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In this simple case the force components are all polynomials. In the general case that is not true.

Use the right nomenclature, please. There are two products defined for vectors in [math]\mathbb R^3[/math], the inner product and the cross product. Neither is denoted with a period.

Where did you get this last equation? It isn't right, so let's drop it. Next, regarding the first three equations: let's get the nomenclature right. The first is better written as [math]F_x[/math] rather than [math]F(x)[/math]. The former indicates the x-component of the force vector; the latter implies that F is a function of x -- and x only. Secondly, you are missing a minus sign. The convention is [math]\vec F = - \nabla \phi[/math]. That said, given that [math]F_x = 2xy+z^3[/math] suggests that the potential is of the form [math]\phi(x,y,z) = -x(xy + z^3) + f(y,z)[/math]. Continuing with the y-component of the force vector, [math]-\,\frac{\partial \phi(x,y,z)}{\partial y} = x^2 - \frac{\partial f(y,z)}{\partial y}[/math] Since [math]F_y = x^2[/math], this says that f(y,z)[/math] has no dependence on y. So let's call it f(z). Doing the same with the z-component, we find that that f(z) can have no dependence on z, either. All that is left is a constant, so [math]\phi(x,y,z) = -x(xy + z^3) + C[/math] If you can construct a potential function you know that the force is a conservative force. The contrapositive is not necessarily true. Failure to construct a potential function might just mean that you couldn't construct a potential function. You would have to prove that no such function could possibly exist to prove that a force is not conservative. One way to prove this is to find a pair of states [math](x_0, y_0, z_0)[/math] and [math](x_1, y_1, z_1)[/math] and two distinct paths between these points. If the change in mechanical energy depends on the path taken the force is necessarily not conservative.

All measures? Of course not. We ration health care right now. Suppose a heart becomes available for transplant, and three people come up as matches: a seventy year old life long smoker, a twenty year old with lots of congenital defects, and an otherwise healthy thirty year old. One guess as to who gets the heart. Unless someone comes up with a way of curing the condition that ails Mr. Skeptic (post #2), we can't always keep the patient alive. Despite all that modern medicine has to offer, that ailment is eventually going to strike all of us because every single one of us is terminally ill.

First off, swaha, do you understand the difference between the inner product between two vectors [math]\vec a \cdot \vec b[/math] and the scalar product [math]\vec a \times \vec b[/math]? There are two products of three vectors in three-space. I named both in post #2, perhaps a bit to tersely. The first is the scalar triple product [math]\vec a \cdot (\vec b \times \vec c)[/math]. Since the inner product is a commutative operation, this is the same as [math](\vec b \times \vec c)\cdot \vec a[/math]. One could eliminate the parentheses in these forms because [math]\vec a \cdot \vec b \times \vec c[/math] has only one viable interpretation. One geometric interpretation of this product is the volume of a parallelepiped with sides specified by the vectors [math]\vec a[/math], [math]\vec b[/math], and [math]\vec c[/math]. Rearrangements (permutations) of the vectors [math]\vec a[/math], [math]\vec b[/math], and [math]\vec c[/math] might change the sign of the result, but never the absolute value. The second triple product is the vector triple product [math]\vec a \times (\vec b \times \vec c)[/math]. Unlike the scalar triple product, those parentheses are essential here. Specifying things in the right order is also essential. In other words, [math]\vec a \times (\vec b \times \vec c)\ne(\vec a \times \vec b) \times \vec c\ne \vec b \times (\vec a \times \vec c)[/math], and so on. One use of the vector triple product is to compute the component of a vector normal to vector. Suppose [math]\hat a[/math] is a unit vector in [math]\mathbb R^3[/math] and [math]\vec b[/math] is some other vector in [math]\mathbb R^3[/math]. The component of [math]\vec b[/math] normal to [math]\hat a[/math] is [math]\hat a \times (\vec b \times \hat a)[/math].

That doesn't make sense.

Do you mean the scalar triple product [math]\vec a \cdot (\vec b \times \vec c)[/math], or the vector triple product [math]\vec a \times (\vec b \times \vec c)[/math]?

Your mistake is treating vectors as numbers. While numbers do form a vector space ([math]\mathbb R^1[/math]), vector spaces have different rules than numbers. In particular, you treated the inner product and scalar multiplication as interchangeable operations. They are not. Given vectors [math]\vec a[/math] and [math]\vec b[/math], [math](\vec a \cdot \vec b)\vec b \ne \vec a (\vec b \cdot \vec b)[/math] in general. That is exactly what you did, and that is invalid.

You missed one possibility: That you misinterpreted/misapplied them, and that is what happened. What you calculated was the orthogonal projection of the acceleration vector a onto the line defined by the radial vector r. There is nothing wrong with that; in fact that is a frequently used operation. Your mistake was assuming that this orthogonal projection is equal to the acceleration vector. That is not the case. As a side exercise, I suggest you investigate [math]d/dt\left(\vec r \times \vec v\right)[/math] , and also the vector triple product, [math]\hat r\times(\vec a \times \hat r)[/math], in the case of non-uniform circular motion. You might also want to re-reread post #20.

What you did correctly, triclino, was to calculate the component of the acceleration vector in the direction of the radial vector. What you did wrong was to assume that this is equal to the acceleration vector. In particular, you calculated [math](\hat r \cdot \vec a)\hat r[/math], and this is equal to the acceleration vector only in the case that the acceleration vector is parallel to the radial vector. The transverse component of the acceleration vector is not zero in the case of non-uniform circular motion.

Let's use that logic on the velocity vector. [math]\hat r \cdot \vec v = 0[/math] Multiply by [math]\hat r[/math] on both sides and: [math]\vec v = 0[/math] The velocity vector is identically zero! Which of course is wrong, and it is wrong for exactly the same reason your derivation was wrong.

Another fallacy, but you have picked at a rather nasty modern problem. Modern technology, and particularly modern biological technology, has wielded WMD capabilities to small nations and even smaller groups. That only a small group of people are all it takes to murder millions is something that should scare you much, much more than do nukes. It scares the bejesus out of policy wonks, and that is policy wonks of all flavors. It takes a fairly large country to make a nuke. It takes a village to make a killer bug.

That is a logical fallacy. Just because nukes do most of their damage in a few seconds and biological weapons take days to months to accomplish their dirty work does not make the biological weapons less harmful. What if the country is North Korea? Obama explicitly left the nuke option on the table regarding North Korea and Iran.

I would suggest splitting classical into mechanics and E&M. For one thing, classical mechanics and E&M don't quite jibe. The same kind of tension that exists today between relativity and quantum mechanics existed more than 100 years ago between classical mechanics and classical electromagnetism. The two were irreconcilable. It is important to remember that the Michelson-Morley experiment was an attempt by classical mechanists to falsify Maxwell's equations.

No, we have laws against that. The military operates under different rules than do we mere civilians, and for good reason. The option to use them should be left on the table, IMO. The cold hard logic of in-kind retaliation coupled with the US classification of biological, chemical, and nuclear weapons as one and the same has acted as a deterrent against the deployment of these weapons by others nations. That deterrent has now vanished. Mad as it may sound, mutually assured destruction kept the US and former USSR from using nuclear weapons against each other -- or against allies. We did not use nukes in Korea or Vietnam because the Soviet Union also had nukes. The Soviet Union did not use nukes against Berlin in the late 1950s/early 1960s, or against various Eastern European uprisings, or against Afghanistan because they too did not want to risk nuclear escalation. Nice sentiment, wrong world. Not near as much potential there as with biological weapons. You are missing the point of the long-standing US policy. Military rules of engagement are essentially the old rule of a tooth for a tooth, and an eye for an eye. You might wish for a different world, but that is just a wish. Our enemies will not see this as a policy of enlightenment. They will see it as a policy of weakness.

You're right. Pound for pound, nukes are not the way to go. Biological weapons are. Critical mass for 22 pounds for 239Pu, 115 pounds for 235U. A nuke comprises multiple subcritical masses, very carefully constructed high-power explosives to force the critical masses to join, precision electronics to make those explosives fire together, and shielding, lots of shielding. A nuke is not a small device. A few ounces of some particularly nasty biological agent is enough to bring down a city, and the delivery device can be very small. A disposable human carrying a disposable aerosol can will do quite nicely.

Yes. That they are one and the same has been the US policy for a long, long time -- and that policy will be reinstated in January 2013. Making this sea-change in policy was not all that bright. Making it a public pronouncement, as opposed to a top-secret executive order, is dumber than dumb.

In Fox's defense, this is quite a sea-change in policy. From the article, The United States pledges not to use or threaten to use nuclear weapons against (those countries)," Gates said -- even in the case of a biological or chemical attack. That is a huge change in policy. The US military policy has been along the lines of an eye for an eye -- to attack in-kind. Use guns, and the US will strike back with guns. Use tanks, the US will strike back with tanks. Use weapons of mass destruction, and the US will strike back with weapons of mass destruction. Chemical and biological weapons are WMD, so per the rules of engagement, use of those kind of weapons justifies (demands!) a counterattack with WMD. We don't have chemical or biological weapons in our arsenal, but we do have nukes.

The magnitude of the velocity vector is constant in uniform circular motion, but not in non-uniform circular motion. Think of a vehicle performing a loop-de-loop on a vertically-oriented circular track. The vehicle might slow down as it climbs toward the top of the track and speed up once it has passed the top. The motion is still circular (the vehicle is following a circular track), but the motion is not uniform.

Not true. In uniform circular motion the magnitude of the velocity vector is, by definition, constant. In non-uniform circular motion the magnitude of the velocity vector is not constant. In terms of polar coordinates, the position vector is [math]\vec r = r \hat r[/math]. Differentiating with respect to time, [math]\vec v = r\dot{\theta}\hat{\theta}[/math]. Note that there is no [math]\dot r \hat r[/math] term because r (the magnitude of the position vector) is constant. Differentiating velocity to obtain the acceleration vector, [math]\vec a = r \ddot{\theta}\hat{\theta} - r \dot{\theta}^2 \hat r[/math]. In uniform circular motion, [math]\dot{\theta}[/math] is constant, so the acceleration vector is purely radial, and its magnitude is equal to [math]v^2/r[/math]. In non-uniform circular motion, [math]\dot{\theta}[/math] varies with time. While the radial component of the acceleration vector is still related to the velocity vector via [math]v^2/r[/math], the acceleration vector by necessity has a non-zero transverse component when the motion is not uniform.

First and foremost, the hypothesis must be well-formed. This is not a well-formed hypothesis: That statement is obviously not true in general (consider the case y=0). Try again, this time stating your hypothesis correctly.

No. That result pertains to uniform circular motion only. In the case of non-uniform circular motion, the acceleration vector must necessarily have some non-zero component normal to the radial vector (or parallel to the velocity vector, same thing).

Then you won't have uniform circular motion. If the motion is still circular, the velocity vector will still be normal to the radial vector because, as I mentioned in post #9, the time derivative of any constant-length vector is necessarily either zero or normal to that constant-length vector.

[math]\omega[/math] is just another name for [math]\dot{\theta}[/math].

The problem at hand is circular motion, so the length of the position vector with respect to the center of rotation is constant. The time derivative of any constant length vector is either zero or normal to the vector.