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Thread: InnerProduct tensor

  1. #1 InnerProduct tensor 
    Forum Isotope Zelos's Avatar
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    a while ago i asked about tensor stuff. since then ive made significan progress. Let A and B be vectors with Both Real and complex components
    Normal scalar product is AB (the dot between is assumed to be, if its cross i´ll have a little x)
    and in tensor you take like
    n=d<sub>cz</sub> A<sup>c</sup> B<sup>z</sup>
    d<sub>cz</sub> is the kronecker delta (spelled right?)
    hope im right this far
    but then when it comes to complex vectors the scalar product is called inner product, right? and look like
    <A|B> = AB*
    where the * after B means it is its complex conjugate.
    My question is, how do you write this in tensor form?


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  3. #2  
    Forum Professor river_rat's Avatar
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    The easiest way I think is to just embed the vector space C^n in R^(2n) and use the standard metric in R^(2n) (ie the krondecker delta)


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  4. #3  
    Forum Isotope Zelos's Avatar
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    could it really be that simple? let me check
    one vector part that is complex
    A=(a+qi)
    B=(s+wi)
    i=imaginary unit, and a/q/s/w is real numbers
    <A|B>=AB*=(a+qi)(s-wi)=as+qw+(qs-aw)i
    if im not misstagen. but i bet i am

    dont seem like a kronecker delta and a 2n tensor is capable to archive that. is it me who is wrong?
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  5. #4  
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    Oops, my idea only works for norms of vectors (as the are the same thing in C and R^2)


    Try [g_ij] = ( 1 -i ; i 1) for the metric tensor of R^2 to obtain what i think is an embedding of C into R^2 with the same inner product structure.
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  6. #5  
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    take that again please
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  7. #6  
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    Well firstly i should have mentioned this earlier that we dont really have a "normal" metric here (the metric is not symmetric, as you can see by the matrix [g_ij] i posted earlier) but lets ignore that and carry on regardless.

    Now every c in C has the form c = a + ib so we associate a vector (a; b) to every complex number. Then if we allow the metric tensor to have the form of the [g_ij] i posted earlier we see that

    c dot d = (c1; c2) dot (d1; d2) = g_ij C^i D^j = c1 d1 + c2 d2 - i c1 d2 + i c2 d1

    We dont have a kronecker delta has this metric is very different to the normal one (as it is not symmetric and is complex valued).
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  8. #7  
    Forum Isotope Zelos's Avatar
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    oh i get it, so the matric g<sub>ij</sub> =
    [[1] [-i]
    [i] [1]]
    ah now i get it, with that matric the imaginary parts disaphere (as wanted)
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  9. #8  
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    Yep, was using ";" to state a new row had started. If you know mathematica notation it would be {{1, -i}, {i, 1}} though i wish we had a latex interpreture here!
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  10. #9  
    Forum Isotope Zelos's Avatar
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    yes, great
    i know that many quantum mechanic formulas use complex vectors. just for the sake of fun, could it be written as tensors most of them?
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  11. #10  
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    Quote Originally Posted by Zelos
    yes, great
    i know that many quantum mechanic formulas use complex vectors. just for the sake of fun, could it be written as tensors most of them?
    I doubt it, as they would not transform properly - tensors are used to handle geometric information and anyway the embedding into R^2n of a complex space C^n loses a lot of the natural algebra if you want to use the standard notations.

    What equations do you have in mind?
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  12. #11  
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    none special. was just thinking about the possebility
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  13. #12  
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    You know why the metric tensor comes into scalar product?
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  14. #13  
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    you mean the kronocker delta?`else it would take all others and add up that isnt i=j
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  15. #14  
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    Well the krondecker delta is just one example of many possible metrics on a real vector space but the actual reason is deeper then that.

    An inner product on the tangent space of a manifold is given by
    U dot V = U^i V_i (ie a contravariant vector "timed" a covariant vector)

    The metric tensor [g_ij] lets you shift between the two different types of vectors as V_i = g_ij V^j

    and thats why U dot V = g_ij U^i V^j
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  16. #15  
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    oh yeah now i remember. well its kinda new hard to remember everything, thx
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  17. #16  
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    lol, im just learning all of this stuff in GR. Im going through "A first course in general relativity" for the third time and it all makes sense this time!
    I demand that my name may or may not be vroomfondel!
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  18. #17  
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    im reading the very same book
    lovly one
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  19. #18  
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    Lol - i have to relearn it everytime i try answer a question! There is just too much to try remember now adays!
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  20. #19  
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    hehe thats what i want
    well while we are add it. how do you derivate tensors? dont quite get it and why its like that
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  21. #20  
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    Okay, thats a long messy routine

    Lets consider a normal contravariant vector field U which we want to take a derivative of in some sense. Now we want a local description of this derivative (as we are actually working in a co-ord. system) so lets suppose take a local chart and suppose U has local co-ords. U(x1, x2, ...)^i wrt some basis e_j.

    Now it woud be dreamy if we could just say that the deriviative for U wrt x1 in this local chart is just the partial deriviative of each U^i wrt x1. But that doesnt work in general (can you see why?)

    Happy so far?
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  22. #21  
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    yeah i see this far, so continue
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  23. #22  
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    Ok, well the problem is that the resulting deriviative does not transform properly between different co-ord. systems unless you started off with a scalar field. Now a rather formal way of motivating the covariant definition is this

    For clarity lets use the notation U<sub>,j</sub> to denote the derivitive of the quality U wrt xj (just the comma notation)

    U<sub>,j</sub> = (U<sup>i</sup> e<sub>i</sub>)<sub>,j</sub>

    But the product rule (as the basis vectors depend on the co-ords as well in general, just think of circular co-ords) gives that

    (U<sup>i</sup> e<sub>i</sub>)<sub>,j</sub> = U<sup>i</sup><sub>,j</sub> e<sub>i</sub> + U<sup>i</sup> e<sub>i ,j</sub>

    Now the derivitives of the basis vectors can be written as
    e<sub>i ,j</sub> = C<sup>k</sup><sub>ij</sub> e<sub>k</sub> where C<sup>k</sup><sub>ij</sub> are the Christoffel symbols. These can be calculated from the metric tensor [g<sup>ij</sup>] by the identity

    C<sub>ijk</sub> = 1/2 (g<sub>jk, i</sub> + g<sub>ki, j</sub> - g<sub>ij, k</sub>) and C<sup>k</sup><sub>ij</sub> = g<sup>ks</sup> C<sub>ijs</sub>

    So we have that

    (U<sup>i</sup> e<sub>i</sub>)<sub>,j</sub> = U<sup>i</sup><sub>,j</sub> e<sub>i</sub> + C<sup>k</sup><sub>ij</sub> U<sup>i</sup> e<sub>k</sub>

    But we can rearrange the entire mess and find that

    (U<sup>i</sup> e<sub>i</sub>)<sub>,j</sub> = (U<sup>i</sup><sub>,j</sub> + C<sup>i</sup><sub>kj</sub> U<sup>k</sup>) e<sub>i</sub>

    So if we define the covariant deriviative as

    U<sup>i</sup><sub>|j</sub> = U<sup>i</sup><sub>,j</sub> + C<sup>i</sup><sub>kj</sub> U<sup>k</sup>

    Then we find that it does transform as a tensor of rank 2, so we have a workable definition of a deriviative. In essence what we have done (if you dont mind imagining your manifold embedded into a larger space) is to kill the components of the derivitive that dont lie in our tangent space (at least thats how i imagine it). You can extend this idea to covariant vectors, (just work out what (e<sup>i</sup>)<sub>,j</sub> = - C<sup>i</sup><sub>jk</sub> e<sup>k</sup>) and as any general tensor is the tensor product of copies of the contravariant and covariant vector spaces we can extend our covariant derivitive to them as well

    For example

    U<sup>i</sup><sub>jk |m</sub> = U<sup>i</sup><sub>jk ,m</sub> + C<sup>i</sup><sub>sm</sub> U<sup>s</sup><sub>jk</sub> - C<sup>s</sup><sub>jm</sub> U<sup>i</sup><sub>sk</sub> - C<sup>s</sup><sub>im</sub> U<sup>i</sup><sub>js</sub>
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  24. #23  
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    river rat, could you please change the previus post so that U_i becomes U<sub>i</sub> and U^i to U<sup>i</sup>
    using < sub> thing < /sub> and < sup> thing < /sup>
    without the space between "<" and "su" easier to read then
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  25. #24  
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    All done
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  26. #25  
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    mr burns: Excellent
    please explain what Christoffel symbol is
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  27. #26  
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    i understand it all except the Christoffel symbol, what is it exacly? what is its purpose?
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  28. #27  
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    Hi Zelos

    Sorry for taking so long to reply, ive been working on my research so i havent had time to post here.

    Well the christoffel symbols are just the coefficients of the derivatives of the basis vectors wrt the different local coordinates. That is because the derivitive of a basis vector wrt a local coordinate is another tangent vector so can be written as a combination of the basis vectors. The christoffel symbols just keep track of that information

    ie. e<sub>i ,j</sub> = C<sup>k</sup><sub>ij</sub> e<sub>k</sub>

    The identity i quoted for the christoffel symbols (which is how they are usually calculated given a metric tensor) comes from Ricci's lemma

    g<sub>ij |k</sub> = 0

    As an exercise derive the identity i gave
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    so what values can it take?
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  30. #29  
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    That depends on the metric. For example in euclidean space the christoffel symbols vanish identically (are zero everywhere). Its a good exercise to calculate the christoffel symbols for the sphere of radius r embedded into R^3 (if you get stuck with this shout, EDIT: gave wrong space, fixed now).
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  31. #30  
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    so then it is nothing and whats the point writing out when its nothing?

    what about in the space einstein used?
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  32. #31  
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    Not every space is flat Zelos, and the vanishing of the christoffel symbols is something very special about euclidean space. You need them to make sense of how the space curves and bends away from being flat. In GR, the christoffel symbols carry the information about the gravitational fields in space in a sense.

    Einsteins equations dont fix the space you are working on, it depends on the problem at hand (though all the spaces have a 4 dimensional tangent space at every point for normal GR the way i understand it)
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  33. #32  
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    Not every space is flat Zelos
    know that

    In GR, the christoffel symbols carry the information about the gravitational fields in space in a sense.
    sweet

    Einsteins equations dont fix the space you are working on, it depends on the problem at hand (though all the spaces have a 4 dimensional tangent space at every point for normal GR the way i understand it)
    okey. guess i´ll have to coem with seperate problems then

    what is best? normal coordinates or polar?
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  34. #33  
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    Normal coordinates? I guess you mean standard cartesian co-ordinates. The coordinates you use depend on the problem at hand. The standard trick is to use coordinates that mirror the symmetry of the problem, for example using spherical coordinates if the problem exhibits spherical symmetry.
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  35. #34  
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    ive notice that C<sup>a</sup><sub>bd</sub> is quite different in polar coordinates. Dont you have to use 4 coordinates in relativity when you deal with polar? i looked in my book and they only use 2
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  36. #35  
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    The christoffel symbols are coordinate specific so they would be rather different.

    What problem are they solving in your book?
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  37. #36  
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    as its now its just understanding the stuff. Im just not understanding why they use 4D cartesian space but 2D polar space.
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  38. #37  
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    Well if its just the introductory stuff then they are using polar co-ordinates because doing the math for general 4-sphere coordinates is god aweful and would cloud the matter more then reveal anything.

    GR is usually done by specifying a possible metric and going from there, so a lot of the anonying grunt work is bypassed.
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  39. #38  
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    here is a problem. Its in 2D.
    i understand the normal e<sub>x</sub> defentions (x taking value ebtween 1-3) but now in polar let "r" be the radius part and "a" the angle.
    i can understand e<sub>r</sub> but the defenition of e<sub>a</sub> how is it aquired?
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  40. #39  
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    Quote Originally Posted by Zelos
    but then when it comes to complex vectors the scalar product is called
    inner product, right?
    Inner, scalar, dot, they all apply
    equally in real and complex spaces.
    and look like
    <A|B> = AB* where the * after B means it is its complex conjugate.
    I don't think this is quite right.

    Although I don't like the bra-ket notation you're using here, let's stick with it. Suppose V is a vector space with |a> and |b> in V. Then by definition the scalar product associates to any pair of this form a number (real or complex) and one writes (in this horrid notation) <a|b> is a number, with <a|b> = (<b|a>)*, i.e. the number given by <b|a> is
    the conjugate of that given by <a|b>. Note that, as the reals are
    a subset of the complex field, the conjugate (*) here is redundant, but
    not wrong.

    If you're interested, the main reason for not using bra-kets (unless
    you know exactly what you are doing - how many of us do?) is that it
    obscures an important distinction between <a|b> as a scalar
    product of two elements in V, and <a| as an element of the dual
    space V* of V, with |b> in V. In a metric space they coincide, of
    course, that's what Dirac had in mind when he introduced the notation.

    But there is a subtle reason why you should distinguish inner product
    spaces from metric spaces.
    My question is, how do you write this in tensor
    form?
    Having skimmed your conversation with river_rat, let me
    suggest this, I hope without giving offence. In my opinion, it is
    almost imossible to really get to grips with tensors without
    first understanding manifolds. River_rat hinted at this when he
    referred to coordinate neighbourhoods. I can try if you like, but it
    would be a longish journey.

    P.S. There is a much more abstract way of looking at this, which I doubt would appeal to you: any rank n covariant tensor can be represented by a degree n polynomial!!
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    Guitarist, i appriciate it. but we have moved on to new subjects
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    Hi Guitarist

    About the bracket representation, if your inner product space is a hilbert space then there is no problem with this notation (i.e. use the riesz representation thm to associate every ket with a bra). Every inner product space is a metric space though, so im not sure about what you are going on about there. Not every metric space is an inner product space (i think the countable product of inner product spaces is the standard counterexample?) is that what you were hinting at?

    Manifolds are scary beasts in general, i would suggest the algebraic treatment of tensors first (which is where i have found most hang ups) as the definition of a tensor field as a section of the tensor product of copies the tangent and cotangent bundle to confuse more at first reading them show anything. In fact i think tensors should be treated in continuum mechanics first, (where everything is embedded in R^3) before handling general manifolds.

    Is your last statement correct? I thought only symmetric covariant tensors had unique polynomial representations?
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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    Quote Originally Posted by Zelos
    here is a problem. Its in 2D.
    i understand the normal e<sub>x</sub> defentions (x taking value ebtween 1-3) but now in polar let "r" be the radius part and "a" the angle.
    i can understand e<sub>r</sub> but the defenition of e<sub>a</sub> how is it aquired?
    Post your definitions (im not quite sure what you are on about)
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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    Quote Originally Posted by river_rat
    About the bracket representation, if your inner product space is a
    hilbert space then there is no problem with this notation (i.e. use the
    riesz representation thm to associate every ket with a bra).
    Indeed, in fact it was in this context that Dirac introduced
    it. I personally don't like to hide the distinction between vectors and
    their duals (which bra-ket certainly does) because this is important to
    (my) understanding the pull-back and push-forward maps. These in turn
    are important to really get a handle on co- and contra-variant
    tensors.
    Every inner product space is a metric space though, so
    im not sure about what you are going on about there. Not every metric
    space is an inner product space (i think the countable product of inner
    product spaces is the standard counterexample?) is that what you were
    hinting at?
    More or less, although I wasn't able to bring a counter-example to mind last night. I was simply pointing out that the two concepts are quite different; but you are right in that it is trivially true that you cannot have an inner product without a metric
    Manifolds are scary beasts in general, i would suggest the
    algebraic treatment of tensors first
    Manifolds scarey? A matter of taste I guess, I rather like them. Let me see if I can find something pithy to say about them.
    I find the algebraic approach highly arbitrary: a tensor is defined as a beast with transforms in a certain way. This may be OK for physicists and engineers, who can no doubt think of loads of situations where this may be self-evident. I was about to say it lacked rigour, but that would be unbearably arrogant of me.
    In fact i think tensors should be treated in continuum mechanics first,
    There you go! As a non-physicist I'm not sure what that is!
    Is your last statement correct? I thought only symmetric
    covariant tensors had unique polynomial representations?
    Hmm. You may be right about symmetry, in fact I'm sure you are. No matter, I suspect it's a bit of a cul-de-sac anyway
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    Quote Originally Posted by river_rat
    Quote Originally Posted by Zelos
    here is a problem. Its in 2D.
    i understand the normal e<sub>x</sub> defentions (x taking value ebtween 1-3) but now in polar let "r" be the radius part and "a" the angle.
    i can understand e<sub>r</sub> but the defenition of e<sub>a</sub> how is it aquired?
    Post your definitions (im not quite sure what you are on about)
    no problem
    e<sub>r</sub> = e<sub>x</sub> cos a + e<sub>y</sub> sin a
    e<sub>a</sub> = -re<sub>y</sub> cos a + re<sub>x</sub> sin a
    i can udnerstand the e<sub>r</sub> (even thoe i dont know how they gor it exacly, could u show that aswell?) but the e<sub>a</sub> i dont get at all
    I am zelos. Destroyer of planets, exterminator of life, conquerer of worlds. I have come to rule this uiniverse. And there is nothing u pathetic biengs can do to stop me

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    Zelos Your implied advice to mind my own business was probably good advice, but, having gone through this thread in rather more detail, I'm puzzled by a couple of things.

    You asked about "normal" vs. polar coordinates. Are you aware that "normal coordinates" has a rather specific meaning in differential geometry? (And certainly doesn't imply Cartesian coordinates - how could it?)

    river_rat Are you using the symbol C for the connection? This could be confusing, as this guy is emphatically not a tensor. If you are using Windows the following might be better. Γ is generated by holding down Alt and typing 226 on the number pad (with NumLock on). The glyph appears when you relaease Alt (I dare say other OS's have similar functions). Actually, this sort of trick generates a surprising numbers of useful symbols. I'll post if you want

    Zelos I can't make sense of your polar coordinates. Are you sure you're not in some tangent space? Even here (in 4-space) we would expect at least 3 entities, I think: one position vector and 2 "angles". Polars are not my strong suit, however, so I may be wrong.

    river_rat You said "GR is usually done by specifying a possible metric and going from there, so a lot of the anonying grunt work is bypassed." Do you like that approach? I think it is highly questionable, but yes, it delivers the metric tensor (via the matrix of the line element) like a dream. Of course it does, as the metric is already assumed!
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    normal is cartesian, i think
    i know that he uses C thing for christoffel symbol and i know its meaning and original symbol

    im working with 2 polar coordinates since, its a introduction to dealing with the stuff.

    You said "GR is usually done by specifying a possible metric and going from there, so a lot of the anonying grunt work is bypassed." Do you like that approach? I think it is highly questionable, but yes, it delivers the metric tensor (via the matrix of the line element) like a dream. Of course it does, as the metric is already assumed!
    prove how to get the matric then. i would love it
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    Zelos
    Do you mean get the metric temsor by assuming a metric? Let's say you
    do. By the way, this isn't a "proof", merely a motivation.

    We are in some 4-space, right? Let's have vectors x, y, z and t. Now,
    the generalized Pythagorean will be x<sup>2</sup> +
    y<sup>2</sup> + z<sup>2</sup> = (ct)<sup>2</sup>, where c is light velocity, included in the quadratic to make dimensions consistent (got that bit?)

    rearranging I find x<sup>2</sup> + y<sup>2</sup> + z<sup>2</sup> - (ct)<sup>2</sup> =0.
    So now I define some s such that
    x<sup>2</sup> + y<sup>2</sup> + z<sup>2</sup> - (ct)<sup>2</sup> = s<sup>2</sup>, and take derivatives thus: dx<sup>2</sup> + dy<sup>2</sup> + dz<sup>2</sup> -d(ct)<sup>2</sup> = ds<sup>2</sup>.

    The term on the RHS is referred to as the line element. But, hey, hang on. Isn't it equal to zero, by construction? Yes, it is. It is a point event in space-time, that is it has zero extenion in space and time.

    Now let's make a matrix, with the dx, dy etc terms as row and column
    headers. First note that we easily recover dx<sup>2</sup> + dy<sup>2</sup> + dz<sup>2</sup> -d(ct)<sup>2</sup> by looking only at like terms. These are found on the principal diagonal i.e. top left to bottom right.

    But where are the dxy, dxz etc terms? And what about dyx, dzx? We may
    suspect antisymmetry, perhaps: each dxy = - dyx, in which case all off
    -diagonal terms cancel. You mentioned the Kroenecker delta, let's apply
    that here to suppress this thought.

    Then we are left with the other possibility, that our vectors x, y, z and ct are linearly independent. That is, neither one can be expressed in terms of the others. This implies that our 4-space is "flat" in the Euclidean sense, and we have what is known as a Minkowski space, whose "signature" is the set {1, 1, 1, -1} of coefficients on the elemets in the principal diagonal.

    Now note that the generalized Pythagorean we started with assumed our
    space has a metric (addition doesn't work otherwise). This is, in general, considered to be very unhygenic, assuming the fact you wish to prove.

    Let's now forget about y, z and ct, and simply use
    x<sup>1</sup>, x<sup>2</sup> etc as our vectors.

    Now let's call the row entries in our matrix i and our column entries j
    (i, j = 1,...,4). Then we may have ds<sup>2</sup> = Σdx<sup>i</sup>dx<sup>j</sup> (i, j = 1,…,4).

    And if we insist our space is everywhere flat we appeal to Herr Kroenecker to stop us from taking unlike matrix entries, thus: ds<sup>2</sup> = δ<sub>ij</sub>dx<sup>i</sup>dx<sup>j</sup>.

    From this we may deduce that, if we don't want to restrict ourselves to flat spaces, there must some other gizmo g<sub>ij</sub> that keeps track of all relative lengths and angles such that ds<sup>2</sup> = g<sub>ij</sub>dx<sup>i</sup>dx<sup>j</sup>.
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    Zelos Have you lost interest in your own thread? Maybe I'm boring you

    But two, no three things. A "normal" coordinate system in differential geometry is emphatically not a Cartesian system. Muddling terms muddles readers!

    Second, I once thought of a Mickey Mouse way of describing the difference between co- and contravariant tensors, which I can show if you want (but it comes with a serious health warning)

    Last, I have a more systematic derivation of the metric tensor, also available if you want.

    Hmm. Sounds too exclusive. So.....

    To All Others The invitation is open to all
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    Guitarist youre not boring me, math can never bore me. i just wait for a answer on my question. how did they get
    e<sub>r</sub> = e<sub>x</sub> cos a + e<sub>y</sub> sin a
    e<sub>a</sub> = -re<sub>y</sub> cos a + re<sub>x</sub> sin a
    i wanna know exacly how they got this
    and i wanna know how you make those mathematical figures in here
    I am zelos. Destroyer of planets, exterminator of life, conquerer of worlds. I have come to rule this uiniverse. And there is nothing u pathetic biengs can do to stop me

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    Zelos Ya, you seem hung up on this. I can't answer this question, and no others have offered either. Maybe if you gave a little more context (from your books) it might help? But, in my limited experience of this subject I have never seen polar coordinates of this form. But my experience is limited.

    Here's my advice: I developed a working understanding of this field without the construction you are offering, so just forget it and move on.
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  52. #51  
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    Hi Zelos

    Been busy so i havent had time to look in here

    To answer your question, a basis for the tangent space of a manifold given a "local" embedding of the manifold into R<sup>n</sup> (ie ambient coordiantes - forgot what they were called yesterday) is given by

    e<sub>i</sub> = y<sup>j</sup><sub>,i</sub> k<sub>j</sub>

    where the k<sub>j</sub> are the unit standard orthonormal basis vectors for R<sub>n</sub>

    So in your case

    y<sup>1</sup> = x<sup>1</sup> cos(x<sup>2</sup>)
    y<sup>2</sup> = x<sup>1</sup> sin(x<sup>2</sup>)

    So taking partial deriviatives (first with respect to x<sup>1</sup> for e<sub>1</sub> and then x<sup>2</sup> for e<sub>2</sub>) and subbing in gives the vectors you gave
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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    then what is y<sup>j</sup><sub>,i</sub>? and how did they got that value?
    and maybe a more basic explination would be better
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  54. #53  
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    y<sup>i</sup><sub>,j</sub> denotes the partial deriviative of the variable y<sup>i</sup> w.r.t the variable x<sup>j</sup>. You can motivate this by the multivariable chain rule.
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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    okey
    still dont know how they got it
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  56. #55  
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    How did you get e<sub>r</sub>? The exact same process gives you e<sub>theta</sub>
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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    lets start this over and imagen im homer simpson and describe it as easy as possible since i dont understand previus explination
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  58. #57  
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    Quote Originally Posted by river_rat
    y<sup>i</sup><sub>,j</sub>
    denotes the partial deriviative of the variable
    y<sup>i</sup> w.r.t the variable x<sup>j</sup>.
    You can motivate this by the multivariable chain rule.
    Ya, well,
    notation is arbitrary, so you can use this if you want, but in
    differential geometry it gets confusing, as this is usually reserved
    for the partial covariant derivative. You showed a few days ago that
    this requires the second Christoffel connection.

    Morever, you say this in a subsequent post:
    To answer your
    question, a basis for the tangent space of a manifold given a "local"
    embedding of the manifold into Rⁿ (ie ambient coordiantes -
    forgot what they were called yesterday) is given by
    ei = yj,i kj
    Notation may be arbitray, but definitions are
    not. "Local embedding" makes no sense (no wonder you were scared by
    manifolds!). The definition of a manifold is that it is locally
    homeomorphic to some Rⁿ (homeomorphism is the
    topological analogue of isomorphism). Embedding refers to the whole
    space as a subspace of some larger space, called the ambient space.

    There is a theorem, can't just remember whose, that says, for an
    n-manifold, the minimum embedding (ambient) space is n + 1, the max is 2n. Homeomorphic spaces, on the other hand, must match dimensions. Or rather, this is the definition of the dimension of a manifold: A point set M of course has no dimension, but if M is locally homeomorphic to some Rⁿ, it is, by definition a manifold, and inherits its dimension from Rⁿ
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  59. #58  
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    Guitarist - I dont know what notation you use but standard mathematical practice is to denote partial derivitives of indexed quantities by commas (so A<sup>i</sup><sub>,j</sub> is the partial derivitive of A<sup>i</sup> wrt x<sup>j</sup> - its called the comma derivative!) with covariant derivitives either denoted by A<sup>i</sup><sub>|j</sub> or A<sup>i</sup><sub>;j</sub>. Never seen A<sup>i</sup><sub>,j</sub> used to denote a covariant derivative.

    Anyway, a local embedding makes sense as well (immersions are easy examples). Let f : M -> N be an immersion. Take a point on the manifold and a neighbourhood U of that point for which f|<sub>U</sub> is an embedding. Then you have a local embedding of M at x into N (with the inverse function thm being a good example of a local embedding thm). Another is the standard param. of a circle in R<sup>2</sup>. The thm you are thinking of is the Whitney embedding thm.

    Where do you get that point sets have no dimension? Dimension is a topological property based on refinements of coverings not a "locally homeomorphic to R<sup>n</sup>" one.

    Oh and manifolds are scary beasts for a whole host of reasons - quotient manifolds, general bundles on manifolds, frolicker spaces etc. Manifolds dont stay the nice fuzzy things you seem to believe - else GR would be a rather easy exercise.
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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