'Symmetrisation Notation' - What is it?

  • Thread starter Thread starter InsertName
  • Start date Start date
  • Tags Tags
    Notation
InsertName
Messages
26
Reaction score
0
I'm reading Hobson, Efstathiou and Lasenby, where on page 490, some notation is used that I am not familiar with.

In the rather long equation (17.62), some terms contain things like

\partial_{(\mu} h^{\rho}_{\nu)}.

The author(s) then says "where we have made use of the symmetrisation notation discussed in Chapter 4."

I've tried looking in Chapter 4 but it's not at all clear what is going on, so I came here :)
 
Physics news on Phys.org
The indices within the parentheses are understood to be totally symmetrized. So

hρν) = ½(∂μhρν + ∂νhρμ)
 
Thanks Bill_K!
 
Thread 'Can this experiment break Lorentz symmetry?'
1. The Big Idea: According to Einstein’s relativity, all motion is relative. You can’t tell if you’re moving at a constant velocity without looking outside. But what if there is a universal “rest frame” (like the old idea of the “ether”)? This experiment tries to find out by looking for tiny, directional differences in how objects move inside a sealed box. 2. How It Works: The Two-Stage Process Imagine a perfectly isolated spacecraft (our lab) moving through space at some unknown speed V...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. The Relativator was sold by (as printed) Atomic Laboratories, Inc. 3086 Claremont Ave, Berkeley 5, California , which seems to be a division of Cenco Instruments (Central Scientific Company)... Source: https://www.physicsforums.com/insights/relativator-circular-slide-rule-simulated-with-desmos/ by @robphy
In Philippe G. Ciarlet's book 'An introduction to differential geometry', He gives the integrability conditions of the differential equations like this: $$ \partial_{i} F_{lj}=L^p_{ij} F_{lp},\,\,\,F_{ij}(x_0)=F^0_{ij}. $$ The integrability conditions for the existence of a global solution ##F_{lj}## is: $$ R^i_{jkl}\equiv\partial_k L^i_{jl}-\partial_l L^i_{jk}+L^h_{jl} L^i_{hk}-L^h_{jk} L^i_{hl}=0 $$ Then from the equation: $$\nabla_b e_a= \Gamma^c_{ab} e_c$$ Using cartesian basis ## e_I...

Similar threads

Back
Top