Recent content by _Matt87_

so for linear acceleration: P(t')=\frac{2}{3}\frac{e^2\dot{\beta^2}}{3c}\gamma^6 right? it's from 8.3 in the pdf you sent. Although at the very end of that chapter there is still the same equation I wrote earlier. And it says 'Note that the radiation power due to linear acceleration is...

I'm not quite sure what is the \delta z or \delta p. It doesn't say, although I think that you guys are right that if f(z,p,t) represents a distribution of particles in a beam then \delta z and \delta p could be the errors. Let suppose that it's acutally (\delta z)^2 and (\delta p)^2 and that...

Homework Statement An electron (e=1.6x10^(-19) Coulomb) of total energy 10.22 MeV (\gamma about 20) is undergoing linear acceleration due to application of a uniform electric field of strenght E_0=1000 MV/m a)how large is the applied force, in Newton? b) Calculate the instantaneous radiated...

oh... right :) thanks. what about 'the hint'? ;]

Homework Statement hi, so I've got this distribution function: f(z,p,t)=\frac{1}{2\pi\partial z\partial p}exp(-\frac{[z-v(p)t]^2}{2\partial z^2})exp(-\frac{[p-p_0]^2}{2\partial p^2}) where: v(p)=v_0+\alpha(p-p_0) v_0=\frac{p_0}{m\gamma_0} \alpha=\frac{1}{m\gamma_0^3} I have to...

light source is about 2 meters from the ccd with a lens 10cm before the camera (100 mm f). software fits a gaussian to an image of the beam and tells me the central coordinates. Now I see that the laser is unstable in a way that the centrum of the Gaussian moves up and down in the focal plane...

hi, I'm measuring my laser pointing stability with a focusing lens and a CCD. Could you guys tell me what would the relationship be between the offset from focal point in CCD and the angle with which the beam exits the laser? CCD is in focus.

Thank you. Only that I am in the same time trying to learn Matlab. And as I've got some other functions similar to mentioned, which I need to solve numerically, I would need someone just to tell me how to write a proper function for that. Or at least tell me why the one I put before do not...

hi all! I've got something! function u = proba(T) k=1.380648813e-23; u0=7.370012199e-19; options=optimset('TolFun',1.0e-16,'TolX',1.0e-16); u=fsolve(@(u)(2*u0^(3/2)/3-quad(@(e)(sqrt(e)./(exp((e-u)./(k.*T))+1)),0,1000000,1e-20)),u0,options); plot(T,u); hold on end only that...

ok ok I started with something like that: function u= chempot k=1.380648813e-23; u0=7.370012199e-19; I=@(e,T,u) sqrt(e)/(exp((e-u)/(k*T))+1); II=integral(I,0,inf); for T=1:1:10000 x0=[-10*u0]; u=fsolve(@(u) 2*u0^(3/2)/3-II,x0); plot(T,u) hold on end doesn't work...

all right. that seems easy. but what if the integral looked like that? (and I wanted to solve it by matlab?): I=\int_{0}^{inf} dε \frac{\sqrt(ε)}{\exp(\frac{ε-μ}{8T})+1}

that 8T is actually under the exp- like that: I=\int_{0}^{inf} dε \frac{\sqrt(ε)}{\exp(\frac{ε-μ}{8T})} does it make a big difference? or I just add it to the code? and what would the analitical solution be?

hi people, I've just recently started using MATLAB (last week) and 've already got plenty of problems. Could anyone help me with this for example [??]: 1) 3=∫(0,inf) (sqrt(ε)/exp[(ε-μ)/(8*T)] dε so this is the equation. I would like to solve in a way that μ=... , and then plot the...

ok. so the Lagrangian with infinitesimal change in the velocity is not equivalent to Lagrangian on the right side of the equation right? becouse the derivative is not linear to v but v^2. does it mean that Lagrangian with some infinitesimal change of velocity is just equal to lagrangian without...

i don't understand one thing. please explain it. if L(v'^2) must differ from L(v^2) only by the total time derivative, then \frac{\partial L}{\partial v^2} 2 v \cdot \epsilon should be equal to the right side of \frac{dL(v^2)}{dt}=\frac{\partial L(v^2)}{\partial v^2} \frac{dv^2}{dt}...