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Relativity Question

  1. Jun 23, 2003 #1
    An observer, moving at a speed of 0.995c relative to a rod as shown measures its length to be 2.00m and sees its length to be oriented at 30.0 degrees with respect to the direction of motion.

    a) What is the proper length of the rod? (Length as measured at rest)

    b) What is the orientation angle in a reference frame moving with the rod? (Again, in rest frame)


    I've found the correct values, however the math seems ambiguous. I was wondering if there was another way to solve it, with cleaner mathematics.

    Do = the length of the rod as measure when the rod is moving a .995c
    Lo = the adjacent side of the triangle formed by the rod moving at .995

    Lo = DoCos(30.0)

    D = the actual length of the rod as observed from the rest frame
    L = the actual length of the adjacent side of the triangle formed by the rod in the rest frame
    @ = the angle formed between L and D.
    h = height of the rest frame triangle

    L = Lo/sqrt(1-v^2/c^2) = DoCos(30)/sqrt{(1-(.995c)^2/c^2)}

    L = 17.34m

    To find D you can use the following equations.

    Tan(@) = h/L
    Sin(@) = h/D
    Cos(@)=L/D
    D^2 = L^2 + h^2

    I ended up getting this to solve for D.

    D = sqrt{L^2 + L^2*Sin^2{Cos^-1(L/D)}^2}

    My calculator solved it and got D = 17.26m or 17.42m.

    The answers are 17.3m and 3.30 degree.

    Is there a cleaner way to do it?

    Am I missing something?
     
  2. jcsd
  3. Jun 23, 2003 #2

    Tom Mattson

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    I would only clean up the notation. Instead of using different letters for each quantity, I would use subscripts and primes.

    L=Length of rod in frame S
    Lx=Lcos(30)=component of length in direction of motion in frame S
    Ly=Lsin(30)=component of length perpendicular to motion in S

    and use L', Lx' and Ly' for the corresponding quantities in frame S'.
     
  4. Jun 23, 2003 #3
    My math is okay then?

    My notation is a lot cleaner on paper, a lot cleaner the second and third time I worked the problem out. How do you make subscripts?

    Thanks for your help!
     
  5. Jun 23, 2003 #4

    Tom Mattson

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    Yes, there's no other way to do it. But, using a more systematic notation will help whoever reads it (like the grader).

    Subscripts:
    If you type this: v[ sub ]0[ /sub ]
    without the spaces you will get: v0.

    Superscripts:
    If you type this: v[ sup ]2[ /sup ]
    without the spaces you will get: v2.

    You can also combine the two as follows: v[ sub ]0[ /sub ][ sup ]2[ /sup ]
    to get v02.
     
  6. Jun 23, 2003 #5
    CoolThanks!
     
  7. Jun 23, 2003 #6

    Tom Mattson

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    Something else: Regarding Greek letters. The ones in the "Smiley" menu (the ones you've been using) are really ugly, and what's worse they don't respond well to being superscripted or subscripted.

    e[mu]
    e[mu]

    See?

    Instead, use the forum code for this, like so:

    & mu ;

    without the spaces gives

    μ

    and for capitals, simply capitalize the name of the Greek letter, like so:

    & Mu ;

    to get

    Μ

    These will super- and subscript nicely, like so:

    eμ
    eμ
     
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