Spacetime Explained

Discussion in 'Pseudoscience Archive' started by lixluke, Jan 3, 2010.

  1. lixluke Refined Reinvention Valued Senior Member

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    An observer has no mass.
    As an observer approaches C, time slows down relative to the observer.
    When an observer travels at C, time stops relative to the observer.
    When an observer is at rest, time is infinite?
    When an observer exceeds light speed, time reverses.
    An observer traveling anywhere in space at light speed will arrive there instantly in referance to the observer.

    Imagine standing on a platfrom in empty space. The platform is accellerating upwards at 9.8m/s2. We will be standing on the platform as if standing on Earth. We are moving upwards at the speed of the platform. If we are standing on Earth, we can pretty much say that we are accellerating upwards at 9.8m/s2. And time flows accordingly.

    The faster an observer travels, the slower time moves in reference to the observer. V = D/T. But V and T have an inverse relationship. Let's imagine that V has a direct inverse relationship with time. Thus, V = 1/T.

    D/T = 1/T
    D = 1

    What does this mean? It just means that D is constant. If D is constant and V is constant (C), then T must be constant.

    It might seem that the existence of matter bends the flow of spacetime around it. But not matter in a solid sense. Matter in the sense of energy bending the spacetime flow around it. However, it might be more accurate to say that the bending of the spactime flow produces energy (matter)?

    Say we have 3 balls in space. They move at different directions and speeds. Each ball observes itself as standing still while the other 2 balls move in space at a particular velocity. Each ball is a ripple of energy. Like a cross section of thread on a fabric. Perhaps there is some point of singualrity that exists as the universal frame of reference for all spactime?
     
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  3. CheskiChips Banned Banned

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    Wow, I think you just solved the Unified Field Theory problem. I can't believe no one thought of D=1!
     
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  5. lixluke Refined Reinvention Valued Senior Member

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    I didn't solve anything. I'm just throwing out facts to lead to a logical solution.
     
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  7. Doreen Valued Senior Member

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    I think there is a problem here, already, but I am not exactly sure what your wording means. The observer does not experience time slowing down, but rather if s/he were to return to the company of those who were not traveling near C, he would find that less time passed for him or her.
    But maybe this is what you meant.
     
  8. Dywyddyr Penguinaciously duckalicious. Valued Senior Member

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    Re-check your "facts".
    Time doesn't slow down for the moving observer, it slows down as seen by a slower/ non-moving observer.
    http://en.wikipedia.org/wiki/Time_dilation
     
  9. lixluke Refined Reinvention Valued Senior Member

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    I don't think so. If a ball is traveling at C. And I am right behind it traveling towards it Time must slow down for me in order for me to perceive the ball in front of me traveling at C.
     
  10. Dywyddyr Penguinaciously duckalicious. Valued Senior Member

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    It doesn't matter what you think.
    Time remains the same for each observer as seen by themselves.
    It doesn't matter how fast I'm going, my clock still runs at the same rate for me, but as seen by observers at different speeds then my clock will vary.

    From the link above.
     
  11. lixluke Refined Reinvention Valued Senior Member

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    Right. From the frame of reference of the observer, spacetime is always at rest. The observer's position in space is always 0. And the observer's perception of time is always constant

    For example, there is a ball at some distance in front an observer, and the ball takes off moving away from from the observer at a velocity of C. Becasue the observer is at rest, he will see the ball moving away at C. At the same time, the ball will see the observer moving away in the opposite direction at C while perceiving itself as being at rest.


    If they both take off at the same time in the same direction at C, the ball in front will be at rest relative to the observer. And the observer will be at rest relative to the ball.

    What they both take off at the same time in the same direction. But the ball takes off at C, and the observer takes off at X < C? Does the observer see the ball in front traveling at C-X?

    What if they both take off at C velocity, but in opposite directions? Do each of them see one another traveling at 2C?
     
  12. Dywyddyr Penguinaciously duckalicious. Valued Senior Member

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    Which contradicts your first post.

    http://en.wikipedia.org/wiki/Faster-than-light
     
  13. lixluke Refined Reinvention Valued Senior Member

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    When an observer travels at C, time stops relative to the observer.
     
  14. Dywyddyr Penguinaciously duckalicious. Valued Senior Member

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    No.
    The clock rate stays the same.
    As posted in #7 and as you yourself said:
     
  15. kmguru Staff Member

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    I thought time moves at the same rate while you are inside a spaceship that is traveling at the speed of light - or better yet, if your space you are in is moving faster than the speed of light. Or did I miss something that is new?
     
  16. lixluke Refined Reinvention Valued Senior Member

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    If an observer departs from a position moving at C, not a split second will pass for the observer when he arrives at his destination no matter how far away it is.
     
  17. CheskiChips Banned Banned

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    Fine lixluke, I'll be nice, Dywyddyr is either dense or pulling your leg.

    It's a matter of semantics. Since neither clock is definite...if you're the one traveling and you want act as if the people not traveling have the clock of perspective then you're "slowing down". If you you have the clock of perspective then you're speeding up. If you have no perspective, you'd sense no change - which is Dywyddyr's unrefined point.
     
  18. AlphaNumeric Fully ionized Registered Senior Member

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    The only velocity an object with no mass can move at is the speed of light. It follows from \(-m^{2} = p^{\mu}p^{\nu}\eta_{\mu\nu} = 0\). So either an observer has mass and moves slower than light or has no mass and moves at the speed of light.

    No, thus \(V \propto \frac{1}{T}\) and if the constant of proportionality is k then \(V = \frac{k}{T}\). Please tell me you're being ironic and not illustrating how you would fail high school physics....
     
  19. lixluke Refined Reinvention Valued Senior Member

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    I don't think Dywyddyr is pulling anybody's leg. He's using Wikipedia.

    Still, the problem is the observer moving at C in relation to what?

    Say there is a ball at a distance from the observer, and the observer travels towards that ball at C. It doesn't matter if the ball is 8 light minutes away or a million lightyears away, he will observe himself hitting that ball instantly.

    But what if the observer was at rest, and the ball was traveling towards him at C? Is that not the same thing? Thus, when the ball departs, not a moment would go by for the observer when the ball hits him.
     
  20. lixluke Refined Reinvention Valued Senior Member

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    The inverse of X is 1/X. In such a case, V = 1/T and T = 1/V.
     
  21. Dywyddyr Penguinaciously duckalicious. Valued Senior Member

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    Neither.
    How fast does MY clock go according to ME if I am travelling at C?

    Re-read his post: he has the observer as the mover.
     
  22. AlphaNumeric Fully ionized Registered Senior Member

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    No, it means \(V \propto \frac{1}{T}\). If you drive twice as fast you complete a set journey in half the time. If you take triple the amount of time you drove one third the speed. BUT that doesn't mean [tec]V = \frac{1}{T}[/tex]. Otherwise distance would not come into the definition of velocity, would it? D = VT so \(V = \frac{D}{T}\). If you have to drive twice as far in the same time you need to drive twice the speed.

    The units don't even match. The units of time is 'seconds'. The units of velocity are metres per second. The units of 1/T is 'per second'. So \(V= \frac{1}{T}\) is wrong by dimensional analysis. An inverse relationship between A and B means \(A = \frac{k}{B}\) for some quantity k which may or may not have units and other dependencies. \(A = \frac{10}{B}\) results in A and B having an inverse relationship yet they do not satisfy \(A = \frac{1}{B}\).

    It would seem you did sleep through physics class.
    Anything with mass.
     
  23. lixluke Refined Reinvention Valued Senior Member

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    It stands still.
     

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