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Dirac delta in curved space

  1. Sep 28, 2012 #1
    The Dirac delta function is defined as:
    [tex]\int_{ - \infty }^{ + \infty } {\delta (x - {x_0})dx} = 1[/tex]
    Or more generally the integral is,
    [tex]\int_{ - \infty }^{ + \infty } {\delta (\int_{{x_0}}^x {dx'} )dx} [/tex]
    But if the metric varies with x, then the integral becomes,
    [tex]\int_{ - \infty }^{ + \infty } {\delta (\int_{{x_0}}^x {\sqrt {g(x')} dx'} )\sqrt {g(x)} dx} [/tex]
    But from the wikipedia.com article we have,
    [tex]\delta (f(x)) = \frac{{\delta (x - {x_z})}}{{\left| {f'({x_z})} \right|}}[/tex]
    where xz is such that [itex]f({x_z}) = 0[/itex].
    And since,
    [tex]\frac{d}{{dx}}(\int_{{x_0}}^x {\sqrt {g(x')} dx'} ) = \sqrt {g(x)} [/tex]
    and
    [tex]f(x) = \int_{{x_0}}^x {\sqrt {g(x')} dx'} = 0[/tex]
    when x= x0 so that xz= x0.
    Then the above becomes,
    [tex]\int_{ - \infty }^{ + \infty } {\delta (\int_{{x_0}}^x {\sqrt {g(x')} dx'} )\sqrt {g(x)} dx} = \int_{ - \infty }^{ + \infty } {\frac{{\delta (x - {x_0})}}{{\left| {\sqrt {g({x_0})} } \right|}}\sqrt {g(x)} dx} [/tex]
    Now obviously if g(x) is constant, we recover the original Dirac delta function. But what if g(x) is not constant, can I still recover the original Dirac delta function? Then the Dirac delta would behave the same in any space.

    Have I evaluated [itex]{\left| {\sqrt {g({x_0})} } \right|}[/itex] wrong? Could there be some sense is saying this should simply by [itex]{\sqrt {g(x)} }[/itex] so that it cancels out? Maybe x never really equals x0, but only approaches it, so that we use some other property of the Dirac delta.
     
    Last edited: Sep 28, 2012
  2. jcsd
  3. Sep 28, 2012 #2
    OK, I think I got it. Starting with
    [tex]\int_{ - \infty }^{ + \infty } {\delta (\int_{{x_0}}^x {\sqrt {g(x')} dx'} )\sqrt {g(x)} dx} = \int_{ - \infty }^{ + \infty } {\frac{{\delta (x - {x_0})}}{{\left| {\sqrt {g({x_0})} } \right|}}\sqrt {g(x)} dx} [/tex]
    I suppose it is safe to assume that the metric g(x) is real and always positive so the squareroot is positive, and we can drop the absolute sign. And since we can bring the denominator outside the integral, we have,
    [tex]\frac{1}{{\sqrt {g({x_0})} }}\int_{ - \infty }^{ + \infty } {\delta (x - {x_0})\sqrt {g(x)} dx} [/tex]
    And then the sifting property lets us bring the function multiplied by the Dirac delta outside the integral with an x value of x0, to get,
    [tex]\frac{1}{{\sqrt {g({x_0})} }}\sqrt {g({x_0})} \int_{ - \infty }^{ + \infty } {\delta (x - {x_0})dx } [/tex]
    But since the squareroots of g cancel, this is just [itex]\int_{ - \infty }^{ + \infty } {\delta (x - {x_0})dx} [/itex]. So,
    [tex]\int_{ - \infty }^{ + \infty } {\delta (\int_{{x_0}}^x {\sqrt {g(x')} dx'} )\sqrt {g(x)} dx} = \int_{ - \infty }^{ + \infty } {\delta (x - {x_0})dx} = 1[/tex]
    Or the Dirac delta function is invariant with changes of coordinates, metric, curvature or spaces.

    But I suppose if we think in higher dimensional terms, the integral inside the Dirac delta is a line integral, and the integral outside the delta is a volume integral, right? So would we still get the cancellation of squareroots of g in higher dimensions? Or would the g(x) term in the line integral not be squarerooted like the volume integral outside the delta?
     
    Last edited: Sep 28, 2012
  4. Sep 28, 2012 #3

    MathematicalPhysicist

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    Well, the Dirac distribution satisifes:

    [tex]\int \delta(x-x_0) f(x) dx = f(x_0)[/tex]

    That pretty much answers your question.
     
  5. Sep 28, 2012 #4

    MathematicalPhysicist

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    BTW, if I am not mistaken we should take minus the squareroot of the det of the metric in the change of variables in integration.
     
  6. Sep 29, 2012 #5
    I'm not satisfied because I'm not sure how this generalizes in multidimensional space. What does
    [tex]\int_{ - \infty }^{ + \infty } {\delta (\int_{{x_0}}^x {\sqrt {g(x')} dx'} )\sqrt {g(x)} dx} [/tex]
    look like in 3D?
    The
    [tex]{\int_{{x_0}}^x {\sqrt {g(x')} dx'} }[/tex]
    part would be a line integral. How does that generalize in 3D with a changing metric?
    The
    [tex]\int_{ - \infty }^{ + \infty } {\delta (......)\sqrt {g(x)} dx} [/tex]
    part would be a volume integral where the squareroot of the metic is commonly used.

    So I guess my biggest question is how to use the metric in a line integral. And since you can have more than one line from x0 to x in 3D, how do you determine which line to take, if it even matters? Would you integrate the line integral as it travels over the geodesic?
     
  7. Sep 30, 2012 #6

    MathematicalPhysicist

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    Well it should be similar to the one dimensional case.

    I assume you know that the n-dimensional Dirac Delta distribution is made up of a product of 1-dimensional Dirac Delta distribution.

    So you do the same trick you did before and get something like:

    [tex] \int_{\mathbb{R}^n} \delta ^n (\bar{x}-\bar{x_0}) \frac{\sqrt{g(\bar{x})}}{\sqrt{g(\bar{x_0})}} d\bar{x}[/tex]

    Ok, in your case I think (though not certain), that th line integral from one dimensional becomes if [tex]\bar{x} =(x^1,x^2,...,x^n)[/tex]

    then the line integral becomes:

    [tex] \int_{x^1_0}^{x^1}...\int_{x^n_0}^{x^n} \sqrt{g(x')}dx'^1...dx'^n[/tex]
     
    Last edited: Sep 30, 2012
  8. Sep 30, 2012 #7
    The differential line segment squared is:
    [tex]d{s^2} = {g_{ij}}(\vec x)d{x^i}d{x^j}[/tex]
    So the line integral in a general curved space is:
    [tex]\int_{{{\vec x}_0}}^{\vec x} {\sqrt {{g_{ij}}(\vec x)d{x^i}d{x^j}} } [/tex]
    which is usually parameterized as:
    [tex]\int_{{t_0}}^t {\sqrt {{g_{ij}}(\vec x(t))\frac{{d{x^i}}}{{dt}}\frac{{d{x^j}}}{{dt}}} } \,\,\,dt[/tex]
    And when we put this inside the Dirac delta to get
    [tex]\delta (\int_{{{\vec x}_0}}^{\vec x} {\sqrt {{g_{ij}}(\vec x)d{x^i}d{x^j}} } )[/tex]
    I don't know how this can be put in the form
    [tex]\delta (f(\vec x)) = \frac{{\delta (\vec x - {{\vec x}_z})}}{{\left| {f'({{\vec x}_z})} \right|}}[/tex]
    where [itex]f({{\vec x}_z}) = 0[/itex].

    Can [itex]{g_{ij}}(\vec x)d{x^i}d{x^j}[/itex] be put in the form of a determinant of the metric time differentials so that we can evaluate [itex]g(\vec x)[/itex] at [itex]{{\vec x}_0}[/itex] and bring it outside the delta in the denominator where it cancels as before?
     
    Last edited: Sep 30, 2012
  9. Sep 30, 2012 #8

    Mute

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    See this section of the wikipedia page. Perhaps it will help?
     
  10. Oct 1, 2012 #9
    In one dimensional space, [itex]\vec x = (x)[/itex], which means [itex]{x^i} = x[/itex] and [itex]d{x^i} = dx[/itex], so that
    [tex]\sqrt {{g_{ij}}(\vec x)d{x^i}d{x^j}} = \sqrt {{g_{ii}}(\vec x)d{x^i}d{x^i}} = \sqrt {g(x)} dx[/tex]
    and then
    [tex]\delta (\int_{{{\vec x}_0}}^{\vec x} {\sqrt {{g_{ij}}(\vec x)d{x^i}d{x^j}} } ) = \delta (\int_{{x_0}}^x {\sqrt {g(x')} dx} ') = \frac{{\delta (x - {x_0})}}{{\sqrt {g({x_0})} }}[/tex]
    And this [itex]{\sqrt {g({x_0})} }[/itex] cancels with the one in the volume integral. And the Dirac delta is invariant in 1D.

    In 2D, we have [itex]\vec x = ({x^1},{x^2})[/itex], and
    [tex]{g_{ij}}(\vec x)d{x^i}d{x^j} = {g_{11}}({x^1},{x^2})d{x^1}d{x^1} + {g_{12}}({x^1},{x^2})d{x^1}d{x^2} + {g_{22}}({x^1},{x^2})d{x^2}d{x^2} + {g_{21}}({x^1},{x^2})d{x^2}d{x^1}[/tex]
    And in flat space, [itex]{g_{12}} = {g_{21}} = 0[/itex], so that the above becomes,
    [tex]{g_{ij}}(\vec x)d{x^i}d{x^j} = {g_{11}}d{x^1}d{x^1} + {g_{22}}d{x^2}d{x^2}[/tex]
    Then if [itex]{g_{11}} = {g_{22}} = g[/itex], we get
    [tex]{g_{ij}}(\vec x)d{x^i}d{x^j} = g{(d{x^1})^2} + g{(d{x^2})^2}[/tex]
    And,
    [tex]\sqrt {{g_{ij}}(\vec x)d{x^i}d{x^j}} = \sqrt g \sqrt {{{(d{x^1})}^2} + {{(d{x^2})}^2}} = \sqrt g ds[/tex]
    So that the [itex]\sqrt g [/itex] cancels out to make the integral of the Dirac delta invariant with changes in rescaling of flat space.

    I wonder what this means. Is the Dirac delta invariant with Lorentz transformation since those are transformation from one flat spacetime to another flat spacetime?

    If we parameterize the line integral with [itex]\vec x = \vec x(t)[/itex] to get,
    [tex]\int_{{{\vec x}_0}}^{\vec x} {\sqrt {{g_{ij}}(\vec x)d{x^i}d{x^j}} } = \int_{{t_0}}^t {\sqrt {{g_{ij}}(\vec x(t'))\frac{{d{x^i}}}{{dt'}}\frac{{d{x^j}}}{{dt'}}} } \,\,\,dt'[/tex]
    Can we show that the time variable must be orthogonal to the flat space dimensions? Can we prove the necessity of Minkowski metric for an invariant integral of a parameterized Dirac delta function?
     
    Last edited: Oct 1, 2012
  11. Oct 2, 2012 #10
    Notice, for example, that in
    [tex]\int_{{t_0}}^t {\sqrt {{g_{ij}}(\vec x(t))\frac{{d{x^i}}}{{dt}}\frac{{d{x^j}}}{{dt}}} } \,\,\,dt[/tex]
    if we make [itex]{g_{ij}}(\vec x) = 0[/itex] whenever i≠j, then we get
    [tex]\int_{{t_0}}^t {\sqrt {{g_i}(\vec x(t)){{\left( {\frac{{d{x^i}}}{{dt}}} \right)}^2}} } \,\,\,dt[/tex]
    Now, in order to make [itex]\sqrt {g(x)} [/itex] come outside the Dirac delta and cancel as before, we need
    [tex]{\left( {\frac{{d{x^i}}}{{dt}}} \right)^2} = 1[/tex]
    Or, we can put this in a different way by saying
    [tex]{(ds)^2} = {(dt)^2} - {(dx)^2} = 0[/tex]
    This looks like the Minkowski metric. But I don't know what it would mean to have ds≠0.
     
  12. Oct 8, 2012 #11
    Now that I think about it, when I write:
    [tex]\int_{ - \infty }^{ + \infty } {\delta (x - {x_0})dx} = \int_{ - \infty }^{ + \infty } {\delta (\int_{{x_0}}^x {dx'} )dx} = \int_{ - \infty }^{ + \infty } {\delta (\int_{{x_0}}^x {\sqrt {g(x')} dx'} )\sqrt {g(x)} dx} = 1[/tex]
    I don't actually know whether the [itex]{g(x)}[/itex] in the volume form of [itex]\int_{ - \infty }^{ + \infty } {\delta (...)\sqrt {g(x)} dx} [/itex] outside the dirac delta is really the same as the [itex]{g(x')}[/itex] in [itex]\delta (\int_{{x_0}}^x {\sqrt {g(x')} dx'} )[/itex] in the line integral inside the dirac delta.

    The volume form [itex]{g(x)}[/itex] outside the delta is
    [tex]\sqrt {g(x)} = \det ({J^T}J)[/tex]
    where
    [tex]J = \left( {\begin{array}{*{20}{c}}
    {\frac{{\partial {u^1}}}{{\partial {x^1}}}}&{\frac{{\partial {u^1}}}{{\partial {x^2}}}}\\
    {\frac{{\partial {u^2}}}{{\partial {x^1}}}}&{\frac{{\partial {u^2}}}{{\partial {x^2}}}}
    \end{array}} \right)[/tex]
    for a change of coordinates from u to x.

    But the [itex]{g(x')}[/itex] in the line integral inside the delta is
    [tex]{g_{ij}}(\vec x)d{x^i}d{x^j} = {g_{11}}d{x^1}d{x^1} + {g_{12}}d{x^1}d{x^2} + {g_{22}}d{x^2}d{x^2} + {g_{21}}d{x^2}d{x^1}[/tex]
    And then when [itex]{g_{12}} = {g_{21}} = 0[/itex] and [itex]{g_{11}} = {g_{22}} = g[/itex], the [itex]{g(x')}[/itex] is just a constant [itex]{g}[/itex].

    Are these two g's actually the same thing... even in curved space? How does that work?
     
    Last edited: Oct 8, 2012
  13. Oct 16, 2012 #12
    Previously, I considered:
    But on second thought, this is probably not useful since I could do the same thing with the dirac delta at the right as I did on the left. I could iterate the process to get another squareroot of g in the denominator making it equal the dirac delta divided by the squarroot of g squared. In fact, I could iterate this an infinite number of times to get a squarroot of g to the power of infinity. This only works if g=1 for flat space, and I have not converted the flat dirac delta to a dirac delta in curved space. I will have to look for something else.
     
  14. Oct 16, 2012 #13

    Mute

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    Perhaps this will help? http://www.fen.bilkent.edu.tr/~ercelebi/mp03.pdf

    It gives the form of the dirac delta in curvilinear coordinates (in 3d). I would guess that to achieve your purposes you would need to generalize to arbitrary dimension and modify the scale factors ##h_i## to account for different metrics.
     
  15. Oct 16, 2012 #14
    Thank you. That's a start. My main goal is to investigate how the properties of the dirac delta function might be used to determine the properties of the background spacetime metric. For example, if the dirac delta describes the density of a mass point, is that enough to determine the properties of the metric? Or, if the dirac delta represents causality, is that enough to determine the Minkowski metric? So I consider equation such as:

    [tex]\int_{ - \infty }^{ + \infty } {\delta (x - {x_0})dx}=\int_{ - \infty }^{ + \infty } {\delta (\int_{{x_0}}^x {dx'} )dx}=\int_{ - \infty }^{ + \infty } {\delta (\int_{{x_0}}^x {\sqrt {g(x')} dx'} )\sqrt {g(x)} dx}=\int_{ - \infty }^{ + \infty } {\delta (\int_{{t_0}}^t {\sqrt {{g_{ij}}(\vec x(t))\frac{{d{x^i}}}{{dt}}\frac{{d{x^j}}}{{dt}}} } \,\,\,dt){\sqrt {g(x)} }dx}=1[/tex]
    Or, if I also apply the first variation to all this and set it equal to zero, is there enough information to specify the metric of Special Relativity or General Relativity? Wouldn't it be interesting if it did?
     
  16. Oct 16, 2012 #15

    pwsnafu

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    My gut reaction would be yes. On ℝ at least, metrics define Borel measures and every Borel measure is a Schwartz distribution. IIRC, Schwartz distributions can be broken up into sums of other singular distributions related to Dirac.

    Honestly, this sounds like the domain of current theory, so ask mathematician who specialise in that?
     
  17. Oct 17, 2012 #16
    God! I hope it's not that complicated.
     
  18. Oct 18, 2012 #17
    Hopefully, it is easier than this. For example, the dirac delta function is a continuous version of the Kronecker delta which is used as the metric in flat space. But the spacetime metric of relativity is also flat spacetime metric. Is it also discrete version of a delta function?
     
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