Why is the phase velocity used in the definition of the refractive index? - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-09-23T18:26:41Z https://physics.stackexchange.com/feeds/question/115152 https://creativecommons.org/licenses/by-sa/4.0/rdf https://physics.stackexchange.com/q/115152 3 Why is the phase velocity used in the definition of the refractive index? Paracosmiste https://physics.stackexchange.com/users/25575 2014-05-30T21:42:53Z 2014-06-08T15:38:46Z <p>I'm aware of the so-called <a href="http://en.wikipedia.org/wiki/Refractive_index#Group_index" rel="nofollow">group index</a> but why is the phase velocity used in the <a href="http://en.wikipedia.org/wiki/Refractive_index#Definition" rel="nofollow">standard definition</a> of the index of refraction? What advantage does this offer?</p> https://physics.stackexchange.com/questions/115152/-/116801#116801 2 Answer by Steve Byrnes for Why is the phase velocity used in the definition of the refractive index? Steve Byrnes https://physics.stackexchange.com/users/3811 2014-06-04T13:24:37Z 2014-06-04T13:24:37Z <p><strong>Refraction</strong> is the phenomenon where light bends when it enters a medium, as described by <a href="https://en.wikipedia.org/wiki/Snell%27s_law" rel="nofollow">Snell's law</a>. The amount of bending depends on the phase velocity. I don't know the real linguistic history, but I bet that is why somebody came up with the term "index of refraction" or "refractive index" to refer to "c / phase velocity".</p> <p>By the way, it is misleading to say that "c / phase velocity" is the "standard definition" of "index of refraction". It is actually the <strong>only</strong> definition of "index of refraction". If someone uses the specific phrase "index of refraction" to refer to anything else, then they are using the phrase incorrectly.</p> <p>Next, the term "group index" is a combination of the word "group" (suggesting the relation with group velocity) and the word "index" (suggesting its analogous mathematical form to refractive index).</p> <p>Rarely, some people use the terms "group index of refraction" or "group refractive index" as a synonym of "group index". Again, I don't know the linguistic history, but I bet it's a <a href="https://en.wikipedia.org/wiki/Back-formation" rel="nofollow">back-formation</a> -- I mean, people probably learned the (correct) term "group index" and then presumed (incorrectly) that it was an abbreviated version of "group index of refraction". The term "group index of refraction" is a bit weird if you think about it, because as I said it has nothing to do with the phenomenon of refraction. But it's no big deal. It doesn't create confusion. It is an unambiguous term, because everyone knows that you are not allowed to shorten the phrase "group index of refraction" by dropping the word "group".</p> https://physics.stackexchange.com/questions/115152/-/116867#116867 2 Answer by John Rennie for Why is the phase velocity used in the definition of the refractive index? John Rennie https://physics.stackexchange.com/users/1325 2014-06-04T18:16:04Z 2014-06-04T18:16:04Z <p>Refraction can be described by the <a href="http://en.wikipedia.org/wiki/Huygens%E2%80%93Fresnel_principle" rel="nofollow">Huygen's-Fresnel principle</a> just like diffraction. In this interpretation it is an interference phenomenon with every point on the wave front acting as a source, and the light propagating only in those directions where the phases reinforce.</p> <p>And this is the key point. The phase change in a distance $d$ is given by:</p> <p>$$\Delta\phi = 2\pi \frac{df}{v_p}$$</p> <p>where $f$ is the frequency of the light (which is a constant) and $v_p$ is the phase velocity. Since the phase velocity determines the phase change, it also determines how the light refracts. That's why it's the phase velocity that determines the refractive index.</p> https://physics.stackexchange.com/questions/115152/-/117409#117409 3 Answer by paisanco for Why is the phase velocity used in the definition of the refractive index? paisanco https://physics.stackexchange.com/users/49105 2014-06-08T15:38:46Z 2014-06-08T15:38:46Z <p>Well at the most fundamental level, the index of refraction of a material is defined as </p> <p>$n = \sqrt {\epsilon \mu}$</p> <p>where $\epsilon$ is the electric permittivity and $\mu$ is the magnetic permeability of the material. </p> <p>This arises from the solution of Maxwell's equations in a medium.</p> <p>Also arising from the wave equation, which can be derived from Maxwell's equations, is that the index of refraction is the speed of light in vacuum, $c$ divided by the speed of light in the material $c_m$.</p> <p>$n ={ c \over c_m}$</p> <p>Worth noting is that since $c_m$ is always less than $c$ , the index of refraction is always greater than 1.</p> <p>Now the phase velocity for an electromagnetic wave of angular frequency $\omega$ is given by</p> <p>$v_p = {\omega \over k}$ where $k= {2\pi \over\lambda_m}$ is the magnitude of the wave vector and $\lambda_m$ is the wavelength in the medium.</p> <p>So after a little algebra, we find that the wave vector and index of refraction are related by</p> <p>$k ={ n\omega \over c}$</p> <p>Where does all this come into play in refraction and Snell's Law? Well, it is the wave vector that comes into play in satisfying the boundary conditions on the electric (and magnetic) fields at the interface between two media.</p> <p>To see this, let's look at the simple case of a plane wave of monochromatic light incident on the interface between two media with indices of refraction $n_1$ and $n_2$:</p> <p><img src="https://i.stack.imgur.com/mVARk.png" alt="enter image description here"></p> <p>In this simple case, considering the boundary conditions on the electric field at the interface is sufficient to derive Snell's Law.</p> <p>The boundary condition is given by equation (1) in the diagram, namely that the incident and reflected electric fields minus the transmitted electric field must be zero, or equivalently that the total electric field at the interface must be continuous.</p> <p>Since we defined y=0 as the plane of the interface, this boundary condition must hold for any value of x. This leads after a little algebra to equation (2) in the diagram. This equation depends only on the angles of incidence $\phi_{inc}$ and refraction $\phi_{tr}$, the wave vector magnitudes in both media $k_1$ and $k_2$, and the transmission and reflection coefficients $T$ and $R$ ( the fraction of energy that is transmitted into the new medium and reflected into the old medium respectively). By symmetry, $\phi_{inc} = \phi_{refl}$.</p> <p>Because the left side of Equation 2 is independent of angle, so must the right side be. This leads to the term in the exponential being zero, which leads directly to Snell's Law, using the relation between wave vector and index of refraction shown above.</p> <p>The group velocity never comes into play in the boundary conditions of refraction. It does come into play in the propagation of energy in the media (as opposed to the fields), but that's another question.</p>