6 ms·

This' something really intriguing to me. Coming from a physics background, this immediately takes me to electrodynamics where the pillar of half of classical ph

This' something really intriguing to me. Coming from a physics background, this immediately takes me to electrodynamics where the pillar of half of classical physics, namely Maxwell's equ, is built on cross product.

nikofeyn 8y ago
it's more that they're built on differential forms and the exterior derivative. for a reference, see section 4.6 and problem 19.13 of an introduction to manifolds by loring tu. this is also covered in gauge fields, knots and gravity by baez and muniain and in much detail in foundations of classical electrodynamics by hehl and obukhov.
ajkjk 8y ago
The 7d cross product is almost certainly not useful for physics. It's not uniquely defined! e_1 x e_2 can equal any of the other basis vectors. The cross product in physics is the wedge product. (which is also featured in 'geometric algebra', as another commenter mentioned, though I would contend that GA is the wrong approach). It produces area vectors from two vectors, and happily extends to any dimensions. For instance the electromagnetic field tensor is a bivector (F = d_u A_v - d_v A_u) which is basically a 4d curl of the 4-potential A, and has bivectorial components for each (uv) plane in spacetime.
- schoen 8y ago
  Isn't the three-dimensional vector cross product also not uniquely defined, since the choice between the right-hand rule and left-hand rule is arbitrary?
  - ajkjk 8y ago
    It is certainly much more uniquely defined, since the direction of the result is determined. But yes, the cross product is defined relative to an orientation and changes coordinates accordingly. That's one of many reasons why the wedge product is more natural. If I could have my way there would not be a single cross product in the entirety of physics.
- yiyus 8y ago
  > I would contend that GA is the wrong approach I just said GA gives the most elegant formulation in another comment. What alternative do you suggest? Seeing that you mention the wedge product and bivectors, may you be thinking in something based in differential forms? I'm fascinated by GA, although I have not had the chance to use it heavily yet. I am very interested if you know something better.
  - ajkjk 8y ago
    I think that the wedge product is fantastic and the geometric product is basically useless -- it has no geometric interpretation, complicates things massively, and conceals intuition wherever it is used. So I'd argue that 'exterior algebra' (which is the language used by differential forms, plus the exterior derivative) is basically the right framework for physics. (The reason the geometric product is used in GA is that it's the extension of complex/quaternion multiplication to higher dimensions, and allows for inverting (non-zero-norm) multivectors . But I don't think that makes it worth including. It's not true a priori that inverting vectors is useful to physics, and it only really makes sense when they're being used as linear transformations -- in which case you can just use linear transformations explicitly.
    - joppy 8y ago
      The exterior algebra by itself is not enough for physics, since it doesn’t “know” about the inner product, and so it cannot tell when vectors or wedges should be perpendicular. Given the inner product, you can define the “hodge star”, which restores this information. Another way of restoring the information is to use a modification of the exterior algebra, where rather than having the product uv = u wedge v, we deform the product by setting uv = u wedge v + <u, v>. This is called a Clifford algebra [1], and when the input vector space is R^3 and the input inner product is the Euclidean dot product, this Clifford algebras is exactly the “geometric algebra”. I prefer to read about Clifford algebras, because the language is much less sensationalistic than many of the articles I’ve seen on geometric algebra. It also generalises to not just inner products, but any symmetric bilinear form, such as the Minkowski form on R^4 which is used in the study of relativity. [1] https://en.wikipedia.org/wiki/Clifford_algebra https://en.wikipedia.org/wiki/Clifford_algebra
yiyus 8y ago
When Maxwell first derived his equations in 1861, he did it without a cross product. As a matter of fact, the vector analysis of Gibbs and Heaviside was developed to formulate Maxwell equations. The most elegant formulation of Maxwell equations is obtained, in my opinion, using Geometric Algebra, which reduces them to a single and very simple equation.

nikofeyn 8y ago

it's more that they're built on differential forms and the exterior derivative. for a reference, see section 4.6 and problem 19.13 of an introduction to manifolds by loring tu. this is also covered in gauge fields, knots and gravity by baez and muniain and in much detail in foundations of classical electrodynamics by hehl and obukhov.

ajkjk 8y ago

The 7d cross product is almost certainly not useful for physics. It's not uniquely defined! e_1 x e_2 can equal any of the other basis vectors. The cross product in physics is the wedge product. (which is also featured in 'geometric algebra', as another commenter mentioned, though I would contend that GA is the wrong approach). It produces area vectors from two vectors, and happily extends to any dimensions. For instance the electromagnetic field tensor is a bivector (F = d_u A_v - d_v A_u) which is basically a 4d curl of the 4-potential A, and has bivectorial components for each (uv) plane in spacetime.

schoen 8y ago
Isn't the three-dimensional vector cross product also not uniquely defined, since the choice between the right-hand rule and left-hand rule is arbitrary?
- ajkjk 8y ago
  It is certainly much more uniquely defined, since the direction of the result is determined. But yes, the cross product is defined relative to an orientation and changes coordinates accordingly. That's one of many reasons why the wedge product is more natural. If I could have my way there would not be a single cross product in the entirety of physics.
yiyus 8y ago
> I would contend that GA is the wrong approach I just said GA gives the most elegant formulation in another comment. What alternative do you suggest? Seeing that you mention the wedge product and bivectors, may you be thinking in something based in differential forms? I'm fascinated by GA, although I have not had the chance to use it heavily yet. I am very interested if you know something better.
- ajkjk 8y ago
  I think that the wedge product is fantastic and the geometric product is basically useless -- it has no geometric interpretation, complicates things massively, and conceals intuition wherever it is used. So I'd argue that 'exterior algebra' (which is the language used by differential forms, plus the exterior derivative) is basically the right framework for physics. (The reason the geometric product is used in GA is that it's the extension of complex/quaternion multiplication to higher dimensions, and allows for inverting (non-zero-norm) multivectors . But I don't think that makes it worth including. It's not true a priori that inverting vectors is useful to physics, and it only really makes sense when they're being used as linear transformations -- in which case you can just use linear transformations explicitly.
  - joppy 8y ago
    The exterior algebra by itself is not enough for physics, since it doesn’t “know” about the inner product, and so it cannot tell when vectors or wedges should be perpendicular. Given the inner product, you can define the “hodge star”, which restores this information. Another way of restoring the information is to use a modification of the exterior algebra, where rather than having the product uv = u wedge v, we deform the product by setting uv = u wedge v + <u, v>. This is called a Clifford algebra [1], and when the input vector space is R^3 and the input inner product is the Euclidean dot product, this Clifford algebras is exactly the “geometric algebra”. I prefer to read about Clifford algebras, because the language is much less sensationalistic than many of the articles I’ve seen on geometric algebra. It also generalises to not just inner products, but any symmetric bilinear form, such as the Minkowski form on R^4 which is used in the study of relativity. [1] https://en.wikipedia.org/wiki/Clifford_algebra https://en.wikipedia.org/wiki/Clifford_algebra

yiyus 8y ago

When Maxwell first derived his equations in 1861, he did it without a cross product. As a matter of fact, the vector analysis of Gibbs and Heaviside was developed to formulate Maxwell equations. The most elegant formulation of Maxwell equations is obtained, in my opinion, using Geometric Algebra, which reduces them to a single and very simple equation.