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Math Help - Can anyone help to recognise the following equation?

  1. #1
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    Question Can anyone help to recognise the following equation?

    Hi,

    It is a long story, and maths really isn't my greatest strength... so I'm here to ask the help of you clever lot

    Basically, I am trying to find out what the following equation represents? Does anyone here recognise it and what it means/refers to?

    Can anyone help to recognise the following equation?-equation.jpg

    Many thanks for the help, if indeed you can.

    Kind regards,

    Kris

    p.s. This has nothing to do with any course I am on, so I'm not trying to cheat on anything.
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  2. #2
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    Re: Can anyone help to recognise the following equation?

    I understand that it could also be a formula/algorithm. I'm an artist, and very much of the right brain... which is why I ask for help. Thanks again.
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  3. #3
    Member Goku's Avatar
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    Re: Can anyone help to recognise the following equation?

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  4. #4
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    Re: Can anyone help to recognise the following equation?

    Right, many thanks for the pointer!

    So just trying to get my head around this. The image I posted up above represents (in part) time, minus (velocity, divided by the squared of the speed of light) which is represented by X.

    Then, the square root...

    l (length?) minus (velocity divided by the speed of light) squared

    and all of this equates to i.

    Sorry, I'm sure I'm very out on this. I know that "i" could represent anything. But I assume that the part that follows is the description that tells you what "i" actually represents.

    So, is this purely another way of expressing observation of a moving object? Like I say, this is a whole different realm to what I am used to. I don't expect a lesson though. Just help in working out the correct description of what this represents. Though I will begin to work through many of the lessons on here. Better late than never huh!

    Thank you for any help.
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  5. #5
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    Re: Can anyone help to recognise the following equation?

    This is a formula of the Lorentz transformation used in the special relativity theory.

    Imagine two frame of references (coordinate systems) F and F'. Suppose they coincide at time t = 0 and F' is moving right at the speed v. Then the same point in space-time has different coordinates in the two frames: (x, y, z, t) in F and (x', y', z', t') in F' . Since F' moves right, i.e., along the x-axis, the y- and z-coordinates coincide in F and F': y' = y and z' = z. Similarly, in Newtonian mechanics, the time coordinate coincide: t' = t because it is assumed that the time flows at the same rate in every frame. As far as the x' coordinate in F', we have x' = x - vt. For example, the origin of F has coordinates (0, 0, 0, t) in F, but (-vt, 0, 0, t) in F' because F' leaves the origin of F further and further behind.

    In the special relativity theory, t' is also different from t, and the difference grows as v approaches the speed of light c. We have t' = \frac{t-\frac{vx}{c^2}}{\sqrt{1-\frac{v^2}{c^2}}}. I think the left-hand side of the formula in the OP is t', not i, and the first symbol under the square root is 1, not l.
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  6. #6
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    Re: Can anyone help to recognise the following equation?

    Many, many thanks. You are a true star. I shall investigate further!

    Yes, I think you are right about the i being a "t". I wish I understood everything you said in your post, but it is a language that I still have much to learn about.

    Anyway, thank you so much for your help folks.

    Kris
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