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Emailanmol's discussion on angular displacement

  1. Mar 25, 2012 #1
    Mod node:

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

    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?
    Last edited by a moderator: Mar 28, 2012
  2. jcsd
  3. Mar 25, 2012 #2
    Your doubts are genuine.
    But as the above post points out, dont 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.


    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
  4. Mar 25, 2012 #3


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    Re: which of the following is vector and which is scalat...

    I don't know what you mean. It IS in fact just a matter of definitions in this case.
  5. Mar 25, 2012 #4

    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'.
    Last edited: Mar 25, 2012
  6. Mar 25, 2012 #5


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    Re: which of the following is vector and which is scalat...

    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?
  7. Mar 25, 2012 #6
    Re: which of the following is vector and which is scalat...


    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
    Last edited: Mar 25, 2012
  8. Mar 25, 2012 #7


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    Re: which of the following is vector and which is scalat...

    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.
  9. Mar 26, 2012 #8

    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.
  10. Mar 26, 2012 #9


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    Re: which of the following is vector and which is scalat...

    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.

    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.

    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.

    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.
  11. Mar 26, 2012 #10
    Re: which of the following is vector and which is scalat...

    It may be treated as scalar or vector.
    thanks yar;...<'-'>
  12. Mar 26, 2012 #11


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    emailanmol, sometimes less is more :redface:

    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??)
    emailanmol, cepheid is obviously correct :smile:

    if you want to discuss this, it would be better to start your own thread, instead of hijacking someone else's homework
  13. Mar 26, 2012 #12
    Re: which of the following is vector and which is scalat...

    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. :-)



    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 whats right.Thanks !

    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.

    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 :-)
    Last edited: Mar 26, 2012
  14. Mar 26, 2012 #13


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    hello emailanmol! :smile:
    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 :redface:)​

    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)
  15. Mar 27, 2012 #14
    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.


    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.
    Last edited: Mar 27, 2012
  16. Mar 27, 2012 #15

    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
    Last edited: Mar 27, 2012
  17. Mar 27, 2012 #16

    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 )

    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 :-)
    Last edited: Mar 27, 2012
  18. Mar 27, 2012 #17


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    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.
  19. Mar 27, 2012 #18
    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


    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?
    Last edited: Mar 27, 2012
  20. Mar 27, 2012 #19


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    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.

    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.
  21. Mar 27, 2012 #20
    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.
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