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Insights Frames of Reference: A Skateboarder's View - Comments

  1. Oct 19, 2016 #1

    kuruman

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  2. jcsd
  3. Oct 20, 2016 #2
    Looking forward to the followup!
     
  4. Oct 20, 2016 #3

    rcgldr

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    Technically, the mass of earth and slide should be included in conservation of mechanical energy in either frame (closed system free from external forces). From a frame where the earth and box are not initially moving, v0 = 0, then vfearth will be slightly below 0 (slightly negative) and vfbox will be slightly below ##\sqrt{2 \ g \ h}##. Since the earth is much more massive than the box, the box gets most of the increase in kinetic energy.

    From the skateboarders frame, v0 will be slightly above (less magnitude) ##-\sqrt{2 \ g \ h}## and vfearth will be slightly below (more magnitude) ##-\sqrt{2 \ g \ h}## . In this frame, the earth gains kinetic energy, while the box loses kinetic energy.

    The problem is similar to a box sliding down a frictionless wedge on a horizontal frictionless surface, except in this case the mass of the wedge (earth + slide) is huge.
     
    Last edited: Oct 20, 2016
  5. Oct 21, 2016 #4
  6. Oct 21, 2016 #5
    I kind of agree with rcgldr:

    Kuruman states:
    "It should be clear to the reader that in either reference frame momentum and energy conservation gives the exact solution while energy conservation only results in an approximation, albeit a good one."

    What happens if you include the kinetic energy of the Earth in the energy conservation calculations?
     
    Last edited: Oct 21, 2016
  7. Oct 21, 2016 #6

    vanhees71

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    As usual, the apparent paradox is much easier to understand when analyzed using the action principle. In the original frame we have
    $$L=\frac{m}{2} \dot{\vec{x}}^2-mgy,$$
    where ##y## points vertically up. Further we have the holonomous constraint
    $$y=f(x)$$
    and thus the action reads
    $$L=\frac{m}{2} [1+f^{\prime 2}(x)] \dot{x}^2-mgf(x).$$
    Now it's clear, why seen from the skater's reference frame the energy is not conserved. We have
    $$x=x'+u t, \quad y=y'$$
    and thus
    $$L=\frac{m}{2} [1+f^{\prime 2}(x'+u t)](\dot{x}'+u)^2-mgf(x'+ut).$$
    Since now the Lagrangian is explicitly time dependent, energy is no longer conserved from the skater's point of view.
     
    Last edited: Oct 21, 2016
  8. Oct 21, 2016 #7
    Great Insight Kuruman!
     
  9. Oct 21, 2016 #8
    Can you clarify what you mean by

    This seems to imply actual conservation of energy is violated in a particular frame of reference... so I must be missing a subtlety of interpretation.

    BTW is the energy of the Earth taken into account? There is energy transferred between the two bodies.

    GENERALLY: Is there any reason we can assume that the Earth can be ignored when dealing with an interaction/force involving the Earth?
     
    Last edited: Oct 21, 2016
  10. Oct 21, 2016 #9
    Kuruman:

    I like the approach with momentum of both the Earth and the block, however, since the locus of interaction is also at the surface of the earth (a point of contact via a shaped ramp) it seems that angular momentum comes into the picture.

    Some of the momentum of the block would indeed be transferred as final linear momentum of the Earth, the Earth would also be imparted angular momentum, since the block end up travelling tangentially to the Earth's surface. Assuming the Earth is rigid, the ratio of the angular to linear momentum transferred involves the moment of inertia of the Earth which of course depends on the density distribution of the Earth.

    I wonder what is the moment of inertia of the Earth and how much of a role does that play in the answer for v0... it may be more complex than what is written in the section "a second opinion"

    EDIT: The concept of "transferring" linear momentum into angular momentum is not correct. The idea that some of the energy is stored in both rotational energy of the Earth and linear kinetic energy of the Earth is likely valid. Sorry for the sloppiness here.
     
    Last edited: Oct 21, 2016
  11. Oct 21, 2016 #10

    rcgldr

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    The reason that earth's energy can be ignored from a earth based frame, is that that v0 = 0, and Δv is very small. The increase in angular energy of the earth in this frame is

    1/5 mearth Δvearth2

    From a different frame, v0 ##\ne## 0, and the increase of energy of the earth is

    1/5 mearth ((v0 + Δvearth)2 - v02) = 1/5 mearth (2 v0 Δvearth + Δvearth2)

    For an "exact" solution, the earth needs to be considered as part of a closed system that includes earth, slide, and box (the skateboarder is effectively an outside observer).

    Using mass of earth = 5.972e24 kg, radius of earth = 6.371e6 m, mass of box 1kg:

    change in velocity of box
    ##\Delta vb = \sqrt{\frac{2(2.3888e24)^2}{2.3888e24+2.3888e24^2}} \ \ \sqrt{g \ h}##
    ##\Delta vb = 0.99999999999999999999999979069 \ \sqrt{2 \ g \ h}##

    change in velocity of earth
    ##\Delta ve = - \sqrt{\frac{2}{2.3888e24 + 2.3888e24^2}} \ \sqrt{g \ h}##
    ##\Delta ve = -0.00000000000000000000000041862 \ \sqrt{2 \ g \ h}##

    For v0 = 0, I get

    Code (Text):

    Δenergyofearth =  0.00000000000000000000000041862 kg g h
    Δenergyofbox   =  0.99999999999999999999999958138 kg g h
    Δenergy        =  1.00000000000000000000000000000 kg g h
     
    for v0 = - 0.99999999999999999999999979069 sqrt(2 g h), I get

    Code (Text):

    Δenergyofearth =  1.99999999999999999999999958138 kg g h
    Δenergyofbox   = -0.99999999999999999999999958138 kg g h
    Δenergy        =  1.00000000000000000000000000000 kg g h
     
     
    Last edited: Oct 24, 2016
  12. Oct 21, 2016 #11

    A.T.

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    The simple approach in the ramp's frame is not based on the conservation of total energy, just on the work theorem applied to the block: Since the ramp does no work on the block, the block's energy should be constant.

    However, this assumes that the ramp's frame is inertial, which it actually isn't as the lack of momentum conservation shows. So even if the ramp does no work on the block in the ramp's frame, the inertial force present in that non-inertial frame does a tiny bit of work on the block.
     
    Last edited: Oct 22, 2016
  13. Oct 21, 2016 #12

    vanhees71

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    Well, I think the "naive-mechanics explanation" has been very well explained in the article. For me it's very difficult to think in terms of forces, but the action-principle approach makes it immediately clear. It's due to the constaint of the block to the ramp which makes the energy non-conserved in the skater's rest frame, because in this frame the Lagrangian becomes explcitly time dependent. The momentum is neither conserved in the restframe of the ramp nor in the skater's restframe, because in neither frame the independent variable #x# is cyclic (see the posting above, where I corrected some typos). Of course nobody would solve this problem in the skater' frame, since it's immediately clear that in the rf. of the ramp the Lagrangian doesn't depend explicitly on time, and thus that the Hamiltonian in these cordinates is conserved.
     
  14. Oct 21, 2016 #13
    What happens if you include the Earth in the Lagrangian? Would it matter that the calculation was made in the skater's rest frame?
     
  15. Oct 21, 2016 #14

    A.T.

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    The skater is an inertial observer. Why wouldn't momentum be conserved in his frame?

    Even if the mass of ramp+Earth was comparable to the block's mass?
     
  16. Oct 22, 2016 #15

    vanhees71

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    Why should momentum be conserved? It's not conserved in either frame, because there are forces acting on the block (gravitation from Earth and the constraining forces from the ramp. The latter forces become obviously time dependent in the skater's rest frame, and thus the forces are not conservative in this frame. Of course the skater's frame is an inertial frame, but that doesn't imply momentum or energy conservation if there are nonconservative forces acting on the block.
     
  17. Oct 22, 2016 #16

    A.T.

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    By "momentum conservation" I meant total momentum of the isolated system, not the momentum of the block only, which obviously not conserved.
     
  18. Oct 22, 2016 #17

    vanhees71

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    Sure, if you have a closed system, the Lagrangian must be Galileo invariant, and all 10 conservation laws must hold. However, in this case it's a very complicated Lagrangian you can't write down to begin with. That's why we use the effective description with constraints (which are idealized too of course) to simplify the problem. I don't see, what's the problem to understand the calculation concerning the Lagrangian I gave above. I thought it might help to understand the problem, expressed in the article in terms of the Newtonian formulation (which I find much more complicated than the Lagrange formalism).
     
  19. Oct 22, 2016 #18

    rcgldr

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    Post #10 includes the Newtonian math assuming a closed (isolated) system.
     
  20. Oct 24, 2016 #19
    The conclusions I get from this discussion is that although particular reference frames may be used to simplify calculations,

    1. there are no incorrect reference frames as such.

    2. When adopting a nonstandard reference frame one must be careful not to ignore the reality of the whole physical system (e.g. the Earth) and one must be careful not to blindly apply the computational shortcuts and assumptions conveniently used in the calculations in the standard frame, which no longer apply when attempting to solve the problem in the nonstandard reference frame.
     
  21. Oct 31, 2016 #20
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