Deriving Multiplication Factor for Geometrized Units: Help Needed!

  • Thread starter Thread starter ofeyrpf
  • Start date Start date
  • Tags Tags
    Units
ofeyrpf
Messages
30
Reaction score
0
Geometrized units use the multiplication factor to convert for example mass to length, in this case the factor would is G/C^2. I know this is to do with setting c=G=1. How does one derive the multiplication factor from this?

Any help greatly appreciated.

Shane
 
Physics news on Phys.org
You have
c = 1 = 3e8 m/s, therefore 1 s = 3e8 m
G = 1 = 6.67e-11 m³/(kg*s²) = 6.67e-11/(3e8)² m/kg, therefore 1 kg = 6.67e-11/(3e8)² m
 
Thank you Ich. I now see where G/c^2 comes from.

Similarily would you be able to show me where \sqrt(G/(4\pi\epsilon_0))^{1/2}/c^2 come from for converting Coulombs to length? I presume it has soemthing to do with the equation for either electric field or electric force as both these contain 4\pi\epsilon_0.

Thanks for your help so far.
 
Yes I have checked out wiki. Thank you. But there they only gave the multiplication factors but not where they came from. But I know where the factor for converting from Coulumbs to length comes from now. Thanks
 
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...
Abstract The gravitational-wave signal GW250114 was observed by the two LIGO detectors with a network matched-filter signal-to-noise ratio of 80. The signal was emitted by the coalescence of two black holes with near-equal masses ## m_1=33.6_{-0.8}^{+1.2} M_{⊙} ## and ## m_2=32.2_{-1. 3}^{+0.8} M_{⊙}##, and small spins ##\chi_{1,2}\leq 0.26 ## (90% credibility) and negligible eccentricity ##e⁢\leq 0.03.## Postmerger data excluding the peak region are consistent with the dominant quadrupolar...
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
Back
Top