Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I once came across the strange, artificial unit "hertz per dioptre", which is dimensionally equivalent to "metres per second". Could this unit, by some stretch of the imagination, be used in some artificial situation where the usage implications of both "hertz" and "dioptre" (frequency of periodic events and refractive power, respectively) would make for the ratio to actually be useful?

share|improve this question
    
For example, a lens causing a crystal to vibrate by refracting light on it (although I don't think light does that). –  Joe Z. Jan 29 '13 at 17:42
    
Units of inverse chromatic aberration? –  dmckee Jan 29 '13 at 17:52
1  
@JoeZeng Light does carry momentum. A crystal could vibrate when exposed to light: on a microscopic scale, this is exactly what occurs when light is travels through materials. The light imparts some of its energy to exciting the crystal atoms as it passes through a medium, which causes the atoms to also emit their own electromagnetic radiation (resulting from relaxing from an excited state to their ground states). It is the sum of the original light wave and the emitted photons that accounts for refraction. –  user44430 Jan 29 '13 at 18:26
1  
I will have my car's speedometer altered to read in units of Hz/D. –  DarenW Jan 30 '13 at 4:04
add comment

2 Answers

up vote 5 down vote accepted

In special relativity, hertz per dioptre is an excellent unit for showing the joint invariance of electromagnetic phenomena in the behavior of all types of lenses, reflective or refractive, under the effects of the Lorentz transformation along the axis of motion. I'm not aware of any other unit that links those two domains in quite that way. In the case of refractive lenses with chromatic dispersion, the invariance turns out to be non-trivial and a bit surprising, since it asserts that the atomic materials in a Lorentz compressed lens must maintain a very specific relationship in how they interact with a spectrum of gamma-shifted light frequencies.

Here's how it works for the easier reflective-lens case. First, imagine a sphere 4 meters across with an $f=280$ THz resonant infrared light wave inside. Why 4 meters? Well, I'm trying to use the correct definition of dioptre. That is the focal length of a refractive or reflective lens, which means the distance it requires to converge parallel light down to a single focal point. In this case, the lens is reflective and has spherical curvature. Looking only at a region small enough (e.g. 2 cm across) to avoid spherical aberration, the focal length of the $d=4$ m sphere is $L=\frac{1}{2}r=\frac{1}{4}d=\frac{1}{4}4=1$ m. So, a 4 m diameter sphere thus correctly gives a dioptre (curvature) of $\delta=1/L=1/1=1$ D, where D $=m^{-1}$.

Next, accelerate the sphere along it X axis to a velocity of $v=\sqrt{\frac{3}{4}}$ c, which gives a Lorentz factor of $\gamma=2$. That means that both the sphere and the resonant light pattern within it will be compressed to $\frac{1}{2}$ their original lengths along the X axis, from the perspective of a viewer "at rest" relative to the moving sphere.

For the small reflective lens regions around either end of where the X axis crosses the sphere, the pre-acceleration curvature was $\delta_0=1D$ (the zero subscript indicates the rest frame). After acceleration to $\gamma=2$ the sphere becomes an oblate spheroid, and the curvatures of the two reflective lens areas have been reduced to $\delta_1=2D$, where higher dioptre numbers indicate flatter curves. (The proof of that is left as an exercise for the reader, but it's not difficult.)

Now let's examine what happens to the frequency of the light within the sphere. The neat thing about special relativity is that physics must remain invariant for both the observer and the observed system. So, if there were n wavelengths of resonant light crossing the sphere along the X axis prior to it being accelerated, there must also be n wavelengths along that same length after the compression. In other words, the wavelengths of the radiation must also be cut in half along X (only), resulting in twice the frequency as before. That transforms the original X-axis $f_0=280$ THz light of the at-rest sphere into $f_1=560$ THz light in the moving sphere. An observer in the rest frame would see this as bright green.

Observant readers may now be saying "Hey, that can't be right! The Lorentz factor also slows time... so shouldn't the light in the moving sphere be slower and thus less energetic?"

While it is true that time will pass more slowly within the moving sphere, it is not correct to think that this same light will be slower when viewed from the rest frame. For that situation the geometry of the wavelengths wins, and the light looks green. However, a simpler way to think of it is that since the light is being emitted and reflected by an object traveling at $\gamma=2$ (or equivalently $v=\sqrt{\frac{3}{4}}$ c), the ordinary Doppler effect will double its frequency.

(@ColinK has correctly noted that the above explanation glosses over some important complications. Please see his excellent comment for more info. I may try to address that soon.)

Now it's time to put this all together.

The original light and sphere had an eta factor of:

$\eta_0=f_0/\delta_0 = (280 THz)/(1 D) = 280\times{10}^{12}$ HpD

where 1 HpD = 1 Hz/D (Hertz per dioptre).

The moving light and sphere has an eta factor of:

$\eta_1=f_1/\delta_1 = (560 THz)/(2 D) = 280\times{10}^{12}$ HpD.

In other words, the eta factor $\eta$, which relates the Lorentz-transformed electromagnetic waves to the Lorentz-contracted physical mirrors from which they reflect, has remained invariant for this example of $\gamma=2$.

It is not an isolated case. It is easy to show that $\eta$ is a universal invariant of special relativity:

$\forall{v_i}(\eta_i=\frac{f_i}{\delta_i}= C)$

where C is a constant in units of HpD = Hz/D = Hertz per dioptre.

Now the remarkable generalization of all of this is that by the same kinds of geometric arguments and application of the "physics must be preserved in both frames" principle, refractive lenses must also fall under the above argument. If a refractive lens has chromatic dispersion (the colored fringes seen in cheap lenses), then the constant C in the above equation will become a frequency-dependent value $C(f)$. Yet the eta invariance remains intact! That is surprising because light dispersion is a pretty complicated phenomenon, yet from the rest frame these messy compressed atoms must nonetheless maintain eta invariance. That is... unexpected.

Thus HpD units not only have real physical meaning, but a meaning that relates directly to the original intent of both the Hertz and dioptre units (versus just being $m/s$ in disguise). This meaning in turn provides an easy way to express an invariant relationship in special relativity that links together the electromagnetic and mechanical Lorentz transformations in an unexpected and non-intuitive fashion.

And finally, despite all the above unexpectedly interesting (to me at least!) SR relationships involved, the HpD unit really did originate as a bit of humor in (as best I could uncover) this xkcd discussion posting back in 2007. So, shrodingersduck from the People's Democratic Republic of Leodensia, wherever you are six years later, I thank you for inadvertently creating an interesting and quite fun opportunity to explore special relativity in a rather unusual context.

Addendum 2013-01-31

The generality of the HpD unit in special relativity can I think be stated even more broadly. So, here goes:

Light frequency, geometric forms, and frequency-dependent refractive indices all change when systems undergo Lorentz transformation, so they are not individually Lorentz invariant. Theorem: If the optical characteristics of an optical system are instead described using HpD (Hertz per Dioptre) and/or its inverse unit DpH (Dioptres per Hertz), the resulting description of its optical properties will remain constant ("eta invariance") regardless of relativistic frame or orientation from which the optical system is analyzed.

That is a theorem only. @ColinK's excellent observation that the Doppler argument I made could be bogus because the shift works differently depending on whether the light is moving with or against the velocity still concerns me. So, I want to look at that a lot more closely and see if I can disprove my own theorem.

Still, wouldn't it be delightful if a unit defined as a joke turned out to be relativistically invariant when the common units for the same phenomena are not?

The other obvious generalization question is this: Does eta invariance (if it exists) apply to other wave phenomena?

And finally, @JoeZeng, I think I misunderstood your question about whether the eta factors (descriptions of optical components using HpD units) are related to the velocity of light. Well, HpD does have dimensional equivalence to a velocity ($m/s$), but if there is a meaningful way to re-interpret an HpD value as a velocity, I sure don't see it. Intriguing question, though...

share|improve this answer
    
Wow. Just... wow. –  Joe Z. Jan 31 '13 at 0:15
1  
Your quite welcome! –  Terry Bollinger Jan 31 '13 at 0:53
    
Now the question is, can you relate this unit to speed? –  Joe Z. Jan 31 '13 at 1:00
1  
Hmm. You can add velocity simply by putting a gamma function $\gamma(v)$ in front of each of the $f_0$ and $\delta_0$ terms. However, if you state it that way the gammas just cancel out. That's why I put expressed it in terms of universal qualifiers from set theory instead, to emphasize that it's an experimental assertion: If you do such experiments, then for any velocity you should see the same constant C that you see at rest. –  Terry Bollinger Jan 31 '13 at 1:53
1  
Heh! All standard SR ideas, but I've of course used some non-standard terminology. This is not from a book; I found your problem challenging and realized there really was an interesting use of it for special relativity (unexpected!). The eta factor $\eta$ is my own notation, and is not related to the speed of light in any direct fashion. It describes the construction of the sphere and the light used in it, and can be set to any value. E.g., $\eta$ increases if you use higher frequency light and/or a smaller sphere, and decreases if you use lower frequency light and/or a larger sphere. –  Terry Bollinger Jan 31 '13 at 2:06
show 7 more comments

It's a joke - to teach people about dimensional analysis

It might make a certain amount of sense to replace the second with the hertz (1/s) since there are a lot more physics equations with a quantity divided by seconds than multiplied - simply because you are often studying the rate of something.

If you follow this logic to its conclusion then you could also use dioptre (1/m) and have speed as hertz/dioptre instead of meters/second.

share|improve this answer
    
What I mean is, is there a unit involving frequency and refractive power that is both dimensionally equivalent to speed and useful in some contrived context? –  Joe Z. Jan 30 '13 at 3:41
    
@JoEZeng not in physics, we don't even use dioptres. –  Martin Beckett Jan 30 '13 at 3:45
    
@martin: We certainly use diopters, we just don't often call them by name. The diopter is the unit of curvature. It's common in optics, and anywhere else you talk about a curved surface. –  Colin K Jan 30 '13 at 3:57
    
@ColinKother than in spectacles/prescriptions does anyone use anything other than meters? Diopters were convenient when you need log tables to find a reciprocal but Zemax manages to do that for you. –  Martin Beckett Jan 30 '13 at 4:01
    
@MartinBeckett: Certainly! Any time somebody describes the curvature of a surface by the reciprocal or its radius, they are using diopters. I believe this is common in many areas of physics and math, but maybe I'm wrong. Regardless, I know first hand that it is very common in optics, both among optical scientists and engineers. –  Colin K Jan 31 '13 at 2:23
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.