
This example simply illustrates that the equation used to calculate ví agrees with Einsteinís 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...  
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