非一般的物理題(2)

First, please note that the definition of force is the same in both classical and relativistic machanic. Force is defined as the rate of change of momentum, i.e.

F = dp/dt ...(1)

where F is force and p is momentum, which is the multiple of mass m and velocity v, i.e.

p = mv

In classical mechanic, m is constant, so we can write F = m dv/dt.
But, in relativity, m is NOT constant, so it is absolutely WRONG to write F = m dv/dt. Rather, we have a variable m

m = mr / √(1 - v²/c²) , where mr is the constant rest mass.

Hence,

p = mr v/ √(1 - v²/c²) ...(2)

Rearranging (2), we can easily get

v = p c / √[ (mr c)² + p²] ...(3)

Now, let us find the solution to the problem by integrating (1).
Let the constant force be -F0 and the initial velocity and momentum be v0 and p0. (By (2), we can easily write p0 in terms of v0.)

∫0t (-F0 ) dt = ∫p0p dp

-F0 t = p - p0

p = p0 - F0 t

Putting this into (3), we have

v = (p0 - F0 t) c / √[ (mr c)² + (p0 - F0 t)²] ...(4)
where p0 = mr v0/ √(1 - v0²/c²)

The v-t graph can be sketched according to
1. when t is small, as p0 is large and so (p0 - F0 t) is almost constant, v decreases very slowly (or almost constant).
2. when v is almost zero, p0 ≈ F0 t , so v ≈ (p0 - F0 t) / mr , where the v-t graph is a straight line with slope - F0 t / mr, consistent with Newtonian mechanic.

Please note that when v << c and F0 t << mr c,
by (2), p ≈ m v
and (mr c)² + (p0 - F0 t)² ≈ (mr c)²
Then (4) becomes

v ≈ (mr v0 - F0 t) c / √[ (mr c)²] = v0 - F0 t / mr , which is consistent with Newtonian mechanic.

For definition of force, please refer to http://en.wikipedia.org/wiki/Force