Emailanmol's discussion on angular displacement

[emailanmol]

Mod node:

This thread was split from this other thread, [thread]590162[/thread].
[/color]Hey vkash.

This is a good one.

See, instantaneous angular displacement is a vector.

Average angular displacement is not a vector.What does this information tell you about instantaneous angular velocity and acceleration.?

What does it tell you about average angular velocity and acceleration?

Do you know why instantaneous angular displacement is a vector while average angular displacement is not?

 

[emailanmol]

Hey
Your doubts are genuine.
But as the above post points out, don't think a lot.These are to be taken more or less as facts , as finding a logical answer would need some practical experiments to be performed.


See,

A vector is something which has magnitude and direction and obeys the laws of vector addition.

Average Angular displacement though has magnitude amd direction, it doesn't obey laws of vector addition.

On the other hand instantaneous angular displacement obeys laws of vector adddition and therefore is a vector.

If someone talks about 50m/s he is talking about speed.Its not right to say velocity is 50m/s if you aren't specifying the direction.

Your velocity of 50m/s has a direction x, and therefore is a vector( plus we also know velocity obeys laws of vector addition so its a vector).

Look at Current.I can say current of 5Ampere flows from point A to B and thus specify it a direction along A to B. But that does't make it a vector as currents add as scalars ( Remember Kirchoff's law)

So basic rule is that it should have magnitude direction and obey laws of vector addition.


When we check average angular displacement it fails to pass the vector addition law, so it isn't a vector.
(you can rotate a book and check)


However, for small displacements it does obey vector addition and therefore Instantaneous angular displacement is a vector.

Since avg angular velocity is avg angular displacement/time it is a scalar.

On the other hand instantaneous angular velocity is instantaneous angular displacement /time and is therefore a vector.

What does this tell you about average angular acceleration and instantaneous angular acceleration?

i don't know <><><>
i think this might be because of these formulas. L=Iω , L[angular momentum] is vector I[moment of inertia] is not a vector so w[angular velocity] must have been a vector quantity.. IT seems like foolish answer but i am not getting any other answer for this.

Applying similar principle,

Instantaneous angular momentum is a vector, average angular momentum is not a vector.


Hope this helps

 

[cepheid]

Hey
Your doubts are genuine.
But as the above post points out, don't think a lot.These are to be taken more or less as facts , as finding a logical answer would need some practical experiments to be performed.

I don't know what you mean. It IS in fact just a matter of definitions in this case.

 

[emailanmol]

I don't know what you mean. It IS in fact just a matter of definitions in this case.

Hey,


It's not just a matter of DEFINITIONS.Usually, we never define a quantity as a vector or scalar in the first place.
We define them as a property of the system.

Like angular displacement was defined as the rotating angle and then put to test against the laws of vector addition. It failed the test.However, small angular displacements passed it . As i said, You can in fact check it by rotating a small book through different angles.

Similarly current and pressure were defined as charge flowing per unit time, and force per unit area.

They were then put to test , and it was seen they add like scalars.

On the other hand, displacement and velocity added as vectors.

At all these places, the quantity was not chosen as a vector in the first place, it was chosen as a physical description of the system and later checked if it added as vectors.

If it did a choice was made if using it as a vector provided any sort a mathematical edge ( or any other advantage), and that model was adopted.

Its seldom we define a quantity straightforward as a vector, and whenever its done, it is usually to have a mathematical edge.

Like the area vector in flux.


It was chosen to point along the outward normal and not the field, so that the flux through a closed surface having no sources and sinks always has a flux of 0.However, before defining it as a vector it was defined and related to a much more physical quantity and aspect 'Area'.

 

[cepheid]

Hey,It's bot just a matter of DEFINITIONS.Usually, we never define a quantity as a vector or scalar in the first place.

Yeah...we do. I'm sorry if my answers seem abrupt, but the mathematical nature of physical quantities comes from the way they are defined.

Like angular displacement was defined as the rotating angle and then put to test against the laws of vector addition.

Similarly current and pressure were defined as charge flowing per unit time, and force per unit area.

They were then put to test , and it was seen they add like scalars.

On the other hand, displacement and velocity added as vectors.

At all these places, the quantity was not chosen as a vector, it was chosen as a physical description of the system and later checked if added as vectors.

I'm sorry, but none of this makes any sense at all. The physical quantities that add like vectors are the ones that are defined to be vectors in the first place. How could we possibly "discover" mathematical properties that they have that aren't already built into their mathematical definitions?

 

[emailanmol]

Ok,

So on what basis, did we come to know that the average angular displacement is not a vector .
Whereas instantaneous angular displacement is a vector.?


We defined pressure as force per unit area.
How did we come to know if the combined pressure at a point added as a scalar and not as vector.?

The basic fact is that we defined these quantities on physical basis and first level and checked if they could act as vectors in the second step.

If they did, we chose the model which best suited the description and scenario.


Chosing pressure as a vector mathematically has no harm.
But it will simply be a waste of a variable and term as it will have no relation what so ever with how a system behaves

 

[cepheid]

Ok,

So on what basis, did we come to know that the average angular displacement is not a vector .
Whereas instantaneous angular displacement is a vector.?We defind pressure as force per unit area.
How did we come to know if the combined pressure at a point added as a scalar and not as vector.?

I don't think that angular displacement is a vector, because it's simply not defined that way. It's defined as a difference between two measured angles. That makes it a scalar.

Pressure is defined as the force that acts per unit area, perpendicular to any surface in a fluid. Since the pressure acts on the surface no matter how it is oriented, fluid pressure has no intrinsic direction of its own. This is built into the definition, which uses the magnitude of the normal force divided by the area.

So the point is, we don't "come to know" anything about the mathematical properties of the physical quantities. We specify them ourselves.

The definitions are not arbitrary, of course. We come up with definitions for physical quantities based on what is most useful in describing physical systems. If a physical quantity, as defined, is not useful for describing nature, then there is no point using it, and we're better off defining some other physical quantity that is useful in this sense.

I don't have anything more to say on the subject.

 

[emailanmol]

I don't think that angular displacement is a vector, because it's simply not defined that way. It's defined as a difference between two measured angles. That makes it a scalar.

Pressure is defined as the force that acts per unit area, perpendicular to any surface in a fluid. Since the pressure acts on the surface no matter how it is oriented, fluid pressure has no intrinsic direction of its own. This is built into the definition, which uses the magnitude of the normal force divided by the area.

So the point is, we don't "come to know" anything about the mathematical properties of the physical quantities. We specify them ourselves.

The definitions are not arbitrary, of course. We come up with definitions for physical quantities based on what is most useful in describing physical systems. If a physical quantity, as defined, is not useful for describing nature, then there is no point using it, and we're better off defining some other physical quantity that is useful in this sense.

I don't have anything more to say on the subject.

Precisely my point.
You CHECKED if the angular displacement is scalar by subtracting the angles.
Not by stating it as a scalar in first place.

In a more accurate way, for angular displacement we rotate objects along some steps and then reverse the chronological order of these steps and SEE/CHECK/FIND OUT if the book is at the same place.
You came to know for large angles it doesn't so its a scalar.
For small angles it does so it's a vector.


Just by defining it as the angle of rotation doesn't make it a vector . YOU HAD TO CHECK IT.


The definition is same mathematically, one is average and the other is instantaneous.
The vector part comes from physical analysis and not by explicity defining small anglular displacements as vectors and large angles as scalars .

I rest my case.

 

[cepheid]

Precisely my point.
You CHECKED if the angular displacement is scalar by subtracting the angles.
Not by stating it as a scalar in first place.

Nonsense. I defined angular displacement as a difference of two angles, and since angles are themselves scalars, that makes angular displacement a scalar by definition. I didn't "check" anything.

And don't forget that just a moment ago you were claiming that it was a vector.

In a more accurate way, for angular displacement we rotate objects along some steps and then reverse the chronological order of these steps and SEE/CHECK/FIND OUT if the book is at the same place.
You came to know for large angles it doesn't so its a scalar.
For small angles it does so it's a vector.

I have no idea what it is that you are trying to accomplish with this proposed experiment. You have to have a precise definition of what an angle is before you attempt to measure one.

Just by defining it as the angle of rotation doesn't make it a vector . YOU HAD TO CHECK IT.

Can't you see that it makes no sense to "check" whether a physical quantity is a vector or a scalar? A physical quantity is something that you measure. In order to measure it, you need to know what it is in the first place. If you ask me to measure the velocity of a car, and I don't know what the definition of velocity is, then I won't be able to make the measurement. I have to know beforehand that velocity is defined as a vector, otherwise I won't know that I'm supposed to: a) use rulers and a clock to measure the distance covered in a unit of time and b) use a compass to determine the direction of motion of the car. If you ask me to measure and report to you the velocity of the car and I don't do part b) because I think it's supposed to be a scalar, then I haven't given you what you asked for. I've given you speed but not velocity.

The definition is same mathematically, one is average and the other is instantaneous.
The vector part comes from physical analysis and not by explicity defining small anglular displacements as vectors and large angles as scalars .

This thing about small displacements being vectors and large ones not...I don't even know where you're getting that from. It's just wrong.

 

[vkash]

Hey,


It's not just a matter of DEFINITIONS.Usually, we never define a quantity as a vector or scalar in the first place.
We define them as a property of the system.

Like angular displacement was defined as the rotating angle and then put to test against the laws of vector addition. It failed the test.However, small angular displacements passed it . As i said, You can in fact check it by rotating a small book through different angles.

Similarly current and pressure were defined as charge flowing per unit time, and force per unit area.

They were then put to test , and it was seen they add like scalars.

On the other hand, displacement and velocity added as vectors.

At all these places, the quantity was not chosen as a vector in the first place, it was chosen as a physical description of the system and later checked if it added as vectors.

If it did a choice was made if using it as a vector provided any sort a mathematical edge ( or any other advantage), and that model was adopted.

Its seldom we define a quantity straightforward as a vector, and whenever its done, it is usually to have a mathematical edge.

Like the area vector in flux.


It was chosen to point along the outward normal and not the field, so that the flux through a closed surface having no sources and sinks always has a flux of 0.However, before defining it as a vector it was defined and related to a much more physical quantity and aspect 'Area'.

It may be treated as scalar or vector.
thanks yar;...<'-'>

 

[tiny-tim]

Hey
Your doubts are genuine.
But as the above post points out, don't think a lot.These are to be taken more or less as facts , as finding a logical answer would need some practical experiments to be performed.


See,

A vector is something which has magnitude and direction and obeys the laws of vector addition.

Average Angular displacement though has magnitude amd direction, it doesn't obey laws of vector addition.

On the other hand instantaneous angular displacement obeys laws of vector adddition and therefore is a vector.

If someone talks about 50m/s he is talking about speed.Its not right to say velocity is 50m/s if you aren't specifying the direction.

Your velocity of 50m/s has a direction x, and therefore is a vector( plus we also know velocity obeys laws of vector addition so its a vector).

Look at Current.I can say current of 5Ampere flows from point A to B and thus specify it a direction along A to B. But that does't make it a vector as currents add as scalars ( Remember Kirchoff's law)

So basic rule is that it should have magnitude direction and obey laws of vector addition.


When we check average angular displacement it fails to pass the vector addition law, so it isn't a vector.
(you can rotate a book and check)


However, for small displacements it does obey vector addition and therefore Instantaneous angular displacement is a vector.

Since avg angular velocity is avg angular displacement/time it is a scalar.

On the other hand instantaneous angular velocity is instantaneous angular displacement /time and is therefore a vector.

What does this tell you about average angular acceleration and instantaneous angular acceleration?



Applying similar principle,

Instantaneous angular momentum is a vector, average angular momentum is not a vector.


Hope this helps

emailanmol, sometimes less is more

if it takes that long to explain something in a homework thread, it's probably clearer to leave it unexplained

(and why did you go on about angular displacement when it wasn't in the question??)

I'm sorry, but none of this makes any sense at all. The physical quantities that add like vectors are the ones that are defined to be vectors in the first place. How could we possibly "discover" mathematical properties that they have that aren't already built into their mathematical definitions?

emailanmol, cepheid is obviously correct

if you want to discuss this, it would be better to start your own thread, instead of hijacking someone else's homework

 

[emailanmol]

Hello cepheid and Tiny Tim,

First of all I would like to point out that any comments I have made (or will make ) so far are just to continue a discussion on a healthy basis and to find out what is actually the right answer to my doubts. My aim is to learn as much as possible from the two of you who are a great asset to the physics community and PF. :-)

Respect!

This thing about small displacements being vectors and large ones not...I don't even know where you're getting that from. It's just wrong.

http://imageshack.us/g/85/img0390nv.jpg/

Kindly refer to these links which are images from the book Resnick Halliday and Krane and see what the authors have to say on this issue. (I hope no one's judging the authenticity and capability of this book and the authors)

my opinions till now have been based on reading this book.

In case you still find my statement and experiments (and all remarks on how a quantity is defined in the first place,) wrong, please explain me where (The Authors) and I
have gone wrong, so that I can improve and know what's right.Thanks !

And don't forget that just a moment ago you were claiming that it was a vector

As i said, its a vector for small angles and scalar when angles are large,( externally validated by the book). Unless the word instantaneous is stated its a scalar.

if it takes that long to explain something in a homework thread, it's probably clearer to leave it unexplained

(and why did you go on about angular displacement when it wasn't in the question??)

You can see in the link why! Angular velocity is angular displacement/ time, so if angular displacement is a vector so is angular velocity and thus in turn angular acceleration.
Am i right?

Also, doesn't everyone have his own way of helping?. I feel the OP gets to know the tone of the message when it's long which plays a crucial role in conveying the actual message.


Also I express myself well when my messages are long.

But I do agree sometimes long message confuse the OP.
So, I would sincerely remould my helping style and try to post shorter messages (if possible).

Thank you so much for pointing it out and going at it like gentlemen.

I appreciate from the bottom of my heart that everywere in your conversations you both aimed at helping me to learn better things, and considered someone as (inexperienced as me who is still in early stages of climbing the ladder of education) as an equal .

I mention yet again, that am having this discussion with a healthy spirit :-)

 

[tiny-tim]

hello emailanmol!

… Angular velocity is angular displacement/ time, so if angular displacement is a vector so is angular velocity and thus in turn angular acceleration.
Am i right?

relative angular velocity isn't a vector …

we can't add relative angular velocities in the same way that we can add relative ordinary velocities (they're not even commutative )​

it's relative angular velocity that is the derivative of angular displacement: neither are vectors

(but of course angular velocity does behave as a vector when we put it in cross-products … loosely speaking, it multiplies as a vector, but it doesn't add as a vector)

 

[emailanmol]

So you mean to say the statement and proof in the book is wrong?


Here's a thread where the last guy speaks of exactly the same thing.

https://www.physicsforums.com/showthread.php?t=279292

I think there are two versions going here.Perhaps treating it as a vector is more of a mathematical manipulation than representing some sort of physical significance.

And I totally agree on the part that angular velocity doesn't add as vectors.

In fact w(A/B)=w(B/A) which violates the commutative law, as you said :-)


However, from what we have been taught (I went to FIITJEE too like Vkash), its treated as vectors for small angles and scalars for large angles.

 

[emailanmol]

It may be treated as scalar or vector.
thanks yar;...<'-'>

Your welcome :-)

In case you still have doubts, open the FIITJEE package or RSM and confirm that the book states and makes exactly the same point as me. I have been a part of FIITJEE for a significant span of time now, and know on what logic the question was based and why it was given as a scalar.

However, in general context I think DH's post was most apt, with exact definition.

For your FIITJEE AITS and phase exams you should follow what the package states

 

[emailanmol]

Can't you see that it makes no sense to "check" whether a physical quantity is a vector or a scalar? A physical quantity is something that you measure. In order to measure it, you need to know what it is in the first place. If you ask me to measure the velocity of a car, and I don't know what the definition of velocity is, then I won't be able to make the measurement. I have to know beforehand that velocity is defined as a vector, otherwise I won't know that I'm supposed to: a) use rulers and a clock to measure the distance covered in a unit of time and b) use a compass to determine the direction of motion of the car. If you ask me to measure and report to you the velocity of the car and I don't do part b) because I think it's supposed to be a scalar, then I haven't given you what you asked for. I've given you speed but not velocity.

I completely agree that once a quantity has been defined , mathematically it includes the information if it is a vector or scalar (or anything else), and in fact my experiment is loosely based on taking advantage of this fact.

You are getting me wrong which is causing a bit of misunderstanding .
You are talking about measuring velocity today , where as I am talking about how it would have been defined and measured the first time(succesfully).

Some logic would have certainly been applied why taking it as a vector poses an advantage, and therefore it was finally chosen as a vector, which is what we both meant when we said it wasn't defined arbitrarily (like a vector mass makes no sense.However why it makes no sense can be easily seen and understood even if we have a vague understanding of how masses behave when kept on one another, and the same principle can be applied for angular displacements )

Yeah...we do. I'm sorry if my answers seem abrupt, but the mathematical nature of physical quantities comes from the way they are defined.

I'm sorry, but none of this makes any sense at all. The physical quantities that add like vectors are the ones that are defined to be vectors in the first place. How could we possibly "discover" mathematical properties that they have that aren't already built into their mathematical definitions?

I meant to say that no one defined displacement and velocity as displacement (which is a vector making velocity a vector) per unit time in the first go.


At that time the person defining it, actually would have had to go through the procedure of determining how the velocity abd displacements would be measured (with which instruments etc) so that it offers the best possible physical description and model .If it wasn't the case he would be making a lucky ( and illogical) guess.

What I (and perhaps the book) wanted to convey is that If OP could retrace the logic involved, he may be able to arrive at which model suits more , or which contradicts and thus be able to arrive at an answer. Plus, he may not know if it's a scalar or vector but he knows it involves angles and can check (exactly same in the link).
He is not finding new properties which mathematical equations don't contain.


And the thing about small displacements being a vector is an exact statement from the book Resnick Halliday Krane (so I hope you see where I get it from and if it's right or wrong!).

And as I said mathematically both small and large displacements have a similar definition, however one is considered a vector(atleast at rotational level) and the other is clearly not.
Can you explain why such discrepancy occurs?

-------

Also I am trying to have a healthy discussion on this topic and not an ARGUMENT,(especially with a man of your STATURE.)

I believe this discussion can help improve my method of looking at vectors and let me know where my ideas have gone wrong, so that I can learn more.

I totally agree with your point that we can't find new mathematical properties which aren't a part of definitions, but I am unable to see where we the book is violating this principle.

From what I can think of, the book is for high school physics, and these terms are defined more clearly at higher physics, so perhaps the book is making an assumption which doesn't cause violations now but may be at higher levels (which you maybe referring to)

Thanks :-)

 

[K^2]

A vector is something which has magnitude and direction and obeys the laws of vector addition.

Average Angular displacement though has magnitude amd direction, it doesn't obey laws of vector addition.

On the other hand instantaneous angular displacement obeys laws of vector adddition and therefore is a vector.

Are we talking in 2D or 3D?

In 2D angular displacement is a vector. The group on addition is cyclic, with 2π=0, but that's not a problem.

In 3D even instantaneous angular displacement is not a vector. Addition group for angular displacements is non-Abelian. In other words, they violate the commutativity axiom of vector space.

Either way, you are mistaken in your assertion.

 

[emailanmol]

Hello K2,

Thanks for your reply :-)

Actually this isn't my statement. It's from Resnick halliday krane where they say that for small angles its true even in 3d

You can view the pages of the book from the links i posted

http://imageshack.us/g/85/img0390nv.jpg/

I think that this problem is arising from the fact that perhaps the book is for high school(and therefore trims the much finer aspects), and clearly the terms we are dealing with are explained and defined in much more detail in the higher branches of physics.

Is this perhaps where I am going wrong?

 

[K^2]

Yes, for small angles it is true in 3D, but that's not the same thing as saying that instantaneous angular displacements are additive. You can say that instantaneous angular velocity is integrable, of course, which is very useful, but it doesn't make displacements themselves vectors.

I think that this problem is arising from the fact that perhaps the book is for high school, and clearly the terms we are dealing with are explained and defined in much more detail in the higher branches of physics.

Is this perhaps why the book is going wrong?

Could be. The vector space is defined sufficiently well here. You can easily test if each of the axioms applies to your quantity to see if it's a vector or not. It's a little hard to say if these are the exact criteria author has had in mind.

 

[emailanmol]

NOTE : I had edited my previous post a bit, but it changes the meaning in no way :-)
----
Ok, thanks K2 . I will go through the wiki page.

 

[D H]

There's quite a few misunderstandings in this thread, which is one of the reasons I separated it from the homework help thread where this thread originated.

First off, Halliday, Resnick, and Krane Physics Volumes I and II is a calculus-based freshman physics text. It is not a high school level text.Rotation
Rotations in ℝⁿ are described by the n×n proper (or special) orthogonal matrices. These special orthogonal matrices form the group SO(n). Rotations are not vectors. They are matrices. Rotation matrices are commutative in ℝ², and ℝ² only. Go beyond ℝ² and rotation A followed by rotation B is generally different from rotation B followed by rotation A. (They are the same if rotations A and B are parallel to the same plane. In this special case, the rotations are just a pair of rotations in ℝ² -- and rotations in ℝ² are commutative.)

Rotations in ℝ³ can be described in terms of aa rotation by some angle \theta about some axis of rotation \hat u. This only works in ℝ³. It doesn't work in ℝ²; rotations in ℝ² are about a point, not an axis. It doesn't work in ℝ⁴ or higher, either. Bottom line: This concept of an axis of rotation is unique to ℝ³.

Since rotations in ℝ³ can be described in terms of an angle (a scalar) and a rotation axis (a direction), does this mean that rotations in ℝ³ are vectors? The answer is no. Having a magnitude and a direction are necessary but not sufficient conditions for some thing to qualify as a "vector." Vectors have to add per vector addition rules. In particular, vector addition is commutative. Rotations in ℝ³ are not commutative, so even though rotations in ℝ³ can be represented by something that superficially looks like a vector in ℝ³, this angle+axis representation is not a vector.

So what is this stuff in Halliday, Resnick, & Krane about small rotations being vectors? That is not what the book says. The book is very clear: "finite angular displacements cannot be represented as vector quantities." The book does some handwaving to arrive at the result "infinitesimal angular rotations can be represented as vectors." Some things to note here:

• Infinitesimal ≠ small. Small is still finite. Small angular displacements cannot be represented as vector quantities.
• There's a lot of handwaving in getting to that result. This is understandable; the underlying math needed to make this rigorous is far, far beyond freshman level math/physics.
• This trick only works in ℝ³. In ℝ², infinitesimal rotations can be represented by a single parameter. Six parameters are needed in ℝ⁴, ten in ℝ⁵, and in general n*(n-1)/2 parameters are needed to describe an infinitesimal rotation in ℝⁿ.
• The n*(n-1)/2 parameters needed to describe an infinitesimal rotation in ℝⁿ act a lot like a vector in ℝ^n*(n-1)/2. They add like vectors, they scale like vectors. In this sense, they are vectors.

Angular velocity
First we need to define what one means by "angular velocity". Since infinitesimal rotations in ℝⁿ can be treated as vectors in ℝ^n*(n-1)/2, one obvious definition of angular velocity is to use this concept of infinitesimal rotations. (Note well: This is handwaving physics math, the kind of stuff that makes mathematicians cringe.)

An even better way to look at these infinitesimal rotations in ℝⁿ is that they are skew symmetric matrices in ℝⁿ. A skew symmetric matrix in ℝⁿ has n*(n-1)/2 independent values -- the exact number of independent values in our infinitesimal rotation. Viewing angular velocity as a skew symmetric matrix is IMO the best way to envision angular velocity in general. This representation is deeply connected to the Lie algebra that generates SO(n).

This leads to an even better way to envision angular velocity. Forget the infinitesimal rotation stuff. The time derivative of a time varying rotation matrix in ℝⁿ can be represented as the matrix product of that rotation matrix and some skew symmetric matrix, or as the matrix product of a different skew symmetric matrix and the rotation matrix. This is very generic. No loosey-goosey, hand waving physics math is needed. Unfortunately, the mathematics is a bit hairy.

Back to ℝ³: A skew symmetric matrix in ℝ³ has three independent parameters which can be represented as a three vector if the resulting three element construct truly does act like a vector. It does. Angular velocities in ℝ³ have a magnitude and direction, they add per the rules of vector addition and they scale per the rules of vector multiplication by a scalar. That is all that is needed to say that these things are "vectors." There is one way that angular velocities differ from things such as position and velocity vectors, and that is how they behave upon reflection. In this sense, angular velocities and angular accelerations are better described by as being pseudovectors or axial vectors.