# Is meters per second equivalent to seconds per meter?

I know this question is probably ridiculous, but bear with me for a moment.

This thought emerged while I was converting between nm and wave numbers ($\rm cm^{-1}$). In order to prove this conversion, one must first take the starting value (let us say nano meters) and flip it:

$$\rm Wavenumber = \frac{1}{wavelength}$$

Starting with $5~\rm{ nm/wave}$ (length per "wave" is $5~\rm nm$), flipping it gives $0.2~\rm waves/nm$ (one fifth of a "wave" per $\rm nm$). Then just proceed forward with normal dimensional analysis ($1\times 10^{7}~\rm nm = 1 cm$) and you get a final answer of:

$$0.2×10^{7}~\rm cm^{-1}$$ or

$$\frac{1×10^7}{5}~\rm cm^{-1}$$

This is when I had the realization: Wait, is this a true "conversion"?

Let's take $\rm 1 m = 1000 mm$. This is a true statement of equivalency because what you do to one side you must do to the other. However, in order to "convert" from nm to wave numbers one must first FLIP the $nm$ side (while keeping the wave number side un-flipped) in order to begin the conversion.

But then I thought, maybe my understanding of equivalency and conversion is fundamentally incorrect. And that you are allowed to flip one side and not the other because the information about the rates remains unchanged.

Therefore, is $10~\rm m/s$ equivalent to $0.1~\rm s/m$? My gut says no which means that nm to wave numbers is not a true "conversion."

• Your guts are right. What you call 'conversion' here, is simply the definition of another quantity which has different units (cm^-1 instead of cm). – xi45 Apr 29 '16 at 0:10
• There is a little bit more here than just a conversion of units. If you study Fourier transforms, especially of periodic functions defined on crystal lattices, then you will find that one can represent them in two different ways, in terms of their spatial representation (that's where the fundamental unit is [$1m$]) and then by their Fourier transform, which leads to the concept of the reciprocal lattice, which is defined in terms of wave numbers with units [$1m^{-1}$] . Both representations have physical meaning and the theory of such systems often switches forth and back between the two. – CuriousOne Apr 29 '16 at 1:17
• Yes, as @CuriousOne indicated, in condensed matter, we take the lattice constant $a=\frac{2\pi}{λ}$ where λ is the wavelength – UKH Apr 29 '16 at 3:22
• "Therefore, is 10 m/s equivalent to 0.1 s/m?" Yes. If you move 10 meters per second, then you spend 0.1 seconds on every meter. – Steeven Apr 29 '16 at 6:18
• consider revising the title of this question which is patently untrue – anon01 Jun 29 '16 at 23:02

In your example, a wavelength and a wavenumber represent in a different way the same information as there is a direct conversion between the two quantities. You can use the same reasoning to answer your question about the m/s and s/m:

You have a speed $v$. Let's define a new quantity based on that speed,

$$\rho = \frac{1}{v}$$

which we conveniently call the inverse speed. Using this formula, you see that $\rho$ is in s/m and that 10 m/s = 0.1 s/m.

If you have a physics problem involving speeds, you could replace all the speeds by their corresponding inverse speed in the equations. And solving the problem with the inverse speeds should lead you to the same result as solving it with the speeds.

The answer to your main question is that you can define an equivalency between m/s and s/m.

But m/s and s/m are not always equivalent, as well as wavenumbers and wavelengths. Besides the considerations on dimensions, if you have a quantity $v_1$ in m/s, and a quantity $\rho_2$ in s/m, you should ask yourself if you are allowed to write

$$\rho_2 = \frac{1}{v_1}$$

Although the dimensional analysis is right, the formula might be wrong. If $v_1$ is the speed I went during my morning jogging and $\rho_2$ is the inverse speed of my cat chasing a mouse, there is no link between the two quantities and the formula is wrong in general.

• Really? I could put an equals sign between 10 m/s and 0.1 s/m? That just seems weird... – Nova Apr 29 '16 at 19:44
• @Nova I did not write about equality between $\rho$ and $v$, but "equivalency". – Roan Apr 29 '16 at 23:42