Recent content by TSny

Yes, good. That's what I get. If @akapm90 is OK with this argument, then they can proceed onward and ignore post #5. Very good point. I agree that the question must be asking for the initial acceleration of M.

@Steve4Physics: Your post made me think of an easier way to see how ##a_y## is related to ##A_x## and ##A_y##. Imagine that the block is lifted vertically upward so that it only has a vertical acceleration ##A_y##. Consider how ##a_y## is related to ##A_y## when the string is still...

I'm not seeing this. If ##\alpha = 0##, so the upper section of the string remains horizontal, then ##a_y = -A_x##. But this is not true for general ##\alpha##.

Even though it starts at rest, we can still consider the configuration after the block has moved some distance. Of course, the block might move down the slope instead of up the slope, but the sign of ##S## handles both directions. (The block might even remain at rest for certain values of the...

The pulley moves with the block ##M##. If the block moves a distance ##S## up the slope, then the pulley moves parallel to the slope the same distance ##S##. Let ##l_{1,0}## represent the initial length of the portion of the string from the pulley ##P## to the fixed point ##A##. Let...

OK. I see now that you already brought up this ambiguity in your first post. Thanks.

I'm wondering if the question is asking for the acceleration ##a## and tension ##T## as functions of time, or if it is asking only for the values of ##a## and ##T## immediately after the system is released (while the upper portion of the string is horizontal)?

I think this might be part of the problem: In calculating the transition rate, you should include a sum over the final polarization states if you are counting photons regardless of their polarization. In this case, you get (similar to Zetili) $$W_{i\to f}=\frac{4\omega_{f\to i}^{3}}{3\hbar...

Sounds right to me.

I think 1.86 is good. Using simple numerical differentiation, similar to your approach, I get 1.83. The approximate formula ##\frac {b}{M} \approx 3^{3/2} + 3.48e^{-\Theta}## yields 1.61. I believe our 1.83-1.86 is probably more accurate than MTW's 1.75. MTW's values of 0.0029 and 0.0000055...

Consdider the case where the test charge moves perpendicularly away from the conductor ("wire") in the lab frame ##S## and suppose the charge has no motion parallel to the wire. Let ##S'## be the frame moving with the test charge. In both ##S## and ##S'##, there is no imbalance of charge...

For online calculation of the incomplete elliptic integral ##F[\phi, m]##, where ##m = k^2##, see this site at WolframAlpha. The link shows the calculation of ##F[0.3, 0.8]##.

I don't see much difference between the MTW (Darwin) calculations and your calculations using ##r_1, r_2, ## and ##r_3##. Darwin shows how the three roots can be expressed in terms of just ##R##, the distance of closest approach. (See post #2). The expressions used in MTW are easily shown to...

The relations between the three root ##(r_1, r_2, r_3)## and the quantities ##R## and ##Q## are given in the paper by C. Darwin referenced in Fig. 25.7 of MTW. You can read this paper online here if you register for a free account. Section 8 of the paper deals with the orbits of light rays...

From ##t = \alpha \sinh(\tau/\alpha)##, we have ##z = (t^2 + \alpha^2)^{1/2} = \alpha \cosh(\tau/\alpha)##. Use these to express the denominator of (3.59) in terms of ##\tau## and ##\tau'##. Note ##\mathbf x = z## and ##\mathbf x' = z'##. Using identities for the hyperbolic sine and cosine...