bei48482_FM

Example 1.5

Find the acceleration of a particle of mass mand velocity vwhen it is acted upon by the con-

stant force F,where Fis parallel to v.

Solution

From Eq. (1.19), since addt,

F (m)m 

m 



We note that Fis equal to ^3 ma, notto ma. Merely replacing mby min classical formulas

does not always give a relativistically correct result.

The acceleration of the particle is therefore

a (1^2 c^2 )^3 ^2

Even though the force is constant, the acceleration of the particle decreases as its velocity in-

creases. As Sc, aS0, so the particle can never reach the speed of light, a conclusion we

expect.

1.8 MASS AND ENERGY

Where E 0 mc^2 comes from

The most famous relationship Einstein obtained from the postulates of special

relativity—how powerful they turn out to be!—concerns mass and energy. Let us see

how this relationship can be derived from what we already know.

As we recall from elementary physics, the work Wdone on an object by a con-

stant force of magnitude Fthat acts through the distance s, where Fis in the same

direction as s,is given by WFs. If no other forces act on the object and the ob-

ject starts from rest, all the work done on it becomes kinetic energy KE, so KE Fs.

In the general case where Fneed not be constant, the formula for kinetic energy is

the integral

KE

s

0

Fds

In nonrelativistic physics, the kinetic energy of an object of mass mand speed is

KE^12 m^2. To find the correct relativistic formula for KE we start from the relativistic

form of the second law of motion, Eq. (1.19), which gives

KE

s

0

ds

m

0

d(m)



0

d



m



 1 ^2 c^2

d(m)



dt

F



m

ma



(1^2 c^2 )^3 ^2

d



dt

^2 c^2

(1^2 c^2 )^3 ^2

1



 1 ^2 c^2





 1 ^2 c^2

d



dt

d



dt

26 Chapter One

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