There seem to be two primary means of "multiplying" vectors in physics, the cross product and the dot product. Assuming the angle between vectors is defined as $(a)$, the dot product between vectors $\vec{A}$ and $\vec{B}$ is $AB\cos(a)$, with no direction assigned, and the cross product is $AB\sin(a)$ with the perpendicular direction assigned. In theory, I don't see why we can't define two more operations or perhaps even more than two by just switching some of these definition parts around. For instance, we could take $AB\cos(a)$ and assign that value to the perpendicular direction and call it the crux product, and we could take $AB\sin(a)$ and say it's a scalar value only, and call it a jimmy product. The current system we use for vector multiplication seems kind of arbitrary to me. Anyone have any thoughts on why the cross product and dot product are in any way superior to the crux product and the jimmy product? Do the cross product and dot product have properties that conform to the physical world better than the crux product and jimmy product?

  • 1
    $\begingroup$ The dot product basically measures angles and the cross product measures areas spanned by two vectors. Both have direct geometric interpretations. An area element has two sides, so you need to decide which one you mean, hence the need for the cross product to be a vector. Angles, on the other hand, do not have two sides, so a single scalar quantity is enough. You would need additional information to make crux unique and jimmy would lose vital information. $\endgroup$ – CuriousOne Jan 4 '15 at 18:38
  • $\begingroup$ Related: physics.stackexchange.com/q/43986/2451 and links therein. $\endgroup$ – Qmechanic Jan 4 '15 at 19:08
  • 1
    $\begingroup$ From a physics perspective, we use dot and cross products because they're really really useful. If you can find a way that crux or jimmy products are useful, physicists will start using them. It's actually the exact opposite of arbitrary! $\endgroup$ – Brionius Jan 4 '15 at 19:20
  • 1
    $\begingroup$ What about the outer, inner, wedge, and "double-dot" products? There are already several ways to multiply, contract, project, rotate, translate, etc. vectors other than just the dot and cross products. $\endgroup$ – honeste_vivere Jan 5 '15 at 20:04

In 3D, the dot and cross product are sufficient. For instance your crux product can be written as $\frac{\vec{A}\cdot \vec{B}}{|\vec{A} \times \vec{B}|}(\vec{A} \times \vec{B})$ (though you were a little vague so it's hard to tell if you meant $\frac{\vec{B}\cdot \vec{A}}{|\vec{B} \times \vec{A}|}(\vec{B} \times \vec{A})$ instead (they differ by a sign). Your vagueness gets worse with the jimmy product, because the angle $a$ for the sine of the angle needs a sign, and if you had $\vec{A}\circ_1 \vec{B}=AB\sin(a)$ and $\vec{B}\circ_2 \vec{A}=AB\sin(a)$, it's hard for me to tell which should be which. Assuming you work out the sign ambiguity, then they'd also be sufficient, but they wouldn't work any better.

But I'd also like to argue against the scalar and vector cross product being the primary ways to product vectors in physics. Firstly they only are defined in 3D. In 2D there is no (nonzero) vector mutually orthogonal to two vectors. And in 4D or higher there are too many. Two vectors really form a plane, and $AB\sin(a)$ tells you the magnitude of the parallelogram they form in that plane, as well as the sign (orientation of that magnitude) and the obvious object to represent that is a plane with an orientation and a magnitude, only in 3D (and maybe 7D) is there a unique way to assign a vector to a pair of vectors. So once you go to 4D (such as in relativity), then you have to accept planes as objects that have to be dealt with as planes, not by working with a vector orthogonal to it.

That said, that math has already been worked out. And for relativistic quantum mechanics no one knows any other math to get the physics right and that math will work for any number of dimensions. The basic idea is that if you multiply something that is orthogonal to everything else you simply get a higher dimensional object. So if you take the unit x vector $\vec{X}$ and multiply it by the unit y vector $\vec{Y}$ then you get the xy plane $\vec{X}\vec{Y}$, if you multiply in the opposite order you get $\vec{Y}\vec{X}$ it is still the xy plane but with the opposite orientation. You can take the x,y, and z vectors and multiply them to get $\vec{X}\vec{Y}\vec{Z}$ to get the xyz 3-volume (which in 4D is useful since there are many 3-volumes).

The other rule is that if you multiply by a unit vector that is contained in something you get something contained in it that is orthogonal to it. That's really so that if you have a unit vector so that $\vec{X}\vec{X}=1$ then you can multiply by $\vec{X}$ again to get back the higher dimensional (original) object. So you made something smaller and got the orthogonal complement so that when you remultiply you get back the original object. To be strictly honest that rule (that $\vec{X}\vec{X}=1$ for unit vectors is the fundamental rule, but it's helpful to know what these new things are, that are really planes, 3-volumes, etc. so I introduce that first to newcomers).

And this is different than dot and cross products because it generalizes. This kind of product is useful to physics because it is absolutely the only known way to do relativistic quantum mechanics and it works for the other stuff too.

Like quantum mechanics itself it has two interpretations, there is the shut-up-and-calculate way that gives the minimum needed to compute answers with no interpretation of what is going on, and that usually goes by the name Clifford Algebra. Then there is a version that produces the same answers but paints a story about what is going on to help you with reasoning about what you are doing and catching your mistakes and making a picture that might be motivating or distracting, but computes the same answers. That version is often called geometric algebra, or geometric calculus. To be fair, in geometric calculus and geometric algebra they do also develop other products but they are based on the fundamental one, so there really are more operations (and they are useful, very useful), it's just that you could technically still just use the fundamental one if all you wanted to to get your results.

Even in 3D, insights can be gained about things that already happened in 3D physics. For instance when you built up a magnetic field out of currents you used the cross product, but with this product we say that a cross product should really be the plane spanned by the vectors, not the vector orthogonal to it. But you notice in the Lorentz force law that we never add a magnetic field (rightly, a plane) to an electric field (rightly, a vector), we only take $\vec{v}\times \vec{B}$ thing as a force (per charge), and that kind of a double cross product turns out to be the orthogonal complement of the projection of the vector v into the B plane, which sounds like a mouthful, but its much more natural than it sounds (because it's the vector that you can combine with the projection to get back the original B-plane)


Your crux and jimmy product do not behave nicely under basis transformations. In particular, if you map $\vec B$ to $-\vec B$ (i.e., a reflection), the crux product would not change sign, so the coordinate system formed by $\vec A$, $\vec B$, and their crux product would change handedness.

Similarly, the jimmy product, which gives a scalar quantity, would change its sign under such a reflection, which scalar quantities don't do.

Probably, your products also fail to satisfy some basic properties of products.

  • $\begingroup$ I just wanted to make sure this wasn't some arbitrary assignment - like minutes are assigned for time and degrees for angles. Because once that arbitrary assignment has been established for a while, it's REALLY tough to get rid of. Even today we still use degrees more often than the more natural radians. I have no idea how old the dot product is, but my thought was, if the dot product was made (or discovered) BEFORE basis transformations became used, then it's more likely to be arbitrary, and transformation rules were built around it instead of the other way around. $\endgroup$ – theboombody Jan 5 '15 at 13:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.