4 ms·

It is important to maintain the spatial structure of the units. Energy and torque are both N.m in SI units, but in one case the vectors are aligned, and in the

It is important to maintain the spatial structure of the units. Energy and torque are both N.m in SI units, but in one case the vectors are aligned, and in the other they are perpendicular. Energy is a scalar, torque is an axial vector.

bjornsing 3y ago
Perhaps I’m stating the obvious, but doesn’t that poke a pretty big hole in Tao’s proposed formalisms?
- r0uv3n 3y ago
  I don't think this distinction is often made in practice, even though it might be useful. As far as I can tell, combining Tao's approach here with a Clifford algebra approach (where energy is a scalar and torque a bivector) would be entirely possible and solve this problem.
  - mikhailfranco 3y ago
    Yes, this is the correct answer. The Gibbs formulation of 3D vector analysis/calculus was certainly useful, but forgets most of the spatial structure. Using Clifford Algebra (Geometric Algebra) fixes everything.
    - ajkjk 3y ago
      daily chance to inject my opinion on GA: Almost all of the value in GA comes from the use of multivectors and the wedge product. It's very hard to find a justification for the "geometric product", and there's no real geometric interpretation of it. Therefore treating it as fundamental is a difficult-to-justify choice. "Exterior Algebra" on its own has everything you want without the objectionable parts.
- contravariant 3y ago
  Dimensional analysis doesn't need to be perfect, it just needs to rule out some options. You can convert all units to mass and still get useful results, but the more diverse you can keep the units the better. I suppose you could even go through the relevant laws of physics and only identify the units for quantities that need to be added together (or to make the quantities inside more complex functions unit-less). In the case of torque and energy you basically run into the problem that torque times angle is an amount of work and angle is traditionally unit-less. You could however give angle a unit by introducing some 'unnecessary' conversion factor like 'degrees per radian', which is only used for trigonometry. In that case you could reformulate torque as 'energy per angle'.
  - cb321 3y ago
    > even go through the relevant laws of physics Possibly of interest, as mentioned in Appendix C of https://www.nist.gov/publications/architecture-software-assisted-quantity-calculus https://www.nist.gov/publications/architecture-software-assi... there is a start to this in https://en.wikipedia.org/wiki/ISO/IEC_80000 https://en.wikipedia.org/wiki/ISO/IEC_80000
largeluke 3y ago
I suppose you can just take the tensor product of the spacial structure with the units? So energy would be R tensored with V^{N.m} (which of course is isomorphic to just V^{N.m}), and torque would be the space of axial vectors tensored with V^{N.m}.
toth 3y ago
That is a valid point but I think that is outside the scope of dimensional analysis. Distinguishing scalars from vectors (or just different rank tensors in general) is usually obvious, because they have a different number of components, and nobody even tries to add them together. If you do try, you will get a result depends on your coordinate system, a big no-no in physics. In some cases, like vectors and axial vectors, it is less obvious because (in 3 spatial dimensions), they actually have the same number of components. But still, if you add them together you get something that does not transform in a sensible way under reflections.
Horffupolde 3y ago
Written that way torque is already the scalar product between force and arm. What you are missing is actually an angular displacement.
pizza 3y ago
That's because technically N.m is the magnitude of either of the energy and torque bivectors; technically the product you should use is the scalar product for the energy, and the wedge product for the torque. You could use the geometric product too and just call both [(N.v1)(m.v2)] for vectors v1 and v2; you would even get as a result [(N.v1)(m.v2)] = [N.v1*m.v2 + N.v1^m.v2] = [(N.v1*m.v2)] + [(N.v1)^(m.v2)] = [Energy part] + [Torque part], which is kind of neat - units that depend on the angles involved but encompassing both cases

bjornsing 3y ago

Perhaps I’m stating the obvious, but doesn’t that poke a pretty big hole in Tao’s proposed formalisms?

r0uv3n 3y ago
I don't think this distinction is often made in practice, even though it might be useful. As far as I can tell, combining Tao's approach here with a Clifford algebra approach (where energy is a scalar and torque a bivector) would be entirely possible and solve this problem.
- mikhailfranco 3y ago
  Yes, this is the correct answer. The Gibbs formulation of 3D vector analysis/calculus was certainly useful, but forgets most of the spatial structure. Using Clifford Algebra (Geometric Algebra) fixes everything.
  - ajkjk 3y ago
    daily chance to inject my opinion on GA: Almost all of the value in GA comes from the use of multivectors and the wedge product. It's very hard to find a justification for the "geometric product", and there's no real geometric interpretation of it. Therefore treating it as fundamental is a difficult-to-justify choice. "Exterior Algebra" on its own has everything you want without the objectionable parts.
contravariant 3y ago
Dimensional analysis doesn't need to be perfect, it just needs to rule out some options. You can convert all units to mass and still get useful results, but the more diverse you can keep the units the better. I suppose you could even go through the relevant laws of physics and only identify the units for quantities that need to be added together (or to make the quantities inside more complex functions unit-less). In the case of torque and energy you basically run into the problem that torque times angle is an amount of work and angle is traditionally unit-less. You could however give angle a unit by introducing some 'unnecessary' conversion factor like 'degrees per radian', which is only used for trigonometry. In that case you could reformulate torque as 'energy per angle'.
- cb321 3y ago
  > even go through the relevant laws of physics Possibly of interest, as mentioned in Appendix C of https://www.nist.gov/publications/architecture-software-assisted-quantity-calculus https://www.nist.gov/publications/architecture-software-assi... there is a start to this in https://en.wikipedia.org/wiki/ISO/IEC_80000 https://en.wikipedia.org/wiki/ISO/IEC_80000

largeluke 3y ago

I suppose you can just take the tensor product of the spacial structure with the units? So energy would be R tensored with V^{N.m} (which of course is isomorphic to just V^{N.m}), and torque would be the space of axial vectors tensored with V^{N.m}.

toth 3y ago

That is a valid point but I think that is outside the scope of dimensional analysis. Distinguishing scalars from vectors (or just different rank tensors in general) is usually obvious, because they have a different number of components, and nobody even tries to add them together. If you do try, you will get a result depends on your coordinate system, a big no-no in physics. In some cases, like vectors and axial vectors, it is less obvious because (in 3 spatial dimensions), they actually have the same number of components. But still, if you add them together you get something that does not transform in a sensible way under reflections.

Horffupolde 3y ago

Written that way torque is already the scalar product between force and arm. What you are missing is actually an angular displacement.

pizza 3y ago

That's because technically N.m is the magnitude of either of the energy and torque bivectors; technically the product you should use is the scalar product for the energy, and the wedge product for the torque. You could use the geometric product too and just call both [(N.v1)(m.v2)] for vectors v1 and v2; you would even get as a result [(N.v1)(m.v2)] = [N.v1*m.v2 + N.v1^m.v2] = [(N.v1*m.v2)] + [(N.v1)^(m.v2)] = [Energy part] + [Torque part], which is kind of neat - units that depend on the angles involved but encompassing both cases