Angle between the orientation of a moving object and its velocity

Discussion in 'Physics & Math' started by Pete, Nov 23, 2011.

  1. Pete It's not rocket surgery Moderator

    Messages:
    10,166
    This is for Tach, who claims that if a surface is moving parallel to itself in one reference frame, it is move parallel to itself in all reference frames.

    First, the easy handwaving rebuttal:

    An observer in a train looks at the ground, and notes that the ground is moving parallel to itself.

    An observer in an elevator looks at the same ground, and notes that the ground is moving perpendicular to itself.

    Therefore, Tach is wrong.
     
  2. Google AdSense Guest Advertisement



    to hide all adverts.
  3. Pete It's not rocket surgery Moderator

    Messages:
    10,166
    And now a more formal approach.

    Please Register or Log in to view the hidden image!


    Consider a flat surface A in inertial motion in a 2+1 spacetime.
    Choose an inertial reference frame S(x,y,t), such that in S:
    • A is at rest.
    • The angle between the surface A and the x-axis is \(\theta_A\).
    • A intersects (x,y,t)=(0,0,0)

    A is thus defined by:
    \(y = x \tan(\theta_A)\)

    An inertial particle P passes through (x,y,t)=(0,0,0) with velocity \(\vec{V_p}\).
    The angle between \(\vec{V_p}\) and the x-axis is \(\theta_P\).
    The worldline of P is thus defined by:
    \(\begin{align} x &= t |\vec{V_p}| \cos(\theta_P) \\ y &= t |\vec{V_p}| \sin(\theta_P) \end{align}\)


    A reference frame S'(x',y',t') has velocity v in the x-direction relative to S, with parallel axes and coincident origin.
    \(\begin{align} \beta &= v/c \\ \gamma &= 1/\sqrt{1-\beta^2} \end{align}\)

    Transforming A, we find:
    \(y' = \gamma\tan(\theta_A) (x' + vt')\)

    The angle between A' and the x'-axis is thus defined by:
    \(\tan(\theta'_A) = \gamma\tan(\theta_A)\)

    Here, I note that if \(\tan(\theta_A) = 0\), then \(\tan(\theta'_A)=0\) as well.

    Transforming the worldine of P, we find:
    \(\begin{align} x' &= t'\frac{|\vec{V_p}| \cos(\theta_P) - v}{1 - v|\vec{V_p}| \cos(\theta_P)/c^2} \\ y' &= \frac{t'}{\gamma}\frac{|\vec{V_p}| \sin(\theta_P)}{{1 - v|\vec{V_p}|\cos\theta_P/c^2} \end{align}\)

    The angle between \(\vec{V'_p}\) and the x'-axis is thus given by:
    \(\tan(\theta'_P) = \frac{|\vec{V_p}| \sin(\theta_P)}{\gamma(|\vec{V_p}| \cos(\theta_P) - v)}\).

    Again, I note that if \(\tan(\theta_P) = 0\), then \(\tan(\theta'_P)=0\) as well, and I also note that this transformation is not the same as for \(\theta_A\).


    The angle \(\phi\) between A and \(\vec{V_p}\) is given by:
    \(\begin{align} \phi &= \theta_A - \theta_P \\ &= \frac{\tan(\theta_A) - \tan(\theta_P)}{1+\tan(\theta_A)\tan(\theta_P)} \end{align}\)

    The angle \(\phi'\) between A' and \(\vec{V'_p}\) is given by:
    \(\begin{align} \phi' &= \theta'_A - \theta'_P \\ \tan\phi' &= \frac{\tan(\theta'_A) - \tan(\theta'_P)}{1+\tan(\theta'_A)\tan(\theta'_P)} \\ \end{align}\)

    Note very well:
    Due to the different transformations of \(\theta_A\) and \(\theta_P\), \(\tan\phi = 0\) does not imply that \(\tan(\phi')=0\), except in special cases such as \(\theta_A = \theta_P = 0\).

    ---------------------------
    Notes:
    Let me know if something needs clarification, doesn't seem to make sense, or is just wrong.
    I have a more general case for a surface moving with arbitrary velocity, but this simpler case is sufficient.
    No, I'm not going to do a (x,y,z,t) derivation. Too hard, no returns.
     
    Last edited: Nov 24, 2011
  4. Google AdSense Guest Advertisement



    to hide all adverts.
  5. Pete It's not rocket surgery Moderator

    Messages:
    10,166
    It is easy to apply the previous analysis to the case of a flat object moving parallel to itself in S:

    A flat object B slides along the surface A.
    P is an element of B.
    Clearly, \(\phi = 0\).

    It is also easy to apply to the rolling wheel where we started:

    Please Register or Log in to view the hidden image!



    In this case, \(\theta_A = \theta_P\), and \(\phi\) is obviously zero.

    Tach claims that in this situation, \(\phi'\) is necessarily zero as well.

    So, is he right? Stay tuned...
     
    Last edited: Nov 23, 2011
  6. Google AdSense Guest Advertisement



    to hide all adverts.
  7. Tach Banned Banned

    Messages:
    5,265
    Rushing much?

    The elevator guy decides to test the flatness of the elevator floor. He sets a laser beam parallel to the floor. The observers decide to call the floor "flat" when the angle between the laser light and the NORMAL to the floor (y- direction) is \(\pi/2\).
    The observer on the train moving with speed V in the x direction concludes that:

    \(cos \theta'=\frac{cos \theta +v/c}{1+ v/c. cos \theta}\)



    1. In the frame of the elevator \(\theta=\pi/2\)
    Now, in the frame of train \(cos \theta' =v/c\), so, \(\theta'< \pi/2\)

    2. Since the two observers know physics, they decide to change the way they measure the flatness of the elevator floor by deciding that the floor is flat if the angle between the laser beam and the TANGENT to the floor is ZERO, i.e. \(\theta=0\)
    Now, the observer on the train measures: \(cos \theta' =1\), so, \(\theta'=0\)

    Now, for the people who want to read a much more straightforward proof using vectors instead of the ugly components, see here. pete was shown this solution before he posted his.

    Next time don't be so quick to declare victory.
     
  8. Tach Banned Banned

    Messages:
    5,265
    Actually the above establishes nothing. You simply put down the formula

    \(tan(\alpha-\beta)=\frac{tan (\alpha) -tan (\beta)}{1+ tan (\alpha) tan (\beta)}\)

    twice, once for the unprimed frame and once for the primed frame. You can't conclude anything from the two trivial formulas.

    Now:

    ...meaning that :

    \(\tan(\theta'_A) = \frac{|\vec{V_A}| \sin(\theta_A)}{\gamma(|\vec{V_A}| \cos(\theta_A) - v)}\)

    But, in the unprimed frame: \(\theta_A= \theta_P\)
    \(\vec{V_p}=\vec{V_A}\)
    so:

    \(\tan(\theta'_A) =\tan(\theta'_P)\)

    meaning that:

    \(\tan(\phi') =0\)

    The end.
     
    Last edited: Nov 25, 2011
  9. Tach Banned Banned

    Messages:
    5,265
    Yep, he's right, \(\phi=0 -> \phi'=0\) . See above.
    In general, \(\phi=k \pi -> \phi'=k \pi\)
     
  10. Pete It's not rocket surgery Moderator

    Messages:
    10,166
    Try again. This time, think about the scenario I actually described:

    The train guy measures the ground to have velocity parallel to the ground.
    Yes or No?

    The elevator guy measures the ground to have velocity perpendicular to the ground.
    Yes or No?

    These are easy questions, and there's no point in continuing if you don't address them.
     
  11. Tach Banned Banned

    Messages:
    5,265
    No more "try again", I used your own math to prove you wrong.
     
  12. Pete It's not rocket surgery Moderator

    Messages:
    10,166
    Sure you can, if you actually follow it through.

    Wrong.
    As shown in the post you didn't seem to read:
    \(\tan(\theta'_A) = \gamma\tan(\theta_A)\)
     
  13. Tach Banned Banned

    Messages:
    5,265
    \(\tan(\theta'_P) = \frac{|\vec{V_p}| \sin(\theta_P)}{\gamma(|\vec{V_p}| \cos(\theta_P) - v)}\)

    \(\tan(\theta'_A) = \frac{|\vec{V_p}| \sin(\theta_A)}{\gamma(|\vec{V_p}| \cos(\theta_A) - v)}\)

    The point A moves exactly the same way as point P. You have no reason for differing formulas.
     
    Last edited: Nov 23, 2011
  14. Pete It's not rocket surgery Moderator

    Messages:
    10,166
    Repeating your mistake doesn't make it correct. How can you critique the maths if you don't bother to work through it?

    From post 2 -
    A is defined by:
    \(y_A = x_A \tan(\theta_A)\)

    Transforming A:
    \(y'_A = \gamma\tan(\theta_A) (x'_A + vt')\)

    The angle between A' and the x'-axis is therefore:
    \(\tan(\theta'_A) = \gamma\tan(\theta_A)\)
     
  15. Tach Banned Banned

    Messages:
    5,265
    Using an incorrect derivation doesn't make things right. Point A moves the same way as point P, so you have no justification for different formulas.
     
  16. Pete It's not rocket surgery Moderator

    Messages:
    10,166
    And I notice you still haven't addressed the exceedingly simple scenario in the opening post:

    An observer in a train watches the ground passing by, and notes that the ground's velocity is parallel to the ground itself. Right?

    An observer in an elevator watches the ground falling away, and notes that the ground's velocity is perpendicular to the ground. Right?
     
  17. Pete It's not rocket surgery Moderator

    Messages:
    10,166
    A is a surface, not a moving point.

    Post 2:
     
  18. Tach Banned Banned

    Messages:
    5,265
    Err , a "surface" that has point coordinates \((x_A, y_A)\)?
     
  19. Tach Banned Banned

    Messages:
    5,265
    Yes, I addressed in great detail in post 4. It explains why co-linearity is a frame-invariant property. Please read it again.
     
  20. AlphaNumeric Fully ionized Moderator

    Messages:
    6,697
    Is Tach actually claiming this? Really? Surely not, there's got to be a line of miscommunication somewhere, isn't there? :shrug:

    It's trivially false. A mirrored line parallel to the x axis and moving in the x direction in frame S will be moving parallel to itself in S. Do a boost which has a non-zero y component and now you have the mirror moving somewhat in both the x' and y' direction but it's still parallel to the x' axis. Contradiction. Case closed.

    Sure I haven't done the actual algebra but all that would tell you is the specific value of things, while Tach's claim is falsified providing the value is non-zero.
     
  21. Tach Banned Banned

    Messages:
    5,265
    This is not what I claim, and pete knows it. It is a mirror that has a zero angle between its velocity and its tangent. So, your counter-example is irrelevant.
     
  22. Pete It's not rocket surgery Moderator

    Messages:
    10,166
    No, post 4 addressed a different scenario regarding the elevator floor and lasers.

    Please, these are simple questions. I just want to know if you agree with the statements as written:

    An observer in a train watches the ground passing by, and notes that the ground's velocity is parallel to the ground itself.
    Do you agree?

    An observer in an elevator watches the ground falling away, and notes that the ground's velocity is perpendicular to the ground.
    Do you agree?
     
  23. Tach Banned Banned

    Messages:
    5,265
    Yes, it showed that your ideas about colinearity are unphysical. It showed that while colinearity is frame-invariant, orthogonality is not. So, try to understand the math and please stop harassing with the same elementary questions.
     

Share This Page