Antisymmetric 4-Tensor: Hodge/Duality Transformation Explained

  • Thread starter Thread starter yukcream
  • Start date Start date
yukcream
Messages
59
Reaction score
0
What is Hodge or duality trnsformation? I just know it is an operation to transform tensor of rank p into rank (n-p), where n is the dimension? I want to know more about it especially how it works in general relativity,can anyone give me some introductory reading for me? Thank you :rolleyes:

yukyuk
 
Physics news on Phys.org
I think the Hodge duality transformation is the most easily understood in terms of Clifford algebras.

There is some introductory reading about Clifford algebras

here

If you are already familiar with the standard vector dot products and wedge products, it should be fairly easy reading. If you are not already somewhat familiar with the wedge product, it may not be so easy.

Anyway, suppose you have three non-collinear vectors in a 4-d space (since we are talking about relativity). (Note that this collection of three vectors is really a three-form. I'm not sure if you are familiar with three-forms or not. The Clifford algebra article will describe three-forms in more detail if you are not already familiar with them).

There is one and only one vector that's orthogonal to all three vectors (the three-form) - you can think of it as the time vector that's associated with the volume element defined by the three non-collinear vectors.

We can make the length of the vector proportional to the volume of the pareallel piped spanned by the three vectors.

This vector is the "hodge dual". In the language of forms, it associates a 1-form with every three-form. (I called it a vector before, but it's not really a vector, its the dual of a vector, a 1-form).

You can do similar transformations with other n-forms.
 
Some suggestions [that worked for me].

First, recognize that the [antisymmetric] cross-product of two vectors, which is most naturally visualized as an oriented plane, can be thought of as a vector in 3-dim Euclidean space... with the help of the Hodge dual operation.

Next, study Hodge duality in electromagnetism.
For example,
http://farside.ph.utexas.edu/teaching/jk1/lectures/node22.html
http://farside.ph.utexas.edu/teaching/jk1/lectures/node23.html

I'd suggest these books:

Schouten - Tensor Analysis for Physicists


Bamberg & Sternberg - A Course in Mathematics for Students of Physics


Burke - Applied Differential Geometry


Schutz - Geometrical Methods of Mathematical Physics


In GR, the Hodge dual shows up when discussing curvature tensors.
Exercise: The Riemann curvature tensor has two pairs of antisymmetric indices. By dualizing each pair, one gets the "double-dual" of Riemann. Take its [nontrivial] trace. What do you get?
The answer is in MTW - Gravitation
 
Last edited by a moderator:
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...
Back
Top