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**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 B_{1} and B_{2} each of mass m_{0} measured in the frame of reference in which they were at rest. Two observers O_{1 }and O_{2} are in motion at very high (relativistic) velocity v in opposite direction parallel to x-axis. Observer O_{1} carries with him ball B_{1} whereas observer O_{2} carries with him ball B_{2}. While passing each other, each observer throws his ball perpendicular to his direction of motion with the velocity v_{0} (where v_{0} << 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,

[Σp_{yi}]_{before collision} → m_{o}v_{o} + m_{o}(-v_{o}) = 0

[Σp_{yi}]_{after collision} → m_{o}(-v_{o}) + m_{o}v_{o} = 0

[Σp_{yi}]_{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:

[Σp_{xi}]_{initial} → m_{o}v + m_{o}(-v_{o}) = 0

[Σp_{xi}]_{final}→ m_{o}v + m_{o}(-v_{o}) = 0

[Σp_{xi}]_{initial} = [Σpxi]_{final} = 0

Therefore for this observer, momentum is conserved.

Now, consider collision from view-point of observer O_{1}. He sees himself to be at rest in his frame of reference and observer O_{2} approaching with velocity v. Remember that v is very large. Collision as seen by observer O_{1} is shown in figure.

Ball moves strictly in ± direction with velocity v_{0} so that v_{1x} = 0 and v_{1y} = 0, v_{1x} and v_{1y }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 v_{2x}and v_{2y} respectively, with v2y reversed after collision. He writes the law of conservation of the y- directed momentum as

[Σp_{yi}]_{before collision }= [Σp_{yi}]_{after collision}

= m_{0}v_{0} + m(-v_{2}y) = m_{0}(-v_{0}) + mv_{2}y

2m_{0}v_{0} = 2mv_{2y}

m = m_{0}v_{0}/v_{2y}

Notice that though he began with equal masses m_{0} for both balls, he reasons that mass of ball B_{2} may have changed and writes m for it as it is moving with relativistic velocity v. He though retains m_{0} for ball B_{1} as it is moving slowly with velocity v_{0} so that its mass is effectively constant. Now, remember that observer O_{2} threw his ball perpendicular to his direction of motion with velocity v_{0} as measured by him. But observer O_{1} measures this as v_{2y} because O_{2} is moving at relativistic speed relative to him. So, we will be justified to use velocity transformation equation, namely Vy = V'y(√1 - β^{2}). In this case, V_{y} = v_{2y}, V'_{y} = v_{0} and V'_{x} = v_{1x} = 0. Making these substitutions, we obtain

v_{2y} = v_{0}(√1 - β^{2})/(1 + (v/c^{2})v_{1x}) = v_{0}(√1 - β^{2})

m = m_{0}/(√1 - v^{2}/c^{2})

Mass m is known as relativistic mass. m_{0} 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^{→} = m_{0}v^{→}/(1 - v^{2}/c^{2})

In terms of components, we can write equation

p_{x }= m_{0}v_{x}/√(1 - v^{2}/c^{2}), p_{y} = m_{0}v_{y}/√(1 - v^{2}/c^{2}) and p_{z} = m_{o}v_{z}/√(1 - v^{2}/c^{2})

Where v^{2} = v^{2}_{x} + v^{2}_{y} + v^{2}_{z}

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[m_{0}v^{→}/√(1 - v^{2}/c^{2})] = 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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