# Delta function in continuation back to Minkowski space

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• jim burns
In summary, the correct scaling for δ(-ix) is δ(-ix)=δ(x), and this can be determined by using the definition and properties of the Dirac delta function.
jim burns
The Green's function for a scalar field in Euclidean space is

$$(2\pi)^4\delta^4(p+k) \frac{1}{p^2+m^2}$$

however when I continue to Minkowski space via GMin(pMin)=GE(-i(pMin)) there's seems to be a sign error:

$$(2\pi)^4\delta^4(-i (p+k)) \frac{1}{-p^2+m^2}=(2\pi)^4\delta^4(p+k) \frac{i}{-p^2+m^2}=-(2\pi)^4\delta^4(p+k) \frac{i}{p^2-m^2}$$

where I used δ(-ix)=(1/(-i))δ(x).

The error seems to be that the scaling of the delta function should instead be δ(-ix)=δ(ix)=(1/(i))δ(x).

But how do we know this? For real 'a' it can be argued δ(ax)=(1/|a|)δ(x) on the grounds that δ is postive, but δ(-ix) is not positive as it's not even a real number.

I would like to offer some insights and clarifications on the sign error that you have encountered.

Firstly, let's look at the definition of the Dirac delta function. The Dirac delta function is a generalized function that is defined as:

$$\delta(x)=\begin{cases} \infty, & x=0 \\ 0, & x\neq0 \end{cases}$$

with the property that

$$\int_{-\infty}^\infty f(x)\delta(x)dx=f(0)$$

for any well-behaved function f(x).

Now, when we have a function of the form δ(ax), where a is a real number, we can use the substitution u=ax to rewrite it as:

$$\delta(ax)=\frac{1}{|a|}\delta(u)$$

This is because when u=0, x=0 and when u≠0, x≠0. Therefore, we can use the substitution to get the correct scaling for the delta function.

However, when we have a function of the form δ(ix), where i is the imaginary unit, we cannot use the same substitution as above. This is because when u=ix, x is a complex number and the substitution does not hold. In this case, we need to use the definition of the delta function and its properties to determine the correct scaling.

Now, let's look at the specific case of δ(-ix). Using the definition of the delta function, we can see that:

$$\int_{-\infty}^\infty f(x)\delta(-ix)dx=f(0)$$

Using the substitution u=-ix, we get:

$$\int_{-\infty}^\infty f(x)\delta(-ix)dx=\int_{-\infty}^\infty f(-iu)\delta(u)du=f(0)$$

Therefore, we can see that the correct scaling for δ(-ix) should be δ(-ix)=δ(x).

In conclusion, the sign error that you encountered is due to the incorrect use of the substitution for the delta function. As scientists, it is important for us to carefully consider the definitions and properties of mathematical functions to ensure the correctness of our calculations.

## What is the Delta function in continuation back to Minkowski space?

The Delta function, also known as the Dirac delta function, is a mathematical function that is used to represent a point mass or an impulse at a specific location. In continuation back to Minkowski space, the Delta function is used to represent the energy density of a system in space-time.

## How is the Delta function related to Minkowski space?

The Delta function is related to Minkowski space through the concept of energy-momentum tensors. In special relativity, Minkowski space is a four-dimensional space-time that combines three dimensions of space and one dimension of time. The Delta function is used to represent the energy density of a system in this space-time.

## Why is the Delta function important in physics?

The Delta function is important in physics because it allows us to describe point-like objects and impulses in a continuous space-time. It is used in many areas of physics, such as quantum mechanics and general relativity, to simplify calculations and represent physical phenomena that cannot be described by regular functions.

## What are the properties of the Delta function?

The Delta function has several important properties, including being infinitely tall and narrow at its peak, having an integral of 1 over its entire domain, and being zero everywhere except at its peak. It also follows the sifting property, meaning that it "filters out" all other values except the one at its peak.

## How is the Delta function used in practical applications?

The Delta function is used in many practical applications in physics, engineering, and mathematics. It is used to represent point sources, such as electric charges or masses, in physical systems. It is also used in signal processing to model impulses in electrical circuits and in probability theory to represent probability distributions with sharp peaks.

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