I got bored and did something with the dielectric constant

[Eclair_de_XII]

Homework Statement

I want to prove or disprove that the dielectric constant $K$ is the projection of $\vec E_0$ onto $\vec E$ using linear algebra.

Homework Equations

$\vec E = \frac{\vec E_0}{K}$

The Attempt at a Solution

$(\vec E)⋅\vec E = (\vec E)⋅\frac{\vec E_0}{K}$
$K(\vec E)⋅\vec E = (\vec E)⋅\vec E_0$
$K \left\| \vec E \right\|^2=\vec E⋅\vec E_0$
$K=\frac{\vec E⋅\vec E_0}{\left\| \vec E \right\|^2}=proj_\vec E \vec E_0$

Since $\left\|\vec E_0\right\| > \left\|\vec E\right\|$, $K>1$. I'm trying to interpret $K$ as the projection of $\vec E_0$ onto $\vec E$ because I was kind of bored one day in physics lecture. I think I remember that $proj_{\vec u} \vec v ≤ |1|$ for some vectors $\vec u$ and $\vec v$, which contradicts what I'm saying. Should I just go with $K=\frac{C}{C_0}$ since it's much simpler? I think I'm reading too much into this...

 

[NFuller]

I think I'm reading too much into this...

I think so too...

$\vec E = \frac{\vec E_0}{K}$

First, what is $E_{0}$ here? I think what you meant to write is $\mathbf{D}=K\epsilon_{0}\mathbf{E}$. Either way the projection of $\mathbf{E}$ and $\mathbf{D}$ or $\mathbf{E_{0}}$ is just
$$\text{proj}\frac{\mathbf{D}\cdot\mathbf{E}}{|\mathbf{E}|^{2}}=\frac{K\epsilon_{0} E^{2}}{E^{2}}=K\epsilon_{0}$$

 

[scottdave]

It looks like you answered your own question, about proving or disproving it.

 

[Eclair_de_XII]

Either way the projection of $\mathbf{E}$ and $\mathbf{D}$ or $\mathbf{E_{0}}$ is just

$\text{proj}\frac{\mathbf{D}\cdot\mathbf{E}}{|\mathbf{E}|^{2}}=\frac{K\epsilon_{0} E^{2}}{E^{2}}=K\epsilon_{0}$

Forgive me if I'm remembering something wrong but aren't projections usually less than or equal to 1 in magnitude? Should it not be: $\text{proj}\frac{\mathbf{E}\cdot\mathbf{D}}{|\mathbf{E}| |\mathbf{D}|}$? Of course, though, it's meaningless anyway, and does not represent the dielectric constant.

 

[NFuller]

aren't projections usually less than or equal to 1 in magnitude?

No, but my notation is messed up. It should be
$$\text{proj}_{\mathbf{E}}\mathbf{D}=\frac{\mathbf{D}\cdot\mathbf{E}}{|\mathbf{E}|^{2}}\mathbf{E}$$