MATT STRASSLER
Of Particular Significance

Zero-Point Energy of Fields: A Warm-Up

Matt Strassler 2/14/2024

As a warm-up to the larger goal of understanding the zero-point energy of a quantum field, let’s learn some important lessons from a simpler problem: the zero-point energy of a guitar string, or any similar string-like object. (You should first read this article on zero-point motion and energy.)

The Basic Idea

A guitar string differs from a ball on a spring, whose zero-point motion and zero-point energy were discussed here, because the latter can have only one type of vibration, while the guitar string has many ways of vibrating. In particular, it has many standing waves. The simplest of these standing waves, and its first three harmonics, which are the next-simplest waves, are shown in the figure below. The simplest wave vibrates at the string’s resonance frequency f, also known as its fundamental frequency, and the harmonics or overtones of that wave, which are standing waves with multiple crests and troughs, have higher frequencies.

Figure from Waves in an Impossible Sea; illustration by Cari Cesarotti.

Each one of these standing waves is like a single ball on a spring; each one acts like a single, independent, vibrating object. So whereas the ball has one “mode” of vibration, the string has many. How many?

Well, the higher the harmonic, the more crests and troughs the corresponding wave will have. You can keep going, to higher and higher harmonics, until the distance between one crest and the next becomes comparable to the internal structure of the string. [For example, you certainly can’t have crests and troughs that are closer together than the atoms that make up the string!] Once you reach that point, the structure of the string itself prevents you from going any further.

Since each of these modes of vibration — each standing wave — is like a single ball on a spring, each one of them has zero-point motion (it can never really entirely sit still, thanks to quantum uncertainty.) It therefore also has zero-point energy, an amount equal to

  • E_{{\rm zero-point}} = \frac12 f h

where f is the mode’s frequency and h is Planck’s constant (which I’ve also called the “cosmic certainty limit”).

Notice that the higher harmonics of the string have higher frequency and therefore higher zero-point energy! This is very interesting. It means that the simpler standing waves of the string contribute only a tiny fraction of its total zero-point energy. Most of the zero-point energy comes from the modes with the highest frequencies — the most frenetic of the standing waves, with crests and troughs microscopically close together, to the point that their details depend on the internal structure of the string.

This is why the total zero-point energy of a string — or of any object — is sensitive to its internal structure. Specifically (as I’ll explain later, using math), suppose the string has length L, and is made from tiny objects that sit a distance \ell apart. Then its zero-point energy grows proportional to its length L, but also grows like the inverse-square of the distance \ell!

  • E_{{\rm zero-point}}^{{\rm total}} \approx hvL/\ell^2

where v is the speed with which traveling waves can move down this type of string.

The fact that the zero-point energy grows proportionally to the length of the string isn’t very surprising. But it’s quite remarkable that the zero-point energy grows even faster if you make the internal structure of the string smaller! If the structure is small enough, that energy could be big — very big — perhaps as big any other form of internal energy that it might have.

You might then wonder whether the string’s vibrational zero-point energy might be the ultimate origin of its rest mass, via the relativity formula E=mc^2. That would be a very clever idea. But… it turns out not to be true, just like many good ideas. [If you thought of this immediately, congratulations for having had a good idea that turned out to be false. For physicists, this is a common experience.]

For an ordinary guitar string, if we use the above formula, take \ell as small as it could possibly be for an ordinary material — the distance between neighboring atoms in the string — and take v to be 500 meters (1600 feet) per second, then we find

  • E_{{\rm zero-point}}^{{\rm total}} \approx 10^{-11}\ {\rm Joules}

which should be compared to the roughly twenty Joules of energy that are required to lift a brick to waist height. While this energy does increase the mass of the string, it does so by about an atom’s-worth — less than a trillionth of a trillionth of its total mass. So for an ordinary string, the zero-point energy of its vibrations cannot be the source of its mass, which must therefore come from something else.

However, let’s imagine we had a string made of some sort of basic material of the universe, one whose structure was found down as far as is believed possible, all the way down to the Planck length \ell_p\sim 10-35 meters, the distance beyond which space and time themselves are suspected to become uncertain. Imagine also that, as we’d expect for the basic material of the universe, waves on this string move at the cosmic speed limit c. Then, if this string were as long as a typical guitar string, its zero-point energy would be

  • E_{{\rm zero-point}}^{{\rm total}} \approx 10^{44} \ {\rm Joules}

WHOA! That’s comparable to the E=mc^2 energy stored in the planet Jupiter!! Quite a bit more than the typical guitar string!

In fact, with so much mass in such a small space, this string would form a black hole.

What do we learn from this? That even in objects of perfectly ordinary size, the amount of vibrational zero-point energy stored within them can be spectacularly large if their internal structure is sufficiently ultra-microscopic. It’s not the case for objects made from atoms; their vibrational zero-point energy is small compared to other forms of energy inside them. But for objects made from the basic ingredients of the universe, their vibrational zero-point energy might be a much bigger deal.

The Simple Math Behind The Ideas

As discussed in this article, quantum uncertainty assures that a ball on a spring has motion and energy even when it is vibrating as little as possible. This is also true for a standing wave on a string; thanks to quantum uncertainty, the wave’s amplitude cannot be strictly zero, and so this mode of vibration has zero-point energy, equal to its frequency times h/2.

For a simple guitar string, the q^{th} standing wave, which has a total of q crests and troughs as you can see from the figure above, has a frequency q f. The total zero-point energy of the guitar string is therefore the sum over the zero-point energies of all the standing waves allowed, from the smallest frequency to the largest. Let’s call the largest frequency q_{{\rm max}} f. Then

  • E_{{\rm zero-point}} = \sum_{q=1}^{q_{{\rm max}}} \frac{1}{2} q f h = \frac{1}{2} fh\sum_{q=1}^{q_{{\rm max}}} q = \frac{1}{4}q_{{\rm max}}(q_{{\rm max}}+1) fh

So to estimate the size of this zero-point energy, we need to know the maximum possible frequency of the modes of a string. How high do the harmonics of a string go?

Naively, the answer is infinite; there’s nothing stopping you from adding more and more crests and troughs, more and more closely spaced, to the vibrating guitar string. However, this is an unphysical answer. At some point you will run into the fact that the guitar string is made from atoms, and so there’s simply not enough material in the string to make the crests and troughs closer together than the atoms are. There’s some distance \ell below which we can’t go, and the maximum q — the maximum number of crests and troughs that can fit on our string — is the length of the string L divided by this minimum distance \ell:

  • q_{{\rm max}} \sim L/\ell

That means

  • E^{{\rm total}}_{{\rm zero-point}} = \frac{q_{{\rm max}}(q_{{\rm max}}+1)}{4} fh \approx \frac{1}{4}\frac{L^2}{\ell^2} fh

Let’s simplify this by using a famous fact about simple waves on a string, also true of light waves inside a mirrored box, that the simplest standing wave on a string of length L has frequency f = v/2L, where v is the speed with which traveling waves would move along this string. This gives us:

  • E^{{\rm total}}_{{\rm zero-point}} \approx \frac{1}{4}\frac{L}{\ell^2} (Lf) h = \frac18{hvL/\ell^2}

This confirms the first formula of this article, up to the factor of 1/8; but when we are making rough estimates of huge or tiny numbers, such details are a distraction, so I omitted them in my more qualitative discussion above.

(Why does a string’s simplest wave have frequency f = v/2L? Click here for some intuition:)

Musical instruments give us some intuition.

  • Larger string instruments generally have longer strings, which make lower notes — i.e. vibrations with lower frequencies. So increasing the length L of a string, leaving everything else the same, makes the frequency smaller.
  • Tightening a string increases its frequency, making its note higher; and tightening a string also increases the speed v with which traveling waves can move along it. In general, anything that increases the waves’ speed will increase the string’s resonant frequency.

Let’s simplify this one more step by writing down the total zero-point energy density , which I’ll call \rho_0, the total zero-point energy divided by the string’s length:

  • \rho_0=E^{{\rm total}}_{{\rm zero-point}}/L\approx \frac18{hv/\ell^2}

Notice

  • the zero-point energy density is independent of L
  • the zero-point energy density increases rapidly as \ell decreases.

When we next turn to fields and their zero-point energies, these two facts will remain true. Click here for the next article in this series.

17 Responses

  1. Great article, thanks. Was there a deliberate intention behind you going straight to the guitar string having a zero point energy based upon its macro frequency f and various modes, rather than treating it as composed of interconnected, interacting atomic springs from the start and working upwards from this?

    I learned about waves, frequencies etc at high school decades ago and recently decided to patiently wade through your various articles on basic physics starting with the ball on a spring, leading to waves on strings such as here:
    https://profmattstrassler.com/articles-and-posts/particle-physics-basics/fields-and-their-particles-with-math/waves-classical-form/

    I’ve found it extremely helpful right from the start to use a physical model of the spring as being composed of atomic balls; interconnected with one another via inter-atomic electrical force springs. Also, I find modeling the propagation of longitudinal waves along a string in one dimension as the obvious first step from understanding how a ball vibrates attached to a spring. From this, then progressing to two then three dimensions. It amazes me how a string manages to vibrate with an overall ordered repetitiveness as the number of dimensions increases.

    Reply

    1. Yes, I chose this approach because it makes it much easier to identify and count the vibrational modes of the string.

      Although one may model a string as a series of balls and springs, one can’t calculate the zero-point energy directly from that model. One would have to first convert their vibrational motion to the modes of the string, after which one would arrive at the starting point of my post. So taking this approach would add a long introduction to my post, one that doesn’t add anything to the calculation.

      Reply

        1. At the moment I have no idea why you would say that. This is how I would teach it in a class for experts, so this *is* the available light. What are you looking for that this presentation does not provide? Are you looking for an explanation as to *why* there is zero-point energy? Or are you looking for an explanation as to why there is so much of it in the theoretical calculation, and so little in the experiments? Are you specifically interested in the zero-point energy of the universe’s fields? Or is it something else? I need you to be more explicit.

          Reply

  2. Quote: “For example, you certainly can’t have crests and troughs that are closer together than the atoms that make up the string!”
    That is, forgive the strong language, utter bullsh*t. By claiming that you deny the existence of liquid or solid matter, ignore the results of spectroscopy, the existence of chemical bonds, quantum entanglement, etc. In other words, you treat the world as consisting of mass points only – a reasonable approach for high energy physics, I admit, but rubbish at every day energy levels. And if you consider a guitar string, the energy involved is only a few eV/c^2, more would tear the steel string apart.
    To start with basic, Newtonian physics, consider how many dimensions does the universe need to have where we live in? Stephen Hawking’s answer was studying a dog: since it does have a mouth, intestines and an anus, a two-dimensional dog would consist of two pieces; to connect these pieces, a third dimension is required. With this argument, Hawking missed out on understanding what time is, and never realised how much nonsense is incorporated in the concept of four-space. Had he started from a self-consistency poit-of-view . . .
    Visualize a Cartesian coordinate system, with axes x,y,z…

    [[Abridged by host]]

    Reply

    1. Thank you for your opinion. However, if you think so little of my scientific presentations, why are you reading them?

      Please note that I will abridge comments that are unreasonably long and/or are insulting to anyone and/or consist of a series of of incorrect physics arguments. And strong language is not appropriate here.

      Reply

    1. The zero-point energy of simple quantum systems is not that hard to measure; for example, tunneling from one such system to another would reveal this upward shift of the energy. I will try to find an example of an experiment that measures this kind of thing very precisely and will post about it.

      Measuring the zero-point energy of large systems is trickier, because it’s not easy to isolate that energy compared to that of other processes that are going on, such as thermal effects. I’m not entirely sure of the experimental state of the art. Zero-point energy can play a role in the phases of materials. But I’ll have to do some reading to get you the best answer.

      For fields of the universe itself, there is a somewhat controversial measurement of what is known as the “Casimir effect”, which is more relevant to the next article. (This can be viewed as a consequence of zero-point energy, but alternative interpretations confuse the issues.) However, the zero-point field fluctuations [the analogue of zero-point motion for a waves on a guitar string] that lead to zero-point energy have all sorts of crucial effects; for instance, there would be no protons or neutrons without them.

      Reply

    2. Zero point energy is why liquid Helium won’t freeze at 1 bar pressure even if you take the temperature down to absolute zero. I would take that observed fact to be an experimental confirmation.

      Reply

      1. I like that answer (and any good condensed-matter physicist probably has quite a few other similar answers) but I do not recall what we can say quantitatively about it; what measurement do you make to verify not only the presence but also the size of the zero-point energy?

        Reply

        1. According to Wikipedia, just about 100 years ago, before Heisenberg proposed the uncertainty principle, “Robert Mulliken[40] provided direct evidence for the zero-point energy of molecular vibrations by comparing the band spectrum of 10BO and 11BO: the isotopic difference in the transition frequencies between the ground vibrational states of two different electronic levels would vanish if there were no zero-point energy, in contrast to the observed spectra. ” I have never studied this but will look into it. The reference is Mulliken, Robert S. (1924). “The band spectrum of boron monoxide”. Nature. 114 (2862): 349-350.

          The year before that (!), “Kurt Bennewitz [de] and Francis Simon (1923), who worked at Walther Nernst’s laboratory in Berlin, studied the melting process of chemicals at low temperatures. Their calculations of the melting points of hydrogen, argon and mercury led them to conclude that the results provided evidence for a zero-point energy. Moreover, they suggested correctly, as was later verified by Simon (1934),[38][39] that this quantity was responsible for the difficulty in solidifying helium even at absolute zero.” The references quoted are: Bennewitz, Kurt; Simon, Franz (1923). “Zur Frage der Nullpunktsenergie” [On the question of zero-point energy]. Zeitschrift für Physik (in German). 16 (1): 183-199. ; Simon, F. (1934). “Behaviour of Condensed Helium near Absolute Zero”. Nature. 133 (3362): 529. Dugdale, J. S.; Simon, F. E. (1953). “Thermodynamic Properties and Melting of Solid Helium”. Proceedings of the Royal Society. 218 (1134): 291.

          (That’s right: Nullpunktsenergie .)

          Clearly there’s more history to read here! This is a big gap in my self-education… So often we physicists learn how things work without having the time or patience to learn the process of how these workings were discovered.

          Reply

  4. Dr.Strassler:
    Check of understanding, if I may. Thermal energy of a “solid” lattice is usually described, in analogy, as the molecules / atoms vibrating back & forth about their equilibrium position, with the “springs” being the intermolecular forces themselves. So, would the zero point energy be the “vibrations” of the atoms themselves? In other words, the internal structure of the atom itself is vibrating, unlike thermal energy where the atoms are vibrating about their equilibrium positions….relative to the other atoms, the zero point energy is the vibration of the internal structure of the atom?
    BTW: I pre- ordered your book on Amazon, anxiously awaiting its release.

    Reply

    1. Actually, no; you’re mixing up two separate points.

      First, ignore the internal structure of the atom. I haven’t addressed this, and for all the discussion so far requires, atoms could be indivisible and have no internal structure. The zero-point energy I’m referring to here has nothing to do with the atoms’ insides.

      Instead, I’m referring to the zero-point energy of the lattice of “springs.” Even when the solid’s lattice is at absolute zero temperature — when it is vibrating as minimally as it possibly can, so there is no thermal energy at all — each of the “springs” in the lattice between one atom and its neighbor contributes zero-point energy. That’s because of quantum uncertainty; if the “spring” between one atom and the next didn’t move at all, then we’d know the positions of all the atoms in the lattice and their motion (i.e. no motion at all.)

      Said another way, the zero-point energy comes from the fact that the atoms can’t be stationary relative to one another, and so there’s quantum jitter going on throughout the solid even when the temperature reaches absolute zero.

      Now, what’s not obvious is how the calculation of the zero-point energy works out. You can’t do it spring by spring, because the springs affect one another (i.e. when one stretches, it causes others to extend.) The actual calculation is a three-dimensional version of what I did for the guitar string here (and the next article in the series will carry it out.)

      Reply

      1. Dr.Stassler: ahh, ok that definitely cleared up my understanding, as I thought ANY jittering between the “springs” would be considered thermal. Is this the reason why we could never actually reach absolute zero?

        Reply

        1. Yes, thermal effects are associated with exciting the springs above their ground state, whereas a small amount of quantum jitter remains even in the ground state.

          The temperature directly tells you how probable it is to find the springs above the ground state. For instance, at a temperature T, the probability of being in the first excited state with energy E1 compared to being in the ground state with energy E0 is exp[-(E1-E0)/(kT)], where k is Boltzmann’s constant. If T goes to zero, this probability goes to zero. But if kT > E1-E0, E2-E0, E3-E0, etc, then the system will rarely be in its ground state and will almost always be found in an excited state.

          The problem of reaching absolute zero temperature has nothing to do with zero-point energy. The problem, as you see from what I just said, is that absolute zero means the system has essentially no probability of being in any of its excited states. But any large system (a) has excited states very close to its ground state, and so it can spend most of its time in its ground state only when kT < E1-E0, an exceedingly small temperature; and (b) it interacts with its environment, which is not at absolute zero (even in deep space), and therefore it is absorbing photons from its environment all the time which easily move it from its ground state to one of its many excited states. You need, therefore, to isolate the system in an extreme way, and then wait a long time as it radiates all its excess energy away. For an atomic nucleus, this takes no time at all; in ordinary life, nuclei behave as though they are at zero temperature. But for a grain of steel, this takes forever. In short, the problem of reaching zero temperature is a quantitative question about how hard it is to isolate systems and how long it takes big systems to radiate their excess energy away. It is not a matter of the zero-point energy, which is the energy that the system still possesses even at zero temperature, and which does not directly affect either the isolating of the system or how long it takes for the system to radiate its excess energy.

          Reply

          1. Dr.Strassler: thank you, that really made the concept understandable. I was going to ask a follow up question, of what about a “molecule” on its own, not part of an ensemble, out in deep space? However, I believe you already answered, that even then, it would be almost impossible to isolate said molecule from photons or even the CMB.

            Reply

            1. The CMB are the photons I was thinking about; they’re the ones at a temperature of 2.7 Kelvin.

              For some molecules, that is in fact cold enough; again it depends on the energy splitting between ground and excited states. But you could always study molecules in a lab that can be made cold enough, using dilution refrigerators. So molecular zero-point energy is not so hard to study.

              I think also I’m not so well-informed about what that can be done with materials and zero-point energy; I’m still learning new things right now.

              Reply

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