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Doubts on rotation

  1. Mar 24, 2013 #1
    I have some doubts on topics related to rotation, so I thought I'd make a single thread rather than multiple ones.

    1. Why are the Coriolis and Centrifugal forces "fictitious"? I think I might be getting confused by the terminology, but do they represent physical forces? My lecture notes say they only appear due to choice of coordinate systems. But I want to develop physical understanding rather than mathematical.

    2. What exactly do Euler angles represent? My lecture notes describe a three step process of rotations of a rigid body to define them, but I don't really understand much.

    Any help will be appreciated.
  2. jcsd
  3. Mar 24, 2013 #2
    Just terminology on your first question. There's nothing fake about the force, it's ordinary and not different at all from the other forces in the chapters you're learning. The only new part is an addition to frame of reference. Depending on where you choose your origin, the force will be pointing in different directions, but 'really,' the force is being applied in a certain way. Enjoy this .gif on wikipedia:

  4. Mar 24, 2013 #3


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    Newton's first law is that objects which are not being acted on by a force travel in a straight line.

    Suppose that you have a kid on a carousel, Carol, and next to the carousel there is a kid standing, Stan. And let's get rid of gravity, the carousel is located in deep space inside a large space ship. Stan throws a ball parallel to the axis of the carousel, and watches it go in a straight line, confirming Newton's first law. Carol also watches the ball, but from her perspective it doesn't go in a straight line, it travels in a helix.

    So, Carol can either say Newton's first law is wrong, objects not acted on by a force accelerate anyway, or she can propose that there must be a force causing the acceleration. Those forces are called "fictitious", but from Carol's perspective, they are essential for doing physics. That is because Carol's reference frame is non-inertial. When you are doing physics in non-inertial reference frames, you must include the fictious forces.
  5. Mar 24, 2013 #4


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    Oh my gosh. This topic comes up approximately once per day on this web site.

    But rather than argue about whether these forces are "real" or "ficititious", let me just list the ways in which they are different from the other forces one is likely to encounter:

    1. "Fictitious forces" can be made to disappear completely through an appropriate choice of coordinates. In contrast, for any other force, if it's present in one coordinate system, then it's also present in another coordinate system.
    2. These forces do not obey Newton's 3rd law. If there is a Coriolis force acting on object A, there is no equal and opposite force exerted by object A on anything else.
    3. All objects, regardless of what they are made out of, or how massive they are, are affected in exactly the same way by these forces.
    4. There is no "source" for these forces. In contrast, an electrical force has a source, which is charged particles.
    5. An object accelerating under these forces, with no non-fictitious forces at work, will feel nothing. It won't be distorted, or stretched or squashed. It's not possible to be crushed by fictitious forces--you can only be crushed by non-fictitious forces.
  6. Mar 24, 2013 #5


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    Thanks, that is a very important point for the OP. You can use fictitious forces to patch the first and second laws, but doing so leaves the third law broken.
  7. Mar 24, 2013 #6
    Thanks all for the replies. So if I am understanding correctly, these forces are only evoked to preserve Newton's second law when viewing a situation from a non-inertial frame.

    Carrying on with carousel analogy, suppose the kid on the carousel rolls a ball. Will she see the ball move in a straight line, and if not, does she need to evoke fictitious forces again?
  8. Mar 24, 2013 #7


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    Carol will see the ball curve and will need to evoke fictitious forces to use Newtons 2nd law to explain it.
  9. Mar 24, 2013 #8
    A "fictitious" force violates the third law of Newton. The third law of Newton breaks the "circular reasoning" seen in the first two laws. A "real" force involves the interaction between at least two bodies. The interaction creates a a pair of forces. According to the third law, the interaction causes a force on one body that is equal in magnitude but opposite in sign to the force on the other body.

    A force can be mathematically defined as mass times acceleration in any frame of reference. It doesn't matter if that frame of reference is inertial or noninertial. The definition is consistent with both the first and second law of Newton.

    Suppose one were to take an inventory of all interactions that cause acceleration in a closed system. For every real force on one body, there has to be an equal and opposite force on another body in the system. If the frame isn't inertial, then there will be residual acceleration in some bodies that is unmatched in other bodies. The unmatched acceleration corresponds to fictitious forces.

    Consider the Coriolus force in the rotating frame where the surface of earth is stationary. The Coriolus force causes acceleration in air masses and pendulums. The air masses don't cause acceleration of the earth in the rotating frame.

    Another way to look at it is in terms of conservation of momentum. In the rotating frame where the surface of the earth is stationary, linear momentum is not conserved. The Coriolus force and the centripetal force can cause objects on the surface of the earth to accelerate. However, the frame of reference has been defined so that there is no opposite change of momentum of the surface of the earth.

    A fictitious force is a force that does not conserve the total momentum of the system. If a fictitious force is observed, then the observer is not in an inertial reference frame.

    A real force is one which conserves total momentum of the system even if it accelerates individual bodies within the system. If all the forces observed are real, then the observer is in a noninertial frame.
  10. Mar 24, 2013 #9
    DaleSpam, does the fact that motion is different in a rotating frame mean we can identify one by means of the fictitious forces? I am thinking in context of the fact that we can distinguish a state of rest from a state of constant motion in a straight line. EDIT: More specifically, I was wondering if identifying a rotating frame has any implications for "absolute rest"

    Darwin123, thanks for the explanation. However, I'm not sure I understand the last part - how can total momentum be conserved if there is a net force on the system? Surely the force would change the momentum.
  11. Mar 24, 2013 #10


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    Yes, you can identify rotating frames by the presence of the centrifugal and Coriolis forces. In general, you can identify some arbitrary non inertial frame by the presence of some arbitrary fictitious force.

    There is no absolute rest.
  12. Mar 24, 2013 #11
    Ok, thanks for clarifying that.

    I guess what I was trying to get to with the question of rest was that is acceleration "absolute" in the sense that we can always identify it?
  13. Mar 24, 2013 #12


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    If you first define a reference frame, which is to be called 'inertial', then other reference frames can be inertial with respect to the first frame. Whether we can always identify if two reference frames have constant motion with respect to each other... That depends on the kinds of experiments you are allowed to do. If you can do whatever experiment you like, then it should always be possible to be able to tell if two reference frames are moving with constant relative motion.
  14. Mar 25, 2013 #13


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    Euler angles are the angle of rotation for each of the three rotations. It's a generic term. If you're talking about attitude control or airplane flight, they're roll, pitch, and yaw. If you're talking about satellite orbits, they're right ascension of ascending node, inclination, and argument of perigee. If you're talking about locations on the Earth, they're longitude, geodetic latitude, and ..... well, sometimes you get lucky and you only need two rotations (or less). It takes a maximum of three rotations - which really only means that if you have more than three, then some of the rotations can be combined until you've reduced them to three.
  15. Mar 25, 2013 #14


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    In general there are two kinds of acceleration. There is coordinate acceleration and proper acceleration. Proper acceleration is the acceleration measured by an accelerometer. It is something that does not depend on the coordinate system, and can always be identified directly through measurement. Coordinate acceleration is the second time derivative of the coordinate position. That is something that obviously does depend on the coordinate system.

    Proper acceleration measures the acceleration due to the sum of all forces except gravity and fictitious forces. So standing on the ground an accelerometer reads a proper acceleration of 1 g upwards, corresponding to the normal force. This similarity between fictitious forces and gravity is the basis of the "equivalence principle" formulated by Einstein in his theory of gravity.
  16. Mar 25, 2013 #15


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    Well, for objects that are macroscopic (much bigger than atoms), the only forces that are relevant are electrical, magnetic, gravitational and contact forces (a contact force is at a microscopic level electromagnetic, too). In the absence of gravity, an inertial (unaccelerated) path is the path taken by a free uncharged particle that is not in contact with any other object. So acceleration can always be measured relative to these inertial paths.

    When you include gravity, there is no way to get a particle to follow an unaccelerated path, since all objects are affected by gravity. The best you can do is try to subtract out the effect of gravity. The approach taken by General Relativity is to measure acceleration relative to freefall paths (paths where the only forces at work are gravity).
  17. Mar 25, 2013 #16
    So is a rotating frame an example of coordinate acceleration?

    But are these angles measured from the same initial orientation? My lecture notes read "∅ and θ are the usual spherical coordinates expressing the direction of the principal axis e3, and ψ expresses the orientation of the object about this axis."

    Then they write the kinetic energy of the body as:
    [itex]T=\frac{1}{2}I_1 \dot{\phi}^2 \sin^2(\theta) + \frac{1}{2}I_2\dot{\theta}^2 + \frac{1}{2}I_3 (\dot{\psi} + \dot{\phi} \cos(\theta))^2[/itex]
    where the I's represent the principal moments of inertia. How does this expression come about?
  18. Mar 25, 2013 #17
    "Net force" means "total force"?

    If the total force on a closed system is zero, then the total linear momentum does not vary in time. "Total momentum" is the sum of the momentum from each body in the system. However, the momentum of individual bodies does not have to be conserved. The change in momentum of one body is cancelled out by the change in momentum of the other body.

    Newton's third law says that if two bodies interact, the <real> force on one body caused by the interaction is equal and opposite to the <real> force on the other body caused by the interaction. Therefore, the change in momentum on one body caused by the interaction has to be equal and opposite to the change in momentum in the other body caused by the interaction. Note that his laws apply to <real> forces rather than <fictitious> forces.

    I put <real> in bracket because both force and momentum vary with the observer. There is no formal definition of "real", so even physicists use the word ambiguously. Observers in different frames of reference will determine different values for force and momentum. Some people use the word "real" as meaning that the quantity is can not vary with the observer. Thus the value of a "real" quantity can not vary with observer. I don't use the word "real" this way because it can be confusing. To me, something can be real even if it varies with the observer.

    I am using the word <real> to distinguish it from <fictitious>. A fictitious force can only exist in a noninertial frame. If you don't know whether or not you are in an inertial frame, then you have to "test" the force. If that force can not be correlated with an interaction between two bodies, then it is <fictitious>. If you determine a force is <fictitious>, then you are not in an inertial frame.

    Newton's in Principia explicitly said that his laws of mechanics are applicable in a hypothetical "absolute space". In a way, this is the definition of "absolute space". An "absolute space" is a set of observers to which all the laws of Newton are valid. His "absolute space" is what we would call an inertial frame. There are no fictitious forces in an inertial frame.

    A little problem His phrase "absolute space" seems to imply that it is unique. "Inertial frames" are not unique. There are different inertial frames. The observers in one inertial frame are moving at a constant velocity (i.e., both speed and direction) with the observers of another inertial frame.

    Important to note: not all reference frames are inertial frames. Obviously, an observer that is stationary with respect to the surface of the earth is not in an inertial frame. In any inertial frame, the surface of the earth is accelerating inward toward the center. Therefore, an observer at rest with respect to the surface of the earth sees fictitious forces. There is no interaction between two bodies that can explain the Coriolis force nor the centrifugal force. That is why they are "fictitious".

    You may have been taught that centrifugal force and centripetal force are different although closely related. Can you see now why "centripetal force" is a <real> force while "centrifugal force" is a <fictitious> force?
  19. Mar 25, 2013 #18


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    Here's the way I look at it: The velocity vector [itex]\boldsymbol{V}[/itex] can be written in terms of noninertial components [itex]x', y', z'[/itex]:

    [itex]\boldsymbol{V} = V^{x'} \boldsymbol{e_{x'}} + V^{y'} \boldsymbol{e_{y'}} + V^{z'} \boldsymbol{e_{z'}}[/itex]

    where [itex]\boldsymbol{e_i}[/itex] is the basis vector in direction [itex]i[/itex]. Now, if we take the time derivative of [itex]\boldsymbol{V}[/itex], we get two parts: the coordinate acceleration and a "correction" due to the use of noninertial basis vectors:

    [itex]\dfrac{D}{Dt} \boldsymbol{V} = \boldsymbol{A_{coord}} + \boldsymbol{A_{corr}}[/itex]


    [itex] \boldsymbol{A_{coord}} = \dot{V^{x'}} \boldsymbol{e_{x'}} + \dot{V^{y'}} \boldsymbol{e_{y'}} + \dot{V^{z'}} \boldsymbol{e_{z'}}[/itex],

    and where

    [itex] \boldsymbol{A_{corr}} = V^{x'} \dot{\boldsymbol{e_{x'}}} + V^{y'} \dot{\boldsymbol{e_{y'}}} + V^{z'} \dot{\boldsymbol{e_{z'}}}[/itex].

    [itex]\boldsymbol{A_{corr}}[/itex] is the "correction" to the coordinate acceleration due to the fact that the basis vectors themselves vary from place to place and from moment to moment.

    For example, if the basis vectors in a rotating coordinate system are related to those in an inertial coordinate system through

    [itex]\boldsymbol{e_{x'}} = \boldsymbol{e_{x}} cos(\omega t) + \boldsymbol{e_{y}} sin(\omega t)[/itex]

    [itex]\boldsymbol{e_{y'}} = \boldsymbol{e_{y}} cos(\omega t) - \boldsymbol{e_{x}} sin(\omega t)[/itex]

    [itex]\boldsymbol{e_{z'}} = \boldsymbol{e_{z}}[/itex]


    [itex]\dot{\boldsymbol{e_{x'}}} = \omega (- \boldsymbol{e_{x}} sin(\omega t) + \boldsymbol{e_{y}}cos(\omega t)) = \omega \boldsymbol{e_{y'}}[/itex]

    [itex]\dot{\boldsymbol{e_{y'}}} = \omega (- \boldsymbol{e_{y}} sin(\omega t) - \boldsymbol{e_{y}}cos(\omega t)) = -\omega \boldsymbol{e_{x'}}[/itex]

    [itex]\dot{\boldsymbol{e_{z'}}} = 0[/itex]

    "Coordinate acceleration" ignores the changes in the basis vectors.
  20. Mar 25, 2013 #19


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    That's the reason that I'm so opposed to the concept of "inertial forces" or "fictitious forces". Mathematically, they are just the time derivative of basis vectors. In general, it's bad practice to think of vectors as ordered tuples, such as

    [itex]\boldsymbol{V} = (V_x, V_y, V_z)[/itex]

    A better way to think about it is in terms of linear combination of basis vectors:

    [itex]\boldsymbol{V} = V_x \boldsymbol{e_x} + V_y \boldsymbol{e_y} + V_z \boldsymbol{e_z}[/itex]

    If you think of it that way, then a change of coordinates does not change a vector, it just rewrites it using different basis vectors. The second advantage of writing it this way is that then you can see that, for a coordinate system with nonconstant basis vectors, you don't just get:

    [itex]\dfrac{D}{Dt} \boldsymbol{V} = (\dot{V_x}, \dot{V_y}, \dot{V_z})[/itex]

    You must also take into account the time-varying basis vectors. You have to include terms like

    [itex]V_x \dot{\boldsymbol{e_x}}[/itex]
  21. Mar 25, 2013 #20
    Acceleration is "absolute" in both special relativity (SR) and Newton's Laws (NL). Although there is no absolute velocity, an absolute acceleration can be defined in terms of "real" forces. A frame of reference is inertial if the total linear momentum is conserved. The acceleration of individual bodies in this inertial frame can be detected by observers in this frame.

    A lot of non-scientists don't understand that velocity is always relative with Newton's Laws. NL are Galilean invariant. However, absolute acceleration can be defined in terms of "real forces". Although SR is different than NL in some ways, this remains the same. An absolute acceleration can be defined in terms of force interactions.

    Acceleration is not absolute in general relativity (GR) because of the "law of equivalence". Acceleration and gravity are conflated by the law of equivalence. However, there are some qualifications to this which I am not prepared to address right now.
  22. Mar 25, 2013 #21
    Thanks for all the replies, I think I understand fictitious forces well now.

    Darwin123, a centripetal force is the force keeping a body in circular motion, so they are real - we can make the action-reaction pairs. For example, if a rope is used to swing a rock in a circle, the tension in the rope is keeping the rock in circular motion, and the rope also feels the reaction force due to the rock.

    Stevendaryl, the discussion using basis vectors was very helpful, thanks.

    I was hoping that we could now tackle the Euler angles.
  23. Mar 25, 2013 #22


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    What? There is definitely a notion of absolute 4 - acceleration in GR: ##a^b = u^{a}\triangledown _{a}u^b## and with ##a^{b}## non - vanishing there is an associated 4 - force ##f^{b} = ma^{b}##. FYI, the (weak) E.P. is something that is present in Newtonian theory as well.
  24. Mar 26, 2013 #23


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    I think what Darwin123 means is that in Minkowski spacetime, we can always choose a 'cartesian coordinate system' in which the twice derivatives of the coordinates of a free particle vanish. But in GR, this is not always possible. (because we can't generally make the Christoffel symbols zero).
  25. Mar 26, 2013 #24


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    So what exactly are you wanting to know about Euler angles? A way to see that there are three angles needed is to consider rotating a globe while keeping its center fixed. You can partly specify the orientation by saying:

    Pick a point [itex]P[/itex] on the globe with latitude [itex]\Psi[/itex] and longitude [itex]\Phi[/itex]. Rotate the globe so that the North pole is moved to the place where [itex]P[/itex] used to be.​

    That doesn't uniquely determine the orientation of the globe, though, because you can still spin the globe on its North-South axis without moving the North pole. So to uniquely determine the orientation, you need two angles (e.g. latitude and longitude) to specify where the North pole goes, and another angle to specify how far to turn the globe on its axis.
  26. Mar 26, 2013 #25
    I want to know how the angles are determined, i.e. what orientation are they describing the body with respect to? And how do we get the formula for kinetic energy in terms of them (which I wrote on the last page.)?
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