I'm reading the book "Advances in the Casimir Effect" by M. Bordag. In section 2.4 it says that the $00$-component of the energy-momentum tensor is given by (eq. 2.59): $$T_{00}(t,x)=\frac{\hbar c}{2}\left(\frac{1}{c^2} [\partial_t \varphi(t,x)]^2 + [\partial_x \varphi(t,x)]^2 + \frac{m^2c^2}{\hbar^2}\varphi^2(t,x)\right)$$

I've tried to get this starting from the general form of the tensor:

$$ T^{\mu\nu} = \frac{\hbar^2}{m}(g^{\mu\alpha}g^{\nu\beta}+g^{\mu\beta}g^{\nu\alpha}-g^{\mu\nu}g^{\alpha\beta})\partial_\alpha\bar{\phi}\partial_\beta\phi-g^{\mu\nu}mc^2\bar{\phi}\phi $$

Where in our case $$\phi = \varphi(t,x) = \sum_n \left[\varphi_n^+ a_n + \varphi_n^- a_n^{\dagger}\right]$$ is the scalar field that satisfies the Klein-Gordon equation, $a_n$ and $a_n^\dagger$ are the anihilation and creation operators, and the metric tensor is given by:

$$g^{\mu \nu}=\begin{pmatrix} -\frac{1}{c^2} & 0 \\ 0 & 1 \end{pmatrix}$$

This way I get: $$T_{00}(t,x)=\frac{\hbar^2 }{m c^4}\left([\partial_t \varphi(t,x)]^2 + c^2[\partial_x \varphi(t,x)]^2\right) + m\varphi^2(t,x)$$

And that's not what the book says. Any idea of what I'm doing wrong?

Appreciate your help.


1 Answer 1


In units $c=\hbar=1$, it's clear the expression you found that the one presented in the book differ by an overall factor of $m$ (and a 2). In principle there's nothing wrong with this as both tensors will still be conserved and this difference can be considered a choice in normalization of the fields and/or of the stress tensor's definition.

However I will point out that it's conventional for the term $m^2\psi^2$ to have units of "energy," and for $T_{00}$ to have units of energy (the 1/2 in the book's expression is also conventional). So, the expression from the book would be the more canonical form for the stress tensor.

The discrepancy here appears to be in the general formula you write down for the stress tensor. No idea where you got this from. The two standard methods for computing the stress tensor involve either an explicit Noether variation (for example, see chapter 2 of Di Francesco et al.'s conformal field theory book) or by taking a metric variation (see that same CFT book or here). Note that the Belinfante tensor and the tensor obtained from Noether's theorem will generically not be identical, but will only differ by an improvement term (a term which does not spoil the conservation).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.