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Why is the cross product used in torque $$t = r\times F ~?$$

I know that the torque is the rate of change of angular momentum and angular momentum is defined as the cross product $$L = r\times P$$ but why is this appearing?

I know that the cross product is a vector of the magnitude of the determinant of the other two vectors and perpendicular. Can someone derive or explain why the cross product is there?

Qmechanic
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ramanujans alkhazarim
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2 Answers2

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I'll use $\vec{\tau}$ to denote torque, and reserve $t$ for time. As you said, torque is the rate of change of angular momentum

$$\vec{\tau} = \frac{d\vec{L}}{dt} = \frac{d}{dt}\left(\vec{r}\times \vec{p}\right)$$

from your definition for angular momentum. Then by the product rule,

$$\vec{\tau} = \left(\frac{d\vec{r}}{dt}\times \vec{p}\right) + \left(\vec{r} \times \frac{d\vec{p}}{dt}\right)$$

The linear momentum, $\vec{p}$, is in the same direction as the velocity, $\vec{v} = d\vec{r}/dt$, so the first term vanishes (the cross product of parallel vectors is zero). So we are left with

$$\vec{\tau} = \vec{r} \times \vec{F}$$

since force is the rate of change of the linear momentum.

Aiden
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    Thanks for your answer however it raises another question why the cross product is used in angular momentum I am loooking for a linear transfromation explanation in the way it justifies the cross product algorithm to be used in this situation – ramanujans alkhazarim Mar 17 '23 at 03:43
  • for example I know that torque in 2d is defined as xFy−yFx how does that translates into 3d – ramanujans alkhazarim Mar 17 '23 at 03:47
  • @ramanujansalkhazarim Can you clarify your question? Are you looking to justify the definition of angular momentum? Or the cross product as a general operation? – Aiden Mar 17 '23 at 03:58
  • Why is the cross product use here instead of the dot product I mean it should be true that at some point of the derivation of torque in 3dimentions the cross product must appear somewhere – ramanujans alkhazarim Mar 17 '23 at 04:00
  • yes justify why the cross product is in the definition of angular momentum – ramanujans alkhazarim Mar 17 '23 at 04:02
  • @ramanujansalkhazarim That looks to be answered here. – Aiden Mar 17 '23 at 04:09
  • @ramanujansalkhazarim You cannot derive the torque in 3D from its expression in 2D. It’s actually the other way around: $xF_y-yF_x$ is the magnitude of the cross product of two vectors in the $x-y$ plane. – Aiden Mar 17 '23 at 04:56
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The cross product is not a vector, it's an axial vector. They are geometrically different. The cross product of 2 vectors, $A_i$, $B_i$, is really:

$$ \bar{C}_{ij} = A_iB_j - A_j B_i $$

which is the antisymmetric part of a rank-2 tensor. Since that has $(3^2-3)/2 = 3$ independent components, we can construct:

$$ C_i = \frac 1 2 \epsilon_{ijk}\bar{C}_{jk} = \epsilon_{ijk}A_jB_k$$

which rotates just like vector, but is even under coordinate inversion, aka parity, (vectors are odd), hence the name: axial vector.

Since Newtons Laws are parity (and time inversion) symmetric, it restricts how they can be formulated.

(Note: the discovery of parity violation in beta decay was because the momentum of the decay electron was aligned with angular momentum of the nucleus. As you pointed out, angular momentum is the cross product 2 vectors, so it is an axial vector. That a vector $\vec p$ was proportional to an axial vector $\vec J$, was Nobel prize level shocking, at the time).

JEB
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