A derivation of the Klein-Gordon equation starts with the following lagrangian for a scalar field ϕ: $$ L=\frac{1}{2}g^{ab}(∇_a\phi)(∇_b\phi)-V(\phi) $$ If we plug this lagrangian in the Euler-Lagrange equations: $$ \frac{∂L}{∂\phi} - ∇_a \left[ \frac{∂L}{∂(∇_a\phi)} \right]=0 $$ we obtain: $$ \frac{∂L}{∂\phi} = -\frac {dV}{d\phi} $$ and $$ ∇_a \left[ \frac{∂L}{∂(∇_a\phi)} \right] = ∇_a \left[ \frac{1}{2}g^{ab} \cdot 2(∇_a\phi) \right] = g^{ab} \cdot ∇_a ∇_b \phi = ∇^a ∇_a \phi=\Box^2 \phi, $$ resulting in a generalised form for the KG equation: $$ \Box^2+\frac {dV}{d\phi}=0 $$ For a potential of the form $$ V(\phi)=\frac{1}{2}m^2\phi^2, $$ we obtain $$ (\Box^2+m^2)\phi=0 $$ This is the KG equation form that I tend to find in QFT books. And since I can derive it myself, I am happy with it.

However, in most books related to General Relativity (Hobson et al. being an illustrative exception), and in numerical relativity papers, I find this other version of the KG equation: $$ (\Box^2-m^2)\phi=0 $$ I am very puzzled by the negative sign on the matter term, and when you try to do some numerical relativity calculations, this sign matters a lot.

Does anyone know what the origin of this discrepancy is?

Is it perhaps a question of different sign conventions? If so, please elaborate.


1 Answer 1


It's just due to a difference in metric convention. $(\square^2+m^2)$ involves a $(+---)$ signature and $(\square^2-m^2)$ involves $(-+++)$. If you are ever confused which is which, note that $$\square^2 e^{i(Et - p\cdot x)} = \pm (E^2-p^2)e^{i(Et - p\cdot x)}$$ with the sign depending on the signature. The sign of the $p^2$ term in the Klein-Gordon equation must match that of the $m^2$ term.

  • $\begingroup$ Thanks for your answer, octonion. I had already considered that option, but I haven't had to specify any metric in my derivation. So I am not convinced yet. $\endgroup$
    – DrD
    Commented Feb 21 at 20:39
  • 3
    $\begingroup$ There is a metric in the first line of your question. The issue is whether the d'Alembert operator (in Cartesian coordinates in flat space) reduces to $\partial_t^2 - \partial_x^2$ or $-\partial_t^2 + \partial_x^2$. When you see someone using $-m^2$ they have the second definition in mind. $\endgroup$
    – octonion
    Commented Feb 21 at 20:45
  • 1
    $\begingroup$ OK, got it :) Thanks a lot $\endgroup$
    – DrD
    Commented Feb 21 at 21:12

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