You know you had a wild vacation when it takes two days to recover from it. You’d think that hanging out with a bunch of neuroscientists would be intellectually stimulating but perhaps a little light on the wild partying. You’d be wrong, at least about the partying part. The social scene surrounding the Marine Biological Laboratory is really something to behold. It was a long weekend, so there were parties every night for four straight nights, and all the parties were too good to leave before the wee hours. I ended up averaging about 5 hours sleep a night, which is nowhere near enough. Somehow biologists simply have better parties than physicists. I think it has to do with the average level of social skills. I know some very socially smooth physicists, but let’s face it – the average physics geek is a little on the dorky side, and a bunch of slightly dorky people all in the same place tend to condense into a big glob of mutually reinforcing dorkiness. Biology dorks don’t undergo the same transition, probably because they are fermion dorks, while physicists are boson dorks. Or something. There’s actually a coherent explanation for why biologist dorks should be fermionic (having to do with the greater degree of distinction between different subfields of biology), but something tells me that it would be better not to go there. Maybe I’m not yet fully recovered from my vacation 🙂
Now *That’s* A Vacation
You know you had a wild vacation when it takes two days to recover from it. You’d think that hanging out with a bunch of neuroscientists would be intellectually stimulating but perhaps a little light on the wild partying. You’d be wrong, at least about the partying part. The social scene surrounding the Marine Biological Laboratory is really something to behold. It was a long weekend, so there were parties every night for four straight nights, and all the parties were too good to leave before the wee hours. I ended up averaging about 5 hours sleep a night, which is nowhere near enough. Somehow biologists simply have better parties than physicists. I think it has to do with the average level of social skills. I know some very socially smooth physicists, but let’s face it – the average physics geek is a little on the dorky side, and a bunch of slightly dorky people all in the same place tend to condense into a big glob of mutually reinforcing dorkiness. Biology dorks don’t undergo the same transition, probably because they are fermion dorks, while physicists are boson dorks. Or something. There’s actually a coherent explanation for why biologist dorks should be fermionic (having to do with the greater degree of distinction between different subfields of biology), but something tells me that it would be better not to go there. Maybe I’m not yet fully recovered from my vacation 🙂
I Want My DNS!
OK, I finally got it working. Sort of. I can ping the LAN. I can ping the internet. I can even get to web sites if I know the IP. But when I ping an internet domain from a client with no IP (even something as simple as yahoo.com) it goes “Huh!” as only computers can do, and sits doing nothing.
Any ideas what I have to do to get ZA (and please, no more stories about what a fool I am to use ZA–those are not helpful at this point) to allow DNS? Or diagnostics I can run to figure out where the problem is?
[Update a few minutes later]
OK, I still don’t know why it’s not doing DNS properly, but I fixed it by assigning some DNS servers manually to the client (Earthlink’s). It seems to work now, but it also seems like a kludge.
Johnnie And Johnnie
…sitting in a tree.
K I S S I N…
I think all this touchy-feely-huggy stuff is going to backfire on Kerry with the non-pansy male demographic.
He Hasn’t Had Time To Be Briefed
Allah is divinely inspired today.
He Hasn’t Had Time To Be Briefed
Allah is divinely inspired today.
He Hasn’t Had Time To Be Briefed
Allah is divinely inspired today.
And The Pope Is Still Catholic
Check out this “Dog Bites Man” headline.
[Update a few minutes later]
I’m told that the link requires registration. It didn’t require it of me–sorry about that. The headline is “Libertarian Seeks Small Government.”
I Wonder?
…if Dick Riordan is starting to lose it?
I Am Quite Disturbed
…at the thought that commenter “Brian” from this post teaches undergraduates.
Scroll down a ways, and be amazed.
[Update a few minutes later]
I should add, that there’s another howler there:
Regarding Newton’s second law of motion, F=ma is just fine for all physics short of things traveling greater than 0.95 the speed of light, or quantum effects.
He’s apparently confused, thinking that I’m referring to Einstein’s Special Relativity version of Newton’s Second Law, in which rest mass is converted to true mass via the factor gamma, which is a function of velocity, or F = dp/dt where p = m*v, or in the Einsteinian version, p = gamma*m*v.
Gamma is a function of velocity. It’s 1/(1-v^2/c^2)^1/2 (or in words, it’s the inverse of the square root of one minus the ratio of velocity squared over the speed of light squared). For low velocities, it’s one divided by the square root of one minus a tiny number, or simply one, so at low velocities, mass equals mass. But for high velocities, you’re starting to divide one by a very tiny number (as the difference between 1 and velocity squared over c squared becomes infinitesimal), so gamma blows up to be a huge number. That’s why mass approaches infinity as its speed approaches that of light.
As I pointed out in the other thread, in the Newtonian case it is simple to take the derivative:
F = dp/dt = d(mv)/dt = m*dv/dt + v*dm/dt. But dv/dt is acceleration, so we get:
F = ma + v*dm/dt.
The Einsteinian case is a much more complicated derivative, because it’s a much more complicated function of velocity. But it’s not relevant, since we’re not talking about near-light speeds. The fact remains that Newton’s Second Law is F = ma + v*dm/dt. The only reason that we always see it as the more simple (and incorrect) F = ma, is that this is a special case in which the mass is constant (the derivative of a constant is zero, and the second term goes away). This is the case for most physics problems, but it certainly isn’t for rocketry, in which the vehicle is ejecting mass (that’s what makes it go).
Anyway, as I said, it’s very disturbing that this person is teaching anyone, let alone undergrads.
[Update at noon Eastern]
Professor Hall, who does teach undergrads as well as grads (and I’m glad of it), expands on his comment via email:
I think Brian’s a bit of a putz in his comments. However, he’s right about F=ma and F=dp/dt. Derivation of the rocket equation is a little tricky to work out, as you say. However, the chain rule does not lead to the correct equation.
In the 2nd law,
F = dp/dt
F is the sum of all applied forces, and p=mv is the linear momentum of the particle of mass m.
If you apply the chain rule to this equation, you get
F = m dv/dt + v dm/dt
as you noted.
However, in order for the chain rule to make any sense here, the two v’s must be the same v. What v is it?
If it’s the velocity of the particle, then this equation can’t apply to a rocket, since it couldn’t lift off the ground. On the ground, v is zero, and initially dv/dt is zero, so F is zero. If F is zero, the linear momentum cannot change, so v remains zero.
If it’s the velocity of the mass leaving the rocket, then initially F = v*dm/dt, which is essentially correct. However, the v in the dv/dt term is clearly not the velocity of the mass leaving the rocket. It’s supposed to be the velocity of the rocket.
The correct derivation of the rocket thrust equation uses a control volume approach, which is essentially a summation of Newton’s 2nd law over a continuum of particles of different velocities (the rocket and the propellant clearly have different velocities).
This leads to the following vector equation for rocket motion
F + ve dm/dt – vehat A (Pe-Pa) = ma
The terms on the left comprise the sum of all external forces acting on the
rocket.F includes all the forces such as gravity, drag, ….
The term ve dm/dt is the thrust due to the rocket, where ve is the exhaust velocity and dm/dt is the mass flow rate (negative number, since m is the mass of the rocket, which is decreasing). The vehat A (Pe-Pa) term is the pressure force. Vehat is a unit vector in the direction of ve, Pe is the exhaust pressure, and Pa is the atmospheric pressure.
I’ll just add that a) I’m glad that at least some of my readers and commenters are smarter than me and b) while I didn’t say that the chain rule led to the rocket equation, I did imply it, and that was a mistake, and c) I had known that at one time, but it’s been a long time.
And we are in agreement that Brian is a putz.