# Virtual length contraction

1. Mar 8, 2014

### ANvH

What should it be, ${sin(ωt-\frac{k}{n}x)}$ or ${sin(ωt-nkx)}$? I am contemplating this with respect to the proper waveform in a medium, following specific rules, particularly when it comes to the dispersion relation and ultimately what this would mean in the context of the Lorentz transformations.

A single wave is described by the expression ${sin(ωt-kx)}$, where ${ω}$ is the angular frequency and ${k}$ is the wavenumber. In transparent media the refractive index should be included into the waveform. Also the speed of the wave is given by the ratio of whatever multiplies ${t}$ to whatever multiplies ${x}$. Taken this as a rule of thumb, the waveform becomes
${sin(ωt-\frac{k}{n}x)}$ (1)​
where ${n}$ can be any real value. Applying the rule of thumb, the propagation velocity, ${u}$, is
${u=\frac{ωn}{k}}=c$ (2)​
In textbooks however, the waveform with a refractive index is given by ${sin(ωt-nkx)}$ and applying the rule of thumb leads to ${\acute{u}=\frac{ω}{nk}}$, which is different from Eq. (2). This form is based on the optical path length (OPL), which is defined as the product of the geometric length of the path that light follows through the medium and the index of refraction, in other words the physical length of the path is multiplied by the refractive index. To make sure ${\acute{u}=\frac{ω}{nk}}$, ${\frac{ω}{k}}$ is equated with ${c}$ and it leads to ${\acute{u}=\frac{c}{n}}$, conforming to Eq. (2). However, Eq. (1) suggests a virtual path length through the medium, as opposed to OPL. The virtual path length can be defined by a transformation where ${\acute{t}=t}$ and ${\acute{x}=\frac{x}{n}}$. Accordingly,
${Δ\acute{x}=\frac{Δx}{n}}$ (3),​
which represents a length contraction. It is the effect of reduced depth conception when observing objects in an aquarium filled with water. The objects look closer (and also larger in size) than they physically are, and vice versa, observing objects in air from water, they will look further away, as can be experienced and deduced by rearranging Eq. (3). Eq. (3) is useful to locate the position of virtual objects. It also suggests that length contraction is an optical illusion.

regards,
Alfred.

2. Mar 8, 2014

### WannabeNewton

No it doesn't, not even in the slightest.

3. Mar 8, 2014

### ghwellsjr

You can't see length contraction. It's different in each Inertial Reference Frame. So how could you see it? It's not optical. So how could it be an optical illusion?

4. Mar 8, 2014

### pervect

Staff Emeritus
Refraction and the speed of light in a media is only loosely related to special relativity. When we write "c" in relativity, it's always the speed of light in a vacuum.

I don't do a lot of optics, but from what I've seen it's always $sin(\omega t - k x)$ with the further provision that $\eta = \frac{c}{v}$, and $v = \frac{\omega}{k}$, v being the so-called phase velocity.

One could solve for k as a function of the other variables, and get $k = \frac{\omega \eta}{c}$. I don't see how to get either of the expressions you write.

Beyond this, I'm not sure what your question was, realtivistic length contraction isn't really related to the optical phenomenon you're apparently trying to connect it to.

5. Mar 9, 2014

### ANvH

For any wave parameterized as $(\omega t - k x)$, traveling in a moving frame of reference with refractive index $n$, the Lorentz transformations
$\acute{t}=\gamma(t-\frac{\beta}{c}x)$​
$\acute{x}=\gamma(x-vt)$​
together with the dispersion relation $\omega n=kc$, leads to
$\omega\acute{t}-k\acute{x}=\gamma[(1+n\beta)\omega t-(1+\frac{\beta}{n})kx]$​
Using the rule of thumb, the phase velocity, $u$, becomes
$u=\frac{(1+n\beta)\omega}{(1+\frac{\beta}{n})k}=\frac{c}{n}\frac{1+ \beta n}{1+\beta n^{-1}}$​
When $\beta$ is small then to first order $u=\frac{c}{n}+v(1-n^{-2})$ where $\beta=\frac{v}{c}$. When $\beta=1$,
$u=c \frac{1+n}{n+1}$​
In both cases, the increase in the propagation velocity should lower the index of refraction, given its definition $n=\frac{c}{u}$. The change in refractive index suggests an optical effect and therefore it is possible that an apparent length contraction, associated with the Lorentz transformations may be an optical effect as well. I understand that such a statement may have a number of ramifications.

First of all, when I write $c$, it is indeed the speed of light in vacuum. I agree with you that the waveform is usually expressed in terms of $\omega t - k x$. In optics $n$ is associated with the wavenumber, because it is said that the frequency of the wave does not change when light enters a transparent medium. By including $n$ in the waveform you would "imprint" the speed of light mathematically. The rule of thumb, I mentioned earlier, does not violate this "imprint"; perhaps it is a matter of semantics, however, your reservations are well taken and could lead to incorrect outcomes.

Second, the response to ghwellsjr is also a response to you, to clarify the connection between relativistic contraction and an optical phenomenon; I apologize for using the word "illusion", perhaps "deception" would have been better.

Last edited: Mar 9, 2014
6. Mar 9, 2014

### WannabeNewton

You're still wrong. Nothing you've done here has anything to do with length contraction. I suggest reading a section on EM waves in (dielectric and/or conducting) media from a standard textbook e.g. Heald and Marion chapters 5-7.

7. Mar 9, 2014

### ANvH

I do not think that this is helpful. The Lorentz transformations and their inverse transformations have everything to do with length contraction and time dilation. The derivation of the speed of light in moving water shows what Lorentz him self could accomplish with the transformations. I quote from "Einstein’s Investigations of Galilean Covariant Electrodynamics prior to 1905", John D. Norton,
Department of History and Philosophy of Science University of Pittsburgh jdnorton@pitt.edu:
http://philsci-archive.pitt.edu/1743/2/Norton.pdf

All I am doing here is what the Lorentz transformations could also mean, i.e., an apparent change of the refractive index. I am not saying that I am correct, but it does strike me, and does me wonder what such a change in refractive index would mean. What would the objects in an aquarium look like when it reaches relativistic speeds?

I hope this clarifies my view point, they say "seeing is believing", but I am not so sure about that ;-)