Debate: Lorentz invariance of certain zero angles

Discussion in 'Formal debates' started by Pete, Nov 25, 2011.

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  1. Pete It's not rocket surgery Moderator

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    Full topic:
    The zero angle between the tangent plane of a microfacet and the tangent velocity as measured in the axle frame transforms into a zero angle through a boost along the x-axis in the ground frame
    For: Tach
    Against: Pete

    Rules are as agreed in the Proposal thread:
    [thread=111135]Proposal: A surface moving parallel to itself in one frame is not doing so in all frames[/thread]
    [thread=111143]Discussion thread[/threaD]

    Tracking list
    1.0 Scenario (Pending)
    1.1 - Coordinate dependence vs. coordinate independence (Resolved)
    1.2 - Definition of rods T1 and T2 ("Active")
    1.3 - Definition of points A and B (Pending)

    2.0 Methodology (not started)

    3.0 Calculations (not started)

    4. Summary and reflection (not started)​
     
    Last edited by a moderator: Nov 28, 2011
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  3. Tach Banned Banned

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    A. Scenario: The zero angle between the tangent plane of any microfacet belonging to a circular wheel and the tangent velocity as measured in the axle frame transforms into a zero angle through a boost along the x-axis in the ground frame


    B. Methodology and Calculations:

    1. In the frame co-moving with the axle, the trajectory of a point on the circumference is:

    \(x=r cos(\omega t)\)
    \(y=r sin(\omega t)\)

    where \(\omega\) is the angular speed of the wheel and r is its radius.

    The tangential speed of a particle \(\vec{v_p}\) is identical to the speed of the tangent plane in any point \(\vec{v_p}=\vec{v_t}=(-r \omega sin(\omega t), r \omega cos (\omega t))\), a well known fact. So,the angle between \(\vec{v_p}\) and \(\vec{v_t}\) is ZERO everywhere in the axle frame.

    2. In the ground frame

    The axle moves with speed \(\vec{V}=(V,0)\) so:

    \(\vec{v'_p}=\vec{v'_t}=(\frac {V-r \omega sin (\omega t)}{1-Vr \omega sin(\omega t) }, \frac{r \omega/ \gamma(V) cos(\omega t)}{1-Vr \omega sin(\omega t)})\)

    so, not only that the two speeds are identical (of course they are), their angle is ALSO ZERO in the ground frame. This is true for both relativistic and "classical" case . It is also true whether the wheel motion has slippage or not.

    The "no slip" situation is a particular case of the above, \(r \omega=V\)

    \(\vec{v'_p}=\vec{v'_t}=(\frac{V(1- sin (\omega t))}{1-V^2 sin(\omega t) }, \frac{V/\gamma(V) cos(\omega t)}{1-V^2 sin(\omega t) })\)

    Of course, the same holds, the zero angle in the axle frame transforms into a zero angle in the ground frame.
     
    Last edited: Nov 25, 2011
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  5. Pete It's not rocket surgery Moderator

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    1.0 Scenario
    To be clear:

    It seems that the scenario you are describing is that of a wheel rolling without slipping along the ground.
    In reference frame S, the axle is at rest at (x,y)=(0,0), the wheel is circular with radius r, and is spinning with angular velocity \(\omega\).

    P is a point on the circumference, located at (x,y)=(r,0) at t=0.
    \(v_p\) is the instantaneous velocity of point P at t=0 in frame S.

    You mention a moving tangent plane, which is not clearly defined.
    I would like add physical realisation of a specific tangent plane to the scenario:
    T is a straight rod
    In frame S:
    At t=0, T is tangent to the wheel at point P.
    T is moving inertially parallel to the y axis, with velocity \(\vec{v_t} = \vec{v_p}\).

    Diagram of frame S at t=0:

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    Does this match what you had in mind, and is it sufficiently well defined?
     
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  7. Tach Banned Banned

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    Yes, but much more general, T is moving in a direction perpendicular on the position vector connecting the center of the wheel and the point P, so \(\vec{v_t(t)}=\vec{v_p(t)}\) for all values of time, t and for all points P on the circumference. This is general stuff, well known so I did not think I needed to get that specific but now we are as specific as we could be.
     
    Last edited: Nov 26, 2011
  8. Pete It's not rocket surgery Moderator

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    1.0 Scenario - Complete

    OK, I'll edit the tracking list to mark the Scenario stage as "Complete", and the methodology stage as active, if you agree.

    I'll prepare a list of the measurements I want us to consider, along with how they will be measured. You do the same (feel free to copy from your first post if you have nothing more to add.)

    Don't forget to add a header to your post, like this one. This might seem anal now, but I think it will really help to keep thing on track later on.
     
  9. Tach Banned Banned

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    Cool.
     
  10. Pete It's not rocket surgery Moderator

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    1.0 Scenario
    Wait. You edited this post as I was responding.

    I think I know what you mean, but I don't think T can be made so general and still be well defined.
    Let's keep to something specific enough that we can actually analyse it.

    We've chosen a specific point on the circumference.
    T is a specific object that is tangent to that point at a specific time.

    It is moving inertially, so it can be easily transformed between frames, and remains straight in all frames.

    Agreed?

    Or, perhaps would you like to add another tangent rod?
    • T1 is a straight rod, moving inertially with \(\vec{v_t} = \vec{v_p}\) at t=0 in S, and is tangent to the wheel at P at t=0 in S.
    • T2 is a straight (in S) rod, tangent and attached to the wheel at P at all times
     
  11. Tach Banned Banned

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    I have big issues with using coordinate-dependent proofs. As such, I prefer the use of coordinate independent entities, like vectors over points. So, for me, the tangent is defined by a pair of time varying points, let's call T_1(t) and T_2(t), or, better by the time varying vector
    \(\vec{v_t}=\vec{T_1T_2}\) which is nothing but the velocity of the tangent plane expressed in the axle frame. Makes sense? Did you notice that none of my proofs contains any coordinate-dependent entities? This approach makes the proofs simpler and less susceptible to error.
     
  12. Pete It's not rocket surgery Moderator

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    1.1 Coordinate dependence

    Sorry, I don't follow what you mean by "coordinate dependent".

    Are you saying that the rods T1 and T2 are a coordinate dependent entities?

    I also don't understand how you are defining T_1(t) and T_2(t).
     
    Last edited: Nov 26, 2011
  13. Tach Banned Banned

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    1.1 - Coordinate dependence vs. coordinate independence




    Coordinate-dependent means writing formalisms in terms of point coordinates , as in (x,y,z). By contrast, a formulation in terms of vectors is considered coordinate - independent.

    No, what I was saying was that choosing T as a point results into a coordinate dependent formulation. Likewise , taking points T1(x1,y1,z1) and T2(x2,y2,z2) as two points defining the tangent results into a coordinate-dependent formulation. On the other hand, working with the vector \(\vec{v_t}=\vec{T_1T_2}\) results into a coordinate-independent formalism. T1 and T2 are points, not rods. Does this make sense now?
     
  14. Pete It's not rocket surgery Moderator

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    1.1 - Coordinate dependence vs. coordinate independence
    I don't understand why using coordinates is a problem, as long as the coordinate system is well defined.
    I note that in your opening post to this thread, you've specified the trajectory of a specific point on the circumference in terms of coordinates:
    This is perfectly clear, given a couple of straightforward assumptions about the origin and orientation of the coordinate system.

    And I didn't notice you having a problem with coordinates in previous threads either (in [post=2847465]shape of a relativistic wheel[/post], for example), so I don't understand why it's a problem now.

    I am not skilled in the formalism of using vectors, so if you insist on going down that path the pace of the discussion will become much slower.
     
  15. Tach Banned Banned

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    Yes, I did just in order to be able to represent the components of the vectors \(\vec{v_p}\), \(\vec{v_t}\)


    ...because that thread was about shapes, more exactly about the implicit representation of a closed curve (the ellipse).


    I don't insist, use what you feel more comfortable with and I will use what I think is more appropriate, including the coordinate-independent formalism. So, at times we may have to reconcile the two.
     
  16. Pete It's not rocket surgery Moderator

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    Ok, thanks for that.

    Definition of T1 and T2

    In post 3, I suggested adding a rod, labelled T, to the scenario.
    In post 7, I suggested adding another rod, labelled T2, and relabelling T as T1:


    • Is there a coordinate dependence problem with the specification of those rods?

      In post 8, you suggested defining T1 and T2 as points:
      But I don't understand exactly how those points are defined.
     
  17. Tach Banned Banned

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    Points T1 and T2 are simply two points defining the tangent in T.
    since you aren't using the coordinate independent formalism, you will need two points to define the tangent. They can be T1 and T2 or T1 and T or whatever.
     
  18. Pete It's not rocket surgery Moderator

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    Definition of T1 and T2
    Sorry, I still don't understand how you are defining T1 and T2 as points. Can you be more explicit?

    Also,
    Do you understand that when T1 and T2 were originally mentioned in post 7, they were defined as rods, not points:
    Is there any problem with that specification?

    Remember, we agreed to answer direct questions in the next post.
     
  19. Tach Banned Banned

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    I am defining them different from you, xcall them A and B, ok?

    I answered the direct question, I am not using any rods, I am defining the tangent through two points. See what happens when you insist in using coordinate-dependent representations? A lot of confusion and wasted time. Please, let's use the tangent vector \(\vec{v_t}\) and be done with this.
     
  20. Pete It's not rocket surgery Moderator

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    Definition of rods T1 and T2
    Agree, we will call your two tangent-defining points A and B.
    We will call the two tangent rods T1 and T2.

    No, you did not.
    The question was whether there is a problem with the specification of the rods, specifically a coordinate dependence problem.

    Is there a problem with the way the rods are specified?

    You seem to imply that they are defined in a coordinate dependent way, but I don't understand why.
     
    Last edited: Nov 27, 2011
  21. Tach Banned Banned

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    Can you do a drawing for each of the two rods and add their equations of motion? Thank you
     
  22. Pete It's not rocket surgery Moderator

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    1.2 - Definition of rods T1 and T2

    T1 is an inertial rod.
    At t=0 in S, T1 is tangent to wheel element P, and moving with the same velocity as P.

    T2 is a rod permanently attached to the wheel at P.
    At t=0 in S, the location of T2 coincides with the location of T1 as illustrated:

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    T2 moves rigidly in S in such a way that it remains tangent to the wheel at P at all times.
    For example, at \(t = \pi/(2\omega)\) in S, T2 has moved around the wheel with P, while T1 has moved inertially, as illustrated:

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    Does that make sense? There is no coordinate dependence in that description that I can see, but here are the equations of motion in vector form anyway:
    Equations of motion
    \(\vec{v_P}(t)\) is the velocity of wheel element P at time t in S.
    Let \(\vec{P}(t)\) denote the position vector of P at time t in S.
    Let \(\hat{P_t}(t)\) denote the unit displacement vector tangent to wheel element P in S at time t.

    Let \(\vec{T_1}(t,l)\) denote the position vectors of the elements of rod T1 at time t in S, where \(l\) is distance along the rod from the element that was in contact with P at t=0.
    Let \(\vec{T_2}(t,l)\) denote the position vectors of the elements of rod T2 at time t in S, where \(l\) is distance along the rod from P.

    Thus:
    \(\vec{T_1}(t,l) = \vec{P}(0) + l\hat{P_t}(0) + t\vec{v_p}(0)\)
    \(\vec{T_2}(t,l) = \vec{P}(t) + l\hat{P_t}(t)\)
     
    Last edited: Nov 28, 2011
  23. Tach Banned Banned

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    1. I can't see any useful role for \(\vec{T_1}(t,l) \)

    2. I would expect:

    \(\vec{T_2}(t,l) = \vec{P}(t) +\hat{P_t}\vec{v_p}(t)\)
     
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