# What is the meaning of purely imaginary physical measures?

I would like to know what is the meaning of working with a physical measure that is purely imaginary. What I mean with "purely imaginary" is $$Z - Re(Z)$$ where $$Z$$ is a complex number.

Furthermore, nowadays are there any physical measures that had been discovered to be purely imaginary?

I am working in a thesis and I have found a wave equation in which velocity is purely imaginary. Did I found a completely theoretical thing, an abstract mathematical situation without physical interpretation, or are there any chances for this purely imaginary velocity be something tangible? Look what I've found below:

$$v = \gamma \cdot k \cdot \pi \cdot i$$

Where $$v$$ stands for velocity, $$\gamma$$ is a constant that depends on the material, $$k$$ is an integer, and $$i$$ is $$\sqrt{-1}$$.

I have rewritten that as the following:

$$v = \gamma k \pi \cdot e^{i\frac{\pi}{2}}$$

This means that my velocity is a vector in the Argand-Gauss plane, directed to the Imaginary axis, with $$\gamma k \pi$$ length, i.e., speed.

In a wave equation, which is not ordinary kinematics, we can, indeed, have an imaginary wave speed: given the ordinary wave equation, taken in one dimension for simplicity,

$$\frac{\partial^2 u}{\partial t^2} = v_w^2 \frac{\partial^2 u}{\partial x^2}$$

if $$v_w = iv$$ for $$v$$ real, then $$v_w^2 = -v^2$$ and the WE looks like

$$\frac{\partial^2 u}{\partial t^2} = -v^2 \frac{\partial^2 u}{\partial x^2}$$

To understand the meaning, note that a wave equation has elementary solution

$$u(x, t) = A \cdot \sin\left(kx - \omega t + \phi\right)$$

where the wave number $$k$$ and angular frequency $$\omega$$ are required to satisfy

$$\frac{\omega}{k} = v_w$$

If $$v_w$$ is imaginary, then that means one of $$\omega$$ or $$k$$ will be so, and you get sine of a complex number, which is also in general a complex number. This will, then, only make sense if the value of the wave, $$u(x, t)$$, itself can make sense as a complex number, too. But it will not involve motion into imaginary directions of space, because the domain of variable $$x$$, the spatial position, is still the real numbers.

ADD 2: I think this should have a real solution, actually - it's a real differential equation after all...

ADD 3: Yes ... I had a hunch but had to work it out, the relevant real-number solution is ... WEEERD! Don't forget the angle sum formula, and you can get, writing as $$kx - \omega t + \phi = (kx + \phi) - \omega t$$, using the identity that $$\cos(ix) = \cosh(x)$$ with imaginary $$\omega$$ as per the relation above,

$$u(x, t) = A\ \sin(kx + \phi)\ \cosh(\omicron t) = [A \cosh(\omicron t)] \sin(kx + \phi)$$

and throw away the imaginary part (justifiable as being a linear combination with a conjugate) as a working real solution, where I've used $$\omicron$$ to denote $$\omega = i\omicron$$. Note that two derivatives in $$x$$ gives you $$-k^2$$ out front but two derivatives in $$t$$ gives you $$\omicron^2$$ out front, so the desired negative sign appears.

Hence, imaginary wave speed means the wave pattern is sitting still but growing exponentially in amplitude!

ADD 4: I believe also there may be a way to get exponential decay, as well.

• I understand that in this context "imaginary" does not mean "fictitious". I just wanted to know if, for example, my purely imaginary velocity (without Real component), is able to cross $R^{3}$ space, i.e., x, y, and z. – Victor Lins Jan 15 at 11:41
• @VictorLins I don't think "does velocity cross space?" is a meaningful question. Do your waves travel through space? – user253751 Jan 15 at 11:45
• @Victor Lins: Added some more details. – The_Sympathizer Jan 15 at 11:46
• Thank you for your explanation, Sympathizer! I will take a more careful look at your explanations and I will try to bring this reality to my paper. – Victor Lins Jan 15 at 11:51
• @Victor Lins : Yes, just make sure to check the maths independently in case I still jinxed something. – The_Sympathizer Jan 15 at 12:12

There is an isomorphism between the field of the complex numbers and the plane $$R^2$$. It means that any theory that has complex numbers can be rewritten without using them. This might be at the expense of introducing a more "complex" (in the sense of complicated) formalism, so using complex numbers many times simplify the math. So, there should not be anything mysterious about imaginary numbers in physics.

I think an intuitive way of approaching imaginary numbers is by looking at the $$i$$ in the same way as we look at the minus ($$-$$) sign in front of a negative quantity. They are both extensions to our range of numbers. The minus sign extends our numbers to allow the subtraction $$a - b$$ where $$b > a$$. This extension can be just as counter-intuitive as complex numbers, we're just more used to it. For example, we would not really be able to imagine what -3 apples are and in earlier times, one might have called you an idiot for using negative velocities, because their car (or maybe horse) and everything else they know can only travel at a positive velocity.

Indeed, what we call a negative number really depends on what we define as being a negative number. In the case of the velocity, we can include directionality and we define the negative number sign as traveling in the opposite direction*. However, we do not need to define the negative numbers for physics to work, it just makes things easier to do so. Expanding on this, we are not able measure the negative number sign itself, we can only measure the absolute value (or, projection along our measurement axis) of a quantity and whether our result is negative or positive ultimately depends on what we choose ourselves to be negative and positive. The minus sign itself is something we cannot measure.

The same really counts for the imaginary unit $$i$$. We use $$i$$ to extend our range of numbers and allow us to compute $$\sqrt{-a}$$ for $$a>0$$. In a similar sense, we do not need to define $$i$$, we can do physics just fine without it. It does however make a lot of things easier to include imaginary numbers.

Now, if we want to measure a complex quantity, what we measure will really depend on our own definition of the $$i$$-axis. For example, if we are doing velocity measurements in $$\cal{R}^3$$, using complex velocities makes no sense because we never defined the meaning of the complex axis. Whether we can measure (the absolute value of) a complex quantity really depends on how we defined the imaginary axis. Also remember that if we have defined the imaginary axis in such a way that allows us to measure what is on it, we cannot measure $$i$$ itself just as we cannot measure the negative number sign.

A good example of using complex numbers to our advantage is the complex frequency and expressing our waves in terms of imaginary exponentials. In the first place, we express our wave (which can for example describe a sound wave) as

$$\cos{\omega x} = \operatorname{Re}(e^{i\omega x}),$$

where we note that $$\operatorname{Re}(e^{ix})$$ returns a real number. Then, using a complex frequency $$\omega = \omega' + i\omega''$$ gives

$$\operatorname{Re}(e^{(\omega' + i\omega'')x}) = \operatorname{Re}(e^{\omega'x}e^{-\omega''x}) = e^{-\omega''x}\cos{\omega x}.$$

So in this case, using a complex frequency allows us to express both the frequency of a harmonic oscillation and the decay constant of this oscillation's exponential decay into one quantity.

*Mathematically speaking, we can really choose any direction. The opposite direction just makes most sense, since we then can compare negative and positive numbers 1 to 1. Regardless of what we exactly define $$a-b$$ for $$b>a$$ as, it would still mathematically work, but it would e.g. force us to distinguish between the subtraction operator and minus sign and also complicate many other things such as making our coordinate system non-Cartesian. So taking the opposite direction makes most sense.