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mathavraj

  • 2 years ago

explain this equation

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

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

  3. mathavraj
    • 2 years ago
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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

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

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

  6. mathavraj
    • 2 years ago
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    there are delta symbols before r and t

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

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

  9. mathavraj
    • 2 years ago
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    oh thank you very much...

  10. mathavraj
    • 2 years ago
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    :)

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

  12. mathavraj
    • 2 years ago
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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

  13. Jemurray3
    • 2 years ago
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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, \[(\Delta r)^2 = (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2 \]

  14. Jemurray3
    • 2 years ago
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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.

  15. Jemurray3
    • 2 years ago
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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.

  16. Jemurray3
    • 2 years ago
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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.

  17. Jemurray3
    • 2 years ago
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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.

  18. Jemurray3
    • 2 years ago
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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 \[s^2 = (c\Delta t)^2 - (\Delta r)^2 \]

  19. mathavraj
    • 2 years ago
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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

  20. Jemurray3
    • 2 years ago
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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.

  21. Jemurray3
    • 2 years ago
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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.

  22. mathavraj
    • 2 years ago
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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

  23. mathavraj
    • 2 years ago
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    @theyatin

  24. mathavraj
    • 2 years ago
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    @mahmit2012 @musicalrose @mukushla

  25. mathavraj
    • 2 years ago
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    @91 @FoolAroundMath

  26. Jemurray3
    • 2 years ago
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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.

  27. mathavraj
    • 2 years ago
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    so no problem abt the unit

  28. Jemurray3
    • 2 years ago
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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.

  29. mathavraj
    • 2 years ago
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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?

  30. mathavraj
    • 2 years ago
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    though its mathematical pls explain...

  31. Jemurray3
    • 2 years ago
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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.

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