Recent content by russdot

  1. R

    Does \(A e^{iA} = e^{iA} A\)?

    Ah yes, so since e^{A} is defined with a power series e^{A} = 1 + A + \frac{A^{2}}{2!} + \frac{A^{3}}{3!} + ... and A commutes with itself then A would commute with e^{A} Thanks!
  2. R

    Does \(A e^{iA} = e^{iA} A\)?

    If A is an operator, is it correct/allowed to say: Ae^{iA} = e^{iA}A Thanks
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    Equating integrands on two equal integrals?

    Homework Statement This is more of a general question. If I have two different integrals that equal the same value, is it valid to equate the integrands? Homework Equations \int P(\theta,\phi)d\Omega = N \int Q(\theta',\phi')d\Omega' = N Where N is a constant and d\Omega = sin\theta...
  4. R

    View from a spaceship at relativistic speeds

    JesseM, Thanks for the reply, those links helped a lot :)
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    View from a spaceship at relativistic speeds

    If one was in a spaceship at rest in frame K and sees an evenly-distributed number of stars around them, what would the distribution of stars look like if you were traveling at relativistic speeds (frame K')? I'm conflicted because I've seen animations online that seem to illustrate the stars...
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    How Do You Simplify Complex Fractions with Nested Radicals?

    Homework Statement \frac{1 + \sqrt{\frac{1}{1 + \left(\frac{s}{u}\right)^{2}}}}{1 - \sqrt{\frac{1}{1 + \left(\frac{s}{u}\right)^{2}}}} Should equal \left(\sqrt{1 + \left(\frac{u}{s}\right)^2} + \left(\frac{u}{s}\right)\right)^{2} Homework Equations above The Attempt at a Solution...
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    Magnetic field above a thin charged disc

    For the magnetic field of each loop (along Z axis), I get: \vec{B}(z) = \frac{\mu_{0}}{4 \pi} \int \frac{\sigma 2 \pi \omega x'^{2} (\hat{\phi} \times \hat{r})dx'd\phi}{x'^{2} + z^{2}} where \hat{\phi} \times \hat{\r} = \hat{k}cos\psi + \hat{R}sin\psi (\psi is the angle between\vec{r} =...
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    Magnetic field above a thin charged disc

    Ah yes, I really should eat some food... but since \vec{I} = \lambda \vec{v} and \lambda = \sigma 2 \pi dr and \vec{v} = r \vec{\omega} then \vec{I} = \sigma 2 \pi \vec{\omega} r dr, correct?
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    Magnetic field above a thin charged disc

    Hi gabbagabbahey, Thanks for the tips, I'm still getting used to the LaTeX notation. marcusl, Ok, and then each infinitesimal loop will have a current I = \sigma 2 \pi R dr.
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    Magnetic field above a thin charged disc

    Homework Statement A thin disc of radius R carries a surface charge \sigma. It rotates with angular frequency \omega about the z axis, which is perpendicular to the disc and through its center. What is B along the z axis? Homework Equations General Biot-Savart law: B(x) =...
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    Potential inside (and outside) a charged spherical shell

    Yes, you're right. The form I used was ambiguous, sorry about that! I guess I just jumped right to using R^{2}sin\theta' d\theta' d\phi as the surface area element. I have solved the problem though, thanks for the assistance!
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    Potential inside (and outside) a charged spherical shell

    Well d^{3}x' turns into r^{2}sin\theta dr d\theta d\phi and since r is constant at R (spherical shell) then an R^2 comes out of the integral and cancels the R^2 in the denominator from the charge density rho = Q / (4 pi R^2). I also figured out the problem, after integration: V(x) =...
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    Potential inside (and outside) a charged spherical shell

    [solved] Potential inside (and outside) a charged spherical shell Homework Statement Use the integral (i) to determine the potential V(x) both inside and outside a uniformly charge spherical surface, with total charge Q and radius R. Homework Equations (i) V(\vec{x}) =...
  14. R

    Mean radius, r.m.s. radius of nucleus

    Great, thanks! I'm assuming if the rms equation is correct, then the mean value equation is also correct..
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    Mean radius, r.m.s. radius of nucleus

    Hello, Given a particular charge distribution p(r) = p_0*exp(-r^{2}/a^{2}), I was wondering if the proper way to calculate the mean radius <r> would be \intp(r)*r*p(r) dV ? Which would make <r^{2}>^{1/2} = (\intp(r)*r^{2}*p(r) dV)^{1/2}, correct?
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