JavaScript is required to use Bungie.net

Destiny 2

Discuss all things Destiny 2.
Edited by SwetiYeti: 3/10/2018 2:25:39 AM
79

The vanguard has advanced tech

Okay. The technology that the vanguard has is beyond incredible. You know when we went to the almighty and destroyed it? Zavala, Cayde, and Ikora were talking to us the whole time. I bet most of you are like: "big whoop Thatch" Well let me remind you that the almighty is NEXT TO THE SUN. And the vanguard? Oh yeah THEY'RE ON EARTH. So if you guys want to learn some basic science, it takes light 8 minutes to get from the sun to earth. Therefore it takes Electromagnetic waves 8 minutes to get from earth to the sun (without a medium) So, to be able to talk to somebody 8 light minutes away AND HAVE AN IMMEDIATE RESPONSE TAKES SOME NEXT LEVEL UNIVERSE BENDING WORMHOLE TRAVELING TECH BECAUSE THAT IS PRETTY MUCH IMPOSSIBLE IF YOU JUST RELATE IT TO "send transmission from point a to point b". So basically, If they have the: -tech to transport radiowaves THROUGH SPACE AND TIME WITHOUT DELAY -the tech to make you feel gravity at 9.8m/s^2 wherever you go (don't even get me started on this) Then they should have the tech to dominate the system if not the galaxy. Come on vanguard, use your noggins. Edit: AND THE SHIPS, THE SHIPS ARE HIGHLY ADVANCED TOO

Posting in language:

 

Play nice. Take a minute to review our Code of Conduct before submitting your post. Cancel Edit Create Fireteam Post

View Entire Topic
  • Well if we look at this: {R_{\mu \nu }-{\tfrac {1}{2}}R\,g_{\mu \nu }+\Lambda g_{\mu \nu }={\frac {8\pi G}{c^{4}}}T_{\mu \nu }\,.} {T_{\mu \nu }^{\mathrm {(vac)} }=-{\frac {\Lambda c^{4}}{8\pi G}}g_{\mu \nu }\,.} {\rho _{\mathrm {vac} }={\frac {\Lambda c^{2}}{8\pi G}}} If the energy-momentum tensor Tμν is that of an electromagnetic field in free space, i.e. if the electromagnetic stress–energy tensor {T^{\alpha \beta }=\,-{\frac {1}{\mu _{0}}}\left({F^{\alpha }}^{\psi }{F_{\psi }}^{\beta }+{\tfrac {1}{4}}g^{\alpha \beta }F_{\psi \tau }F^{\psi \tau }\right)} -is used, then the Einstein field equations are called the Einstein–Maxwell equations (with cosmological constant Λ, taken to be zero in conventional relativity theory): Additionally, the covariant Maxwell equations are also applicable in free space: Fαβ;β=0 F[αβ;γ]=1/3(Fαβ;γ+Fβγ;α+Fγα;β)=1/3(Fαβ,γ+Fβγ,α+Fγ, β)=0 {{\begin{aligned}{F^{\alpha \beta }}_{;\beta }&=0\\F_{[\alpha \beta ;\gamma ]}&={\tfrac {1}{3}}\left(F_{\alpha \beta ;\gamma }+F_{\beta \gamma ;\alpha }+F_{\gamma \alpha ;\beta }\right)={\tfrac {1}{3}}\left(F_{\alpha \beta ,\gamma }+F_{\beta \gamma ,\alpha }+F_{\gamma \alpha ,\beta }\right)=0.\end{aligned}}} where the semicolon represents a covariant derivative, and the brackets denote anti-symmetrization. The first equation asserts that the 4-divergence of the two-form F is zero, and the second that its exterior derivative is zero. From the latter, it follows by the Poincaré lemma that in a coordinate chart it is possible to introduce an electromagnetic field potential Aα such that {{\begin{aligned}{F^{\alpha \beta }}_{;\beta }&=0\\F_{[\alpha \beta ;\gamma ]}&={\tfrac {1}{3}}\left(F_{\alpha \beta ;\gamma }+F_{\beta \gamma ;\alpha }+F_{\gamma \alpha ;\beta }\right)={\tfrac {1}{3}}\left(F_{\alpha \beta ,\gamma }+F_{\beta \gamma ,\alpha }+F_{\gamma \alpha ,\beta }\right)=0.\end{aligned}}} F αβ = A α ; β − A β ; α = A α , β − A β , α {F_{\alpha \beta }=A_{\alpha ;\beta }-A_{\beta ;\alpha }=A_{\alpha ,\beta }-A_{\beta ,\alpha }} in which the comma denotes a partial derivative. This is often taken as equivalent to the covariant Maxwell equation from which it is derived. However, there are global solutions of the equation which may lack a globally defined potential. First, the determinant of the metric in 4 dimensions can be written: det ( g ) = 1 24 ε α β γ δ ε κ λ μ ν g α κ g β λ g γ μ g δ ν {\displaystyle \det(g)={\tfrac {1}{24}}\varepsilon ^{\alpha \beta \gamma \delta }\varepsilon ^{\kappa \lambda \mu \nu }g_{\alpha \kappa }g_{\beta \lambda }g_{\gamma \mu }g_{\delta \nu }\,} using the Levi-Civita symbol; and the inverse of the metric in 4 dimensions can be written as: g α κ = 1 6 ε α β γ δ ε κ λ μ ν g β λ g γ μ g δ ν det ( g) . {\displaystyle g^{\alpha \kappa }={\frac {{\tfrac {1}{6}}\varepsilon ^{\alpha \beta \gamma \delta }\varepsilon ^{\kappa \lambda \mu \nu }g_{\beta \lambda }g_{\gamma \mu }g_{\delta \nu }}{\det(g)}}\,.} Substituting this definition of the inverse of the metric into the equations then multiplying both sides by det(g) until there are none left in the denominator results in polynomial equations in the metric tensor and its first and second derivatives. The action from which the equations are derived can also be written in polynomial form by suitable redefinitions of the fields. In differential geometry, the Einstein tensor (named after Albert Einstein; also known as the trace-reversed Ricci tensor) is used to express the curvature of a pseudo-Riemannian manifold. In general relativity, it occurs in the Einstein field equations for gravitation that describe spacetime curvature in a manner consistent with energy and momentum conservation. So, basically Space Magic.

    Posting in language:

     

    Play nice. Take a minute to review our Code of Conduct before submitting your post. Cancel Edit Create Fireteam Post

    19 Replies
    You are not allowed to view this content.
    ;
    preload icon
    preload icon
    preload icon