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5

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


3

The theorem does not apply, as we do not have spherical symmetry. All we have is rotational symmetry about a preferred axis. In fact, the gravitational field outside a rotating object will be Kerr, which only reduces to Schwarzschild in the case of no rotation. Otherwise, there will be time-space terms in the metric, making it not static. Still, Kerr is ...


3

There is a relatively fast approach to computing the Riemann tensor, Ricci tensor and Ricci scalar given a metric tensor known as the Cartan method or method of moving frames. Given a line element, $$ds^2 = g_{\mu\nu}dx^\mu dx^\nu$$ you pick an orthonormal basis $e^a = e^a_\mu dx^\mu$ such that $ds^2 = \eta_{ab}e^a e^b$. The first Cartan structure ...


3

The singularities are in a sense mere artefacts of your coordinate system, but do describe important surfaces. The reason your metric was ill defined up on those surfaces is that you picked a metric that was nice when you are far far from the black hole (all those $1/r$ terms get small and it looks like the SR metric for a spherical coordinate system). ...


3

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 ...


2

If we restrict ourselves to special relativity then the form of the Minkowski metric is an assumption. You can argue whether it is derived from the Einstein postulates or whether the Einstein postulates are derived from it, but this is really a philosophical nicety as you end up having to make equivalent assumptions either way. If you consider general ...


2

$g^{\mu\nu}$ is the inverse of the metric $g_{\mu\nu}$; and the expression is for a general metric, not for the Minkowski metric. In 4 dimensions, if $g = \det(g_{\mu\nu})$ $$\det(\sqrt{-g} g^{\mu\nu}) = (\sqrt{-g})^4\det(g^{\mu\nu}) = \frac{g^2}{\det(g_{\mu\nu})} = g$$


2

The notion of derivative requires a notion of comparison. In a general manifold, tangent vectors at different points belong to totally different vector spaces (see footnote 1), so we must define a way of mapping one tangent vector to another tangent space that we shall take, by definition to be the the "invariant image" of the vector in the new tangent space ...


2

The short answer is that calculating the Riemann Tensor is a grind. It will take a while, no matter what way you do it. Presumably you're doing the Schwarzschild metric in the standard (Schwarzschild) coordinates, so you're aided by the fact that the metric tensor is diagonal. This means that $R^\alpha_{\beta \gamma \delta} = g^{\alpha \alpha}R_{\alpha ...


2

The Schwarzschild metric describes the geometry of the spacetime containing a time independant spherically symmetric mass and nothing else. In other words the spacetime has to have existed unchanged for an infinite time and continue to exist unchanged for an infinite time, and there must be nothing else in the universe. Obviously there is no object in the ...


1

Just to complement John Rennie's answer, one can always perform a Lorentz transformation to a coordinate system such as the particle is at rest for a given time. It's called instantaneous rest frame (IRF). This frame changes point to point, unless the particle's velocity is constant. In such a frame, we have $ ds^2 = -c^2d\tau^2, $ where $\tau$ is the ...


1

In general relativity, the Minkowski metric plays a privileged role because it is the unique asymptotically flat solution to the vacuum Einstein equations that has zero ADM energy. The positive energy theorem in general relativity says all asymptotically flat spacetimes satisfying the dominant energy condition have non-negative ADM energy. Thus, one can ...


1

By "derived the Schwarzschild metric" I assume you mean calculating the exterior solution of the form $$ds^2 = -\left(1-\frac{2M}{r}\right)dt^2 + \left(1-\frac{2M}{r}\right)^{-1}dr^2 +r^2\Omega^2$$ Applications: Describing deflection of light by the sun Precession of the perihelia of the orbits of the inner planets Schwarzschild singularity and ...


1

You set $\rho$ equal to one for no reason. In detail the expression for $\Gamma$ is: $$\Gamma^1_{01}=\frac{1}{2}\sum_\rho g^{1\rho}\left[\frac{\partial g_{1\rho}}{\partial x^0} + \frac{\partial g_{\rho 0}}{\partial x^1} - \frac{\partial g_{01}}{\partial x^\rho}\right].$$ And since $g_{01}$ equals zero (since your metric is diagonal), all four partial ...


1

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 ...



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