Changes
Added derivations of force
== Force in Relativity ==
== Force in Relativity ==
In [[Theory of Relativity|special relativity]], these prescriptions must be modified so as to be Lorentz invariant, which among other things, means that all [[inertial reference frame]]s stand on equal footing and have the same prescription for all physical and dynamical quantities, though observers in different [[inertial reference frame]]s will measure different values for many of them, each observer being correct for his own frame. The relationship between force and [[acceleration]] in [[special relativity]] can be derived using Newton's second law:
In [[Theory of Relativity|special relativity]], these prescriptions must be modified so as to be Lorentz invariant, which among other things, means that all [[inertial reference frame]]s stand on equal footing and have the same prescription for all physical and dynamical quantities, though observers in different [[inertial reference frame]]s will measure different values for many of them, each observer being correct for his own frame. The relationship between force and [[acceleration]] in [[special relativity]] can be derived using Newton's second law using relativistic momentum:
<math>F = \frac{dP}{dt} = \frac{d \gamma}{dt} m_0 v + \gamma \frac{dm_0}{dt} v + \gamma m_0 \frac{dv}{dt}</math><br/>
<math>F = \frac{dP}{dt} = \frac{d \gamma}{dt} m_0 v + \gamma \frac{dm_0}{dt} v + \gamma m_0 \frac{dv}{dt}</math><br/>
Since <math>m_0 </math> is the [[rest mass]] and therefore constant, it's derivative is zero.
Since <math>m_0 </math> is the [[rest mass]] and therefore constant, it's derivative is zero. The derivative of [[speed]] is [[acceleration]] Hence,
<math>F = \frac{d \gamma}{dt} m_0 v + 0 + \gamma m_0 a</math><br/>
Now consider a force that is parallel to motion and perpendicular to motion.
=== Force parallel to motion ===
To differentiate the [[Lorentz factor]], <math>\gamma</math>, use the chain rule:
<math>\frac{d \gamma}{dt} m_0 v + \gamma m_0 a = \frac{d \gamma}{dv} \frac{dv}{dt} m_0 v + \gamma m_0 a</math><br/>
Differentiating the {Lorentz factor]] with respect to speed, we find
<math>\frac{d \gamma}{dv} = \frac{{\gamma}^3 v}{c^2}</math>
<math> F_{\parallel} = \frac{{\gamma}^3 v}{c^2} a m_0 v + \gamma m_0 a = (\frac{{\gamma}^3 v^2}{c^2} + \gamma) m_0 a </math>
After a bit of rearranging, we find that the term in the bracket simplifies to <math>{\gamma}^3</math>. Hence we find for a force acting parallel to an objects motion is:<br/>
<math> F_{\parallel} = {\gamma}^3 m_0 a</math>
=== Force perpendicular to motion ===
A force perpendicular to an objects motion cannot do any [[work]] on that object. Hence the [[kinetic energy]] of the object and so its [[speed]] must remain constant. Therefore <math>\gamma</math> must be constant and its derivative zero. Hence we derive the expression for a perpendicular force as:
<math>F_{\perp} = \gamma m_0 a</math><br/>
=== Direction of Force and Acceleration ===
Rearranging the two forces for acceleration, we find:
<math> a_{\parallel} = \frac{F_{\parallel}}{{\gamma}^3 m_0} </math><br/>
<math> a_{\perp} = \frac{F_{\perp}}{\gamma m_0} </math>
Note that since all forces can be resolved into a component parallel and perpendicular to motion, and for all velocities except zero, the parallel and perpendicular accelerations will be '''different'''. Hence we can conclude that in the general case (where the force is not completely parallel or perpendicular to motion) that the [[acceleration]] [[vector]] will have a '''different''' direction to the force [[vector]].
== Fundamental Forces ==
== Fundamental Forces ==
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