Relativistic Mechanics - Force

Force

In special relativity, Newton's second law does not hold in its form F = ma, but it does if it is expressed as

where p = γm0v is the momentum as defined above and m0 is the invariant mass. Thus, the force is given by

Derivation

Starting from

Carrying out the derivatives gives

using the identity

,

gives

If the acceleration is separated into the part parallel to the velocity and the part perpendicular to it, one gets

\begin{align} \mathbf{F} & = \frac{\gamma^3 m_0 v^{2}}{c^2} \, \mathbf{a}_{\parallel} + \gamma m_0 \, (\mathbf{a}_{\parallel} + \mathbf{a}_{\perp})\\ & = \gamma^3 m_0 \left( \frac{v^2}{c^2} + \frac{1}{\gamma^2} \right) \mathbf{a}_{\parallel} + \gamma m_0 \, \mathbf{a}_{\perp} \\ & = \gamma^3 m_0 \left( \frac{v^{2}}{c^2} + 1 - \frac{v^{2}}{c^2} \right) \mathbf{a}_{\parallel} + \gamma m_0 \, \mathbf{a}_{\perp} \\ & = \gamma^3 m_0 \, \mathbf{a}_{\parallel} + \gamma m_0 \, \mathbf{a}_{\perp} \end{align}\,

Consequently in some old texts, γ3m0 is referred to as the longitudinal mass, and γm0 is referred to as the transverse mass, which is numerically the same as the relativistic mass. See mass in special relativity.

If one inverts this to calculate acceleration from force, one gets

The force described in this section is the classical 3-D force which is not a four-vector. This 3-D force is the appropriate concept of force since it is the force which obeys Newton's third law of motion. It should not be confused with the so-called four-force which is merely the 3-D force in the comoving frame of the object transformed as if it were a four-vector. However, the density of 3-D force (linear momentum transferred per unit four-volume) is a four-vector (density of weight +1) when combined with the negative of the density of power transferred.

Read more about this topic:  Relativistic Mechanics

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