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What specifically about the torque vector is perpendicular? Is the torque vector like this only so that it works smoothly with linear algebra?

The only explanation I get usually is "because it's just like that" or "because cross product". But why is it a cross product? Just for convenience?

I am a first year physics student. Do I have to be an expert in calculus and linear algebra before I can understand this first year physics concept?

I just want to understand physics ... :(

Isaac Sechslingloff
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  • I answered a similar problem a couple of days ago. Maybe it will help a bit. – garyp Jun 15 '22 at 16:43
  • I think the short answer is that, yes, while you do not need to be "expert" in calculus and linear algebra, those things are the language in which the discussion makes best sense. And it's not quite basic linear algebra, but "geometric algebra", which is not a usual topic in undergrad math classes, but is not difficult, and does appear in physics contexts. – paul garrett Jul 22 '22 at 00:56

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It's neither calculus nor linear algebra; rather, it's geometry, or perhaps representation theory, or Clifford algebra... each of which can lead to a long an in-depth answer. That will not be presented here.

Clifford algebra is the most complete. In that, torque ($\tau$) is not a vector, it's a bi-vector, as is angular acceleration ($\alpha$). The bi-vector in 3 dimensions can be characterized by the cross product.

From a geometric perspective, it is au currant to express physical laws in a coordinate free manner that relates similar geometric objects on both sides of the equal sign. Hence:

$$ \tau = I\alpha$$

relates two bi-vectors.

If you'd rather stick with plain old Euclidean objects, then torque is really the 3 independent components of an antisymmetric rank 2 tensor:

$$ T_{ij} = r_iF_j - F_jr_i $$

This can be converted into an axial-vector via:

$$ \tau_i = \frac 1 2 \epsilon_{ijk}T_{jk}$$

which rotates like a vector, but it's different (e.g., it's parity even).

JEB
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