Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Minkowski norm

  1. Oct 19, 2009 #1
    From my reading of introductory texts on special relativity, I've seen this defined in various ways, and I'm curious about whether any of these definitions are preferable to others, for example because they're more convenient, consistent, logical, clearer, more widely used or more easily extended to general relativity. Most often, I've seen a Minkowski norm is defined as

    [tex]\left||\mathbf{v}|\right| = \sqrt[+]{t^{2} - x^{2} - y^{2} - z^{2} }[/tex]

    or

    [tex]\left||\mathbf{v}|\right| = \sqrt[+]{-t^{2} + x^{2} + y^{2} + z^{2} }[/tex]

    depending on the sign convention the author happens to be using. This "norm" fails to meet a number of the criteria of a true norm. It's not necessarily real valued (it can be real or imaginary), and the norm of a lightlike vector is always 0, even if the vector isn't 0). I'm not sure about the triangle inequality...

    In The Geometry of Spacetime, Callahan defines the Minkowski norm ||v|| of a vector v thus:

    [tex]\left||\mathbf{v}|\right| = \sqrt[+]{\left|t^{2} - x^{2} - y^{2} - z^{2} \right|}[/tex]

    ensuring that the value is always real and positive, although I think there's still the problem with lightlike vectors. Taylor & Wheeler in Spacetime Physics seem to fluctuate between these two approaches. They offer the following definition of "the interval":

    [tex]\left(interval \right) = \left[ \left(\Delta t \right)^{2} - \left(\Delta x \right)^{2} \right]^{1/2},[/tex]

    yet go on to define timelike and spacelike intervals (proper time and proper distance) in exactly the same way as Callahan defined his Minkowski norm:

    [tex]\Delta \tau = \left[ \left(\Delta t \right)^{2} - \left(\Delta x \right)^{2} \right]^{1/2};[/tex]

    [tex]\Delta \sigma = \left[ \left(\Delta x \right)^{2} - \left(\Delta t \right)^{2} \right]^{1/2}.[/tex]

    They mention Minkowski's idea of multiplying the time coordinate by the square root of minus one, i, but dismiss it on the grounds that it serves no purpose except to obscure a genuine physical difference between time and space. Lawden does use this method in An Introduction to Tensor Calculus, Relativity and Cosmology, at least for the early chapters which deal with Cartesian tensors. But Kip Thorne in the Introduction to general relativity section (part II) of this lecture series (I can't remember which one exactly)

    http://elmer.tapir.caltech.edu/ph237/CourseOutlineA.html

    says there are problems with extending this method to general relativity (adding that the concept of "imaginary time" which Stephen Hawking refers to in A Brief History of Time is something else entirely). Instead he introduces an orthogonal basis for Minkowski spacetime where the spatial basis vectors each coincide with their corresponding reciprocal basis vectors, as in a Euclidean orthonormal basis, while the temporal basis vector is equal in length but opposite in sign to the corresponding reciprocal basis vector for time, so that

    [tex]\mathbf{e}_{0} \cdot \mathbf{e}_{0} = -1[/tex]

    [tex]\mathbf{e}_{0} \cdot \mathbf{e}^{0} = 1.[/tex]

    I've written in my notes: "The dot product is defined the same as for R3, and so is the length (the square root of the dot product of the vector with itself."

    [tex]\left||\mathbf{a}| \right| = a^{i}a_{i}[/tex]

    In response to a student's question, Kip Thorne says he won't be using the concept of the dual space in these lectures. If I've understood and correctly represented the ideas of this lecture in my notes, it seems that he's defined a kind of norm for Minkowski spacetime much more like the Euclidean norm that those other authors: always real, always positive except when zero... Does it fulfill all the requirements of a true norm? Does it sort of allow us to treat Minkowski spacetime as, in some ways, like R4? Does it, in fact, effectively accomplish what multiplying the time coordinate by i was inteded to do? This "dot product" seems to be equivalent to

    [tex]g^{\alpha}_{\beta} = \left( \begin{matrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix} \right)[/tex]

    but not

    [tex]g^{\alpha}^{\beta}[/tex] or [tex]g_{\alpha}_{\beta}[/tex]

    defined by the same matrix. But would all of these be called different forms of the Lorentz metric (metric tensor)?

    Finally, I see that the term Minkowski norm is also used for a rather different concept, synonymous with Hölder norm:

    http://books.google.co.uk/books?id=PDjIV0iWa2cC&pg=PA17&lpg=PA17

    the generalisation of the Euclidean norm which Wikipedia calls a p-norm:

    http://en.wikipedia.org/wiki/Norm_(mathematics)#p-norm
     
  2. jcsd
  3. Oct 19, 2009 #2

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I prefer to talk about the bilinear form g defined by

    [tex]g(x,y)=x^T\eta y[/tex]

    where x and y are 4x1-matrices and

    [tex]\eta=\begin{pmatrix}-1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}[/tex]

    The alternative convention is to define [itex]\eta[/itex] with the opposite sign. Neither convention has a significant advantage over the other. My convention seems to be more common in GR books and the other convention seems to be more common in QFT books. (Weinberg's QFT book is an exception).
     
  4. Oct 20, 2009 #3
    the triangle inequality is reversed. the sum of the 2 sides must be less than the 3rd.

    if 2 points are separated by time then the interval is imaginary (timelike). if they are separated by distance then the interval is real (spacelike). what do you mean the norm of a lightlike vector is always 0, even if the vector isn't 0? (nevermind I get it)

    is the Minkowski 'norm' the same as the 'interval'?
     
  5. Oct 20, 2009 #4
    Yeah, I realised eventually!

    I don't know. Would the term interval only apply to position vectors or displacement vectors? I couldn't find any examples of it being used of the energy-momentum vector, but Google shows plenty of references to the energy-momentum vector's length, magnitude or norm (including one reference with scare-quotes to its Minkowski "norm"). Maybe magnitude is the safest and most neutral term.

    If a true norm is defined

    [tex]\left(1 \right) \;\left||\mathbf{v}| \right| \text{is real}[/tex]

    [tex]\left(2 \right) \; \left||\mathbf{v}| \right| \geq 0[/tex]

    [tex]\left(3 \right) \; \left||\mathbf{v}| \right| = 0 \Leftrightarrow \mathbf{v} = \mathbf{0}[/tex]

    [tex]\left(4 \right) \; \left|| \lambda \mathbf{v}| \right| = \lambda \left|| \mathbf{v} ||[/tex]

    [tex]\left(5 \right) \; \left|| \mathbf{u} + \mathbf{v} | \right| \leq \left|| \mathbf{u} | \right| + \left|| \mathbf{v} | \right|[/tex]

    where v and u are vectors and lamba a scalar, then Callahan's Minkowski norm meets conditions 1, 2 and 4. It conforms to the reverse of the triangle inequality, 5, and meets condition 3 only for non-lightlike vectors. The definitions which make the norm simply a positive square root of plus-or-minus the time component minus the space components are as norm-like except that they fail to meet condition 1. One of those books defined a seminorm as a function which meets all the conditions for a norm except 2, but I don't think any of these Minkowski "norms" is even that. Multiplying the time component (or the space components) with i and summing the squares of the components has the same effect as subtracting either the squares of the time component or space components.
     
  6. Oct 20, 2009 #5

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I think a seminorm is usually defined differently. Here's the definition from "A course in functional analysis" by John Conway:

    (Here [itex]\mathbb F[/itex] means "either [itex]\mathbb R[/itex] or [itex]\mathbb C[/itex]").

    If [itex]\mathcal H[/itex] is a vector space over [itex]\mathbb F[/itex], then a seminorm is a function [itex]p:\mathcal H\rightarrow[0,\infty)[/itex] having the properties

    (a) [itex]p(x+y)\leq p(x)+p(y)[/itex] for all [itex]x,y\in\mathcal H[/itex]

    (b) [itex]p(\alpha x)=|\alpha|p(x)[/itex] for all [itex]\alpha\in\mathbb F[/itex] and all [itex]x\in\mathcal H[/itex]

    It follows from (b) that [itex]p(0)=0[/itex]. A norm is a seminorm such that

    (c) [itex]x=0[/itex] if [itex]p(x)=0[/itex].

    Conway also defines a "semi-inner product" in a similar way. An inner product is a semi-inner product u such that x=0 if u(x,x)=0.

    So the words "seminorm" and "semi-inner product" don't seem to be good choices for the functions we're dealing with in SR. I think the terms "non-degenerate non-negative linear functional" and "non-degenerate symmetric bilinear form" are understood by mathematicians to mean exactly what we want, but I don't think there are any terms for these things that aren't long and awkward.
     
    Last edited: Oct 20, 2009
  7. Oct 20, 2009 #6

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The triangle inequality holds whenever the Cauchy-Schwarz inequality does, and it does for spacelike vectors. Formula 3 in this article is the Minkowski space version of Cauchy-Schwarz.
     
  8. Oct 20, 2009 #7
    Thanks for this, Fredrik, and for the link to Giulini's article. For (4), I should have written:

    [tex]\left(4 \right) \; \left|| \lambda \mathbf{v}| \right| = | \lambda | \left|| \mathbf{v} ||[/tex]

    And for the seminorm, I meant to write "except for 2 and 3".

    Would "non-degenerate" mean that

    [tex]\left||\mathbf{v}| \right| = 0 \Rightarrow \mathbf{v} = \mathbf{0}[/tex]?

    And if so, wouldn't the existence of lightlike vectors prevent this condition from being met?
     
  9. Oct 20, 2009 #8

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Yes and yes. The maps we're dealing with are degenerate. I'm actually not sure how "standard" that term is. I would expect a mathematician to understand what I mean, but I would still explain it, just in case.
     
  10. Nov 22, 2009 #9
    I see Sean Carroll takes yet another approach in his Lecture Notes on General Relativity where he defines the norm as the "inner product of the vector with itself":

    [tex]\eta_{\mu\nu} V^{\mu} V^{\nu}[/tex]

    and the spacetime interval as the quantity

    [tex]-(c \Delta t)^{2} + (\Delta x)^{2} + (\Delta y)^{2} + (\Delta z)^{2}[/tex]

    rather than its square root.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Minkowski norm
  1. Minkowski metric (Replies: 8)

  2. Minkowski force (Replies: 4)

  3. Minkowski Metric (Replies: 36)

  4. Minkowski space (Replies: 28)

  5. Minkowski Force (Replies: 3)

Loading...