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Let me get this straight

  1. Jan 12, 2013 #1
    Okay I am new to relativity. I have been reading some books. Let me see if I understand this correctly. Consider the following scenario. There are 2 observers or inertial fames. They are moving apart at a constant velocity. Each ovserver can consider itself at rest relative to the other. Observers move through spacetime at the speed of light. In each relative frame the following can be considered true. Motion is converted from the time dimension into space through adjacent frames of symmetry, so that each observer can consider the other as falling over a sort of horizon as frames transate. These distortions are from the perspective of each observer a sort of fleshing out of an underlying symmetry in which mirror images or probability zones can be considered equal even though they are kind of negative opposite. Sorry if I am not very coherent, but am I on the right track?
     
    Last edited: Jan 12, 2013
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  3. Jan 12, 2013 #2

    Drakkith

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    I may not be up to speed on SR and GR like many here, but I think I know a little bit about the basics, and I literally have no idea what you are trying to say after about halfway down. Start with JUST basic SR and use terminology JUST from there.
     
  4. Jan 12, 2013 #3
    Sorry I don't know all the terms, I am hoping to learn by being corrected. I have only read a few layperson targetted books that use analogies to describe the theories. I am not even sure I understand the difference between SR and GR. SR is a kind of 'freeze frame' or contour line of symmetry (sorry only basic math here) within the greater framework of GR in which the speed of light remains constant no matter where it is observed from in those frames? Sorry if this is confusing.

    Edit: nvm I know I am getting this wrong because the speed of light remains constant from all frames in SR and GR right?
     
    Last edited: Jan 12, 2013
  5. Jan 12, 2013 #4

    Drakkith

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    Yeah, light always travel at c in any inertial frame.
     
  6. Jan 12, 2013 #5
    And any rotating frame?
     
  7. Jan 12, 2013 #6

    Vanadium 50

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    This is a very bad strategy and we discourage it. It's inefficient, and it tends to annoy the people doing the correction. It also puts all the effort on their part, as opposed to yours.
     
  8. Jan 12, 2013 #7

    Dale

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    No, if they are moving apart at a constant velocity then they are movinig relative to the other. They can each consider themselves at rest, but not at rest relative to the other.

    This is completely unintelligible. Are you perhaps not an english speaker and using some sort of machine translation tool?
     
  9. Jan 12, 2013 #8
    am i correct in answering this with:
    No, a rotating frame is not inertial thus c will change constantly in this rotating frame
     
  10. Jan 12, 2013 #9

    Dale

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    Yes. This is essentially correct, although I would say the speed of light is changing, rather than c.
     
  11. Jan 12, 2013 #10
    yes, ofcourse, c is an agreed upon value describing the speed of light in vacuum.
    It is the speed of light that is constantly changing in a rotating frame, not the value of c.
    thank you.
     
  12. Jan 13, 2013 #11

    Dale

    Staff: Mentor

    No worries, it was a very minor point, and it seems like you understand the major points.
     
  13. Jan 13, 2013 #12
    No. I'm sorry, but what you wrote is simply word salad. And there's no need for it. Relativity can be described and understood using everyday comprehensible language. I suggest taking a look at the book "It's About Time" by N. David Mermin.
     
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