Rasalhague
- 1,383
- 2
http://hyperphysics.phy-astr.gsu.edu/HBASE/relativ/airtim.html#c5
I don't understand the approximation T0 = -TS that they make in the final step of the section "Kinematic Time Shift Calculation". From this, and the other equations in this section, I get
-T_0=T_0\left ( 1+\frac{R^2\omega^2}{2c^2} \right )
-1=1+\frac{R^2\omega^2}{2c^2}
c^2=\frac{R^2\omega^2}{-4}
c=\pm \frac{R\omega}{2i}
but how can this be when c is a constant positive real number, not dependent on the product of the rotation of the Earth with its radius? And
T_A=T_S-T_S\left ( \frac{2R\omega v+v^2}{2c^2} \right )
T_0\left ( 1+\frac{R^2\omega^2}{2c^2} \right )=-T_S\left ( -1+ \frac{2R\omega v+v^2}{2c^2} \right )
T_0\left ( 1+\frac{R^2\omega^2}{2c^2} \right )=T_0\left ( -1+ \frac{2R\omega v+v^2}{2c^2} \right )
4c^2= 2R\omega v+v^2 - R^2\omega^2
c=\frac{\sqrt{(R\omega+v)^2-2R^2\omega}}{2}
which can't be right, since c doesn't depend on these arbitrary variables: radius of the earth, etc.
I don't understand the approximation T0 = -TS that they make in the final step of the section "Kinematic Time Shift Calculation". From this, and the other equations in this section, I get
-T_0=T_0\left ( 1+\frac{R^2\omega^2}{2c^2} \right )
-1=1+\frac{R^2\omega^2}{2c^2}
c^2=\frac{R^2\omega^2}{-4}
c=\pm \frac{R\omega}{2i}
but how can this be when c is a constant positive real number, not dependent on the product of the rotation of the Earth with its radius? And
T_A=T_S-T_S\left ( \frac{2R\omega v+v^2}{2c^2} \right )
T_0\left ( 1+\frac{R^2\omega^2}{2c^2} \right )=-T_S\left ( -1+ \frac{2R\omega v+v^2}{2c^2} \right )
T_0\left ( 1+\frac{R^2\omega^2}{2c^2} \right )=T_0\left ( -1+ \frac{2R\omega v+v^2}{2c^2} \right )
4c^2= 2R\omega v+v^2 - R^2\omega^2
c=\frac{\sqrt{(R\omega+v)^2-2R^2\omega}}{2}
which can't be right, since c doesn't depend on these arbitrary variables: radius of the earth, etc.