Particle motion equations

Derivation of the particle coordinate path equation.

The derivation of the particle path is somewhat more involved and the reader is referred to the mathpages derivation by Kevin Brown. As far as I can tell the derivation is correct except that Kevin Brown succumbs to the usual temptation to insert an ad-hoc absolute function so that the result agree with the accepted traditional physical interpretation. These are the final corrected steps shown below.

Starting with:

$t = \sqrt{{R \over 2m} -1}\left[\left( {R \over 2} +2m \right)\alpha + \left(R \over 2\right) sin(\alpha)}\right] - 4m\sqrt{-1}\,\,arctan\left(cos(\alpha)-1 \over sin(\alpha)\sqrt{1-R/2m} \right)$

Substituting

the equation becomes:

$t = \left( {R \over 2} +2m \right)Q\alpha + \left(R \over 2\right)Q \,sin(\alpha)}\right] -4m\sqrt{-1}\,\,arctan\left(cos(\alpha)-1 \over sin(\alpha)\sqrt{-1}\,Q \right)$

Substituting the identity

we get

$t = \left( {R \over 2} +2m \right)Q\alpha + \left(R \over 2\right)Q \,sin(\alpha)}\right] -4m\sqrt{-1}\,\,arctan\left(-tan(\alpha/2) \over \sqrt{-1}\,Q \right)$

Now the final substitution of

gives:

$t = \left( {R \over 2} +2m \right)Q\alpha + \left(R \over 2\right)Q \,sin(\alpha)}\right] -4m \, ln\left(Q+tan(\alpha/2) \over Q-tan(\alpha/2) \right)$

and not

$t = \left( {R \over 2} +2m \right)Q\alpha + \left(R \over 2\right)Q \,sin(\alpha)}\right] -4m \, ln\left(\left| Q+tan(\alpha/2) \over Q-tan(\alpha/2)\right|\right)$ as claimed by Kevin Brown.

The alpha variable that appears in the particle equation is a parametric variable. It can be removed to give a normal Cartesian equation by noting that mathpages gives the radial part of the proper motion of a particle as r=R(1+cos(alpha))/2 which is easily solved to give alpha = acos(2r/R-1) which is the form used in the applets.

The derivation by Kevin Brown of the particle motion in terms of proper time appears to be correct and can be seen in the link given above.

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dt = dT - 2m dr/(r-2m) in equation 13.4