Green function, Delta and Heaviside

[Reingley]

Homework Statement

Show that the explicitly covariant expression:

G_R(x-y) = θ(x⁰-y⁰)δ(($\vec{x}$-$\vec{y}$)²)/2$\pi$

agrees with the retarded Green function:

δ(x⁰-y⁰-|$\vec{x}$-$\vec{y}$|) / (4$\pi$|$\vec{x}$-$\vec{y}$|)

Homework Equations

N/A

The Attempt at a Solution

I know that the derivative of the Heaviside function is the delta function, though I'm completely unsure how to apply that to this problem.
My other instinct was to do some Laplace transform, but that seemed futile as well.

Is there some simple property I am missing that would help get me started?

 

[mark726]

Thank you for your post. It is always exciting to see someone tackling a problem in theoretical physics. I am a scientist and I would be happy to help you with this problem. Let's start by looking at the two expressions that we need to show are equal:

GR(x-y) = θ(x0-y0)δ((\vec{x}-\vec{y})2)/2\pi

and

δ(x0-y0-|\vec{x}-\vec{y}|) / (4\pi|\vec{x}-\vec{y}|)

The first thing to notice is that both expressions involve the delta function, which is defined as the derivative of the Heaviside step function. This means that we can use the properties of the delta function to simplify the expressions. In particular, we can use the property that

δ(ax) = δ(x)/|a|

where a is a constant. This will come in handy when we compare the two expressions.

Let's start with the first expression. We can write it as

GR(x-y) = θ(x0-y0)δ((\vec{x}-\vec{y})2)/2\pi = θ(x0-y0)δ((x-y)2)/2\pi

since the norm of a vector squared is the same as the scalar product of the vector with itself. Now, using the property above, we can write

GR(x-y) = θ(x0-y0)δ((x-y)2)/2\pi = θ(x0-y0)δ(x-y)/2π

We can also write the second expression in a similar form. We can start by using the property

δ(x0-y0-a) = δ(x0-y0)/|a|

where a is again a constant. This means that we can write the second expression as

δ(x0-y0-|\vec{x}-\vec{y}|) / (4\pi|\vec{x}-\vec{y}|) = δ(x0-y0)/(|\vec{x}-\vec{y}|4π)

Now, we can see that both expressions involve the Heaviside step function θ(x0-y0), which is equal to 1 when x0 > y0 and equal to 0 otherwise. This means that we can write both expressions as

GR(x-y) = θ(x0