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I apologize if this question is dumb, but I've looked all over for a straightforward answer and either I can't find one or the terms are too complex for me to understand. I have only a rudimentary knowledge of Mechanics, but I do understand basic Linear Algebra.

So torque, mathematically, is the cross product of the radial distance vector and a force vector. This cross product gives another vector that is orthogonal to both vectors and it points either outside or towards the "page" (in the context of a two dimensional diagram).

Assuming this is correct, I do not understand what it pointing in or out means. Does it even have a phyisical, intuitive meaning?

The best answer I've been able to come up with is that it's just a mathematical convention with no actual phyisical meaning, meant to provide a framework within which operations between torque vectors, such as addition and substraction, make sense.

Am I correct or way off the mark here?

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  • $\begingroup$ Yes, it's a convention. The 'interpretation' is that the rotation will be curling around the axis defined by the torque in the same way the RIGHT hand curls around the thumb (right-hand rule). $\endgroup$ – Danu Oct 30 '13 at 22:00
  • $\begingroup$ @Danu So, if I understand correctly, it means that if your right hand would hold the torque vector, which is the axis of rotation, with it's thumb pointing outwards from what is being rotated, and you would bend your wrist forward, it would rotate in a certain direction. And if the torque vector was facing the opposite direction, the forward motion of the right hand would produce a movement in the opposite direction due to the hand itself having changed direction to mantain the rule of the thumb pointing outwards. Would this be correct? Thank you very much. $\endgroup$ – xxe Oct 30 '13 at 22:28
  • $\begingroup$ Yes, I believe so. This right-hand rule is something that pops up quite often when dealing with cross products (for instance, the direction of the magnetic field due to a current). $\endgroup$ – Danu Oct 30 '13 at 22:32
  • $\begingroup$ To really get to the bottom of this, you must grok that torque is actually a pseudo-vector. Indeed, in the relativistic context, angular momentum is rank 2 tensor, not a vector: en.wikipedia.org/wiki/Relativistic_angular_momentum So... to get a handle on the physical meaning of this, you've got some homework to do. $\endgroup$ – Alfred Centauri Oct 30 '13 at 23:21
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As in the comments, there's certainly something of a convention at work here and it's to do with the "co-incidence" that we live in three spatial dimensions.

As in Greg's answer, torque is intimately linked with angular momentum through Euler's second law. That is, torque and angular momentum are about rotational motion. And rotations, in general, are characterized by the planes that they rotate together with the angles of rotation for each of these planes. In three dimensions, the plane of rotation can be defined by a single vector - namely the vector orthogonal to the plane. So we have the concept of the "axis" of rotation, but this is not general, its simply that a line happens to be the subspace of a three dimensional vector space that is orthogonal to the plane of rotation. In four and higher $N$ spatial dimensions, the concept of an axis is meaningless: not only does an axis not specify a plane (the space orthogonal to a plane is of dimension $N-2$), but also a general rotation rotates several planes (up to and including the biggest whole number less than or equal to $N/2$).

So the "true" information specifying a three dimensional rotation is the "bivector" $A\wedge B$, where $A, B$ are linearly independent vectors defining the plane, and a bivector is an abstract directed "plane" just like a "vector" is an abstract directed "line". Cross products in three dimensions are actually bivectors, not vectors, but we can get away with thinking of them as such in three dimensions.

Some further reading to help you out: the Wikipedia pages Plane Of Rotation, Rotation Matrix and Orthogonal Group (rotation matrices form the group $SO(N)$, the group of orthogonal matrices with unit determinant).

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While the notion of the 'meaning' is somewhat subjective so this may not provide any more meaning but, you should think about the origin of the torque. Namely as the rate of change of the angular momentum: $$\textbf{N} = \frac{d\textbf{L}}{dt},$$ Now the fact is orthogonal to the plane containing the position vector and force vector is a direct result of the definition of angular momentum $$\textbf{L} = \textbf{r} \times \textbf{p} $$ where $\textbf{p}$ is the linear momentum. So one can see the torque as the vector acting on the angular momentum.

This is where I claim that it is not a convention, but a requirement that the angular momentum and hence torque lie perpendicular to the plane. Imagine you could define it in some other way, where it didn't lie perpendicular to this plane, then for a constant angular momentum (e.g no torque) the angular momentum vector would have to rotate with the system and would not be constant! For this reason the choice of orthogonality, is not really a choice or convention but it describes the system.

This is my own personal interpretation so I don't claim it as exactly true in any sense.

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In electrical engg, itis very clear in that the current carrying conductor and the magnetic field produced by it are mutually perpendicular to each other.In mechanics it is very difficult to assign a physical meaning for directions of angular velocity,angular momentum and torque.

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