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In the framework of special relativity (SR), why is it that only energy differences are observable?

Is it simply because, starting our with some action $$S=\int d^{4}x\,\mathcal{L}(x)$$ one can always add some constant $M^{4}$ to the Lagrangian resulting in a "shifted" action $$S'=S+M^{4}\int d^{4}x$$ such that the equations of motion are unaffected by this new contribution, since $\delta\left(M^{4}\right)=0$ and so $$\delta S'=\delta S$$ Hence the dynamics of the theory are unaffected by this constant shift. In particular, the corresponding Hamiltonians of the "shifted" and "un-shifted" theories are equal up to an arbitrary constant, which we can set to zero.

Also, when one takes into account gravity, is the reason why it is sensitive to absolute energies because in the this case general covariance (or diffeomorphism invariance) demands that $$S'=S+M^{4}\int d^{4}x\sqrt{-g}$$ and as the metric is a dynamical quantity, this additional constant contribution $M^{4}\int d^{4}x\sqrt{-g}$ does affect the dynamics of the theory, since under a variation of the metric $$\frac{1}{\sqrt{-g}}\frac{\delta}{\delta g_{\mu\nu}}\left(M^{4}\sqrt{-g}\right)=\frac{1}{2}M^{4}g^{\mu\nu}$$ and hence makes a measurable contribution - since its inclusion affects the dynamics of the metric and hence the local curvature of spacetime.

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Energy is like height, which is also meaningful relative to an arbitrarily placed zero level.

Hamiltonian dynamics involves derivatives of the Hamiltonian (total energy), hence a shift has no effect. Thermodynamical observables are derivatives of free energy, hence a shift has no effect. Spectral lines are differences of energy levels, hence a shift has no effect.

$\int dx^4$ is not a well-defined term since it is infinite. Thus your argument is devoid of meaning. (In renormalized relativistic quantum field theory, the renormalization procedure guarantees that the vacuum energy is always exactly zero, obviating the need for such a term.)

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  • $\begingroup$ In QFT, when quantized a classical field, the vacuum energy enters in exactly in the form $E_0 \int d^4x$. Do you think that this renders all of QFT invalid? $\endgroup$
    – tparker
    Commented Oct 23, 2017 at 16:04
  • $\begingroup$ @tparker: No, since the vacuum energy is always made exactly zero in the renormalization procedure! $\endgroup$
    – Arnold Neumaier
    Commented Oct 24, 2017 at 7:47
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    $\begingroup$ Actions are frequently only defined over a finite region of spacetime, in which case a term like $\int d^4 x$ is perfectly well-defined. And if you're going to exclude terms from the Lagrangian because their integral is infinite, what would you do with a term like $\vec{\nabla} \phi \cdot \vec{\nabla} \phi$, whose integral over all of space is infinite for a plane-wave solution? $\endgroup$
    – Michael Seifert
    Commented Oct 24, 2017 at 14:27
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    $\begingroup$ Adding a constant term as an energy shift is typically done by adding a term like $M^4f(x)$, where $f(x)\sim 1$ for $|x|\ll R$ and $f(x)\to 0$ for $|x|\gg R$ for some $R$ (in a Euclidean sense). Given this compact support, the analysis in the question (if performed a little more carefully), seems perfectly right. $\endgroup$
    – Bob Knighton
    Commented Oct 24, 2017 at 14:45
  • $\begingroup$ @BobKnighton: But then the energy term added is not constant but position dependent. In the renormalization procedure, the terms are carefully constructed so that the contributions to the vacuum energy exactly cancel in the limit $R\to\infty$ and $f\to 1$. $\endgroup$
    – Arnold Neumaier
    Commented Oct 24, 2017 at 15:32

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