Wednesday, August 02, 2006

All of it


We sell this shirt in the NAP bookstore, where I am working for half of today (making three times what I used to make for working in a bookstore, so I can't really complain). The bookstore manager told me that people ask what it is all the time, so I wrote it down and brought it home to Dan, who confirmed that it was F=ma written in the most obscure way possible. (The marketing people didn't seem to get why that was funny... but then, that's what they need me for.) A guy came in today and was delighted that I could tell him what it was and could recognize some of the symbols, but he too wasn't sure exactly what formulation of F=ma he was looking at (he said the acceleration of the Earth was in there somewhere).

Eventually, since this is the National Academies, a prominent physicist will undoubtedly come in and recognize it at a glance. Meanwhile, though, Dan explained volume integrals and R-double-dot, but we're otherwise stumped. Can anyone help unpack the equation?

6 Comments:

Julie said...

I think you have angular frequency and angular velocity in there as well as coordinates and triple integrals. I flipped through a few books (quantum mechanics and my CRC Guide) and none of my quant books seem to include triple integrals. However, complex integrals always make me suspicious of quantum mechanics because you are integrating the probability that an electron will occupy a particular place in its orbit path. Therefore, you always factor angular momentum, velocity and the wave function that predicts the electrons path.
Is that what this is? Well, it is a measur of force - this usually implies an energy calculation is taken somewhere because force directly relates to energy. The most logical thing it is measuring to me is an electron. You always have to use quantum mechanics when you measure the position of an electron because you can only predict its probability. (I think this is the poly-exclusion principle or something like that- an electron can only be pinpointed by one of two definite coordinates but not by both at once)

8/02/2006 10:52 PM  
jess said...

I think we're pretty sure they're volume integrals...

Of course by "me" I almost always mean "Dan" when I'm talking about matters of physics notation.

8/02/2006 11:47 PM  
Nick said...

Wow, that one is a doozie. So, here's how I read it. Generally, as you said, it's supposed to be Newton's 2nd law, but for a general massive fluid (or collection of particles) in a non-inertial frame. From left to right I'd interpret it as follows:

The first term F is clearly force, as you know.

The next two terms are supposed to be acceleration, I'm pretty sure. The 2nd term I can't immediately identify. Let's come back to that. The integrand of third term is the general formula for acceleration of a point particle in a non-inertial frame multiplied by the mass density, so this is part of adding up all the accelerations. R is the position of the origin of that frame of reference (with respect to some inertial frame of reference), V_xyz (that's supposed to be xyz in subscript) is the velocity of the element of mass at point (x,y,z), omega is the angular velocity vector of rotation of the reference frame (with respect to an inertial one), and rho is the mass density at each point.

I think the next thing is an equal sign (right?), but it's a little blurry for me. Anyway, after that I think is the expression for the change in momentum, since Newton's second is F=ma or F=dp/dt (i.e. rate of chance of momentum). The 4th term is the rate of momentum flow over the boundary of the region considered. You can see that because V_xyz*rho is like momentum/volume, taking the dot product with dA tells you about what component is headed out of (or into) the surface, and then the velocity tells you how much of that momentum escapes in a given time. In a closed system, this term would equal zero. The 5th term is the change in momentum at each point inside the volume, since it's just the time derivative of a term that's adding up the momentum/volume at each point.

Getting back to term number two, it's presumably some added piece to the acceleration that you miss by just adding up the point particle formula at each point. Presumably, it has something to do with the effect of the change in mass density at each point. I've never seen the notation beta_rho before, but I think you could work it out from terms 4 and 5, since term 3 comes from term 5 when you take the derivative of V_xyz with respect to t. Term 2 then encompasses the rest, which comes from differentiating term 5, and from term 4. You could obtain it by carrying through the time differentiation on term 5, using Greens' theorem on term 4, and applying local conservation of mass.

And it's convenient that I said YOU and not ME, because I'm going to bed. :-) I'm not sure if that clears anything up anyway.

8/08/2006 1:56 AM  
jess said...

Actually, that was pretty helpful... not least because I hadn't noticed how blurry the signs were! It sounds like Dan's right that it's sort of nonsensical, kind of like someone trying to rewrite a cliche with as many words as possible, but it's useful to know vaguely what the building blocks are and what they signify. Thanks Nick.

8/08/2006 9:42 AM  
Julie said...

Nick- Is there nothing to do with quantum mechanics? I suspected it was related to the path of an electron but it sounds like no. Am I understanding you correct?

8/19/2006 2:30 PM  
Nick said...

Julie: Doh! I missed your comment when you actually made it. Well, in case anyone actually reads this, here is my reply.

Yes, I'm fairly certain this equation has nothing to do with quantum mechanics. It's certainly true that when you're finding observable values using quantum mechanics of a particle in 3D space you'll usually have integrals over volume of expressions involving the wave function.

In this case, the integrals involve the greek letter rho, which is usually used for the mass density function (the mass per volume at each point in space). So, I think it's mass density that's being integrated over space (with appropriate factors of acceleration) to give force. If one were to interpret it as having to do with quantum mechanics, one would have to adopt a very different interpretation. It don't bare a close resemblence to any quantum equation I'm familiar with, and the notation would have to be pretty unusual.

10/13/2006 11:38 AM  

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