Undergrad Is photon energy negative in a negative-index metamaterial?

[jeast]

Negative-index metamaterials are engineered to have a negative relative electric permittivity $\epsilon_r$ and negative relative magnetic permeability $\mu_r$ so that the index of refraction $n$ is negative:
$$n=-\sqrt{\epsilon_r\mu_r}.$$
The dispersion relation for photons travelling in a medium with refractive index $n$ is:
$$\omega=\frac{c}{n}k.$$
The photon energy E is given by
$$E=\hbar \omega=\frac{\hbar c k}{n}.$$
If the refractive index $n$ is negative then is the photon energy $E$ negative?

You can see from this simulation of an EM plane wave entering a negative-refractive index material that the phase velocity becomes negative.

 

[Cthugha]

If the refractive index $n$ is negative then is the photon energy $E$ negative?

No, when you do the math from scratch using negative permeability, you find that the direction of the flow of energy associated with the Poynting vector is directly opposite to the direction of wave propagation (which is associated with k). Therefore, using the definitions and coordinate systems applied here, k will become negative, which in turn results in positive energy.

 

[jeast]

No, when you do the math from scratch using negative permeability, you find that the direction of the flow of energy associated with the Poynting vector is directly opposite to the direction of wave propagation (which is associated with k). Therefore, using the definitions and coordinate systems applied here, k will become negative, which in turn results in positive energy.

But the energy density $\rho$ of the EM wave inside the negative index metamaterial is
$$\rho=\epsilon_r\epsilon_0E^2$$
where $E$ is the magnitude of the electric field.

If $\epsilon_r$ is negative then the energy density $\rho$ is negative.

 

[PeterDonis]

But the energy density $\rho$ of the EM wave inside the negative index metamaterial is
$$\rho=\epsilon_r\epsilon_0E^2$$
where $E$ is the magnitude of the electric field.

Where does this formula come from?

 

[jeast]

Where does this formula come from?

The energy density of an electromagnetic field inside a medium with relative permittivity $\epsilon_r$ and relative permeability $\mu_r$:
$$\rho=\frac{\epsilon_r\epsilon_0|\mathbf{E}|^2}{2}+\frac{|\mathbf{B}|^2}{2\mu_r\mu_0}.$$
For an EM wave in a medium with refractive index $n$ we have:
$$
\begin{eqnarray*}
|\mathbf{E}|&=&\frac{c}{n}|\mathbf{B}|,\\
n&=&\sqrt{\epsilon_r\mu_r},\\
c&=&\frac{1}{\sqrt{\epsilon_0\mu_0}}.
\end{eqnarray*}
$$
By combining the above expressions we find:
$$\rho=\epsilon_r\epsilon_0|\mathbf{E}|^2.$$
If $\epsilon_r$ is negative in the metamaterial then the EM energy density $\rho$ is negative.

 

[PeterDonis]

The energy density of an electromagnetic field

I didn't ask you to explain what your formula means, I asked you where you got it from. That means you need to give a reference.

 

[jeast]

I didn't ask you to explain what your formula means, I asked you where you got it from. That means you need to give a reference.

The electrostatic energy density inside a dielectric medium:
https://farside.ph.utexas.edu/teaching/em/lectures/node72.html
Eqn. 855

The magnetostatic energy density inside a linear magnetic material:
https://farside.ph.utexas.edu/teaching/em/lectures/node78.html
Eqn. 894

 

[PeterDonis]

The electrostatic energy density inside a dielectric medium:
https://farside.ph.utexas.edu/teaching/em/lectures/node72.html
Eqn. 855

The magnetostatic energy density inside a linear magnetic material:
https://farside.ph.utexas.edu/teaching/em/lectures/node78.html
Eqn. 894

As the pages you reference note, these are "textbook" formulas for normal linear dielectric media. What justifies applying these formulas to the very different kind of material you are talking about?

Also, as even you explicitly say in your own post, quoted above, these are formulas for electrostatic and magnetostatic energy density. But an EM wave is neither electrostatic nor magnetostatic. What justifies applying these static formulas to an EM wave?

 

[Demystifier]

But the energy density $\rho$ of the EM wave inside the negative index metamaterial is
$$\rho=\epsilon_r\epsilon_0E^2$$
where $E$ is the magnitude of the electric field.

If $\epsilon_r$ is negative then the energy density $\rho$ is negative.

The formula above is derived for the case $\epsilon_r>0$, so you cannot blindly apply this formula when $\epsilon_r<0$. Instead, you must start over from first principles, and derive a more general formula which is valid for any sign of $\epsilon_r$. @Cthugha gave a more fundamental reason why energy is positive, which indicates that the formula valid for any sign of $\epsilon_r$ is
$$\rho=|\epsilon_r\epsilon_0|E^2$$

Physics is not just a bunch of formulas. Instead, there is a hierarchy between them, it is important to understand which formula arises from which.

 

[renormalize]

If $\epsilon_r$ is negative then the energy density $\rho$ is negative.

Not true. Here's a link to a paper discussing the EM properties of a homogeneous metamaterial medium consisting of arrays of linear wires and split-ring resonators (SRRs):
Boardman & Marinov-Electromagnetic energy in a dispersive metamaterial
The authors show that, for operating frequencies $\omega$ near the "plasma" frequency $\omega_{p}$ of the wires and the resonant frequency $\omega_{0}$ of the SRRs, the effective relative permittivity and permeablility of the metamaterial (ignoring loss) take the form:$$\varepsilon_{\text{eff}}\left(\omega\right)=1-\frac{\omega_{p}^{2}}{\omega^{2}}\quad,\quad\mu_{\text{eff}}\left(\omega\right)=1+\frac{F\omega^{2}}{\omega_{0}^{2}-\omega^{2}}$$where $F$ is a positive, dimensionless constant that characterizes the geometry of the SRRs. Clearly, there is a range of frequencies where $\varepsilon_{\text{eff}}, \mu_{\text{eff}},$ or both are negative. Nevertheless, for time-harmonic fields $\boldsymbol{E}\left(t\right)=\boldsymbol{E}_{0}e^{-i\omega t},\boldsymbol{H}\left(t\right)=\boldsymbol{H}_{0}e^{-i\omega t}$, they show that the time-averaged EM energy density in the medium is (sensibly) positive semi-definite:$$\left\langle w_{EM}\right\rangle \equiv\left\langle w_{E}\right\rangle +\left\langle w_{H}\right\rangle =\frac{\varepsilon_{0}}{4}\left(1+\frac{\omega_{p}^{2}}{\omega^{2}}\right)\left|\boldsymbol{E}_{0}\right|^{2}+\frac{\mu_{0}}{4}\left(1+F\frac{\omega_{0}^{2}\left(\omega_{0}^{2}+\omega^{2}\right)}{\left(\omega_{0}^{2}-\omega^{2}\right)^{2}}\right)\left|\boldsymbol{H}_{0}\right|^{2}$$

 

[Demystifier]

For pedagogic purposes, and for fun, let me construct another example where one might naively think that energy is negative. The gravitational force on Earth is $F=-mg$, where the negative sign indicates that it acts downwards. Due to buoyancy, however, the net force in a fluid may act upwards, which behaves as if the mass $m$ is negative in the fluid (Archimedes law). Then using the formula for the kinetic energy
$$E=\frac{mv^2}{2}$$
one concludes that kinetic energy is negative in the fluid. The conclusion is of course wrong, but its purpose is to make an analogy, because the error is of the same type as the error in this thread.

 

[PeterDonis]

The OP question has been answered, and this thread is now closed. Thanks to all who participated.