Variation of Mass with Velocity:
Consider effect of very high speeds on mass and consequently momentum and force acting on object. Talking about momentum, reminds of Newton's second law of motion. This law describes relationship between two quantities. It says rate of change of momentum is directly proportional to applied force. In mathematical terms we write this as F→ = d(mv→)/dt. For constant mass m, we get F→ = ma→ or a→ = F→/m. But we know that v = u + at so that as t → ∞, v → ∞. Hence, classically, if the constant force is applied on the object for sufficiently long time, its velocity increases infinitely. This clearly opposes theory of special relativity as velocity of the object is prevented from reaching or exceeding velocity of light in relativistic mechanics. Hence at very high velocities, something is wrong with F→ = ma→
Consider the elastic collision of two identical balls B1 and B2 each of mass m0 measured in the frame of reference in which they were at rest. Two observers O1 and O2 are in motion at very high (relativistic) velocity v in opposite direction parallel to x-axis. Observer O1 carries with him ball B1 whereas observer O2 carries with him ball B2. While passing each other, each observer throws his ball perpendicular to his direction of motion with the velocity v0 (where v0 << v) such that glancing collision of balls occurs at point D, which is mid-way between two observers and rebounds with same velocity and at equal angles to x-axis. As seen by each observer, path of motion of his ball is strictly ±.
Therefore, as seen by observer at rest at D in xyz frame, paths of balls are as shown in the figure. For him,
[Σpyi]before collision → movo + mo(-vo) = 0
[Σpyi]after collision → mo(-vo) + movo = 0
[Σpyi]before collision = [Σpyi]after collision = 0
In x-direction, he sees balls approaching each other with equal velocity before collision and move away from each other after collision. Therefore he writes:
[Σpxi]initial → mov + mo(-vo) = 0
[Σpxi]final→ mov + mo(-vo) = 0
[Σpxi]initial = [Σpxi]final = 0
Therefore for this observer, momentum is conserved.
Now, consider collision from view-point of observer O1. He sees himself to be at rest in his frame of reference and observer O2 approaching with velocity v. Remember that v is very large. Collision as seen by observer O1 is shown in figure.
Ball moves strictly in ± direction with velocity v0 so that v1x = 0 and v1y = 0, v1x and v1y being the x and y components of velocity of ball. On the other hand, he sees B2 ball move with velocity whose x and y components are v2xand v2y respectively, with v2y reversed after collision. He writes the law of conservation of the y- directed momentum as
[Σpyi]before collision = [Σpyi]after collision
= m0v0 + m(-v2y) = m0(-v0) + mv2y
2m0v0 = 2mv2y
m = m0v0/v2y
Notice that though he began with equal masses m0 for both balls, he reasons that mass of ball B2 may have changed and writes m for it as it is moving with relativistic velocity v. He though retains m0 for ball B1 as it is moving slowly with velocity v0 so that its mass is effectively constant. Now, remember that observer O2 threw his ball perpendicular to his direction of motion with velocity v0 as measured by him. But observer O1 measures this as v2y because O2 is moving at relativistic speed relative to him. So, we will be justified to use velocity transformation equation, namely Vy = V'y(√1 - β2). In this case, Vy = v2y, V'y = v0 and V'x = v1x = 0. Making these substitutions, we obtain
v2y = v0(√1 - β2)/(1 + (v/c2)v1x) = v0(√1 - β2)
m = m0/(√1 - v2/c2)
Mass m is known as relativistic mass. m0 is known as rest mass.
Momentum and Force in Relativistic Mechanics:
At very high velocities force could no longer be represented by Newton's second law as F = ma as m is not constant but differs with velocity. We stress here that F = ma is not correct in relativistic range. Best we can do is to leave relation in form F→ = d(mv→)/dt. This means F→ = m(dv/dt) + v(dm/dt) and (dm/dt) doesn't vanish if velocity of body differs with time. Resultant force is always equal to rate of change of momentum.
Then, provided m is given by equation, momentum can be written as p→ = mv→. Therefore, for very high (relativistic velocity), momentum is correctly represented by
p→ = m0v→/(1 - v2/c2)
In terms of components, we can write equation
px = m0vx/√(1 - v2/c2), py = m0vy/√(1 - v2/c2) and pz = movz/√(1 - v2/c2)
Where v2 = v2x + v2y + v2z
Velocity v which appears in denominator of these expressions is always velocity of object as measured from the inertial frame of reference. It is not velocity of the inertial frame itself. Velocity in numerator can be any of components of velocity vector.
Newton's second law can then be written in form
F→ = dp→/dt = d/dt[m0v→/√(1 - v2/c2)] = d/dt(mv→)
If net force acting on the system of particles is zero, momentum in relativistic range is conserved provided the mass of particle is given by equation.
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