Recent content by physics_student123

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...

Yes, I managed to do it, TSny. $$\dot{p} = \dot{\gamma}mc\beta + \gamma mc\dot{\beta}$$ $$\dot{p} = \frac{\beta^2\dot{\beta}mc}{(1-\beta^2)^{3/2}} + \gamma mc\dot{\beta}= \frac{\dot{\beta}mc}{\sqrt{1-\beta^2}}\left(\frac{\beta^2}{1-\beta^2}+1\right) = \frac{\dot{\beta}mc}{(1-\beta^2)^{3/2}}$$...

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...