...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.

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.

Actually, Newton's second law, F=dp/dt, only applies to a point mass particle with constant mass. The chain rule applied here appears, erroneously, in Bate, Mueller and White. It may also appear in other sources.

I have seen it used with the hand-waving trick of saying that the v multiplying dm/dt must have to do with the dm/dt, so it must be the exhaust velocity, and magically, that gives the equation for thrust due to a rocket engine with given exhaust velocity and mass flow rate. However, the flaw in that statement is that the v in the chain rule is clearly the velocity appearing in the vehicle momentum.

The correct application of Newton's 2nd law to a variable mass system is a bit more complicated, requiring the use of a control volume to account for the mass flow. See, for example, Hill and Peterson.

Right you are, Chris. That's one more layer of complexity to the issue.

The fact remains, though, that F = ma does not ineluctably lead to the conclusion that all supersonic flow creates shock waves...

The way I first was taught was as follows...

Consider the rocket at rest in vacuum, far from any other object. The total momentum is zero. Once the rocket begins firing, for a rocket of constant exhaust velocity, you can bookkeep the total momentum in the initial instant (making the usual physicist type approximation that the ship greatly outmasses the parcel of exhaust) -- Mship*Vship - Me*Ve. Because the momentum is conserved, d(Ptotal)/dt = 0. In the initial instants, the mass of the ship is approximately constant, so this results in the usual Mship d(Vship)/dt = Ve * d Me/dt.

I know, in rigorous terms, the approximations made sorta short-cut Chris' more complete derivation, but it served the purpose of getting us students past the hurdle of thinking that rocket science was, well, rocket science. Kind of like dipping a toe in the water slowly. The more complete version could then follow.

BTW, why is everyone so surprised about the whole idea of shockless (or as a practical engineering matter, more likely greatly reduced shock) supersonic travel? It's been known for something like 50 years that reshaping a fighter's fuselage to have a narrowed 'waist' would reduce drag and improve transonic performance, so why should shaping an aircraft to improve things be such a surprise? (I always figured that the one that would be tough to conquer would be the detachment shocks, rather than the leading edge shock.)

- Eric.

Eric: I hate to break this to you, but area-
ruling does not result in shock-free supersonic
flight as our friend Rand "might" be claiming.
Im not really sure what he's claiming since
he's resorting to name calling. Class.

Then again, who really knows. He still has not
produced a drawing/link of these shock-free
supersonic sniper bullets Im so eager to buy.

Everyone should read Anderson's "Modern
Compressible Flow" Ill save you some time:
start at page 90. Is that in the System's
Engineering program at Big Blue? Anderson's
an Ohio St. guy: probably not. No, Ive
never been to Columbus.

I have not gone to the comments section of the
earlier post, so I could be proved wrong about
the drawings. Going there now.

that last post was mine.

Rand: so I went to the previous post comments
section: had no idea you and Brian locked horns
so bad (Ive known you were a Wolverine for a long
time). Nevertheless, he's got a point (about
the physics). This statement of yours should
give everyone pause:

"The answer is nowhere, Greg. They are not for
sale on the retail
market (like much military equipment),
they would probably require a
gun designed to shoot them, and as far as
I know they're not now, nor
have they ever been manufactured for
operational use. But it's not
because they defy physics."

Do you have aluminum foil in your baseball cap
to keep the mind control waves out?

Eric: I hate to break this to you, but area-ruling does not result in shock-free supersonic flight as our friend Rand "might" be claiming. Im not really sure what he's claiming since he's resorting to name calling. Class.

I am not claiming that. If you'd read the original piece, you'd realize that I was in fact claiming that it's not a solution at all, and is being overhyped.

It's OK for Brian to call me a "science lightweight", and to talk about "math that I couldn't possibly understand," when he obviously knows nothing about me, but not for me to agree with Professor Hall's characterization of him as a result of that?

Do you have aluminum foil in your baseball cap to keep the mind control waves out?

Why no, not at all. Why do you ask?

It certainly wouldn't seem to follow from your quote.

Are you claiming that there is no military ordnance that is unavailable, over the counter, to the general public? Or is it your claim that everything that is theoretically possible is put into military production?

Just what is your point?

Greg,

Fer crying out loud, read the actual words I posted. I never asserted that area-ruling results in shock-free supersonic flight. The point was that body shaping (and by implication other seemingly unconventional approaches) can result in unexpected (at least at the time) improvements of aero characteristics. I therefore find the whole idea of nearly shock-free supersonic flight to be totally unsurprising.

BTW, most texts' discussion of shocks seemed (to me) to follow a certain circular logic. Given the (oft unstated) assumption that certain conditions *result* in a shock, the text sets out to mathematically describe the situation. All Rand is doing (I think) is pointing out that the above starting assumption ain't necessarily so.

- Eric.