Normalization of a delta function in curved spacetime

In summary, the first statement (\int d^4 x \delta^4(x - x_0) = 1) is the usual definition in curved spacetime, while the second statement (\int d^4 x \sqrt{-g} \delta^4(x - x_0) = 1) is incorrect due to the transformation properties of the delta function and the measure. However, the delta function can be tensorized by dividing by the square root of the metric determinant.
  • #1
jdstokes
523
1
Which of the following are true in curved spacetime?

[itex]\int d^4 x \delta^4(x - x_0) = 1[/itex] (1)

[itex]\int d^4 x \sqrt{-g} \delta^4(x - x_0) = 1[/itex] (2)

I think the first one is incorrect in curved spacetime, or in general when the metric is non-constant. I would argue this by saying that the delta function does not transform, whereas the fourth-order differential transforms in the opposite way to [itex]\sqrt{-g}[/itex], so the whole thing transforms as a scalar as it must.

I've also heard that [itex]\delta^4[/itex] is not a scalar, which suggests that (1) is the correct statement. However, this seems strange to me as I would think that (1) will fail to hold in curvilinear coordinates e.g.
 
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  • #2
actually the right difinition is
[tex]\int_M F(x^{\mu})[\frac{\delta^{(4)}(x^{\sigma}-y^{\sigma})}{\sqrt{-g}}]\sqrt{-g}d^4x=F(y^{\sigma})[/tex]
 
  • #3
The delta function is a scalar density. You don't need curved space or four dimensions to see this, it follows from the identity δ(f(x)) = (1/|f '(x0)|) δ(x-x0) where f(x0) = 0. For example δ(3x) = (1/3) δ(x). Or take plane polar coordinates: δ(x) ≡ δ(x) δ(y) = (1/r) δ(r) δ(Θ).
 
Last edited:
  • #4
jdstokes said:
Which of the following are true in curved spacetime?

[itex]\int d^4 x \delta^4(x - x_0) = 1[/itex] (1)

[itex]\int d^4 x \sqrt{-g} \delta^4(x - x_0) = 1[/itex] (2)

I think the first one is incorrect in curved spacetime, or in general when the metric is non-constant. I would argue this by saying that the delta function does not transform, whereas the fourth-order differential transforms in the opposite way to [itex]\sqrt{-g}[/itex], so the whole thing transforms as a scalar as it must.

I've also heard that [itex]\delta^4[/itex] is not a scalar, which suggests that (1) is the correct statement. However, this seems strange to me as I would think that (1) will fail to hold in curvilinear coordinates e.g.

I'd say that

[itex]\int d^4 x \delta^4(x - x_0) = 1[/itex]

is the usual definition. The RHS is trivially a scalar. The measure on the LHS is a density. So the delta distribution is also a density, as was mentioned by others here.

You can "tensorize" the delta distribution by defining

[tex]
\delta^4(x - x_0) \rightarrow \frac{\delta^4(x - x_0) }{\sqrt{g}}
[/tex]
 

FAQ: Normalization of a delta function in curved spacetime

What is a delta function?

A delta function, also known as a Dirac delta function, is a mathematical function that is defined as zero everywhere except at a single point, where it is infinite. It is commonly used in physics and mathematics to represent a point-like source or to model a very narrow impulse or spike in a system.

How is a delta function normalized?

In curved spacetime, the normalization of a delta function is a bit more complicated than in flat spacetime. It involves taking into account the effects of gravity and the curvature of spacetime. The normalization factor is determined by the metric tensor, which describes the curvature of spacetime, and is often represented as a function of the coordinates of the spacetime.

Why is normalization important in curved spacetime?

In curved spacetime, the concept of distance and volume changes due to the effects of gravity. This means that the simple definition of a delta function as a point-like source in space no longer holds. Normalization takes into account the changes in distance and volume, allowing for a more accurate representation of a delta function in curved spacetime.

Can a delta function be normalized in any curved spacetime?

The normalization of a delta function in curved spacetime depends on the specific metric tensor of the spacetime. Different metrics can result in different normalization factors, so it is not a universal process. However, in many cases, a delta function can still be normalized in curved spacetime, making it a useful tool in various theoretical physics models.

How does normalization of a delta function affect its properties in curved spacetime?

The normalization factor of a delta function in curved spacetime affects its properties, such as its height and width. This means that the delta function will appear differently in curved spacetime compared to flat spacetime. For example, a normalized delta function in curved spacetime may have a different peak value and spread out over a larger area compared to its flat spacetime counterpart.

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