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friend

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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 x

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= x

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 x

[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 x

_{z}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= x

_{0}so that x_{z}= x_{0}.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 x

_{0}, but only approaches it, so that we use some other property of the Dirac delta.
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