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(1) Of course bosons can be coupled to gravity. The graviton is a spin 2 field. (2) We want to use vierbien to couple fermions to gravity because the Dirac equation is formulated in Minkowski space and we want the behavior of fermions in a general spacetime to be locally like their behavior in Minkowski space. Therefore we use the local frame ...


You can always transform to the coordinates of the traveler which experiences no motion in space but only in time (with proper time!). That way $ d\tau=dt $ and $ dx=dy=dz=0 $ . $ |u|^2 = \eta_{\alpha \beta} u^{\alpha} u^{\beta}=-1 $ - which is a scalar. Scalars are invariant under any coordinate transformation, so even after you return to your original ...


Use: $$ u_{\alpha}= g_{\alpha\beta}u^{\beta} $$ where $g_{\alpha\beta}$ is the metric tensor.


No, that transformation is not legit. Your final statement (A) says that the covariant derivative of the stress-energy tensor vanishes for every component. It has three free indices which means it represents 4$\times$4$\times$4=64 real-valued equations. Your initial statement (B) states only that four-at-a-time linear combinations of those derivatives ...


The metric times the Kronecker delta gives $$g_{ab} \delta^a_c = g_{cb}$$ Since the Kronecker delta tells us to replace the $a$ indice with $c$. We do this for both terms in your equation, $$\frac{\partial L}{\partial \dot{x}^c} = \frac{1}{2} g_{cb} \dot{x}^b + \frac{1}{2} g_{ac} \dot{x}^a$$ and then rename the dummy indices (the indices that are summed ...


Rewrite the Shouten identity by pulling down $\nu$: $$ 0 = \eta_{\mu\nu} \epsilon^{\rho\sigma\tau\lambda} + \delta_\mu^\rho \eta_{\chi\nu} \epsilon^{\sigma\tau\lambda\chi} + \cdots$$ Then plug this into the left hand side of the equation you marked with (?). You will get four terms, all having the correct structure of the right hand side (one free index at ...


I was just making a slight mistake: $ \nabla_a(u^b u_b)=\nabla_a (c^2)=0=u_b\nabla_a(u^b )+u^b\nabla_a(u_b )$ Using the raising and lowering properties of the metric: $ 0 = u_b\nabla_a(g_{bc}u_c )+u^b\nabla_a(u_b ) = g_{bc}u_b\nabla_a(u_c )+u^b\nabla_a(u_b ) = u^c\nabla_a(u_c )+u^b\nabla_a(u_b)= 2u^b\nabla_a(u_b) $ From which the result follows.

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