4
$\begingroup$

I have been searching about the cross product and still can't grasp the physical intuition of it. As far as I know, mathematically the cross product is a tool that creates a new vector perpendicular to the two given independent vectors. This is useful for describing a plane with a single vector instead of two. Some definitions include that the magnitude of the cross vector is the area of the parallelogram defined by the two vectors of which I don't see the relation or how it makes sense since it's just a result and not how it was derived, hence why is there a sine in that formula.

But the main thing I don't get is, since the cross is a vector as well then in physics it has a magnitude and direction, so it should describe a phenomenon. What phenomenon is it? And then most of its uses or application are in circular motions (angular momentum, torque, etc), what about linear motions? Wouldn't a cross product of two vectors in linear motion give us a perpendicular vector as well? What is that vector representing? What would be the cross product of two force vectors?

$\endgroup$
5
  • 1
    $\begingroup$ Somewhat related to physical interpretation: Pseudovector $\endgroup$
    – DanDan0101
    Nov 10, 2021 at 23:27
  • 9
    $\begingroup$ Just because an abstract mathematical operation is defined on some quantity or quantities, that does not always mean that it expresses an important truth. E.g., I can compute the arctangent of the price of a gallon of gas, but do I learn anything by it? I don't know of any reason to find the cross product of two forces, but then, I am not a physicist, so I won't write that as an actual answer. Meanwhile, here's some things that Wikipedia thinks you can do with cross products: en.wikipedia.org/wiki/Cross_product#Applications $\endgroup$ Nov 10, 2021 at 23:40
  • 2
    $\begingroup$ Do not break your head about it, it is just a useful definition that simplifies the description of many things. $\endgroup$
    – user65081
    Nov 11, 2021 at 0:06
  • $\begingroup$ The cross product of two vectors $\vec a$ and $\vec b$ is an antisymmetric second rank tensor, namely $A_{ij} = a_i b_j - a_j b_i$. In 3D $\bf A$ has 3 components and transforms like a vector, except under inversion. It is also known as a pseudovector. $\endgroup$
    – my2cts
    Nov 11, 2021 at 8:47
  • $\begingroup$ A question on Mathematics that may help answer your question: Cross Product definition $\endgroup$ Nov 11, 2021 at 19:53

2 Answers 2

2
$\begingroup$

The cross product of two vectors is really a bivector. It has a magnitude and a direction, but the magnitude is an area instead of a length, and the direction is a plane instead of a line.

Like two vectors can point in opposite directions while lying on the same line, two bivectors can "point" in opposite directions while lying in the same plane. You can think of the directions as clockwise and counterclockwise, though which of those is which depends on which side of the plane you're on.

Bivectors are useful for things that lie in a plane and have a clockwise/counterclockwise direction and a magnitude, like angular velocity.

In three dimensions (and only in three dimensions), you can identify a bivector with a vector perpendicular to the plane of the bivector, whose length is the bivector's area. Because of this, bivectors are usually not taught as such. Instead, you have a cross product that produces another vector, whose direction is given by the right hand rule.

$\endgroup$
1
  • $\begingroup$ thanks benrg this really helps by seeing it from that view if we see it as a plane and an area we can describe any kind of motion of 2 vectors in the 3d with one vector , could you please explain that for the force of lorentz where the cross product appear , and it's a force and not only a motion description $\endgroup$ Nov 11, 2021 at 13:00
2
$\begingroup$

From a physics perspective, the cross product is nothing but a mathematical tool that just so happens to fit to som real-world application. This is the case for all math by the way.

  • Push on a train along the tracks and you move the train, meaning you supply energy via work on the train. Push sideways to the tracks and the train doesn't move. No energy supplied, no work done. It seems clear that on only parallel forces to the displacement do work. Push at an angle and again only the parallel part of the force does work. The magnitude of the work turns out to be the product of those parallel components. Let's invent a mathematical tool that multiplies just the parallel components - we can call it a dot product: $$W=r\cdot F.$$

  • In another scenario you are turning a screw with a wrench. Pulling or pushing the wrench causes no turning. Only pulling perpendicular to the wrench arm causes turning. Pull at an angle and still only the perpendicular portion matters. The magnitude of the torque created turns out to be the product of those perpendicular parts. Let's invent a new mathematical tool that multiplies just the perpendicular components of two vectors - we can call it a cross product: $$\tau=r\times F$$

There is not really any significance to the cross product nor any other mathematical tool in itself. It just turns out to be useful for certain applications.

Now, in the former case we need nothing more than the magnitude of the work done. There is no doubt about any directions or anything like that and work is not defined as a vector quantity in the first place. So we are good here.

In the latter case the magnitude is also fine for many applications. But there is actually a directionality involved - which way is the torque turning the screw? Since this might be relevant, why don't we invent the cross product to not just give magnitude but also direction. Meaning, why don't we invent it to be a vector. It would have to be a "curving" vector to point along the actual rotation direction (around the screw), which doesn't match our usual Euclidian vector definitions. Let's instead invent the cross-product vector form to be perpendicular to both input vectors. Then we invent the right-hand-rule to indicate how this vector's direction corresponds to a turning direction around the screw, and then all is good.

Again, there is not really any significance to neither the cross product magnitude nor its direction mathematically. But if it happens to work in some applications in physics, then we can choose to use it and interpret the direction in the real world. Many other mathematical tools and relations could also have been invented - had they ever become relevant in physics.

$\endgroup$

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.