The Photon Impulse Equation

By using of the Newton formula $F=\frac{dp}{dt}=\frac{d(mc)}{dt} $ (1) together with the Einstein formula $E=mc^2$ the following equation can be received : $F=\frac{d(mc)}{dt}=\frac{1}{c}\cdot \frac{d(mc^2)}{dt}=\frac{1}{c}\cdot \frac{dE}{dt}$ (2). In [1,2] was shown: $-{dE} \mathord{\left/ {\vphantom {{dE} {dt}}} \right. \kern-\nulldelimiterspace} {dt}=P=hf^2$ (3). The solution of the equation system (2,3) delivers the expression for the photon force:$F=-\frac{1}{c}\cdot hf^2=-\frac{hc^2}{c\lambda ^2}=-\frac {hc}{\lambda ^2}=-\frac{hf}{\lambda }$ (3). With Eq. (2) and (3) the following relationship can be presented:$\frac{d(mc)}{dt}=-\frac{E} {\lambda }=-mc\frac{c}{\lambda }$ (4). The Eq. (4) let us to derive the photon impulse equation finally: $\frac{dp}{p}=\frac{d(mc)}{mc}=-f\cdot dt$ (5). [1] About the calculation of the photon power. S. Reissig, Bulletin of the APS, March Meeting 2004, Part I, Montreal, Vol. 49, No.1, p. 255 [2] The Photon Power and Stefan-Boltzmann Radiation Law. S. Reissig, Bulletin of the APS, March Meeting 2004, Part I, Montreal, Vol. 49, No.1, p. 255.