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Some questions on tensors

  1. Feb 21, 2006 #1
    Why do we say that the position 4-vector, [tex]x^{\mu}[/tex], is naturally contravariant and that the del operator, [tex]\partial_{\mu}[/tex], is naturally covariant?

    The only thing I could come up with is that the contravariant del components [tex]\partial^{\mu} = (-c^{-1}\partial_t,\nabla)[/tex] have a negative sign in front of the c. Is this 'unnatural'?

    The relationship between time, [tex]t[/tex], and proper time, [tex]\tau[/tex], is simply [tex]t = \gamma\tau[/tex]. Which makes [tex]t[/tex] a function of velocity. Now the velocity 4-vector:


    is definately a 4-vector right? Im sure it is. But is


    a 4-vector? Could the fact that [itex]t[/itex] is not a scalar have an influence on this question?
    Last edited by a moderator: Feb 21, 2006
  2. jcsd
  3. Feb 21, 2006 #2


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    The itex tags (inline tex) don't work on PF; we just use tex for inline formulas. I took the liberty of editing them.

    I think you are making too heavy weather over this. The covariant/contravariant distinction is not a function of anything like a minus sign; both covariant and contravariant tensors form an algebra.

    The distinction is rather in the definitions of the two kinds of tensor. A covariant vector (to just illustrate with rank 1 tensors) takes a tangent vector and produces a number. That's what a derivative does; it gives the slope, a number, of a tangnt, right?

    A contravariant vector on the other hand takes a tangent vector and produces another tangent vector. You can think of the position vector as taking a unit vector in the right direction and extending it to the position.

    It is a consequence of these definitions that when you make a diffeomorphic change of coordinates, the components of a covariant vector are inner producted with the partials of the old coordinates with respect to the new ones, while the components of a contravariant vector are producted with the partials of the new coordinates woth respect to the old ones. "CO, LOW, PRIMES BELOW" as Lillian Lieber taught, Covariant has its index subscript, and the partials have the primed (new) coordinates in the denominators. Contravariant has the opposites.
  4. Feb 21, 2006 #3


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    sure they do. but i wouldn't use \frac{}{} for inline. [itex] \alpha = e^2/(\hbar c 4 \pi \epsilon_0) [/itex] might be more readable than [itex] \alpha = \frac{e^2}{\hbar c 4 \pi \epsilon_0} [/itex].

    unfortunately, that sometimes floats your equation way above or way below the text baseline. sometimes i use tex instead if itex when it looks better.
  5. Feb 21, 2006 #4


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    You could add a \displaystyle to get
    [tex] \alpha = \displaystyle\frac{e^2}{\hbar c 4 \pi \epsilon_0} [/tex]
    but not with itex
    [itex] \alpha = \displaystyle\frac{e^2}{\hbar c 4 \pi \epsilon_0} [/itex]
  6. Feb 21, 2006 #5
    Where did you get this term "naturally" from? It is not needed.

    The 4-position is a Lorentz 4-vector, and not a general tensor such as a tangent vector. The term "Lorentz tensor" can, and often is, defined in two equivalent ways. The first way is is to say that a Lorentz tensor maps Lorentz 4-vectors and Lorentz 1-forms to scalars. The other way is by stating the transformation properties on the components of the Lorentz tensors.

    A Lorentz 4-vector is a set of 4 quantities which transform in the same way as the coordinates of an event relative to a well defined origin. This makes finite spacetime displacements 4-vectors.

    A general 4-vector is defined in a similar way, i.e. the coordinates of a general 4-vector transform in the same way as the coordinates of an infinitesimal spacetime displacement.
    No. The components don't transform in the same way as the components of a 4-vector does.
    Yes. Absolutely.

  7. Feb 23, 2006 #6
    I've been playing around with some Lorentz transformations and I've reached a problem. In SR Lorentz transformation matrices, [itex]L^{\mu'}{}_{\nu}[/itex] are 4x4 matrices which obey the Lorentz transformation property

    [tex][g_{\mu\nu}] = [L^{\mu'}{}_{\nu}]^{T}[g_{\mu\nu}][L^{\mu'}{}_{\nu}][/tex]

    The problem I have regards the [itex]L^4_4[/itex] entry. Why can it never be zero? Why must we have [itex]L^4_4 \geq 1[/itex] or [itex]L^4_4 \leq -1[/itex]? What happens if [itex]L^4_4 = 0[/itex]?
  8. Feb 23, 2006 #7
    I just had a thought. Could it be that if [itex]L^4_4 = 0[/itex] then the [itex]x^4[/itex] coordinate is not invariant under Lorentz transformations?
  9. Feb 23, 2006 #8


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    With L44=0, does the inverse transformation exist?
  10. Feb 23, 2006 #9

    George Jones

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    Let [itex]L[/itex] be a Lorentz transformation, so that [itex]g\left( Lu,Lv\right) =g\left(u,v\right)[/itex] for all 4-vectors [itex]u[/itex] and [itex]v[/itex], and let [itex]\left\{ e_{1},e_{2},e_{3},e_{4}\right\}[/itex] be an orthonormal basis for Minkowski space. Then, [itex]\left\{ e_{1}^{\prime},e_{2}^{\prime},e_{3}^{\prime},e_{4}^{\prime
    }\right\}[/itex], with [itex]e_{\mu}^{\prime}=Le_{\mu}[/itex] is also an orthonormal basis. In particular, [itex]e_{4}^{\prime}=Le_{4}[/itex] is a unit-length timelike vector, and this is what your condition guarantees:

    e_{4}^{\prime} & =Le_{4}\\
    & =L_{4}{}^{\mu}e_{\mu}\\
    & =L_{4}{}^{1}e_{1}+L_{4}{}^{2}e_{2}+L_{4}{}^{3}e_{3}+L_{4}{}^{4}e_{4},


    1 & =g\left( e_{4}^{\prime},e_{4}^{\prime}\right) \\
    & =-\left( L_{4}{}^{1}\right){}^{2}-\left( L_{4}{}^{2}\right) ^{2}-\left(L_{4}{}^{3}\right) ^{2}+\left( L_{4}{}^{4}\right) ^{2}.

    Allowing only the [itex]+1[/itex] restricts to orthochronous Lorentz transformations. Note that because [itex]L[/itex] is not symmetric, indices should be staggered, but also note that I might have used a non-standard staggering convention.

    Last edited: Feb 23, 2006
  11. Feb 24, 2006 #10
    It appears as if you've chose the LT between two inertial frames which are in the standard relationship, i.e. S' moves relative to S in the +x direction. When this is the case then [itex]L^4_4 = \gamma[/itex]. You can rotate the spatial axes of the S frame so that [itex]L^4_4 = 0[/itex].

  12. Feb 26, 2006 #11
    Co means it transforms like basis vectors. Contra means it transforms inverse to basis vectors.
  13. Feb 27, 2006 #12
    Hi Rob!

    Yes. In fact the inverse transformation always exists.

  14. Feb 27, 2006 #13


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    I assume that the '4' here is the time axis?

    From your mathematical description, you are looking for the set of general linear coordinate transorms that leave the metric invariant.

    Your transform basically defines a new coordinate system [itex]x'^a = L^a{}_b x^b[/itex], and we demand that the metric coefficients be identical in the new primed coordinates.

    The sort of coordinate transformation that does this includes Lorentz boosts, reflections, and rotations.

    Thinking about this, we can eliminate spatial diagonal elements (i.e. [itex]L^3{}_3[/itex]) via a spatial reflection We can swap a time coordinate with a space coordinate, and this will make [itex]L^4{}_4[/itex] zero, but unfortunately the metric will change - it will still be Minkowskian, but the time coordinate is distinguished by a minus sign in the metric.

    In other words, we can't replace a time coordinate with some superposition of purely space-like coordinates and maintain a minkowski metric unchanged, i.e. we can't write

    t' = a*x + b*y + c*z + 0*t

    and keep t' as being a time coordinate (distinguished by a negative sign in the metric).
  15. Feb 27, 2006 #14
    For purposes of searching the index there are two terms which apply here. When L00 > 0 or L00 < 0 the two terms "orthochronos" and "antichronos" are used to define which one applies to the Lorentz transformation. Which one belongs with which will depend on the convention used for the metric.

    Try searching the internet using those terms and see you find.

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