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Basis vectors under a Lorentz transformation

  1. Jan 11, 2010 #1
    Hello, I am new to the forums and I hope this fundamental topic has not been previously treated, as these forums don't seem to have a search function. I am studying general relativity using S. Carroll's book (Geometry and Spacetime) and I am having a fundamental problem with basis vectors under a Lorentz transformation.

    Consider a four-vector [tex]V[/tex] having the components [tex]V^{\mu}[/tex] with respect to some basis vectors [tex]\hat{e}_{(\mu)}[/tex] associated with a coordinate system [tex]x^\mu[/tex]. According to the book (and using his notation) the coordinate system transforms to the primed coordinate system according to

    [tex]x^{\mu'}=\Lambda^{\mu'}_{\nu}x^\nu[/tex]

    for some Lorentz transform [tex]\Lambda^{\mu'}_{\nu}[/tex]. Similarly the vector components transform in the same way according to

    [tex]V^{\mu'}=\Lambda^{\mu'}_{\nu}V^\nu[/tex]. (1)

    As I understand it the vector components are with respect to some basis and the primed components with respect to a corresponding primed basis. The (invarient) vector can thus be expressed with respect to either basis as

    [tex]V=V^\mu\hat{e}_{(\mu)}=V^{\mu'}\hat{e}_{(\mu')} [/tex] (2)

    where, in going from the unprimed to the primed coordinate system, the bases are related by the inverse transform:

    [tex]\hat{e}_{(\nu')}=\Lambda^{\mu}_{\nu'}\hat{e}_{(\mu)} [/tex]

    where [tex]\Lambda^{\mu}_{\nu'}[/tex] (subtly) denotes the inverse transform (because the prime is on the second index).


    This is all well and good but when I try to apply it to a concrete example it seems as though the basis vectors should transform (from unprimed to primed) via the transform, not its inverse. To construct an example, consider a rotation in the [tex]x^1x^2 [/tex] ([tex]xy [/tex]) plane, which I believe is Lorentzian and is accomplished via the transformation matrix

    [tex]\Lambda=\left[\begin{array}{cccc}
    1 & 0 & 0 & 0\\
    0 & \cos{\theta} & \sin{\theta} & 0\\
    0 & -\sin{\theta} & \cos{\theta} & 0\\
    0 & 0 & 0 & 1
    \end{array}\right]
    [/tex].

    To make this even more concrete let us set the angle at 45 degrees, i.e. [tex]\theta=\pi/4[/tex], in which case the transform becomes

    [tex]\Lambda=\left[\begin{array}{cccc}
    1 & 0 & 0 & 0\\
    0 & \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} & 0\\
    0 & -\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} & 0\\
    0 & 0 & 0 & 1
    \end{array}\right]
    [/tex],

    which has an inverse transform of

    [tex]\Lambda^{-1}=\left[\begin{array}{cccc}
    1 & 0 & 0 & 0\\
    0 & \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} & 0\\
    0 & \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} & 0\\
    0 & 0 & 0 & 1
    \end{array}\right]
    [/tex].

    If we then let the unprimed basis be orthonormal and lie on each axis with unit length then we can express a vector lying at a 45 degree angle from the x-axis and in the xy plane can be expressed as [tex]V=\hat{e}_{(1)}+\hat{e}_{(2)}[/tex] and has components

    [tex]V^0 = 0[/tex]
    [tex]V^1 = 1[/tex]
    [tex]V^2 = 1[/tex]
    [tex]V^3 = 0[/tex].

    A diagram would be useful here but it should be easy enough to visualize. Anyway, the components with respect to the primed coordinate system (and the yet-to-be-determined primed basis?) are, by (1),

    [tex]V^{0'} = 0[/tex]
    [tex]V^{1'} = \sqrt{2}[/tex] (3)
    [tex]V^{2'} = 0[/tex]
    [tex]V^{3'} = 0[/tex],

    which makes pefect sense if one visualizes the vector and the primed and unprimed coordinate axes. According to the book the basis vectors change by the inverse transform and are found to be

    [tex]\hat{e}_{(0')} = \hat{e}_{(0)}[/tex]
    [tex]\hat{e}_{(1')} = \frac{\sqrt{2}}{2}(\hat{e}_{(1)}-\hat{e}_{(2)})[/tex] (4)
    [tex]\hat{e}_{(2')} = \frac{\sqrt{2}}{2}(\hat{e}_{(1)}+\hat{e}_{(2)})[/tex]
    [tex]\hat{e}_{(3')} = \hat{e}_{(3)}[/tex],

    Which, already at this point, do not correspond to the primed coordinate axes. So the vector in terms of its primed components and primed basis vectors is

    [tex]V=\sqrt{2}\hat{e}_{(1')}=\sqrt{2}\frac{\sqrt{2}}{2}(\hat{e}_{(1)}-\hat{e}_{(2)})=\hat{e}_{(1)}-\hat{e}_{(2)}[/tex],

    which is clearly different from the original vector, and is rather its rotation in the wrong direction.

    I am convinced that I am doing something wrong here or misinterpreting things but I cannot find the source of error and am hoping that another pair of eyes might see it clearly. Any help with this is much appreciated.
     
    Last edited: Jan 11, 2010
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  3. Jan 11, 2010 #2

    Fredrik

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    Welcome to Physics Forums. There's a search link at the top of the page. :smile: I'm sure this has been answered before, but it can be difficult to find the right post. I answered a similar question in the quantum physics forum a few days ago. You might find it useful even though it's about a unitary transformation in a complex vector space, instead of a Lorentz transformation. Link.

    I don't understand what you're doing here. Did you decide what the primed coordinate axes were before you found the primed basis vectors? How? Also, I don't know how you arrived at the first equality. Your definition of V and the primed basis vectors you found implies that [tex]V=\sqrt{2}\hat{e}_{(2')}[/tex], so why did you write [tex]=\sqrt{2}\hat{e}_{(1')}[/tex]?
     
  4. Jan 11, 2010 #3
    There were a couple errors in my post which I have corrected and I have added some more equation references, which I'll need here.

    First, I guess I assumed that the primed basis vectors would correspond to the primed coordinate system in the same way that the unprimed basis vectors correspond to the unprimed coordinate system, i.e. they lie on the latter's axes. This is apparently not the case at all as my example shows (unless there is something wrong there), and my assumption may be unfounded. This correspondence is however evidenced by (2), according to which the primed components, which are correct with respect to the primed coordinate system, are with respect to the primed basis vectors.

    Regarding the first part of the equation you pointed out above, this comes from applying the primed components given in (3) to the last part of (2) while the latter part comes from the substitution of the primed basis vector given by (4). Though [tex]V=\sqrt{2}\hat{e}_{(2')}[/tex] is the correct vector given the primed basis, I don't see how this is obtained.
     
  5. Jan 11, 2010 #4

    atyy

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    If you let the inverse act on the transformed coordinates, do you get back the original coordinates?
     
  6. Jan 11, 2010 #5
    Yes, which makes perfect sense.
     
  7. Jan 11, 2010 #6

    atyy

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    Hmmm, well if you write the primed basis vector as the transpose of [0 1 0 0] and apply the inverse transform matrix, you get the transpose of [0 sqrt(2) sqrt(2) 0] which is what you want?
     
  8. Jan 11, 2010 #7
    Yes if you do that (which is to say applying the Lorentz transform to the basis vector) you get the correct result. The problem is that, according to the text, the basis vectors transform using the inverse transformation, which seems to be the cause of my woes as, if they are transformed, using the original transformation matrix, everything works out.

    Again though, I am now not certain that the primed basis vectors correspond to the primed coordinate system in the same way that the unprimed basis vectors correspond to the unprimed coordinate system. I may be confusing the relation of the coordinate system to the basis vectors.
     
  9. Jan 11, 2010 #8

    atyy

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    Well, no - what we did was apply the transform to the *components* of the basis vectors, so that's fine.

    I have no idea how you apply the transform written as a matrix to a vector itself, as opposed to its representation in coordinates. Probably have to do very careful index gymnastics ...
     
  10. Jan 11, 2010 #9
    I don't think the basis vectors technically have components in themselves (as they are abstract and invariant geometric entities) without reference to some coordinate system. A vector can be expressed through it's components with respect to some coordinate system and/or with respect to some basis. In my example the basis chosen in the unprimed coordinate system happens to line up with the coordinate system axes, though this does not necessarily have to be the case. The problem is that, because they transform inversely, the primed basis vectors do not line up with the primed coordinate system. Again, I'm not sure they need to, but things work out if we refer the primed components of the example vector to the primed coordinate system but not if we refer them to the primed basis vectors. But, according to (2), these components should be with reference to the primed basis vectors. The fundamental conflict seems to be why the coordinates transform one way while the basis vectors transform inversely.
     
  11. Jan 11, 2010 #10

    atyy

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    The basis vectors do have components. V=v0e0+v1e1+v2e2+v3e3, ie. V has components [v0 v1 v2 v3] in the unprimed basis, so e0=1e0+0e1+0e2+0e3, so e0 has components [1 0 0 0] in the unprimed basis.
     
  12. Jan 11, 2010 #11
    Yes, I see, though I'm not sure how it helps resolve the original problem.
     
  13. Jan 11, 2010 #12

    Fredrik

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    I think I got it. Took me some time because I'm slower than usual this time of night. (No, I didn't work on it this whole time. I took a break to watch Chuck :smile:). In every 4x4 matrix below, the row index is upstairs to the left, and the column index is downstairs to the right.

    [tex]V=V^\mu e_\mu=V'^\mu e'_\mu=\Lambda^\mu{}_\nu V^\nu e'_\mu[/tex]

    [tex]e_\nu=\Lambda^\mu{}_\nu e'_\mu[/tex]

    [tex](\Lambda^{-1})^\nu{}_\rho e_\nu=(\Lambda^{-1})^\nu{}_\rho\Lambda^\mu{}_\nu e'_\mu=\delta^\mu_\rho e'_\mu=e'_\rho[/tex]

    What you're doing is equivalent to

    [tex](\Lambda^{-1})^\rho{}_\nu e_\nu=e'_\rho[/tex]

    and that's not equivalent to the correct transformation that I just derived.
     
  14. Jan 12, 2010 #13
    Yes, I understand perfectly now, I was summing over the row when I should have been summing over the column. Correcting this, the primed basis vectors line up perfectly with the primed coordinate system in the same way the unprimed basis vectors line up with the unprimed coordinate system and the primed vector components are then in terms of both the primed coordinate system and the primed basis vectors. Thanks a lot Fredrik! The index notation is going to take some getting used to and I'll have to be sure to be more careful about the meaning of the equations. I have a feeling though that, later on, they'll make things much simpler. Now on to dual vectors! Thanks again!
     
  15. Jan 12, 2010 #14

    Fredrik

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    You're welcome. This post might help with the annoying index notation. I actually recommend that you avoid that notation when you can. I don't know why anyone would prefer to write [itex]\eta_{\rho\sigma}\Lambda^\rho{}_\mu\Lambda^\sigma{}_\nu=\eta_{\mu\nu}[/itex] when they can write [tex]\Lambda^T\eta\Lambda=\eta[/tex].
     
  16. Jan 12, 2010 #15

    atyy

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    When using the index notation on the basis vectors as opposed to the coordinate representation of the basis vectors as matrices, does the index equation actually have a matrix multiplication interpretation?
     
  17. Jan 12, 2010 #16

    Fredrik

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    A formula like

    [tex](\Lambda^{-1})^\nu{}_\rho e_\nu=e'_\rho[/tex]

    can be interpreted as

    [tex]\begin{pmatrix}e_0 & e_1 & e_2 & e_e\end{pmatrix}\Lambda^{-1}=\begin{pmatrix}e'_0\\ e'_1\\ e'_2\\ e'_3\end{pmatrix}[/tex]

    i.e. you allow the components of the 4x1 matrix and the 1x4 matrix to be 4-vectors instead of numbers. But some people would just find this confusing, so it's probably better to avoid it, and use the first version of the formula instead (with indices).

    What I'm objecting to is that there are plenty of calculations that don't mention basis vectors at all, that still use the annoying index notation. For example, if I want to prove that the inverse of a Lorentz transformation is a Lorentz transformation, there's no need to involve indices. It's just a simple excercise in matrix multiplication, and to do it with indices is to insert the definition of matrix multiplication for no apparent reason, as if we need to explain what matrix multiplication is every time we multiply two matrices.
     
  18. Jan 12, 2010 #17

    atyy

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    I see. That's almost like notating v=v1e1+v2e2=[e1 e2][v1 v2]T, which could be confusing as you say since I normally have in mind row/column matrices as representations of one forms/vectors, rather than basis vectors/components of vectors.

    Agreed. I suppose the SR authors that do this are trying to prepare readers for GR, where the index notation helps keep track of order of the tensor.
     
  19. Jan 12, 2010 #18
    Yes I find the matrix notation simpler but I think when dealing with higher ranked tensors later, the matrix analogy becomes difficult if not impossible since, for example, you might be dealing with a 4D matrix (in the case of a (0, 4) tensor) and multiplying such things with other tensors and vectors in different ways. I'm getting more used to it and I think tensors will be simper to manipulate but it's very easy to lose track of what's really going on mathematically and physically.
     
  20. Jan 12, 2010 #19
    I've been doing battle with metric tensors over here using matrices. Admittedly they're winning so far... I've sometimes been treating a basis as a 1xn matrix and find that idea far less confusing than some aspects of the subject, such as Carroll's "somewhat subtle notation".

    It seems that the component matrix for the Riemannian metric tensor in plane polar coordinates can be got by

    [tex]g = B^T \, B[/tex]

    [tex]g_{ab} = \frac{\partial x^a}{\partial x^m'} \frac{\partial x^b}{\partial x^n'} \, \delta_{m' n'}[/tex]

    (and don't know how to fix the indices) where B is the change of basis matrix such that

    [tex]B e_{column} = e'_{column}.[/tex]

    But the components of the pseudo-Riemannian metric tensor seem to be defined the other way around, by the component transformation, as in these examples of Lorentz transformations:

    [tex]g = (B^{-1})^T \eta B^{-1}[/tex]

    [tex]g_{\alpha \beta} = \frac{\partial x^\mu'}{\partial x^\alpha} \frac{\partial x^\nu'}{\partial x^\beta} \eta_{\mu \nu}[/tex]
     
  21. Jan 12, 2010 #20
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