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The first definition transforms as a four-vector: $\dfrac{dx^{'\mu}}{d \tau} = \Lambda^{\mu}{}_{\nu} \dfrac{dx^{\nu}}{d \tau}$. The second definition transforms not quite as a four-vector: $\dfrac{dx^{'\mu}}{d t'} = \dfrac{dt}{dt'} \Lambda^{\mu}{}_{\nu} \dfrac{dx^{\nu}}{d t}$. This makes sense, since in the first definition you divide the differentials of ...

6

@Milan already answered the technical problems of your definition. I would like to point out conceptual problems. We would like the 4-velocity to somehow characterize the movement of an object through spacetime. Conceptually it makes sense to demand, that such quantity should depend only on the quantities that have direct relation to that movement. So ...

4

No. The right side of your equation is meaningless. You have to contract indices two at a time, not four at a time. Otherwise you don’t get a Lorentz-invariant result.

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As a third, but equivalent, perspective to the existing answers: When in doubt, you can always put the summation back explicitly. (Just take it back out once you understand the answer and before getting evil looks from your peers for not having used Einstein summation!) In your case, you would have $$\sum_a \sum_b g^{ab} g_{ab} = ? = \sum_a \sum_a g^{aa} ... 4 The negative curvature is the output, not the input. To calculate the curvature invariant (which is called the Kretschmann scalar K) you have to compute$$K=R_{\mu\nu\lambda\rho} \ R^{\mu\nu\lambda\rho}$$For a rotating black hole (here with a=M) you get regions with positive and negative curvature: The bold black lines show the hard crossings between ... 3 The Hamiltonian is not always the sum of potential and kinetic energies. Like the Lagrangian, it is a theoretical construct that is not unique for any given physical system, and you can always construct a time-reparametrization invariant formulation of a system where the Hamiltonian vanishes. See this answer of mine for the construction and the first part of ... 3 Eq. (3) is a consequence of the symmetry imposed by the principle of relativity. I hope my proof is what you are looking for. First of all, we assume (x'-ct') can be expressed as a function of (x- ct),$$x' - ct' = \Phi(x-ct) \quad ,$$with \Phi being at least invertible and C^1. Side remark : From the technical point of view this follows from ... 3 As @G.Smith pointed out in his answer, g^{aa}g_{aa} doesn't make sense. You can calculate g^{ab}g_{ab} as a special case of raising and lowering indices. Contracting the b indices first you get$$g^{ab}g_{ab}=\delta^a_a=...$$I leave out the final step as an exercise for you. 3 i am wondering if saying that we “move through spacetime at the speed of light” is something genuinely derived or just a matter of definition I would say that the truth is somewhere in the middle. The four-velocity is defined as$$\frac{dx^{\mu}(\tau)}{d\tau}$$We can look at this expression from the perspective of Newtonian physics. The top is a change in ... 3 This is an issue that confused me when I was learning general relativity. In special relativity, we put so much emphasis on the physical meanings of coordinates, while in general relativity we treat them as being almost completely arbitrary. As you know, the point is that once you have spacetime curvature, you can't define the usual network of clocks and ... 2 Other answers already stated that in standard covariant formulation of general relativity there is only one metric. Nothing in MTW quotes suggests otherwise, only different notation for spacelike / timelike separation. At the same time I would like to mention that in the nonrelativistic, Newtonian limit, spacelike and timelike displacements become ... 2 Your first equation is a bit of an abuse of notation (although it is sometimes used). A better expression would be$$t_A\cdot t_B$$if it is clear that t_A, t_B are vectors or$$(t_A)^\mu (t_B)_\mu$$using Einstein's summation convention (as you already did). In the same vein your second equation is an abuse of notation. However, there it is even worse ... 2 This question is quite old, but let me give it a shot. In general, as you pointed out, all coordinate charts are "equally good", as long as the physical processes that you're interested in take place in the domain of your chart. If you want to calculate the orbit of body around a black hole, the precession or mercury, the gravitational redshift and so on, ... 2 This is because you work in first order in \varepsilon. So you need to consider only the linear terms in \varepsilon:$$ \delta g^{\mu\nu}(x)=\bar g^{\mu\nu}(x)-g^{\mu\nu}(x)=-\frac{\partial{ \bar{g}^{\mu\nu}}}{\partial x^{\alpha}}\varepsilon^{\alpha}+ \frac{\partial \varepsilon^{\mu}}{\partial x_{\nu}}+\frac{\partial \varepsilon^{\nu}}{\partial x_{\mu}}...

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For fundamental questions about time you must refer to the fundamental notion of proper time instead of coordinate time. One essential difference between general relativity and Newton's system of space and time is the fact that instead of one absolute time concept there are two time concepts - coordinate time and proper time. Both time concepts are linked ...

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The Schwarzschild interior solution is a non-physical model that that gives the correct qualitative prediction for the interior metric of astrophysical bodies only within a certain parameter range. It is constructed by assuming that the density of the material of the "Schwarzschild star" is constant and the pressure profile is assumed to automatically ...

2

The correct relation is $$\partial_\mu \left(\frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)}\right)= \partial_0 \left(\frac{\partial \mathcal{L}}{\partial(\partial_0 \phi)}\right)+\partial_i \left(\frac{\partial \mathcal{L}}{\partial(\partial_i \phi)}\right).$$ Essentially you're asking if $$A_\mu B^\mu \overset{?}{=} A_0 B^0 + A_i B^i,$$ or $$A_\... 2 Only diffeomorphism invariant quantities are measurable. The metric does not have this property. ''every observer has a local proper time that he/she can measure on their clock, and proper distances that they can measure with a meter stick much shorter than the radius of curvature. If I insist on using these coordinates to describe my space'' These are not ... 2 I think your confusion is most concisely caught in this part of your question: If I insist on using these coordinates to describe my space, I don't see how there can be any remaining gauge freedom. Without getting into the argument that's emerged in comments to your question about whether your particular choice of coordinates makes sense, it is generally ... 1 If there was a material that could resist collapse so that r_g=r_s were possible it would mean that this material would have a divergent bulk modulus. The bulk modulus$$ B=-V\frac{dP}{dV} $$for an isentropic process is PV^\gamma=k a constant and so B=\gamma P. However, if the collapse stops at the horizon it means there is a large outward ... 1 The interior Schwarzschild metric models a fluid in equilibrium, and it is clearly static. It doesn't represent a collapsing body; for that you need to take a look at, for example, the Oppenheimer-Snyder solution, which models a ball (or was it a shell?) of presureless dust collapsing to a black hole. 1 OP's Hamiltonian of the form$$ H~=~\frac{p^2+m^2}{2}, \qquad p^2~:=~p_{\mu}g^{\mu\nu}(x) p_{\nu}~\leq~0,\tag{1} $$is called a super-Hamiltonian in e.g. MTW, cf. e.g. this Phys.SE post. (For a massless particle like the photon m=0.) The super-Hamiltonian (1) is the e=1 gauge of the Hamiltonian$$H~=~ \frac{e}{2}(p^2+m^2)\tag{2}$$for a relativistic ... 1 It is often said that the event horizon of a Schwarzschild black hole is lightlike. Is this correct and, if so, what exactly does this mean? It means that only lightlike geodesics can stay stationary at the horizon, all other geodesics must fall in (or escape, assuming there are space-like geodesics as well) and the proper time intervall they can spend at ... 1 To account for the minus sign, one can e.g. use the following trick$$ -\det g~\stackrel{(2)}{=}~+\det M, \tag{1}$$where the matrix$$M^{\mu}{}_{\nu}~:=~(\eta^{-1})^{\mu\lambda} g_{\lambda\nu}\tag{2} has positive determinant, and \eta_{\mu\nu} is a fixed fiducial reference metric of determinant -1, e.g. the Minkowski metric. Then the infinitesimal ... 1 Using indices can be confusing. You are actually calling g_{\mu \nu} the metric tensor as a matrix, not a particular element of this matrix. Your derivation is correct until the second last step. It becomes clearer if you use the substitution/notation \begin{align} g_{\mu \nu} &\to g \\ g^{\mu \nu} &\to g^{-1} \\ \delta g_{\mu \nu} &\to \... 1 Use the chain rule to evaluate the variation of the logarithm: let F[X] be a functional of X = X(\tau). Then\delta \log(F[X]) = \frac{1}{F[X]} \delta F[X]$$In your example (suppressing indices on the metric):$$ \delta \log(-\det(g)) = \frac{1}{\det(g)} \delta \det(g) $$You should then use Jacobi's formula:$$ \delta \det(g) = \det(g)g^{\mu\nu}\...

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In GR, the proper distance is a property of curves connecting two points, not of the points by themselves. If two points are causally disconnected, then you can define a "distance" between them as the minimum proper distance over all the spacelike curves that connect them (which will necessarily be attained by a spacelike geodesic). But this doesn't really ...

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Let me first write down my interpretation of the notation you've used, then I'll proceed to write down my solution. I guess $t_A=t^\mu_A {\partial \over \partial x^\mu}={\partial x^\mu \over \partial \xi^A}{\partial \over \partial x^\mu}={\partial \over \partial \xi^A}$. So, $t_{A,B}={\partial^2 \over {\partial \xi^A \partial \xi^B}}$ and \$t_{A,B}\cdot t_D=...

1

Yes, you can talk about there being two metrics, which are coincident for a proper relativistic spacetime - but whose distinction becomes important when considering the classical limit. This is similar to the case for quantum mechanics, where the classical limit requires us to bifurcate the wave function to separate positional and momental wave functions, so ...

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