The General Lorentz Transformation

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  • Опубликовано: 17 дек 2024

Комментарии • 47

  • @gauravtulsiani9870
    @gauravtulsiani9870 2 месяца назад +1

    Thank you Sir. Have been scratching head and pages to find an elegant general Lorentz boost. Your work is truly elegant.

  • @gabzitcho
    @gabzitcho 4 года назад +19

    Actually, is very hard to find this demonstration in regular EM books. In most of them, the boost matrix is just shown for one axis. Thank you!

  • @rikteembhowmick3432
    @rikteembhowmick3432 4 года назад +4

    The easiest explanation yet elegant. Thanks sir

    • @TensorCalculusRobertDavie
      @TensorCalculusRobertDavie  4 года назад

      Hello Rikteem and thank you for your comment. It is great to hear you found the video helpful.

  • @geez6666
    @geez6666 5 лет назад +3

    Now this is how physics is explained!!! I don't know why a coordinate free, geometric, approach is not often encouraged. Or better, best of both worlds like your explanation demonstrated. Thank you so much!!!

    • @TensorCalculusRobertDavie
      @TensorCalculusRobertDavie  5 лет назад

      Thank you for your encouraging comment, I am glad that you found the video useful. Cheers!

  • @deigivan8595
    @deigivan8595 6 лет назад +8

    Thank you very much. It's so useful.

  • @sirawutnatyanun9497
    @sirawutnatyanun9497 6 лет назад +7

    You save my life from Relativistic EM homework.

  • @lamequemaciel6412
    @lamequemaciel6412 4 года назад +2

    Thank you so much, Sir.

  • @joaoaugusto3361
    @joaoaugusto3361 7 лет назад +2

    Thanks, Robert! It was very difficult to find this demonstration. The main problem that me and my friends found is that you couldn't deduce this general transformation using the same ideas to demonstrate only for a x Boost, for example. We got frustrated, and then recognized that it should have another way to do this, probably using vectors, and it would be a lot 'cleaner'. Anyway, thanks again!

  • @chiragkshatriya9486
    @chiragkshatriya9486 4 года назад +4

    Sir, Can you please explain the term v/|v| in the equation at 8:11. Sir shouldn't it be unit vector r.

    • @TensorCalculusRobertDavie
      @TensorCalculusRobertDavie  4 года назад +1

      Hello Chirag. Note v/|v| is just a unit vector like any other unit vector that points in the same direction as v. Vectors are geometric objects and so all unit vectors have a magnitude of one and point in some direction. We just need to choose the one that points in the direction we are interested in.
      In the video r_(parallel) points in the same direction as v and so when we want to write r_(parallel) as a vector we can take its magnitude |r_(parallel)| and multiply it by a unit vector that points in the direction we are interested in, which is v. So we have, r_(parallel) = |r_(parallel)| v/|v|

  • @mathsclinicbymuhammadasads4756
    @mathsclinicbymuhammadasads4756 7 лет назад +2

    Thank you so much Sir, you made my assignment easy

  • @rathindranathdas7879
    @rathindranathdas7879 5 лет назад +2

    Thank you very much. It was very helpful.

  • @yihe7133
    @yihe7133 2 года назад +2

    This is very nice! I do have one question: the position vector r' in S' frame is expressed in terms of velocity vector v and position vector r in S-frame. To compare both sides (at 10min-ish of the video), one has to assume the unit vector ex'=ex, ey'=ey, ez'=ez, etc. What is the justification here?

  • @prashantshukla4759
    @prashantshukla4759 Год назад +1

    Is it necessary for r parallel Vector to be parallel to velocity vector? 3:10

    • @TensorCalculusRobertDavie
      @TensorCalculusRobertDavie  Год назад

      Yes it is because we will need to consider the Lorentz contraction which is more easily achieved if the parallel part of the vector is parallel to the velocity vector.

  • @erdemmk62
    @erdemmk62 6 лет назад +3

    Thank you!

  • @mohamedmouh3949
    @mohamedmouh3949 11 месяцев назад +1

    thank you so much 🤩

  • @vishnuakundi4766
    @vishnuakundi4766 4 года назад +2

    amazing

  • @amirmohammadtavakkoli6222
    @amirmohammadtavakkoli6222 6 лет назад +2

    Thanks. Very useful.

  • @Reidemeistermoves
    @Reidemeistermoves 6 лет назад +2

    thanks! very useful

  • @wanigaming4055
    @wanigaming4055 4 года назад +2

    Thanks

  • @naveedhussain1
    @naveedhussain1 4 года назад +2

    what is its inverse transformation?

    • @TensorCalculusRobertDavie
      @TensorCalculusRobertDavie  4 года назад

      Hello Naveed and thank you for your comment. I didn't include the inverse transformation because the focus of the video was on deriving the forward transformation and not its inverse which can be easily found using a software package.

  • @hossainahd
    @hossainahd Год назад +1

    Great

  • @pradeepshekhawat1296
    @pradeepshekhawat1296 3 года назад +2

    Can you suggest me a graduate level book for EM plz ??

    • @TensorCalculusRobertDavie
      @TensorCalculusRobertDavie  3 года назад +1

      Can I suggest you have a look at these links,
      physics.stackexchange.com/questions/142419/recommended-books-for-advanced-undergraduate-electrodynamics
      www.physicsforums.com/threads/graduate-electrodynamics-books.337872/

  •  6 лет назад +3

    This derivation is slightly different from that proposed by Albert Einstein himself in his book "relativity" : ruclips.net/video/koPnW0mXcvI/видео.html

    • @TensorCalculusRobertDavie
      @TensorCalculusRobertDavie  Год назад

      This is the general case where motion is in any direction and not aligned with a single axis.

  • @aritradey8112
    @aritradey8112 3 года назад

    Which book can i get this?

    • @TensorCalculusRobertDavie
      @TensorCalculusRobertDavie  3 года назад +3

      Hello and thank you for your question. I haven't seen this derivation in any textbook.