# Physical meaning of Vector product

1. Oct 14, 2009

### collectedsoul

Can someone please explain to me why it is that the direction of a cross product is perpendicular to the plane of the original vectors? What is the physical significance of this? I mean, if I take torque as an example, what the hell does it mean to say that the direction of the torque is perpendicular to the plane of 'action'. What is the torque doing in that plane?

2. Oct 14, 2009

### Gear300

It is becoming erect.

3. Oct 14, 2009

### Klockan3

Because a torque can't be described in any simpler way than a vector perpendicular to the plane of action. It do not signify anything else except that you can see it as the axis the torque is turning around, it is just the simplest notation.

4. Oct 14, 2009

### symbolipoint

The most helpful part of Klocken3's answer is, "...you can see it as the axis the torque is turning around...".

Still, the cross-product is confusing, physically. Not sure how it represents a paralellogram (am I confused with something else?).

5. Oct 14, 2009

### lurflurf

The main reason is so the cross product is rotationally invariant. If the cross product pointed any direction other than perpendicular to the plane, rotating the three vectors would change there relative position. It would be a quite badly defined torque if it depended on the direction you were looking. It is often desired to find a perpendicular to a plane. There is also the relationship with reciprocal basis.

Last edited: Oct 14, 2009
6. Oct 14, 2009

### Phrak

You are right to be skeptical. Nothing is happening perpendicular to the plane. It's mathamagic. It also happens to be useful mathamagic. We can change coordinate systems so that the force and the distance are pointing along some other directions and it still works as intended.

If space were 4 dimensional or more we would have a hard time defining a unique direction for torque. It all works out nicely only in three dimensional space.

In some higher mathematics we might rather define torque as the "dual of the wedge product of force times displacement." But this is a lot harder to wrap your head around.

7. Oct 15, 2009

### collectedsoul

I had a feeling it has no real significance. I should have clarified that I was bothered about the direction per se of the torque. So, to those of you who said that it denotes the rotational axis, it still isn't necessary that the rotational axis be 'directed' in the up or down direction, is it? You could just as well take the right hand rule and turn it around - its merely a matter of convention. I say do away with it completely. The torque has no business having a direction at all.

I had a further doubt about all this. If the torque's direction has any meaning, then is it that if there are two forces acting on a body that are parallel and opposite and separated by a certain distance (enough to cause rotation basically), the torque gives the resultant of these two forces in a perpendicular plane BECAUSE if the body rotates in a particular direction (say clockwise) then it also MOVES in the downward direction? I hope that sentence made sense. I'm trying essentially to visualize the action of a screw and asking if the torque has 'real significance' because the screw moves in or out depending on direction of rotation. The obvious problem that comes to me is that in the screw's case it is the twisted shape that drives it inward. Am I right about all this?

8. Oct 15, 2009

### Klockan3

No, there is no movement along the direction of the vector, it just states if you are turning clockwise or anti clockwise. As you see you can just rotate in two ways around an axis, and you have two directions for any given axis. Therefore the set of vectors in the axis's vector space can describe any kind of torque around that axis you can imagine.

The most important thing to remember is that you shouldn't add a lot of things to the definition. Like thinking that it would be moving in that direction just because the imagine in your head do that.

9. Oct 15, 2009

### lurflurf

I bet you have lots of fun screaming at the weather report "You fools that 30 inches mercury may just as well be -30 inches!@#!". Conventions are not superfluous, two choices may work just as well, but you still need to differentiate between two cases.

10. Oct 15, 2009

### Phrak

On the other hand, you can accept everything in science and mathematics as immutable tools without need of improvement or further consideration. Then it's not science, it's engineering.

11. Oct 15, 2009

### GRB 080319B

I'm just learning this concept myself, so feel free to correct any mistakes I've made. The vector (cross) product physically means how much of an applied force is going into rotating an particle/object about an axis (xy plane will be action plane and z-axis will be axis of rotation). In the example of using a wrench to rotate a bolt, the cross product $$\vec{r}$$ X $$\vec{F}$$ is the torque $$\tau$$, which is how hard the bolt is being twisted by your applied force (your push on the wrench). The $$\vec{r}$$ is the position vector from the bolt to the end of the wrench (along the x-axis), and the $$\vec{F}$$ is the applied force you pushing on the end of the wrench (at the terminal point of $$\vec{r}$$ and at some angle $$\theta$$ from the x-axis). $$\tau$$ = $$\vec{r}$$ X $$\vec{F}$$ = rFsin$$\theta$$$$\vec{n}$$, where $$\vec{n}$$ is the unit normal vector that is perpendicular to the action plane ($$\vec{n}$$ is along z-axis) and r (length of wrench) and F are the magnitudes of their respective vectors. What rFsin$$\theta$$$$\vec{n}$$ tells us is that for a longer wrench (bigger r) there is more torque on the bolt, and also the component of the force acting to rotate the wrench (Fsin$$\theta$$) is greater when $$\theta$$ is closer to $$\left|\pm90|$$. When pushing wrench at an angle 0$$\leq$$$$\theta$$$$\leq$$90, you will be pushing the wrench and rotating the bolt in the anticlockwise direction from the x-axis, which is represented by the $$\vec{n}$$ pointing toward you along the z-axis. Also, since sin(-$$\theta$$) = -sin$$\theta$$, when pushing the wrench at an angle -90$$\leq$$$$\theta$$$$\leq$$0, then you will be pushing the wrench and rotating the bolt in the clockwise direction from the x-axis. The negative sign from -sin$$\theta$$ represents the direction of the $$\vec{n}$$, which this time is pointing away from you along the negative z-axis. So, the angle of the applied force not only determines how much of your push is going into twisting the bolt, but also whether sin$$\theta$$, and therefore $$\vec{n}$$, is positive or negative, and the sign of $$\vec{n}$$ tells us which direction (anticlockwise or clockwise) you are rotating the bolt. $$\vec{n}$$ doesn't tell you if the bolt is moving out or in the z direction (tightening or loosening), all it tells you (by it's sign) is if the bolt is being rotated anticlockwise or clockwise wrt the action plane. The magnitude of torque (how much of your push on the wrench is going into rotating the bolt) is represented by rF$$\left|$$sin$$\theta$$|, and is independent of the direction of rotation. Hope this helps.

12. Oct 16, 2009

### Feldoh

You're right about the convention. The whole "right-hand rule" is just an arbitrary choice, however that does not mean that the direction is insignificant, it just give us a way to describe torques relative to one another.

13. Oct 16, 2009

### lurflurf

I see I did not realize the cross product was being improved, I would like to see that. It will be difficult to improve since the cross product is unique (up to multiplication by a constant). collectedsoul has suggested the improvement of taking the constant to be -1 (aka the left hand rule), which I guess is better if by better one means more left handed. One does not really need to have faith in the cross product, it is not secret. Anyone can experience it for herself/himself. As far as torque involving a net force, it does not. It is true that a torque is the same as a force couple, but the forces cancel. I would like a definition of 'real significance'. If it has anything to do with "is a vector" the cross product is a pseudovector. A bag of potatoes is the same is we say it weighs 10# or -10#.

14. Oct 16, 2009

### collectedsoul

Did I say that the right hand rule should be replaced by the left hand rule? What am I - a fanatical left wing activist?? What I did say was that the torque need not have a direction at all. To those who say that the torque's gives us the direction of rotation, I say we already know that from resolving the forces on the body. I don't need the torque to tell me that.

Where I can see use for having direction assigned to the torque is if there is direct comparisons of torques and problems dealing with multiple torques. In principle I see how the extra information of direction being included in the torque would be useful, although I don't know how such situations of 'torque comparison' actually arise.

Basically what I understand from this discussion is that the vector product doesn't have much of a physical sense, its more a useful abstraction. Does everyone agree with that? Or is that also a leftist interpretation?

15. Oct 16, 2009

### lurflurf

The direction of a torque is not only needed to compare two torques, but to know the effect of the torque. You are talking about the magnitude of torque. Sure sometimes you only need to no the magnitude, and other times the direction is available from other information. The same is true of any vector or pseudovector. This is like saying "why should I have a velocity vector like 5 knots north east? If I want to know what direction it is pointing I will just look at it". The advantage of a vector is it collects the information into one object. If we change coordinates or units all aspects change together. A vector is abstract in the sense that is a mathematical representation of (in these cases) a physical effect. A vector is not any more abstract than a scalar. It is a more physical view to think of a torque as being a directed quantity than to thing of it as a number and some direction In simple terms it matters very much what direction a torqe is applied. The wrong choice could lead to something breaking, something moving the wrong (ie dangerous or not helpful) way, of much more effort being used than needed. For example if I asked someone at the controls to rotate me horizontally I would not like to be rotated vertically by mistake. I still do not know what you mean by real significance or physical sense.

16. Oct 16, 2009

### collectedsoul

Take your example. Would the person who was rotating you stop and think, 'Oh, should I create a torque which will be directed in .... plane or .... plane'? Or would he think, ' Should I apply a force in ... direction or ... direction?'

17. Oct 16, 2009

### slider142

I believe the intention of the torque vector proponents is to create a mathematically manipulable object that is isomorphic to the former formulation of the question. As long as we can calculate with it in complicated situations without worrying about fallable human intuition, it is a good thing. Intuition is great motivation, but it's good to have something (empirical or mathematical) to check your intuition against.

18. Oct 16, 2009

### Phrak

Yes, well I'll take this as sarcasm. Feel free to study other mathematical systems. Quarternians, wedge products, skew symmetric matrices, and more.

Last edited: Oct 16, 2009
19. Oct 18, 2009

### collectedsoul

I get that. That is what the use of an abstraction is, which, as I stated, I understood to be the sole use of including the direction in the vector product. Btw, its fallible, not fall-able. But I'm sure you'd say that was a typo. ;)

20. Oct 19, 2009

### lurflurf

Are we using torque as in physics or as in engineering?
I don't think you would like the moment of inertia tensor, generalized coordinates, or much of the rest of physics.
An abstraction retains the important aspects of a concrete situation and discards the unimportant aspects. So you are quite right that the sole use of including the direction in a torque isbecuse important. As you say we can often work with forces (in fact force couples are often needed) instead of torques, besides being quite tedious that would require consideration of unimportat aspects. It is quite natural that in dealing with rotational situations we formulate a rotational version of our equations. If we view a rotation as a type of complicated linear motion we increase our confusion and effort. If rx[\sup]F is what we need why would we want to seperate them? One situation where torque is particularly helpful is conservation of angular momentum. In simple situation conservation of angular momentum follows from conservation of conservation of linear momentum. In a polar fluid for example we have angular momentum due to linear momentum and internal angular momentum as well.