Deriving the General Lorentz Transformation | Special Relativity

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  • Опубліковано 24 гру 2024

КОМЕНТАРІ • 23

  • @Darkness_7193
    @Darkness_7193 6 днів тому

    Matrices in this case is absolutely evil, it needs a lot of efforts to obscure such a simple idea from 0:56 to 3:00
    Thanks for great video

  • @redandblue1013
    @redandblue1013 Рік тому +2

    Very nice, my course's lecture notes absolutely sucked and I needed some clarification on this. Finally found the full matrix

  • @michael_zaki6903
    @michael_zaki6903 Рік тому +1

    Thank you this was the only video I could find involving the derivation of this specific matrix. You just helped me a ton.

  • @Abon963
    @Abon963 Рік тому +1

    Thanks a lot!

  • @sathyanarayananviswanathan3770

    great video, really enjoyed it

  • @HomoLamarckiens
    @HomoLamarckiens Рік тому +1

    Awesome video that I wanted for long

  • @lowersaxon
    @lowersaxon Рік тому

    Well, thats correct of course. There are many ways to picture the LT and by applying the economics of Occam‘s razor I think we should use the easiest way to do.

  • @NH-zh8mp
    @NH-zh8mp Рік тому

    I still don't get the point at 1:39, may you explaine more detail on why you got that second term on the right in that formula ?

    • @deepbean
      @deepbean  Рік тому +1

      Hi!
      So basically, the idea is that the current velocity vector can be separated into two components, one of which is parallel to the velocity, and one of which is perpendicular. The perpendicular component remains unchanged while the parallel component is modified according to the position Lorentz transform (derived in the video before this one in the series).

    • @NH-zh8mp
      @NH-zh8mp Рік тому

      @@deepbean I got it. I still got some questions.
      (1) can you explaine more about the equality at 2:22 ?
      (2) how do you conclude that it is proj_(vec v) of vec r ?
      (3) how about the equaliy at 2:47 ?

    • @deepbean
      @deepbean  Рік тому +1

      @H N Hi, apologies for the late reply. So this is just the idea that the position vector can be decomposed into components perpendicular to and parallel to the velocity vector, the latter being just the projection of the position vector on the velocity vector, which you can calculate using the dot product. That equation at 2:47 is just a way of finding the projection via the dot product. Hope this answers it.

  • @Benjamin-gx8ho
    @Benjamin-gx8ho 11 місяців тому

    around 2:16, why do you make the velocity have an arrow in the second equation?

    • @premanav
      @premanav 11 місяців тому

      Good Question. Hope we got the answer soon.

    • @premanav
      @premanav 11 місяців тому

      My guess is with an arrow on velocity make it free from parallel component in this context and direction of velocity vector is now is the direction of relative velocity direction of inertial frames in Lorentz transformation happening.

    • @deepbean
      @deepbean  8 місяців тому

      That's because it's now a vector. So, on the first line, we have a unit vector on the second RHS term, which we then get rid of by vectorizing the two terms in the bracket.

  • @NH-zh8mp
    @NH-zh8mp 10 місяців тому

    This is a general Lorentz boost in arbitrary direction. What about the more general case where the boost is applied for an inertial frame (t''',z) whose 3 spatial axises are a rotation of of a initial frame (t,x) ? Is the time coordinate in this frame (t''',z) coincide with the time coordinate in the frame (t',y) which is a boost of (t,x) ?

  • @NH-zh8mp
    @NH-zh8mp 11 місяців тому

    what about using Lie algebra to find the form of the general Lorentz transformation ?

  • @milessodejana2754
    @milessodejana2754 8 місяців тому

    Wait, in case of t --> t', if t = 0, does that mean that negative value of t' is valid?

  • @NH-zh8mp
    @NH-zh8mp Рік тому +1

    So, is there a linear algebra way to gain this general Lorentz transformation ?

    • @deepbean
      @deepbean  Рік тому +3

      Yup! If you wanted, you could also do this by using rotation matrices; rotating the velocity vector such that it lies completely on the x-axis, then applying the 1-D Lorentz transform, then rotating back again.