Energy Momentum Tensor - General Properties

TSC
Messages
39
Reaction score
1
It is often stated and proved in textbooks that the momentum density is also the energy flux.
The explanation is often done using the dust model.
However, it is possible that in a real fluid, there is heat conduction via particle collision. There is energy flux, but since no molecules are ever transported in the IRF, how come there is momentum density?
 
Last edited:
Physics news on Phys.org
Energy and momentum in special relativity are related by the fact that their combination ##(E, \vec{P})## transforms as a 4-vector, via the Lorentz transform.

This means that what appears as "just energy" in one frame of reference, appears as energy and momentum in another. A transfer of energy in one frame of reference is a transfer of energy and momentum in another frame of reference, moving relative to the first.

This is similar to the way that time and space form a 4-vector, Time and space 'mix together" in just the same way as energy and momentum due, the mathematical formalism that describes this mixing together is the 4-vector formalism.
 
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