15.20: Acceleration
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Figure XV.33 shows two references frames, Σ and Σ′, the latter moving at speed v with respect to the former.
A particle is moving with acceleration a′ in Σ′. (“ in Σ′ ” = “referred to the reference frame Σ′ ”.) The velocity is not necessarily, of course, in the same direction as the acceleration, and we’ll suppose that its velocity in Σ′ is u′. The acceleration and velocity components in Σ′ are a′x′,a′y′,u′x′,u′y′.
What is the acceleration of the particle in Σ? We shall start with the x-component.
The x-component of its acceleration in Σ is given by
ax=duxdt,
where
ux=u′x+v1+u′xvc2
and
t=γ(t′+vx′c2)
Equations 15.16.2 and 15.5.19 give us
dux=duxdu′xdu′x=du′xγ2(1+u′xvc2)2
and
dt =∂t∂t′dt′+∂t∂x′dx′=γdt′ + γvc2dx′
On substitution of these into Equation ??? and a very little algebra, we obtain
ax=a′γ3(1+u′xvc2)3
The y-component of its acceleration in Σ is given by
ay=duydt,
We have already worked out the denominator dt(Equation ???). We know that
uy=u′y′γ(1+u′x′vc2)
from which
duy=∂uy∂u′x′+∂uy∂u′y′∂u′y′=1γ(−vu′y′c2(1+vu′x′c2)2du′x′+11+vu′x′c2du′y′).
Divide Equation ??? by Equation ??? to obtain
ay=1γ2(−vu′y′c2(1+vu′x′c2)2a′x′+11+vu′x′c2a′y′).