# What can space-time bend

Here is indirect evidence that "photons" bend space-time. Look at that * Peres metric * (I use $ c \ equiv 1 $ span> and the $ (1, -1, -1, -1) $ span> convention): \ begin {equation} \ tag {1} ds ^ 2 = dt ^ 2 - dx ^ 2 - dy ^ 2 - dz ^ 2 + F (x, y, t - z) (dt - dz) ^ 2, \ end {equation} span> where $ F (x, y, u) $ span > any function of three independent variables ($ u = t - z $ span>). Plug this metric into Einstein's equation. First: without stress tensor (and without cosmological constant): \ begin {equation} \ tag {2} G _ {\ mu \ nu} = 0. \ end {equation} span> After some algebra you will get a restriction for $ F (x, y, u) $ span>: \ begin {equation} \ tag {3} \ frac {\ partial ^ 2 F} {\ partial x ^ 2} + \ frac {\ partial ^ 2 F} {\ partial y ^ 2} = 0. \ end {equation} span> Therefore $ F $ span> must be a * harmonious function * in $ x $ span> and its $ y $ span>. The simplest non-trivial solution (with space-time curvature) is a linear superposition of quadratic functions (there are two independent polarization states for the gravitational wave): \ begin {equation} F (x, y, u) = \ mathcal {A} (u) (\ , x ^ 2 - y ^ 2) + \ mathcal {B} (u) \, x \, y, \ end {equation} span> where $ \ mathcal {A} (u) $ span> and $ \ mathcal {B} (u) $ span> are secondary functions of $ u = t - z $ span>. The metric (1) then describes a * planar gravitational wave *that spreads in a vacuum.

Then add a * planar monochromatic electromagnetic wave * with energy impulse added \ begin {equation} \ tag {4} T _ {\ mu \ nu} = \ Phi (x, y, u) \, k _ {\ mu} \, k _ {\ nu}, \ end {Equation} span> where $ k ^ {\ mu} = (\ omega, 0, 0, \ omega) $ span> is the wave number and $ \ Phi (x, y, u) $ span> is arbitrary. Einstein's equation then becomes \ begin {equation} \ tag {5} G _ {\ mu \ nu} = - \, \ kappa \, T _ {\ mu \ nu}. \ end {equation} span> Of course $ \ kappa \ equiv 8 \ pi G $ span>. A lot of algebra has the following restriction: \ begin {equation} \ tag {6} \ frac {\ partial ^ 2 F} {\ partial x ^ 2} + \ frac {\ partial ^ 2 F} {\ partial y ^ 2} = 2 \ kappa \ omega ^ 2 \, \ Phi. \ end {equation} span> I am looking at a planar monochromatic electromagnetic wave that propagates in space-time with a circular polarization (this is a classic field that comes closest to a "quantum photon" with an angular frequency $ \ omega $ span>) : \ begin {equation} \ tag {7} A ^ {\ mu} (x, y, u) = \ varepsilon_1 ^ {\ mu} \, \ mathcal {F} (u) + \ varepsilon_2 ^ {\ mu} \, \ mathcal {G} (u), \ end {equation} span> where $ \ varepsilon_ {1, \, 2} ^ {\ mu} $ span> are the four vectors of the space-like polarization, orthogonal to $ k ^ {\ mu} $ span> and \ begin {align} \ tag {8} \ mathcal {F} (u) & = a_0 \ cos {(\ omega \, u)}, & \ mathcal {G} (u) & = a_0 \ sin {(\ omega \, u)}. \ end {align} span> The amplitude $ a_0 $ span> is just a constant. It is easy to check whether (7) and (8) (4) with $ \ Phi (x, y, u) = \ text {cste} \ propto a_0 ^ 2 $ result in span>. Then (6) can be solved to get a simple, non-trivial solution (the Riemann curvature tensor is not 0): \ begin {equation} \ tag {9} F (x, y, u) = \ frac {\ kappa \, a_0 ^ 2 \, \ omega ^ 2} {8 \ pi \ alpha} \, (\, x ^ 2 + y ^ 2). \ end {equation} span> ($ 4 \ pi \ alpha $ span> is the electromagnetic coupling constant that occurs in the energy-momentum tensor. It depends on your preferred units for the field amplitude $ a_0 $ span>. I use the Fine structure constant $ \ alpha \ approx \ frac {1} {137} $ span>). The metric (1) with the function (9) then describes a circularly polarized EM wave (and the associated gravitational wave) that moves in space-time The Riemann curvature is not 0 (its components are constants in this case, since the wave energy momentum is homogeneous).

The * not localizable * "Photon" bends space-time in a non-trivial way. Due to the circular polarization, the curvature is homogeneous (but not isotropic, since the wave propagation defines a privileged orientation)

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