Consider two observers, A and B* moving towards each other as
shown here. 



Suppose we are told that A moves at 3ms^{1}
and B moves at 2ms^{1}. 

Hearing this we will assume that, in one second, A moves
3m measured along the ground
and similarly that B covers 2m
in the same time. 

In other words we are assuming that these
velocities are the velocities of the two observers relative to
the ground. 

What is the velocity of B relative to
A? 

To answer this question, imagine yourself to be A. What do you
observe? 

In one second you move
3m forwards and B moves
2m in the opposite sense, so B
moves 5m closer to you (that is, in
the negative sense) in one second. 

So, if you could make a measurement** of the velocity of B, you
would find 5ms^{1} in the
negative sense or 5ms^{1} 



Also, if we imagine ourselves to be B, using the same logic we conclude that the
velocity of A relative to B is +5ms^{1}. 



Now consider the following case. 



Suppose that A and B are both moving at
3ms^{1} 

Imagining yourself to be A, you will conclude that you will
never catch up with B. In other words, the velocity of B relative to
A is zero. 



By imagining yourself to be one of the observers you are giving
yourself zero velocity relative to that observer. In other words,
mathematically, you are removing (subtracting) the velocity
of that observer from your measurement. 

So, mathematically 

v_{B} relative to A = v_{B}
relative to the ground  v_{A} relative to the ground 

and, of course, there is nothing special about "the ground", it
could be relative to any other body/reference point and, to save a
bit of writing, I will use the following notation: 

v_{B(A)} = v_{B(C)}  v_{A(C)} 

where C represents a third body (the ground in this case). 

Using the figures for the first situation above we have 

v_{B(A)} = 2 
(+3) = 5ms^{1} 



If all this seems pretty trivial, be warned... Einstein found
that it is not quite true! 

However, the errors don't become a problem until the relative
velocities involved are very large. 

More of that on other pages. 



One of Einstein's aims in developing the theory of relativity
was to be able to take measurements made by one observer and
transform them into measurements made by another observer. 

In other words, if one observer says that a body is moving at
vms^{1}, what will another
observer say about the same body? 

The answer to this question obviously depends on the relative
motion of the two observers. 



We now introduce a third body***, p (for
pelican, as you can see). 







Suppose that we do not know how fast A and B are moving relative
to the ground but we do know that their relative velocity has
magnitude u = 4ms^{1} 

Suppose also that observer A has measured the velocity of p
(relative to A). 

Let this velocity be v. 

If A tells B that p moves at 10ms^{1}
relative to A then B (observing that he is moving in the same sense
as p) will conclude that p is moving at 6ms^{1}
in the negative sense, relative to him. 

Let this be v'. 

Clearly, this result is again obtained by a simple subtraction: 

v_{p(B)} = v_{p(A)} 
v_{B(A)} 

or 

v' = v  u 

v' = 10 (4) = 6 





* I was going to use the letters A and G for obvious reasons: A
for Albert and G for... ok, who knows? Yes, the great Groucho but
eventually settled on boring old A and B. 

**How might this measurement be made in practice?
See Measuring Relative Velocity 

*** Of course, we already had a "third
body" before; the ground. 
