Page:BumsteadContraction.djvu/14

The acceleration along the radius vector is

$\scriptstyle {f_{n}\sin \phi +f_{t}\cos \phi ={\frac {{\text{V}}^{2}e}{m_{0}r^{2}}}(1-\beta ^{2})^{\frac {1}{2}}(1-\beta ^{2}\cos ^{2}\phi )}$

The acceleration perpendicular to the radius vector is

$\scriptstyle {f_{n}\cos \phi -f_{t}\sin \phi ={\frac {{\text{V}}^{2}e}{m_{0}r^{2}}}(1-\beta ^{2})^{\frac {1}{2}}\sin \phi \cos \phi \cdot \beta ^{2}}$

If we take the earth as a numerical example, this perpendicular acceleration is very small. Its maximum value will occur when the earth is at the extremities of the minor axis of its orbit; at this point

$\scriptstyle {\cos \phi =\epsilon ;\ \sin \phi ={\sqrt {1-\epsilon ^{2}}};}$

where $\scriptstyle {\epsilon }$ is the eccentricity of the orbit. Taking $\scriptstyle {\epsilon =1\cdot 7\times 10^{-3}}$ and $\scriptstyle {\beta ^{2}=10^{-8}}$ we find

Acceleration along $\scriptstyle {r={\frac {{\text{V}}^{2}e}{m_{0}r^{2}}}(1-\beta ^{2})^{\frac {1}{2}}~[1-2\cdot 9\times 10^{-12}]}$

Acceleration perpendicular to $\scriptstyle {r={\frac {{\text{V}}^{2}e}{m_{0}r^{2}}}(1-\beta ^{2})^{\frac {1}{2}}~[1\cdot 7\times 10^{-10}]}$

I am not sufficiently familiar with the details of astronomical calculations to be able to say with entire confidence whether or not such an acceleration perpendicular to $\scriptstyle {r}$ could be detected. It seems, however, unlikely. The maximum effect is of the same order as would be produced by a perturbing body at a distance equal to that of the sun, and whose mass was only $\scriptstyle {\frac {1}{200,000}}$ that of the earth. The perturbation, moreover, would be periodic, vanishing at perihelion and aphelion and accelerating the earth's motion in one-half the orbit, retarding it in the other half.

When the sun is also moving, the problem becomes more complicated. For the present purpose it will be sufficient to obtain the order of magnitude of the acceleration perpendicular to the radius vector. Let $\scriptstyle {\mathbf {v} }$ be the velocity of the sun, and $\scriptstyle {\mathbf {u} }$ that of the planet relative to the sun. Then the force on the planet is $\scriptstyle {(\mathbf {E} )=\mathbf {E} +(\mathbf {v} +\mathbf {u} )\times \mathbf {H} }$ where $\scriptstyle {\mathbf {E} }$ is given by equation (1), in which $\scriptstyle {\beta }$ is now the ratio of the velocity of the sun to the velocity of light, and $\scriptstyle {\theta }$ is the angle between the radius vector and the sun's path. The magnitude of $\scriptstyle {\mathbf {H} }$ is given by equation (2). The force $\scriptstyle {\mathbf {E} }$ is along the radius vector; the force $\scriptstyle {(\mathbf {v} +\mathbf {u} )\times \mathbf {H} }$ is normal to the resultant path of the planet. Let $\scriptstyle {\psi }$ be the angle between $\scriptstyle {r}$ and the tangent to the resultant path of the planet, then