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Consider a particle which, in its instantaneous rest frame , has
mass and constant acceleration in the direction .
Let us transform to a frame , in the standard configuration with respect
to , in which the particle's instantaneous
velocity is . What is the value of , the
particle's instantaneous acceleration, in
S?
The easiest way in which to answer this question is to consider the
acceleration 4vector [see Eq. (1429)]

(1541) 
Using the standard transformation, (1397)(1400), for 4vectors, we obtain
Equation (1542) can be written

(1544) 
where is the constant
force (in the direction) acting on the particle in .
Equation (1544) is equivalent to

(1545) 
where

(1546) 
Thus, we can account for the ever decreasing acceleration of a particle
subject to a constant force [see Eq. (1542)] by supposing that the
inertial mass of the particle increases with its velocity according to
the rule (1546). Henceforth, is termed the rest mass, and
the inertial mass.
The rate of increase of the particle's energy satisfies

(1547) 
This equation can be written

(1548) 
which can be integrated to yield Einstein's famous formula

(1549) 
The 3momentum of a particle is defined

(1550) 
where is its 3velocity. Thus, by analogy with Eq. (1545),
Newton's law of motion can be written

(1551) 
where is the 3force acting on the particle.
The 4momentum of a particle is defined

(1552) 
where is its 4velocity. The 4force acting on the particle
obeys

(1553) 
where is its 4acceleration. It is easily demonstrated that

(1554) 
since

(1555) 
Next: The force on a
Up: Relativity and electromagnetism
Previous: Fields due to a
Richard Fitzpatrick
20060202