Question

\frac{d}{d\tau}U^\alpha + \Gamma^\alpha_{\mu\nu}U^\mu U^\nu = 0. The particle is initially at rest, the initial value of its acceleration is $(dU^\alpha/d\tau)_0 = -\Gamma^\alpha_{00} = -\frac{1}{2}\eta^{\alpha\beta}(h_{\beta0,0} + h_{0\beta,0} - h_{00,\beta}).

          \frac{d}{d\tau}U^\alpha + \Gamma^\alpha_{\mu\nu}U^\mu U^\nu = 0.

The particle is initially at rest, the initial value of its acceleration is

$(dU^\alpha/d\tau)_0 = -\Gamma^\alpha_{00} = -\frac{1}{2}\eta^{\alpha\beta}(h_{\beta0,0} + h_{0\beta,0} - h_{00,\beta}).

(d)/(dτ)U^α+ Γ^αμνU^μU^ν= 0.

The particle is initially at rest, the initial value of its acceleration is

(dU^α/dτ)0 = -Γ^α00 = -(1)/(2)η^αβ(hβ0,0 + h0β,0 - h00,β).

Added by Bryan M.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Adi S

06/13/2022

Step 1

(9.22) which states that if a particle is initially at rest, the initial value of its acceleration is given by (dU"/dt) = -r^2 = -n(h0,t + h0,t - h0,t). Show more…

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Give a more rigorous proof that Eqs. (9.22) and (9.23) imply that a free particle initially at rest in the TT gauge remains at rest. Eq. (9.22): If a particle is initially at rest, the initial value of its acceleration is (dU"/dt) = -r^2 = -n(h0,t + h0,t - h0,t). Eq. (9.23):