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The Derivation of the Energy–Momentum Relation in Relativistic Classical Mechanics |
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Special and General Relativity have been tested and accepted as correct theories of the physical world. Classical Newtonian mechanics is an adequate description of the physical world for velocities small compared to the speed of light. It is often alluded that relativistic mechanics applies only for velocities approaching the speed of light. That is not correct. Relativistic mechanics is the correct description at all velocities, but Newtonian mechanics is an adequate approximation at low velocities. However, some of the hallowed formulas such as the kinetic energy of a particle being equal to one half of the product of its mass and its velocity squared are only approximations.
The key to the exact formulas for relativstic mechanics is the formula for relativistic mass. Let m_{0} be the rest-mass of a particle and v its velocity. Velocity relative to the speed of light c is denoted as β. Then relativistic mass m is given by
where, as noted previously, β=v/c. Mass is a nonlinear function of velocity.
Momentum, which in Newtonian mechanics is a linear function of velocity m_{0}v, is in relativistic mechanics mv, a nonlinear function of velocity.
For a free particle the total energy E is
The first approximation of this function is
The Maclaurin series expansion of (1-(v/c)^{2})^{-1/2} is:
Thus the series expansion for energy is:
The kinetic energy is E-m_{0}c^{2} and thus while (1/2)m_{0}v^{2} is the first order approximation of kinetic energy it is not the precise value. The total kinetic energy E-m_{0}c^{2} is larger than (1/2)m_{0}v^{2} by a factor of
When v/c=0.1 this factor is equal to about 1.00813. Thus even at a speed of 18,600 miles per second the Newtonian formula for kinetic energy is in error only by less than one percent. But for v/c=0.5 the error is approximately 20 percent and at v/c=0.8 the errror is more than 50 percent.
The formula is
where E is the total energy.
The standard practice in textbooks for the derivation of this formula starts with the creation of a four dimensional momentum vector P given by:
The squared magnitude of P is then said to be
It is then argued that the squared magnitude of P is constant and furthermore that that constant is m_{0}².
This derivation has some perplexing transitions so it is worse than hand waving; it is more like theoretical sleight of hand. There is no explanation for how the sign of the sum of squares of the three momenta is opposite of the sign of the square zeroeth component. An alternate derivation is considered below based on the formulas for the terms in the relation.
Note that relativistic momentum p is given by
Note also that (m_{0}c²)² is equal to m_{0}²c^{4}.
Thus
On the other hand
Thus
(To be continued.)
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