Ah, very well then. I'm not used to Gaussian units.
TSny, thank you SO MUCH for taking the time and effort to help me with this problem. I simply could not have done it without you. Your help was invaluable.
Also, thank you hutchphd for your answers.
The units are indeed proving to be a problem, TSny. From the relation ##F = -\alpha v##, one arrives at the conclusion that the units of ##\alpha## are ##kilogram/second##. But from the expression that I derived above with your help, the units of ##\frac{2e^2\alpha}{3cm}(\gamma_0-1)## are...
Ok, so we have ##dt = -\frac{d\beta}{\frac{\alpha}{m}\beta(1-\beta^2)^{3/2}}## which implies
$$\int_{t=0}^{t=t_f}Pdt = -\frac{2e^2\alpha^2}{3cm^2}\int_{\beta_0 = \frac{v_0}{c}}^0\beta^2\frac{1}{\frac{\alpha}{m}\beta(1-\beta^2)^{3/2}}d\beta =...
$$P = \frac{2e^2\gamma^6\alpha^2\beta^2}{3cm^2}(1-\beta^2)^3 = \frac{2e^2\gamma^6\alpha^2\beta^2}{3cm^2}\frac{1}{\gamma^6} = \frac{2e^2\alpha^2\beta^2}{3cm^2}$$
Indeed, hallelujah!
Power is the time derivative of work, which is energy transferred, so in order to find the total energy emitted...
So we have $$\frac{\dot{\beta}mc}{(1-\beta^2)^{3/2}} = -\alpha c\beta$$ which implies $$\dot{\beta} = -\frac{\alpha\beta}{m}(1-\beta^2)^{3/2}$$
Using this in the relativistic Larmor formula, we get $$P = \frac{2e^2\gamma^6\alpha^2\beta^2}{3cm^2}(1-\beta^2)^3$$ but this is the power radiated at a...
Thank you for your replies, hutchphd and TSny.
TSny, the rate of change of relativistic momentum, which is equal to the net force F, is given by
$$\dot{\vec{p}} = \gamma m\vec{a} + \gamma^3m\left(\frac{va}{c^2}\right)\vec{v}$$
but I don't understand how to input the information that the...
Honestly, folks, I don't even know how to start. I included in the Relevant Equations section the relativistic generalization of the Larmor formula according to Jackson, because that's the equation for the power emitted by an accelerated particle, but I don't see how that gets me very far.
The...