Cap'n Refsmmat

Oh. What about general news aggregation?

Is it worth subscribing to Google News RSS feeds? I'm afraid of being swamped by news.

Yeah, that's basically it. The machine is there to prevent it from becoming a question of sexual abuse ("what if the prisoner is gay?") or anything else -- my question is purely about pleasure.

Right, but (done correctly) there are no harmful side effects. No long-term damage. Just pleasure. Is the addiction a significant torture in itself?

It's hard to find really good news sources on the Internet -- CNN's website usually makes me want to hit someone, for example ("do I really care about Tiger Woods?!??!"). I subscribe to the Times (of London, not New York) on my Kindle, but I think I could be getting more. What sites/devices/formats do you use to get your world news? Newspaper, web site, RSS feed, electronic gizmo...? Merged post follows: Consecutive posts mergedI might add that I'm specifically looking for global news, not just one country's politics and celebrity gossip. The Times is nice, but I think I could be getting more. I can get online access to the Economist through my parents, which may be a good idea.

Even when performed under proper anasthesia, with painkillers provided and good standards of care? If that's still a problem, what if I do this through some sort of transcranial magnetic stimulation, so all I have to do is put an electronic helmet on the guy? (Or electrodes applied to, er, "strategic" locations.)

Suppose there's some intelligence organization that's captured a high-level enemy and needs to extract information to prevent possible attacks on innocent civilians and prevent many deaths. Now, this organization has rules against torture and cruel punishment. So, instead, it subjects the prisoner to surgery (with painkillers and proper treatment) to implant an electrode in his brain, so the interrogators can immediately stimulate it and cause pleasure. Not pain, but pleasure. They have a little dial that can turn up the level of pleasure farther than is imaginable. They press the button a few times to show the prisoner what it can do, and inform him that they'll press it again if he gives them information. He's placed in a boring ol' prison cell and gets no other pleasures -- crappy food, little human contact, etc. Nothing abusive, just nothing very nice. Is this "reverse torture" unethical? (Considering the obvious objection, "it's just like bribery" -- well, I assume they could tinker with the knob and button to make the guy dependent on that pleasure. Give strong enough pleasure, then deny it, and you have a powerful bargaining tool.)

By the magic of definite integration! Ooh, basic calculus makes me all tingly inside. [math]\frac{\mu_0 I}{4\pi R} \int_{\theta_1}^{\theta_2}\sin \theta \, d\theta = \frac{\mu_0 I}{4\pi R} \left[- \cos \theta\right]_{\theta_1}^{\theta_2} = \frac{\mu_0 I}{4\pi R} \left[ -\cos \theta_2 + \cos \theta_1 \right][/math] Similarly, we can do the same thing to u, integrating with respect to theta. [math]ds_1 = r \, d\theta[/math] and so on. (Basic angles there.) The rest is explained in post #72, with the trig substitutions. Because I divided by length, as you can see by the conspicuous absence of L in the second equation. Incidentally, BIPM gives results in Newtons per meter: Page 113 of that PDF you're so fond of.

What physical quantity does dl represent? Indeed. But if you set the length to infinite (for infinitely long wires), you will get an infinite force, which is clearly not very useful. So we divide by length to get force per unit length. Merged post follows: Consecutive posts merged I did this already.

Indeed. Note that the scan uses sine, while I use cosine -- but I choose a different angle to be theta, namely the opposite angle in the triangle. Since [imath]\sin (\frac{\pi}{2} - \theta) = \cos \theta[/imath], it works.

Perhaps this is a bit clearer. [math]\mathbf{F}_{12} = \frac {\mu_0} {4 \pi} I_1 I_2 \oint_{C_1} \oint_{C_2} \frac {d \mathbf{s_2}\ \mathbf{ \times} \ (d \mathbf{s_1} \ \mathbf{ \times } \ \hat{\mathbf{r}}_{12} )} {r_{12}^2} [/math] [math]\mathbf{F}_{12} = \frac {\mu_0} {4 \pi} I_1 I_2 \oint_{C_2} d\mathbf{s_2} \times \oint_{C_1} \frac{d\mathbf{s_1} \times \hat{\mathbf{r}}_{12}}{r^2_{12}}[/math] (Reordering multiple integrals is not a problem AFAIK.) Now, let's do this chunk by itself: [math]\mathbf{u} = \oint_{C_1} \frac{d\mathbf{s_1} \times \hat{\mathbf{r}}_{12}}{r^2_{12}}[/math] (so [math]\mathbf{F}_{12} = \frac {\mu_0} {4 \pi} I_1 I_2 \oint_{C_2} d\mathbf{s_2} \times \mathbf{u}[/math]) Let's consider only its magnitude: [math]u = \oint_{C_1} \frac{ds_1 \sin \theta}{r^2_{12}}[/math] Using the same steps as in post #72 in this thread, we can turn this integral into: [math]u = \frac{\cos \theta_1 - \cos \theta_2}{R}[/math] where R is the distance between the wires. And since [math]\theta_1 = 0[/math] and [math]\theta_2 = \pi[/math] for an infinite wire, as in post #72 again, we get: [math]u = \frac{2}{R}[/math] Now, plug in to our original double integral... (I know we only found the magnitude of u, but we're only interested in the magnitude of the force anyway.) [math]\mathbf{F}_{12} = \frac {\mu_0} {4 \pi} I_1 I_2 \oint_{C_2} d\mathbf{s_2} \times \mathbf{u} = \frac {\mu_0} {4 \pi} I_1 I_2 \oint_{C_2} d\mathbf{s_2} \times \frac{2}{R}[/math] [math]d\mathbf{s_2}[/math] is always perpendicular to the force it experiences, so we can turn this into an easy magnitude: [math]F_{12} = \frac {\mu_0} {4 \pi} I_1 I_2 \oint_{C_2} \frac{2}{R} \, ds_2[/math] Integrating: [math]F_{12} = \frac {\mu_0} {4 \pi} I_1 I_2 \frac{2s}{R}[/math] (s is simply a length, so let's call it L for consistency) [math]F_{12} = \frac{\mu_0 I_1 I_2 L}{2\pi R}[/math] so [math]\frac{F_{12}}{L} = \frac{\mu_0 I_1 I_2}{2\pi R}[/math] Done! Merged post follows: Consecutive posts merged --1--------------------------------------------------------2-- theta \ ----------------------------A--------------------------------- Imagine here that theta is the angle between the wire, A, and point 1. If each wire is infinitely long, theta will approach 0. Now look at the angle between the wire, A, and point 2. As the wire becomes infinitely long, it approaches [imath]\pi[/imath].

For additional verification: [math]\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi R}[/math] Using the values ambros gives... [math]\frac{F}{L} = \frac{4\pi \times 10^{-7} \times 1 \times 1}{2 \pi \times 1} = 2 \times 10^{-7}[/math] Checking units, next. [imath]\frac{F}{L}[/imath] should be in units of Newtons/meter. [math]\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi R} = \frac{\mbox{N} \mbox{ A}^2}{\mbox{A}^2 \mbox{ m}} = \frac{\mbox{N}}{\mbox{m}}[/math] Physics works!

s1 and s2 are not necessarily directly across from each other. This is a double integral, not a single integral of two wires at a time, so s1 and s2 can have different values. Hence [math]\hat{\mathbf{r}}_{12}[/math] is not constant. Imagine integrating [math]\int_0^4 \int_{-2}^7 xy\, dx \, dy[/math]. [math]x \neq y[/math], clearly.

http://en.wikipedia.org/wiki/Proof_verification http://us.metamath.org/index.html I think you'd enjoy the latter site.

At the moment, the Physics forum isn't very well-organized. For example, where do I go to post a question about electromagnetism? It doesn't belong in "classical physics," because that forum says it's about "facets of mechanics," but electromagnetism isn't exactly "modern and theoretical physics" either. Is there a better organizational strategy that would improve the Physics section?

[math]\mathbf{F}_{12} = \frac {\mu_0} {4 \pi} I_1 I_2 \oint_{C_1} \oint_{C_2} \frac {d \mathbf{s_2}\ \mathbf{ \times} \ (d \mathbf{s_1} \ \mathbf{ \times } \ \hat{\mathbf{r}}_{12} )} {r_{12}^2} [/math] [math]\mathbf{F}_{12} = \frac {\mu_0} {4 \pi} I_1 I_2 \oint_{C_2} d\mathbf{s_2} \times \oint_{C_1} \frac{d\mathbf{s_1} \times \hat{\mathbf{r}}_{12}}{r^2_{12}}[/math] (Reordering multiple integrals is not a problem AFAIK.) Now, let's do this chunk by itself: [math]\oint_{C_1} \frac{d\mathbf{s_1} \times \hat{\mathbf{r}}_{12}}{r^2_{12}}[/math] Let's consider only its magnitude: [math]\oint_{C_1} \frac{ds_1 \sin \theta}{r^2_{12}}[/math] Using the same steps as in post #72 in this thread, we can turn this integral into: [math]\frac{\cos \theta_1 - \cos \theta_2}{R}[/math] where R is the distance between the wires. And since [math]\theta_1 = 0[/math] and [math]\theta_2 = \pi[/math] for an infinite wire, as in post #72 again, we get: [math]\frac{2}{R}[/math] Now, plug in to our original double integral... [math]\mathbf{F}_{12} = \frac {\mu_0} {4 \pi} I_1 I_2 \oint_{C_2} d\mathbf{s_2} \times \oint_{C_1} \frac{d\mathbf{s_1} \times \hat{\mathbf{r}}_{12}}{r^2_{12}} = \frac {\mu_0} {4 \pi} I_1 I_2 \oint_{C_2} d\mathbf{s_2} \times \frac{2}{R}[/math] [math]d\mathbf{s_2}[/math] is always perpendicular to the force it experiences, so: [math]\frac {\mu_0} {4 \pi} I_1 I_2 \oint_{C_2} d\mathbf{s_2} \times \frac{2}{R} = \frac {\mu_0} {4 \pi} I_1 I_2 \oint_{C_2} \frac{2}{R} \, ds_2[/math] Integrating: [math]\frac {\mu_0} {4 \pi} I_1 I_2 \frac{2s}{R}[/math] (s is simply a length, so let's call it L for consistency) [math]F = \frac{\mu_0 I_1 I_2 L}{2\pi R}[/math] so [math]\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi R}[/math] Done!

I think if you "up the ante" even more, people will think you're BSing them on the resume. According to everyone I've spoken to, "Hi, I have four papers published already, here are copies" already gives you a huge advantage over every other applicant.

This is, roughly, like saying this: [math]\int a \, dx = a \, dx[/math] which is patently false. What physical quantity does [math]d\mathbf{l}[/math] represent in your final equation? And I'm still waiting on those sources. I asked for sources that explicitly state and derive "your" equation, not ones that just state the Biot-Savart law. I'd like so see how they derive it so I can work this out correctly. edit: hmm. I'm attempting to evaluate the integral. Does anyone know if [math]\int \mathbf{a} \times \mathbf{x} \, dx = \mathbf{a} \times \int \mathbf{x} \, dx[/math]? (That integral probably doesn't work, I'm just curious if constant cross product terms can be pulled out like constant scalar multiples can be.)

This is where you've screwed up. Here's how the double integral is working: ----1------2--------3----4------5--- -------------------A---------------- For each point on the upper wire, calculate the force caused by the current there on point A, and sum them. You will notice that r varies. Next, move A along the entire lower wire, repeating the sum each time. This is how the integral works. That's the only way you're going to find the complete influence of one wire on the other -- remember, magnetic fields extend an infinite distance, so the force on point A depends on the field from every point on the upper wire, and the force on the entire lower wire depends on the forces at every point on the lower wire. That is why your integral disagrees with every other source I have pointed you to. It's also why I don't yet know how to perform the integration correctly -- that's one nasty integral. But every source I point to shows that my formula ([imath]\frac{\mu_0 I}{2\pi R}[/imath]) works, so I think the integral will work out. Now, again, can you cite sources to support your claim? Ones that directly state that your formula is the correct formula for the force experienced by a wire?

So Wikipedia and the three sources it cites are wrong, yes? To say that the equation I used tells us the force that we're looking for? Just curious. I'll work on doing the integral; I have other work to do at the moment.

http://en.wikipedia.org/wiki/Ampere%27s_force_law It's still Ampere's force law. I have shown you earlier how I derived "my" formula. Can you find me sources that agree with you explicitly, and state the magnetic field created by a wire as you do? I can provide many that agree that you're wrong. http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magcur.html http://www.pa.msu.edu/courses/1997spring/PHY232/lectures/ampereslaw/wire.html http://webphysics.davidson.edu/physlet_resources/bu_semester2/c14_long.html And my textbook, as cited earlier. Any you have that agree with you? Merged post follows: Consecutive posts merged You took the words right out of my mouth.

Also, dl is an infinitesimal value (hence the "d"). It cannot be 1m.

Someone would have to get on Google Adsense to reject the Scientology ads. I don't have access, though.

Now, if drugs were legalized, it would also be easy to subsidize pharmaceutical companies to come up with aids to break addiction, just like nicotine patches. I wonder how feasible that is.

And, of course, I bet the addiction rate would be far different if everyone were trying it, vs. just people who have already used various drugs trying it.