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Why is the stress-energy tensor for electromagnetic radiation traceless?

A photon gas obeys the equation of state $\rho=P/3$ and hence $T^{\mu}_{\quad\mu}=3P-\rho=0$. (Can also be seen by expressing the stress energy tensor in terms of of the electromagnetic tensor as ...
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Derivation of Raychaudhuri equation - Trace

In Wald (Wald: General Relativity on page 218, equation 9.2.10) is stated that $$v^c∇_cB_{ab}=−B^c_bB_{ac}+R^d_{cba}v^cv_d$$ and to continue in order for the equation to be derived one needs to take ...
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This is related to a previous question of mine. We consider a density matrix $\sigma(t)$ operating on a Hilbert space $\mathscr{H}_{s}\otimes \mathscr{H}_b$ with Hamiltonian $H = H_s \otimes \mathbb{... 0answers 46 views Traceless stress tensor What does it mean, when the viscous (or viscoelastic) stress tensor is traceless$\tau_{rr}+\tau_{\theta \theta}+\tau_{\phi \phi}=0$? Why if the viscoelastic model is linear it is traceless and if ... 1answer 69 views Integrating of von Neumann equation for density matrix Suppose we are given the Hamiltonian $$H=f \frac{\text{Tr}\sigma_x \rho}{\text{Tr}\rho}\sigma_x,$$ where$\rho$is the density matrix, and$\sigma_x$is the Pauli matrix $$\sigma_x= \begin{... 1answer 84 views Is tracing out a subsystem always akin to discarding all information about it? Suppose we have some quantum system with sub-systems A and B. It could be, for example, two qubits or groups of qubits. Is it fair to say that tracing out the sub-system A is always akin discarding ... 1answer 85 views Why can the partial trace be written as \text{Tr}_B(\rho)= \sum_k (1 \otimes \langle k|) \rho (1 \otimes |k \rangle)? I don't really understand a notation that I stumbled upon regarding a partial trace. According to the definition I have, partial trace is$$\rho_A=\text{Tr}_B(\rho_{AB}):= \sum_k (1_A \otimes \... 0answers 31 views Diagrammatic expansion of an operator insertion in path integral for Trace Anomaly calculation Starting with a scale invariant classical field theory, we can prove that the energy-momentum tensor will be traceless. \begin{equation} \Theta^\mu_{\ \mu }=0 \end{equation} In the context of the ... 1answer 85 views Relation between the trace anomaly and the energy-momentum tensor being off-shell Let's say we have a massless QED theory with a Lagrangian \begin{equation} L=i\bar{\psi}\not{D}\psi-\frac{1}{4}F_{\mu\nu}F^{\mu\nu} \end{equation} The symmetric energy-momentum tensor is \begin{... 1answer 77 views What is the trace in the Chern-Simons action I have been looking at the Chern-Simons Lagrangian in$(2+1)$-dimensional spacetime$M$in terms of a gauge field$A: $$S[A] = \frac{k}{4 \pi}\int_M \text{Tr}(A \wedge \text{d}A+ \frac{2}{3}A \... 1answer 185 views How can I prove that the partial trace is well-defined? When I define the partial trace as below, how can I prove it well-defined? I understand that I have to indicate Tr_k(\rho) does not depend on how to take the ONB of \mathbb{C}^2$$n\in \mathbb{Z}_{... 0answers 55 views QCD Trace Anomaly and Mass In the paper in equations 4 and 5, some of the mass of the nucleons comes from the "trace anomaly" of the QCD energy-momentum tensor (as described in the paragraph following these equations). Is there ... 1answer 103 views Physics Meaning of Trace Technology in QED [closed] As it pointed out on page 133 of Peskin and Schroeder, any QED amplitude involving external fermions, when squared and summed or averaged over spins, can be converted to traces of products of Dirac ... 1answer 119 views reduced density matrix of state [closed] given a multi particle state I have to calculate the reduced density matrix where I trace out the third particle $$|\psi\rangle = \frac{1}{\sqrt{3}} \left ( |\uparrow \uparrow \downarrow\rangle + \... 1answer 215 views What is a definition of the trace norm? I have found that (one?) definition of the trace norm is$$\mid\mid A\mid\mid = \sqrt{A^*A} \tag{1}$$but now I am reading this paper where (on page 4) it says In particular, we will restrict ... 1answer 60 views Proving identity \DeclareMathOperator{\Tr}{Tr} \Tr\left[\gamma^{\mu}\gamma^{\nu}\right] = 4 \eta^{\mu\nu} In the lecture notes accompanying a course I'm following, it is stated that$$\DeclareMathOperator{\Tr}{Tr} \Tr\left[\gamma^{\mu}\gamma^{\nu}\right] = 4 \eta^{\mu\nu} $$Yet when I try to prove this,... 1answer 138 views Double-trace operators in CFT? What is the conceptual difference between so called "single-trace" and "double-trace" (or "multi-trace") operators e.g., in a Conformal Field Theory? 0answers 36 views How to Explicitly Calculate z-Component of Berry Curvature? While numerically playing with the 2-level Haldane model recently, I wondered how I could analytically calculate the z-component of the Berry curvature F. I framed my problem as needing an ... 1answer 29 views Evaluating a trace with two factors of \gamma^5 In the process of calculating a spin-averaged square amplitude in QFT, I came across the following expression:$$ \text{Tr}\left[\gamma^\mu\gamma^5\gamma^\alpha\gamma^\nu\gamma^5\gamma^\beta\right] ... 1answer 50 views Confusion with trace of gamma matrices Using \{\gamma^\mu, \gamma^\nu\} = 2 \eta^{\mu\nu} \mathbf{1}, it is easy to show that: \begin{align*} \operatorname{tr} \gamma^\mu \gamma^\nu = 4\eta^{\mu\nu} \end{align*} Now, it is also true that ... 1answer 99 views Trace over configuration basis Let us take a many-body quantum system, whose phases in the configuration basis are labeled by \mathbf {\hat q}=(q_1,\cdots, q_N) and momenta \mathbf {\hat p}=\left(-i\frac{\partial}{\partial \hat ... 2answers 52 views Alternate definitions of Thermal states The definition of thermal states I'm used to is:\tau_{\beta} = \frac{1}{Z}\,e^{-\beta H}$$where Z is the partition function defined as Z= \mathrm{Tr}(e^{-\beta H}), \beta the inverse ... 1answer 89 views Why can we write lagrangian for gauge theory without the traces? I understand that trace is needed in order to preserve gauge invariance of the lagrangian equation by using the cycling property. But I fail to see why the following equation holds true:$$-\frac{1}{2}... 2answers 433 views Electromagnetic stress tensor is only traceless in 4D? The electromagnetic stress tensorF_{\mu \nu}$is as we all know traceless in 4 dimensions. With$F_{\mu \nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$and$A = (A_0,A_1,A_2,A_3)= (\phi, A_1, A_2, ...
In most textbooks (Georgi, for example) a scalar product on the generators of a Lie Algebra is introduced (the Cartan-Killing form) as $$tr[T^{a}T^{b}]$$ which is promptly diagonalised (for compact ...