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Invarients for 4-vector quantites.

  1. Jun 10, 2008 #1
    I was summarizing for myself the various four-vectors of mechanics:

    x &= ct + \mathbf{x} \\
    V &= \frac{dx}{d\tau} = \gamma(c + \mathbf{v}) \\
    P &= m V = E/c + \gamma\mathbf{p} \\
    f &= m\frac{d^2 x}{d\tau^2} = m\frac{d V}{d\tau} \\


    \gamma^{-2} &= (1 - {\lvert \mathbf{v}/c \rvert}^2) \\
    d\tau &= {\left(\frac{dx}{d\lambda} \cdot \frac{dx}{d\lambda}\right)}^{1/2} d\lambda \\
    x \cdot x = {\lvert x \rvert}^2 &= c^2t^2 - {\lvert \mathbf{x} \rvert}^2 \\
    E &= \int f \cdot (c d\tau) \\
    \mathbf{v} &= \frac{d\mathbf{x}}{dt} \\
    \mathbf{p} &= m\mathbf{v} \\

    Invarients for the first three four vectors are:

    {\lvert x \rvert}^2 &= c^2 t^2 - {\lvert \mathbf{x} \rvert}^2 = c^2 \tau^2 \\
    {\lvert V \rvert}^2 &= \gamma^2 (c^2 - {\lvert \mathbf{v} \rvert}^2) = c^2 \\
    {\lvert P \rvert}^2 &= m^2 {\lvert V \rvert}^2 = m^2 c^2 \\

    Is the minkowski norm of the four vector force:

    f = m\frac{d^2 x}{d\tau^2}

    also an invarient? I think it has to be. Assuming that is the case, what would the value (and significance if any) of this be?
    Last edited: Jun 10, 2008
  2. jcsd
  3. Jun 10, 2008 #2
    It ought to be invariant in the sense that it is the same in every reference frame. But as far as I know, it doesn't have to take on any particular value. Also, what you have written,
    [tex]{\lvert x \rvert}^2 &= c^2 t^2 - {\lvert \mathbf{x} \rvert}^2 = c^2 \tau^2[/tex]
    is only true for propagation at constant velocity. Also, it is a convention to sometimes leave the bars and bolds off of 4-vectors, so write [tex]P^2[/tex].
  4. Jun 10, 2008 #3
    Alright. Thanks (and for pointing out my error).

    Re: conventions.

    One book (New foundations for Classical Mechanics) I was using is using [itex]P\bar{P}[/itex] (using a conguage geometric product to define the norm). Baylis' "Relativity in Introductory Physics" paper uses this too.

    Another (Geometric Algebra for Physicists) uses [itex]P^2[/itex] ... but that book uses a geometric product on a four vector with an explicit unit vector for the time component. There the product is defined in such a way that the square is the norm, so omitting the bars can be done without ambiguity.

    So after seeing so many notations for the same thing I picked one that I thought would be understood without ambigity;)
  5. Jun 10, 2008 #4
    Mmm... well it's probably ok. I would shy away from learning geometric algebra and special relativity at the same time. The convention I mentioned is usually used by relativists and field theorists. Geometric algebra is nice, but is not standard fare for physicists.
  6. Jun 10, 2008 #5
    I was actually studying electromagnetism, which can be expressed very nicely with GA, but the two are so interrelated I can't avoid relativity too.
  7. Jun 10, 2008 #6
    Well, physicists usually just stick to differential forms for that. I'm not so sure how GA handles higher dimensions (i.e,. > 4)
  8. Jun 11, 2008 #7
    GA works uniformly with any number of dimensions. An example is how it is being used by the CS folks to model 3D space using an explicit origin for the viewport and point at infinity. Another example is application to linear solutions. All the problems of intersecting subspaces of any dimension (lines, planes, hypervolumes), and minimal distances between these and so forth, have a very elegant solution in this context.

    I actually started with differential forms trying to see how you get from maxwells equations to the retarded time equations. That looked like a 4D stokes/greens/... theorm type boundary value calculation to me and none of my old Calculus III worked once that extra dimension was added to the mix.

    I had some trouble with the forms ideas because I kept getting distracted trying to understand what I thought to be the more simple ideas first. ie: how the wedge product fit in regular geometry without introducing all the additional manifold calculus ideas. I went looking for that, and found a very satisfactory answer in the GA infrastructure.

    Even if the ideas aren't part of mainstream physics I've found them very helpful trying to understand many of the traditional physics ideas. Maybe I'm just dumb, I couldn't get my head around a lot of the tensor ideas, with raised and lowered indexes, before seeing how this fit in a GA context. Two examples. The completely antisymetric tensors like the:

    R^{\text{T}} R'

    of rigid body motion, or the electromagnetic tensor


    have a very natural bivector representation in GA. To me the forms or tensor approaches, even though they may be conventional, seem less intuitive.

    In the long run it's more work for me to learn things this way because I don't really have a choice but to learn this and the traditional ways, but if I'm able to learn the traditional ways easier that way and have a better toolbox, I think I'm better off in the long run. However, I also have the luxury of learning multiple approaches since I'm not under deadlines for exams. This is a completely self study project for fun. I've been out of school for a long time (now doing boring, but well paying, computer programming for a database product). My work unfortunately provides absolutely no justification for any sort of physics or mathematics.
  9. Jun 11, 2008 #8


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    These equations aren't valid notation. You can't use "+" like that. In this context, I would write them as

    [tex]\textbf{X} = (ct, \textbf{x})[/tex]
    [tex]\textbf{V} = \frac{d \textbf{X}}{d\tau} = \gamma(c, \textbf{v})[/tex]
    [tex]\textbf{P} = m \textbf{V} = (E/c, \gamma \textbf{p})[/tex]
    [tex]\textbf{F} = m \frac{d^2 \textbf{X}}{d\tau^2} = m \frac{d \textbf{V}}{d\tau}[/tex]​

    Different authors use different notation conventions, but a common one is to use uppercase for 4-vectors and lowercase for 3-vectors. (In GR, most authors would use [itex]X^\mu[/itex] instead of the [itex]\textbf{X}[/itex] often used in SR.)

    Note that as F is orthogonal to V (which can be proved by differentiating ||V||^2), then in the rest frame (co-moving inertial frame) of the accelerating object the "time" component of F is zero (because in that frame the "space" component of V is zero). Thus ||F|| must be the magnitude of the "proper force", the force experienced by the object in its own rest frame.

    If Q is a genuine 4-vector (i.e., in the context of SR, its components transform via the Lorentz transform) then its norm ||Q|| must be invariant. The derivative of a 4-vector with respect to a scalar invariant is also a 4-vector. So once you know [itex]d\textbf{X}[/itex] is a 4-vector then [itex] d\tau = ||d\textbf{X}|| / c [/itex] must be invariant and so V, P and F must be 4-vectors, and their norms all invariant.
  10. Jun 11, 2008 #9
    It is valid to do so in geometric algebra which defines a product (clifford algebraic product) of vectors that produces a scalar and bivector term, incorporating both dot and cross product like parts (really wedge product) behaviour respectively.

    Similarily vector/bivector products produce vector trivector parts. I figured what was written would be understood independent of any knowledge of this algebra.

    Thanks for the note about the F and V orthogonality. I'll try that calculation.
  11. Jun 12, 2008 #10
    fyi. A paper that discusses this algebra applied to spacetime is available here (my standalone description is probably not too coherent if you don't already know the ideas).


    See: RelEasyPs.pdf (it has a long Lotus notes url that probably won't cut and paste well).

    ps. When I wasn't at work I sat down to do demonstrate to myself that F and V are orthogonal as you said, but saw how immediately from V \cdot V and was left with nothing interesting to calculate;(
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