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The value of the fine structure constant is given as $$ \alpha = \frac{e^2}{4\pi\varepsilon_0\hbar c} = \frac{1}{137.035\,999..} $$ It's value is only dependent on physical constants (the elementary charge $e$, speed of light $c$, Plancks constant $\hbar$), the vacuum permitivvity $\varepsilon_0$) and the mathematical constant $\pi$, which are considered to be constant under all cirumstances.

However the Wikipedia article Coupling constant states

In particular, at low energies, α ≈ 1/137, whereas at the scale of the Z boson, about 90 GeV, one measures α ≈ 1/127.

I don't understand how this can be possible, except that one of the physical constants above or even $\pi$ are actually not constant, but dependent on the energy scale. But that seems nonsense.

So what do physicists mean when they say that the fine structure constant $\alpha$ increases with energy? Can you perhaps reformulate the quoted sentence above so that it makes more sense?

asmaier
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    The above question (v1), which basically asks What is renormalization?, should not be confused with the following physics.SE question with a similar sounding title http://physics.stackexchange.com/q/2725/2451 – Qmechanic Jun 15 '11 at 12:47
  • I believe this question was already essentially answered here: http://physics.stackexchange.com/q/9706/ – Marek Jun 15 '11 at 14:40
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    or perhaps are you asking about general phenomenon of running coupling of quantum field theories that every coupling constant depends on the scale -- http://en.wikipedia.org/wiki/Coupling_constant#Running_coupling – Marek Jun 15 '11 at 15:04

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Expanding on what Vladimir said: the thing that is changing with energy is $e$ (the others are not constants so much as conversion factors between length and time, time and energy, etc.). The reason the charge can vary is that the vacuum is not entirely empty. Sloppily speaking, near a charge, the electric field interacts with virtual (electron/positron) pairs and the effect is that the virtual pairs screens the "raw" electric field. Thus, if you're far away, you see one value, but as you get closer the electric field raises faster than $1/r^2$. With scattering experiments, how close you get to a charge is directly related to the in-going energy of the particles. Now, in modern physics, we account for this by saying that the charge $e$ changes with energy scale; this sounds bizarre in the form I just explained (since you might expect that we just declare the force to be not $1/r^2$), but it turns out that this is the neatest way intellectually to understand it, due to a convergence of issues to do with wanting to preserve observed symmetries in the theory at all scales.

Incidentally, for things like colour charges in QCD, the vacuum anti-screens, which is to say that the observed field strength increases as you get further away. Heuristically, this is what leads to confinement of quarks in the "normal" phase.

genneth
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    But why can't we say $\varepsilon_0$ changes with energy scale? I could argue that due to the interaction of the virtual particles with the charge the vacuum permittivity changes locally near that charge and that's why the coupling changes with energy. What are the observed symmetries we want to preserve at all scales that force us to vary the elementary charge with energy? – asmaier Jun 15 '11 at 13:54
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    @asmaier: well, you always have a factor $e^2 / \epsilon_0$ for every interaction with two vertices, so whether you scale $e$ or $\epsilon_0$ is just a matter of convention. In fundamental physics we don't bother writing $\epsilon$s $c$s and $\hbar$s at all, so we can only change $e$ :) The symmetry in QED is a gauge symmetry with $U(1)$ group. Don't get scared by these words, it essentially says that the theory behaves like Maxwell's electrodynamics and in particular there will be familiar Coulomb $1/r^2$ interaction in long distance limit. – Marek Jun 15 '11 at 14:50
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    But I don't really subscribe to @genneth's point of view of $1/r^2$ at all distances and forces. At high eneries (and equivalently short distances) it makes no sense to talk about forces. Well, unless you are talking about effective forces due to charge screening but these definitely don't behave like classical Coulomb ones. So I'd like genneth to clarify what he means by preservation of $1/r^2$ at small distances. – Marek Jun 15 '11 at 14:55
  • @Marek: it's sloppiness on my part, meant only to convey an image (since the OP probably isn't geared up on renormalisation). I can't (as you noticed) really mean $1/r^2$ force literally. If you pushed me to make it precise, I might say that we can try and sum up tree diagrams and call that classical (so produces in some sense $1/r^2$), and consider other diagrams to be quantum corrections. – genneth Jun 15 '11 at 15:33
  • @asmaier: What Marek said :-) – genneth Jun 15 '11 at 15:33
  • @genneth: fair enough :) – Marek Jun 15 '11 at 15:57
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    @asmaier: The vacuum permittivity $\epsilon_0=\frac{1}{\mu_0c_0^2}$ is in the current SI convention not a statement about the Vacuum that could be experimentally measured, but rather it is a man-made definition, see http://en.wikipedia.org/wiki/Vacuum_permittivity#Value In particular, $\epsilon_0$ does not run. – Qmechanic Jun 15 '11 at 18:46
  • @Marek and @Qmechanic: Your arguments why $\varepsilon_0$ cannot change with energy seem to boil down to the argument that one wants the speed of light $c$ not to change with energy to not break Lorentz invariance. Am I right? But what about Planck's constant? Couldn't one instead of rescaling $e$ rescale $\hbar$ with energy and achieve the same result of a running $\alpha$ with energy? – asmaier Jun 16 '11 at 08:29
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    @asmaier: you should distinguish between fundamental constants of theory and between coupling constants of a concrete model. $\hbar$ and $c$ are very fundamental to QFT (irrespective of the concrete model!) and that's why we work in Planck units and set them equal to one. It is masses and interaction "constants" that you are really interested in when studying some theory and its predictions. – Marek Jun 16 '11 at 11:31
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    @Marek Isn't it purely for historical reasons that the Planck units are set to one in QFT and thereby classified as fundamental? One could also use http://en.wikipedia.org/wiki/Stoney_units and then $\hbar$ wouldn't be fundamental. Isn't the differentiation between fundamental constants and non-fundamental constants completely arbitrary? – asmaier Jun 16 '11 at 13:58
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    @asmaier: no it isn't. Planck units are unique, all other systems are not as fundamental. Fundamental physics is based on few very deep theories like special and general relativity and quantum mechanics and each of these theories carries with it one fundamental constant. Besides these we have dozens of non-fundamental constants, like masses of particles, their charges, interaction strengths, etc. It is very easy to construct models where you add 20 new particles and 10 new interactions. But I dare you to construct something as fundamental as quantum mechanics or special relativity ;) – Marek Jun 16 '11 at 15:14
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Do not worry, it is a constant.

There is a sloppiness and misunderstanding in physics when the energy dependence of the cross sections is wrongly attributed to the "fundamental constant" whereas it is a cross section feature.

Vladimir Kalitvianski
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  • This is a pet peeve of mine! People invest so much time studying perturbative renormalisation they confuse the couplings that appear in the low energy effective lagrangian with "the" fine structure constant, which if QED doesn't require a UV completion is a perfectly fine thing to define. –  Nov 26 '18 at 04:56
  • Just to be clear, is the accepted answer wrong? or is it just a matter of language/interpretation? – Quillo Aug 27 '22 at 15:18