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© The scientific sentence. 2010


Adding velocities



1. With Galilean transformations:

Let's consider an object moves from x1 at the time t1 
to x2 at the time t2 in the inertial frame xOy. (x1,t1) 
and (x2,t2) represente the space-time coordiantes of 
two events measured by this observer. The velocity 
ux mesured by this observer 
is:
ux  = (x2 - x1)/(t2 - t1)

Similarly, for an observer in the inertial frame x'Oy' moving at 
the speed v relative to xOy, the two events are (x'1,t'1) and 
(x'2,t'2), and the velocity measured by this observer is: 

ux'  = (x'2 - x'1)/(t'2 - t'1)

Using the newtoniam concept of an universal absolute time, 
we write:
t'1= t1 and t'2 = t2. Therefore:

ux'  = (x'2 - x'1)/(t'2 - t'1)  = (x'2 - x'1)/(t2 - t1) 

Using the Galilean transformations:
x'1 = x1 - vt1
x'2 = x2 - vt2 , we get:

ux'  = (x2 - vt2 - x1 + vt1)/(t2 - t1) =  
(x2 - x1)/(t2 - t1) - v = ux - v


ux' = ux - v


This the galilean velocity transformations that is 
not correct for high speeds .

2. With Lorentz transformations:

We have:
x'1 = γ(x1 - vt1)
t'1 = γ (t1 - v x1/c2)
and
x'2 = γ(x2 - vt2)
t'2 = γ (t2 - v x2/c2)
We get:

ux'  = (x'2 - x'1)/(t'2 - t'1)) =  
[γ(x2 - vt2) - γ(x1 - vt1)]/[γ (t2 - v x2/c2) - γ (t1 - v x1/c2)] 
= [(x2 - x1)/(t2 - t1) - v ]/[1 - v/c2((x2 - x1)/(t2 - t1))]
= [ux - v]/[1 - v ux/c2]

Similarly:
ux  = [ux' + v]/[1 + vux/c2]

ux'  =  (ux - v)/(1 - vux/c2)
ux  = (ux' + v)/(1 + vux/c2)


For "y" and "z" components, we have:

uy' = γ uy/ (1 - vux/c2)
uz'  = γ uz/ (1 - vux/c2)

ux'  =  (ux - v)/(1 - vux/c2)
uy' = γ uy/ (1 - vux/c2)
uz'  = γ uz/ (1 - vux/c2)



Remark that, it the moving observer in x'Oy, referential 
frame measures the velocity of light in this frame and finds 
V'x = c , the first relation gives also Vx = c. The conclusion 
is the speed of light is the same in these two referential 
frames. Just a verification!


Example:

If we observe two spacecraft S1 and S2 going away at 
speed of v1 = 0.7c and v2 = 0.8 c respectively, what 
is the speed v21 of S2 relative to S1? 

Consider S1 as the inertial frame xOy, so:
v21 = ux', v2 = ux and v1 = v.
Therefore:
v21 = [v2 - v1][1 - v1 v2/c2] = c[0.8 - 0.7]/[1 - 0.8 0.7] 
= 0.223 c

  


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