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    if youve read the book, can you please explain to me why at pg 77 how the hypotenuse can be s^2=(ct)^2-x^2 ???? :eek:
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    (Original post by swagadon)
    if youve read the book, can you please explain to me why at pg 77 how the hypotenuse can be s^2=(ct)^2-x^2 ???? :eek:
    Can you scan the page and post?

    Would be much easier to understand the context of the question. Thanks.
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    Because it is not Euclidean space here... it's hyperbolic/Minkowski space.
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    The length of the hypotenuse ^2 represents the magnitude of the four vector in Minkowski space^2, or the dot product of a four vector with itself
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    (Original post by KeyFingot)
    Because it is not Euclidean space here... it's hyperbolic/Minkowski space.
    i know but i wanted to know how they derived the equation s^2=(ct)^2-x^2
    and i researched it and it was because they added i^2 infront of x,which is an imaginary number and i^2=-1 which made -x^2,but i didnt see why they added i
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    (Original post by swagadon)
    i know but i wanted to know how they derived the equation s^2=(ct)^2-x^2
    and i researched it and it was because they added i^2 infront of x,which is an imaginary number and i^2=-1 which made -x^2,but i didnt see why they added i
    I'm not sure as such how it's derived (whether it was derived, or just a nice result of special relativity).

    Consider the lorentz transformations for an x direction boost,

     t=\gamma\left(t'+\frac{vx'}{c^2}  \right),\ x=\gamma\left(x'+vt'\right), \\ \mathrm{with\ }\gamma\equiv\dfrac{1}{\sqrt{1-v^2/c^2}}.

    I've omitted the y and z ones, as we are only considering x-direction boosts. If you combine (ct)^2-x^2, you can see that it is identical to (ct')^2-x'^2 which is an incredible result. This quantity is called s^2 which is the distance in spacetime^2 and it is the same for all observers. That's where the formula comes from.
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    Does anyone else read this in Brian Cox's voice or is it just me?
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    (Original post by SkaGirl9)
    Does anyone else read this in Brian Cox's voice or is it just me?
    Yep, I do too! I'm not very far into it yet though, only in the second chapter Bu reading it atm


    This was posted from The Student Room's iPhone/iPad App
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    (Original post by SkaGirl9)
    Does anyone else read this in Brian Cox's voice or is it just me?
    hahaha yeah! i do it with all authors of all books
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    (Original post by KeyFingot)
    I'm not sure as such how it's derived (whether it was derived, or just a nice result of special relativity).

    Consider the lorentz transformations for an x direction boost,

     t=\gamma\left(t'+\frac{vx'}{c^2}  \right),\ x=\gamma\left(x'+vt'\right), \\ \mathrm{with\ }\gamma\equiv\dfrac{1}{\sqrt{1-v^2/c^2}}.

    I've omitted the y and z ones, as we are only considering x-direction boosts. If you combine (ct)^2-x^2, you can see that it is identical to (ct')^2-x'^2 which is an incredible result. This quantity is called s^2 which is the distance in spacetime^2 and it is the same for all observers. That's where the formula comes from.
    i doont know what the lorentz transformations for an x direction boost is, im only an AS level student lol, but thanks for trying to help! , are you an undergraduate? and where did you learn that anyway?
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    (Original post by swagadon)
    i doont know what the lorentz transformations for an x direction boost is, im only an AS level student lol, but thanks for trying to help! , are you an undergraduate? and where did you learn that anyway?
    Oh sorry, I had no idea! Yeah I'm studying currently a first year Physics student, it was in my special relativity module (which was awesome!).

    I don't really know how to explain it without the maths, sorry I can't help you further.
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    (Original post by KeyFingot)
    Oh sorry, I had no idea! Yeah I'm studying currently a first year Physics student, it was in my special relativity module (which was awesome!).

    I don't really know how to explain it without the maths, sorry I can't help you further.
    what university do you go to? and how is it?
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    (Original post by swagadon)
    what university do you go to? and how is it?
    Manchester and it's great!
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    Just saying, the negative sign comes from the Minkowski metric.

    Without getting into too deeps in mathematics about the Minkowski metric (you can just treat it as an example of a pseudo-Riemannian manifold), the hypotenuse is actually the arc length, and if we reduce the interval, we get a differential arc length ds

    Relativity tells you that the trajectory of an object can be obtained from the equation ds^2=g_{ij}dx^i dx^j (by principle of least action), which in special relativity aka Minkowski metric (The g_{ij}, which is also a matrix) gives ds^2=c^2 dt^2 - dx^2-dy^2-dz^2. For simplicity, we just state s^2=c^2 t^2 - x^2 -y^2-z^2=c^2 t^2 - \vec{x}\cdot\vec{x}
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    (Original post by agostino981)
    Just saying, the negative sign comes from the Minkowski metric.

    Without getting into too deeps in mathematics about the Minkowski metric (you can just treat it as an example of a pseudo-Riemannian manifold), the hypotenuse is actually the arc length, and if we reduce the interval, we get a differential arc length ds

    Relativity tells you that the trajectory of an object can be obtained from the equation ds^2=g_{ij}dx^i dx^j (by principle of least action), which in special relativity aka Minkowski metric (The g_{ij}, which is also a matrix) gives ds^2=c^2 dt^2 - dx^2-dy^2-dz^2. For simplicity, we just state s^2=c^2 t^2 - x^2 -y^2-z^2=c^2 t^2 - \vec{x}\cdot\vec{x}
    thanks for trying to explain but im just an AS level student haha
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    (Original post by swagadon)
    thanks for trying to explain but im just an AS level student haha
    Never underestimate yourselves. Physics is always open to everyone.
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    (Original post by agostino981)
    Never underestimate yourselves. Physics is always open to everyone.
    thanks i just didn know what a pseudo-Riemannian manifold is
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    Sorry about gate crashing this thread but was wondering if any of you could help me with the following question attached?
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