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mathavraj

explain this equation

  • one year ago
  • one year ago

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  1. mathavraj
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    s^2=/_\r^2-c^2/_\t^2

    • one year ago
  2. mathavraj
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    it is the euclidean spacetime interval equation

    • one year ago
  3. mathavraj
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    @agentx5 @sami-21 ...i dont know how to write this equation there are no keys of scientific operators in my keyboard

    • one year ago
  4. Jemurray3
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    you type \[ \text{\[ s^2 = r^2 - c^2 t^2 \] } \]

    • one year ago
  5. Jemurray3
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    It will become \[s^2 = r^2 - c^2 t^2 \]

    • one year ago
  6. mathavraj
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    there are delta symbols before r and t

    • one year ago
  7. Jemurray3
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    \[\text{ \[ s^2 = \Delta \r^2 - c^2 \Delta t^2 \] } \]

    • one year ago
  8. Jemurray3
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    \[s^2 = \Delta r^2 - c^2 \Delta t^2 \]

    • one year ago
  9. mathavraj
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    oh thank you very much...

    • one year ago
  10. mathavraj
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    :)

    • one year ago
  11. Jemurray3
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    Spacetime isn't Euclidean, by the way. Did you have a question about that equation?

    • one year ago
  12. mathavraj
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    yes found that in wikipedia when i typed spacetime what are those equations and wat do u mean by it is not euclidean?http://en.wikipedia.org/wiki/Spacetime

    • one year ago
  13. Jemurray3
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    In Euclidean three-dimensional space, the distance between two points is defined as \[ \Delta r = \sqrt{ (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2 } \] or equivalently, \[(\Delta r)^2 = (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2 \]

    • one year ago
  14. Jemurray3
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    That distance, which can be thought of as the separation between two points in space, is invariant under translation and rotation. That means that an observer moving through Euclidean space would observe the distance between two objects to be exactly the same no matter which way they were turned and no matter how they happened to be moving.

    • one year ago
  15. Jemurray3
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    However, when we combine the dimension of time with the dimensions of space in a four-dimensional manifold called spacetime, we label our points not as points in space but as events in space and time. However, relativistically it turns out that observers moving at different speeds would not measure the distance above to be the same, and therefore \[(\Delta r)^2 = (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2 \] is no longer invariant.

    • one year ago
  16. Jemurray3
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    However, it was found that there is an analogous invariant interval that will be measured to be the same by all observers regardless of velocity. That interval is called s^2, and is written as \[s^2 = (\Delta r)^2 - (c\Delta t)^2 = (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2 - (c\Delta t)^2 \] This plays the role of the invariant interval in four-dimensional spacetime.

    • one year ago
  17. Jemurray3
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    Lastly, spacetime is not Euclidean because that invariant interval has a minus sign in it. If it were a plus sign, it would be Euclidean space, but it's not. It's usually referred to as Minkowski space.

    • one year ago
  18. Jemurray3
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    As an FYI, sometimes the minus sign gets reversed. For reasons I'm not going to go into because it requires knowledge of special relativity, I prefer to write the spacetime interval as \[s^2 = (c\Delta t)^2 - (\Delta r)^2 \]

    • one year ago
  19. mathavraj
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    (Note that the choice of signs for\[s^2\] above follows the space-like convention (-+++). Other treatments reverse the sign of \[s^2\] Spacetime intervals may be classified into three distinct types based on whether the temporal separation (\[c^2\Delta t^2\]) or the spatial separation (\[\Delta r^2\]) of the two events is greater what does this mean please explain it @Jemurray3 ...thank you very much

    • one year ago
  20. Jemurray3
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    if \[ (c\Delta t)^2 > r^2 \] then the interval is called timelike. If \[ (c\Delta t)^2 < r^2\] then the interval is called spacelike and if they're equal, the interval is called lightlike.

    • one year ago
  21. Jemurray3
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    Events that are separated by spacelike intervals cannot be causally connected because the time it would take for information to travel from one point in space to another (at the speed of light) is greater than the separation of those two points in time. I.e. if I clapped my hands on earth and an alien in the andromeda galaxy fell out of his chair, there couldn't possibly be a causal connection between these two events because it would take years for the signal to traverse the space between us.

    • one year ago
  22. mathavraj
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    dont be mad at me...could you please explain what this invariant interval actually mean..if it is the distance it should be S right but it is \[S^2\] HEREwhich means \[m^2\] right

    • one year ago
  23. mathavraj
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    @theyatin

    • one year ago
  24. mathavraj
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    @mahmit2012 @musicalrose @mukushla

    • one year ago
  25. mathavraj
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    @91 @FoolAroundMath

    • one year ago
  26. Jemurray3
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    It is the spacetime equivalent of distance, yes. Take the square root if you'd like, but if s is invariant then s^2 is as well, and because it can be negative you can get into trouble. It's just more useful to refer to s^2, in general.

    • one year ago
  27. mathavraj
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    so no problem abt the unit

    • one year ago
  28. Jemurray3
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    The units are consistent. Honestly the invariant spacetime interval is not of much use without the rest of special relativity, and the geometry and topological properties of flat spacetime (Minkowski space) require more mathematics to adequately understand, so I don't really think you should worry about it until you have a more thorough grounding in the math and physics.

    • one year ago
  29. mathavraj
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    one last question how did we incorporate that term \[c \Delta t^2\]? why do we subtract it from the variant euclidean distance why not plus it?

    • one year ago
  30. mathavraj
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    though its mathematical pls explain...

    • one year ago
  31. Jemurray3
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    It didn't just pop out of nowhere. If there was a plus sign, rather than a minus sign, then the interval would change depending on the observer. Once you derive the Lorentz transformation equations of special relativity, it can be shown that the quantity s^2 as written above is the same for all inertial observers.

    • one year ago
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