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I know this comes from cross product rule but what is the meaning of Angular velocity and Angular momentum directing in upward direction ?

Is it literally mean directing in upward direction but then why objects doesn't move upward ?

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- Thread starter akashpandey
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- #1

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I know this comes from cross product rule but what is the meaning of Angular velocity and Angular momentum directing in upward direction ?

Is it literally mean directing in upward direction but then why objects doesn't move upward ?

- #2

anuttarasammyak

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Can you please explain in more detail ?

Why it is expressed conventionally upward direction?

Why it is expressed conventionally upward direction?

- #4

Nugatory

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Vectors are mathematical objects that have both a magnitude and a direction so they are a natural choice for describing angular momentum, where we need to specify both the magnitudeof the rotation and the direction of the axis of rotation. So when we say that the angular momentum vector points up, we’re saying that’s the direction of the axis of rotation.

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It means counter-clockwise rotation about a vertical axis as you are looking down from above the horizontal plane of rotation.what is the meaning of Angular velocity and Angular momentum directing in upward direction ?

Because it is angular velocity and not linear velocity. It literally means rotating about a vertical axis.Is it literally mean directing in upward direction but then why objects doesn't move upward ?

It doesn’t mean moving in a vertical direction. That is what linear velocity means. If angular velocity were the same thing as linear velocity then we wouldn’t need both of them. They must be different from each other for them to both be useful.

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- #6

anuttarasammyak

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Angular momentumCan you please explain in more detail ?

Why it is expressed conventionally upward direction?

[tex]L_{ij}=r_ip_j-p_ir_j[/tex]

have two indeces instead of one as vector has. Conventionally we can introduce L of one index, e.g.

[tex]L_k:=\epsilon_{ijk}L_{ij}[/tex]

where

[tex]\epsilon_{ijk}=1[/tex] for {ijk}={123},{231},{312}

[tex]\epsilon_{ijk}=-1[/tex] for {ijk}={132},{321},{213}

[tex]\epsilon_{ijk}=0 [/tex] otherwise. It is the cross product rule. You see independent three components survive as if it is a vector perpendicular to plane of rotation.

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Halc

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I always thought it more logical to think of it as clockwise when viewed in the direction of the vector. Sure, it goes counter-clockwise if you look in the opposite direction, but why choose to look that way?.It means counter-clockwise rotation about a vertical axis as you are looking down from above the horizontal plane of rotation.

Right hand rule trumps any view preference. If your fingers curl in the direction of rotation motion, the thumb will point in the direction of the vector.

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Because that is the way that my fingers and thumb go naturally when I look at them. I have to twist my hand uncomfortably to get the fingers of my right hand to go clockwise.Sure, it goes counter-clockwise if you look in the opposite direction, but why choose to look that way?.

AgreedRight hand rule trumps any view preference. If your fingers curl in the direction of rotation motion, the thumb will point in the direction of the vector.

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Usually you start to use vectors in (Euclidean) geometry in the sense of "polar vectors". They always have to do with translations or a direction along something is moving. The most direct example is a vector defined by two points in Euclidean space ##\overrightarrow{AB}##, which you can use to depict the translation of the point ##A## to the point ##B## along the straight line connecting the two points. That's where the name "vector" comes from (in Latin "vehere" means "to drag" or "to move").

Now you describe the motion by defining some "origin" ##O## and each point ##P## is then uniquely defined by the position vector ##\vec{x}=\overrightarrow{OP}##, and thus you can describe the trajectory of a point particle as a function of time, and that's what we are after in classical point-particle mechanics. Time is mathematically just a parameter which parametrizes the trajectory (but with a physically well-defined measure which is given by some clock). This already implies that also time derivatives of ##\vec{x}## all are polar vectors. Particularly useful in mechanics are of course velocity ##\vec{v} = \dot{\vec{x}}## and acceleration ##\vec{a}=\dot{\vec{v}} = \ddot{\vec{x}}##, and both are polar vectors.

As was already stated in the answers above, there are also another kind of vectors, the socalled "axial vectors", where "axial" refers to the idea of a axis around which something rotates. For this by convention we use the right-hand rule: To describe a rotation of a point around an axis you point the thumb of your right hand in direction of a unit vector ##\vec{n}##. Then the (naturally) curled fingers give the sense of the rotation of the point around the axis. Now you also need the angle of the rotation ##\phi \in [0,2 \pi)##. In this sense you can define the rotation by the vector ##\vec{\phi}=\phi \vec{n}##, but here indeed the vector doesn't have anything to do with dragging something from one place to another but rather it gives a precise description in which way something rotates around the axis.

Now take as a simple example a particle moving on a circle of radius ##a## around the ##z## axis of a Cartesian right-handed basis system (i.e., the basis vectors ##\vec{e}_i## are perpendicular to each other and of unit length, i.e., ##\vec{e}_i \cdot \vec{e}_j=\delta_{ij}## and oriented relative to each other such that if you point the thumb of your right hand in direction of ##\vec{e}_1##, the index finger in direction of ##\vec{e}_2##, then the middle finger points in direction of ##\vec{e}_3##).

This motion is obviously described by

$$\vec{r}(t)=\vec{e}_1 a \cos[\phi(t)] + \vec{e}_2 \sin[\phi(t)],$$

where the angle of rotation is an arbitrary function of time. Now calculate the velocity (a polar vector!)

$$\vec{v}(t)=\dot{\vec{r}}(t)=a \dot{\phi}(t) \{-\vec{e}_1 \sin [\phi(t)] + \vec{e}_1 \cos[\phi(t)] \}.$$

On the other hand

$$\vec{e}_3 \times \vec{r}(t)= a \{-\vec{e}_1 \sin [\phi(t)] + \vec{e}_1 \cos[\phi(t)] \}.$$

So we can write

$$\vec{v}(t)=\dot{\phi}(t) \vec{e}_3 \times \vec{r}(t)=\dot{\vec{\phi}}(t) \times \vec{r}.$$

From this it's clear that the cross product describes infinitesimal rotations, because the infinitesimal change of the position due to the rotation of the point around the ##z## axis is given by

$$\mathrm{d} \vec{r} = \mathrm{d} \vec{\phi} \times \vec{r},$$

and that's why we call ##\vec{\omega} = \dot{\vec{\phi}}## the angular velocity, and it's clear that it is an axial vector as is the case for ##\vec{\phi}##.

The last question now is, how you can distinguish mathematically a polar from an axial vector. The answer is, how the vectors behave under spatial reflections. A spatial reflection at the origin of the coordinate system for a position vector is defined by ##\vec{r} \rightarrow \vec{r}'=-\vec{r}##. Since of course time has nothing to do with the reflection, by definition it doesn't change. Thus also ##\vec{v}=\dot{\vec{r}}## and ##\vec{a}=\ddot{\vec{r}}## just flip their sign under spatial reflections.

For the rotation of the particle around the ##z## axis we got

$$\vec{v}=\vec{\omega} \times \vec{r}.$$

Now ##\vec{v}## and ##\vec{r}## are polar vectors and thus both flip sign under rotations. For that to be consistent with this formula, you must define that the axial vector ##\vec{\omega}## does NOT flip its sign, i.e., under spatial reflections ##\vec{\omega} \rightarrow \vec{\omega}'=\vec{\omega}##.

So the axial vectors simply don't change at all under spatial reflections while polar vectors flip their sign.

Another way to see this is to look again at the formula for the velocity of the particle rotating around the ##z## axis. We just take the cross product with ##\vec{r}##:

$$\vec{r} \times \vec{v}=\vec{r} \times (\vec{\omega} \times \vec{r}) = r^2 \vec{\omega}-\vec{r} (\vec{r} \cdot \vec{\omega})=a^2 \vec{\omega}.$$

Now we have expressed the axial vector ##\vec{\omega}## in terms of a cross product of two polar vectors, and obviously under space reflections we have

$$\vec{r} \times \vec{v} \rightarrow \vec{r}' \times \vec{v}'=(-\vec{r}) \times (-\vec{v})=\vec{r} \times \vec{v},$$

i.e., the cross product of two polar vectors is an axial vector.

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