Recent content by buttersrocks

No worries. If you do manage to get in that form, I'd appreciate you setting me in the right direction. I can get almost there, but I can never get rid of terms involving \rho_2 that are not inside of the LHS sine. Mainly, is it the second term of the Maclaurin series where I substitute in the...

you probably have the first printing. If you check the errata: http://astro.physics.sc.edu/goldstein/1-2-3To6.html you'll see that the expression typed above is indeed correct and it is what is printed in my sixth printing goldstein.

Homework Statement (Goldstein 3.3) If the difference \psi - \omega t in represented by \rho, Kepler's equation can be written: \rho = e Sin(\omega t + \rho) Successive approximations to \rho can be obtained by expanding Sin(\rho) in a Taylor series in \rho, and then replacing \rho...

Thanks cmos. That last part of the last post really solidified it for me. You are correct, I was not using "direct relationship" in the mathematically rigorous sense; I should have seen that. I think that this discussion really helped to destroy a few subconscious misconceptions. Much appreciated.

Okay, I see where it is coming from now. Thanks.

Bear with me for a moment... Okay, I understand the combinatorial example, but does it apply? For each particular frequency there is one and only one wavelength that it relates to. You don't have 100hz and 1000hz both corresponding to the same wavelength. Hence, my confusion. Also, how...

Can you please elaborate? I'm certain that the formula c=\lambda\nu, does not relate the peak of the frequency curve to the peak of the wavelength curve. You can check by attempting to calculate the temperature of the sun given the form found using wavelength and the form using frequency. If...

I hope this is the right place. Since the law is derived from statistical mechanics, I assumed this was the place to post. I understand how the Wien's Displacement law is derived. What I don't understand, other than mathematically, is why there is a difference between the peak intensity as...

Okay, changed my mind, that doesn't make sense, does it? Does this work because we started with a function for a full cycle and it just got reduced when we took the time average?

Got it. Okay, so basically, how this works is that the work done is periodic with half the period of period of full oscillation. So, the time average of the power leaves us with the average power for 1/2 cycle. So, the power absorbed over a full cycle is twice that. Since the photons are...

Don't think so. I'm actually close to having the result I think. It just hit me that the \frac{dW}{dt} is periodic with freq \frac{\omega}{2}, whereas the E-field is periodic with frequency \omega I feel like it comes from there. Gonna work through it now.

Hi, I originally posted this question in the homework section, but I really don't need any help calculating anything, my answers are right. I'm having conceptual trouble, so I figured that this question belongs here. So, let's say there is a field driving a single atomic oscillator (hydrogen...

For anyone following this thread, mapes was right, the answer should contain \frac{dC}{dT}

Yes, that's the same as I had, just to save time, I drew by analogy; q ~ V , -\phi ~ P. I just got confused with the - sign, but when I worked it out, I did what you did above and got the same relation. I get the F=\frac{q^2}{2C} because this operation is carried out at constant temp, so I set...

Yes, that negative sign should certainly be there.