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Transformation of solid angle

  1. Mar 17, 2015 #1

    ShayanJ

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    I'm trying to figure out how the element of solid angle transforms under a transformation between two inertial frames moving with velocity v w.r.t. each other under an arbitrary direction. But I should say I disappointed myself! Anyway, some books which contain a brief discussion on this(which doesn't include an actual derivation!), simply state that ## d\varphi'=d\varphi ##, but this is not at all trivial to me! What should I do?
    In my desperate actions toward the above goal, I managed to get myself confused about something else too! How curvilinear coordinates transform under Lorentz transformation?
    I'll appreciate any suggestion about the above problems.
    Thanks
     
  2. jcsd
  3. Mar 17, 2015 #2

    BiGyElLoWhAt

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    Does psi represent an angle element? Or does it represent the measure of the whole angle?
     
  4. Mar 17, 2015 #3

    ShayanJ

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    If you mean ## \varphi ##, then its the last coordinate in the spherical coordinate system(## (r,\theta,\varphi) ##).
    Then the solid angle element becomes ## d \Omega=\sin \theta d\varphi d\theta ##.
     
  5. Mar 17, 2015 #4

    Orodruin

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    The angle ##\varphi## represents a rotation in the plane perpendicular to the motion, thus it is given by the ##y## and ##z## coordinates (in standard configuration) which do not change. Therefore ##\varphi = \varphi'## and also ##d\varphi = d\varphi'##.
     
  6. Mar 17, 2015 #5

    ShayanJ

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    But this is an assumption about the relative motion. It doesn't work if we consider an arbitrary direction!
     
  7. Mar 17, 2015 #6

    Orodruin

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    You can always chose your coordinates in such a way that this is the case (as long as you satisfy yourself with rotation free Lorentz transformations). Just define ##\theta## as the angle between the direction of the transform and the differential solid angle.
     
  8. Mar 17, 2015 #7

    ShayanJ

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    Thanks.
    So what about the other question? How can we transform curvilinear coordinates between inertial frames moving uniformly w.r.t. each other?
     
  9. Mar 17, 2015 #8

    Orodruin

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    Usually when you deal with this kind of problems, only one event is of real importance (i.e., the decay or interaction). The ##\varphi## angle is unchanged and you can typically compute the ##\theta## transformation using the abberation formula.
     
  10. Mar 17, 2015 #9
    "The angle φ represents a rotation in the plane perpendicular to the motion, thus it is given by the y and z coordinates (in standard configuration) which do not change. Therefore φ=φ′ and also dφ=dφ′."

    This is true for transformation properties of the longitude, φ, which NOT a solid angle. The solid angle, usually denoted by Ω actually does involve directions parallel to the direction of relative velocity. In arbitrary dimensions, say n-space (plus time), the solid angle is defined in terms of cartesian components by the differential form,

    dΩ= ∑i=1,n [(-1)(i-1) xi dx1 ∧...dxi-1 ∧ dxi+1....dxn] /rn.

    You can check for test cases for n=2 and 3 that this general (n-1)-form expression indeed reduces to familiar planar polar angle element, dθ and solid angle element, dΩ= sinθ dθ ∧ dφ respectively. When cast In this cartersian form, the solid angle element does not appear to me to be Lorentz invariant due to "dt ∧ dz or dt∧dy" terms. I checked explicitly by plugging Lorentz transformation for n=2 case.
     
  11. Mar 17, 2015 #10

    PAllen

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    If you really want to play with explicit transform between spherical coords in one inertial frame to same in another, you can get them (in ugly form) by a straightforward process. Starting from the Lorentz transform between Cartesian coordinates. Then, substitute standard expressions of r',θ',φ' for for x',y',z' on the left; and the same for r,θ,φ for x,y,z on the right. Then simplify per φ'=φ. With just trivial rearrangement you end up with (for example, depends on conventions chosen):

    φ' = φ
    tanθ' = rsinθ/ (γ(rcosθ-vt))
    r'sinθ' = rsinθ
    t' = γ (t- vrcosθ)

    I am sure someone, somewhere, has derived a much more clever way of writing these. This is just a quick scribble to show the ideas. Never, in my life, have I had occasion to use such an explicit set of relations, so I had no idea what they might look like before writing these down.
     
  12. Mar 18, 2015 #11

    ShayanJ

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    Its clear now, thanks guys!
     
  13. Mar 18, 2015 #12

    Orodruin

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    I never claimed anything else ...
     
  14. Mar 18, 2015 #13
    "I never claimed anything else ... "

    In that case, my apologies...
     
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