- #1
ExtravagantDreams
- 82
- 5
This should be very simple, but I can find a simple example that violates my general conclusion. I clearly must be doing something wrong with my integration by parts. Any suggestions would be great.
Define a distribution such that the density;
[tex]\eta(\vec{x})=\int d\vec{k} f(\vec{x},\vec{k})[/tex]
which is a function ONLY of x.
Now, let me arbitrarily calculate
[tex]\int d\vec{k} k_{x} f(\vec{x},\vec{k}) = \int dk_{x} k_{x} \int dk_{y} \int dk_{z} f(\vec{x},\vec{k})[/tex]
I should be able to integrate over kx by parts;
[tex] k_{x} \int dk_{x} \int dk_{y} \int dk_{z} f(\vec{x},\vec{k}) - \int dk_{x} \frac{d k_{x}}{d k_{x}} \int dk_{x} \int dk_{y} \int dk_{z} f(\vec{x},\vec{k}) = k_{x} \int d\vec{k} f(\vec{x},\vec{k}) - \int dk_{x} \int d\vec{k} f(\vec{x},\vec{k})[/tex]
Using the stipulation at the beginning this is simply;
[tex]k_{x} \eta(\vec{x}) - \int dk_{x} \eta(\vec{x})[/tex]
Since, the density, eta, is only a function of x, this is zero.
There must be something wrong with this because I can come up with a simple example that contradicts this;
Let
[tex]f(\vec{x},\vec{k})=\eta(\vec{x})\delta (\vec{k}-\vec{k'})[/tex]
It is easy to see that
[tex]\eta(\vec{x})=\int d\vec{k} f(\vec{x},\vec{k})[/tex]
Now calculating the original quantity
[tex]\int d\vec{k} k_{x} f(\vec{x},\vec{k}) = \int dk_{x} k_{x} \eta(\vec{x})\delta (k_{x}-k'_{x}) = \eta(\vec{x}) k'_{x}[/tex]
What gives? Where have I gone wrong?
Define a distribution such that the density;
[tex]\eta(\vec{x})=\int d\vec{k} f(\vec{x},\vec{k})[/tex]
which is a function ONLY of x.
Now, let me arbitrarily calculate
[tex]\int d\vec{k} k_{x} f(\vec{x},\vec{k}) = \int dk_{x} k_{x} \int dk_{y} \int dk_{z} f(\vec{x},\vec{k})[/tex]
I should be able to integrate over kx by parts;
[tex] k_{x} \int dk_{x} \int dk_{y} \int dk_{z} f(\vec{x},\vec{k}) - \int dk_{x} \frac{d k_{x}}{d k_{x}} \int dk_{x} \int dk_{y} \int dk_{z} f(\vec{x},\vec{k}) = k_{x} \int d\vec{k} f(\vec{x},\vec{k}) - \int dk_{x} \int d\vec{k} f(\vec{x},\vec{k})[/tex]
Using the stipulation at the beginning this is simply;
[tex]k_{x} \eta(\vec{x}) - \int dk_{x} \eta(\vec{x})[/tex]
Since, the density, eta, is only a function of x, this is zero.
There must be something wrong with this because I can come up with a simple example that contradicts this;
Let
[tex]f(\vec{x},\vec{k})=\eta(\vec{x})\delta (\vec{k}-\vec{k'})[/tex]
It is easy to see that
[tex]\eta(\vec{x})=\int d\vec{k} f(\vec{x},\vec{k})[/tex]
Now calculating the original quantity
[tex]\int d\vec{k} k_{x} f(\vec{x},\vec{k}) = \int dk_{x} k_{x} \eta(\vec{x})\delta (k_{x}-k'_{x}) = \eta(\vec{x}) k'_{x}[/tex]
What gives? Where have I gone wrong?