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we have some independent means of accurately determining this mass — which seems a very remote possibility indeed — any motion of the perihelion of Mercury within reasonable limits can be so explained. It will be remarked that the value (37) of $e\delta \varpi$ for Mercury is only little larger than the part of the motion (1″.203), which Seeliger explains by a rotation of the astronomical system of coordinates with respect to the "inertial system," i.e. the system to which the equations of motion are referred. If this rotation were rejected, and the term (37) were added, most of Seeliger's residuals would be very little changed. There would then be appreciable residuals only for the nodes of Mercury and Venus, and the perihelion of Mars. The node of Venus could be put right by increasing the density of Seeliger's second ellipsoid to about three times the value adopted by Seeliger (which would still leave it less than 1/2000 of that of the inner ellipsoid). Seeliger's rotation is practically determined from the motions of the node of Mercury and the perihelion of Mars, the excesses of which over their theoretical values, as determined by Newcomb, are 1.2 and 2.1 times their respective probable errors. If the density of the second ellipsoid were so determined as to represent the node of Venus exactly, the residual in the perihelion of Mars would be reduced to 1.6 times its p.e. The other residuals would not be much affected,^[1] and the representation would thus be on the whole very satisfactory.

14. We now come to the effect of a Lorentz-transformation on the elements of the orbit. First consider the orbital plane. In the system ($x,y,z,t$), of which the origin is in the Sun, or in the mean centre of gravity if the mass of the planet is not negligible, this plane is—

$ax+by+cz=0$

We must transform to simultaneous coordinates by (6) or (7). Since $x_{2}=y_{2}=z_{2}=0$ and $\xi _{2}=\eta _{2}=\zeta _{2}=0$ we have $\Delta _{q}=r_{q},\ x_{1}-x_{2}=x,$ etc. The transformed plane is thus—

$a\left(x'_{1}-x'_{2}\right)+b\left(y'_{1}-y'_{2}\right)+c\left(z'_{1}-z'_{2}\right)+{\frac {q_{1}}{1-q_{1}}}r'_{q}(a\alpha +b\beta +c\gamma )=0,$

or

$a\left(x'_{1}-x'_{2}\right)+b\left(y'_{1}-y'_{2}\right)+c\left(z'_{1}-z'_{2}\right)=0,$

where

$a'=a\left(1-q_{1}\right)+q_{1}\mathrm {C} \alpha ,$ $b'=b\left(1-q_{1}\right)+q_{1}\mathrm {C} \beta ,$ $c'=c\left(1-q_{1}\right)+q_{1}\mathrm {C} \gamma ,$

⁠$\mathrm {C} =a\alpha +b\beta +c\gamma .$

1. The residual for the node of Mars, which is found by Seeliger about equal to the probable error, would almost disappear.