# Tag Info

109

Nice discovery! The formula for time dilation outside a spherical body is $$\tau = t\sqrt{1-\frac{2GM}{c^2r}}$$ where $\tau$ is the proper time as measured by your object at coordinate radius $r$, $t$ is the time as measured by an observer at infinity, $M$ the mass of the spherical body, and $G$ and $c$ the gravitational constant and the speed of light. ...

17

There seem to be several confusions here. Massive and massless particles behave qualitatively differently, even if the massive particle is traveling very fast. The minimum radius for a stable orbit for a massive particle is $3 r_s$. Circular orbits above this radius are all stable. Massless particles only have circular orbits at the photon sphere, $(3/2) ... 7 Thanks to the hint given by knzhou I figured out that if one wants to give the particle a proper initial velocity of$v_0$, the initial velocity in terms of Schwarzschild coordinates$v_i$would then be $$v_i = \frac{v_0}{ \color{green}{\sqrt{ 1-v_0^2/c^2}}\cdot \color{blue}{\sqrt{1-r_s/r_0}}}$$ for the transversal component, and the same without the blue ... 6 The metric reads, restoring$c$, $$\mathrm{ds}^2 = -(1+gz/c^2)^2 c^2\mathrm{dt}^2 + \mathrm{dz}^2 + \mathrm{dx}^2\:.$$ The metric can be used to determine the geodesics by means of the associated quadratic Lagrangian $$L = -(1+gz/c^2)^2 c^2\dot{t}^2 + \dot{z}^2 +\dot{x}^2\tag{0}$$ where the dot denotes the derivative with respect to the affine parameter$s$. ... 5 The spacetime interval is a relativistic invariant, and is proportional to the travelers proper time. So in a since you are traveling one second per second, per your own wrist-watch. Every other measurement would be the speed of some other inertial reference system, measured with your clock. Let$s^2 = x^2 + y^2 +z^2- (ct)^2$, where$x$,$y$,$z$are ... 4 It's because you have to replace the erroneous$\gamma$by$\gamma^2$(and similarly$m$by$m^2$) in the inner product and because $$\gamma^2 m^2 c^2 - \gamma^2 m^2 v^2 = \gamma^2 m^2(c^2-v^2)=\dots$$ and $$\gamma^2 = \frac{1}{1-v^2/c^2} =\frac{c^2}{c^2-v^2}$$ and$c^2-v^2$from the explicit factor cancels against the denominator of$\gamma^2$, while ... 4 To find points where local flatness breaks down, a general strategy is to calculate the curvature tensor$R_{\mu\nu\rho\sigma}$, and find the locus of various singularities (points where$R_{\mu\nu\rho\sigma}$becomes unbounded). Except in rare cases (i.e. unusual cancellation), singular behavior of$R_{\mu\nu\rho\sigma}$is apparent in the Ricci scalar ... 3 Indeed,$f$is a symmetric form, since$\omega$and$\omega '$are Grassmann-even: $$(\text dx \wedge \text d y)\wedge (\text d z \wedge \text d t)=(\text d z \wedge \text d t)\wedge(\text dx \wedge \text d y)$$etc.. Now, to calculate the signature, you should find a basis which diagonalizes$\omega$, the dimension of the space is$6$. A basis is given ... 2 A general diffeomorphism does not map geodesics to geodesics. Some simple counter examples You can a build diffeomorphism on the Euclidean plane by imagining putting one finger on a tablecloth at point$x$and dragging it. This map is clearly smooth, a smooth inverse is constructed by dragging your finger back. Any geodesic on the plane (a line) passing ... 2 There are lots of ways of approaching special relativity. My own preferred approach is the invariance of the line element. Suppose you move a small distance in spacetime$(dt, dx, dy, dz)$then the length of the line element$ds$is defined by: $$ds^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2 \tag{1}$$ This equation is known as the metric equation and is derived ... 2 Yes the Schwarzschild metric describes the spacetime geometry around the Earth, and I describe how to use the geodesic equation to describe objects falling in Earth's gravity in How does "curved space" explain gravitational attraction?. An example of how the Schwarzschild metric describes the Earth's gravitational field is the time dilation of GPS ... 2 The acceleration should be $$a = \frac{G\cdot M}{r^2 \cdot \sqrt{1-r_s/r}}$$ with$r$as the height above the center of mass and the Schwarzschildradius $$r_s = \frac{2\cdot G\cdot M}{c^2}$$ The force to hold the ball at rest is $$F=m\cdot a$$ As one can see it now takes an infinite force and energy to keep a body at a fixed height when$r=r_s$. 2 I will answer this with a simple example. Let us consider the metric for weak gravity, $$ds^2 = \left(1 - \frac{2GM}{rc^2}\right)c^2dt^2 - dr^2 -r^2d\Omega^2.$$ The$g_{tt}$metric element is largest by a factor of$c^2$and we have $$\Gamma^r_{tt} = \frac{1}{2}g^{rr}\partial_r g_{tt} = \frac{GM}{r^2}.$$ Now let us work with the geodesic equation that is ... 2 The metric is usually defined this way because of the speed limit (c). Imagine that: If you are centered at the origin of a coordinate frame (x,y,z) a emit a signal, this signal would never travel a distance bigger than the distance the light would travel. So, $${|r|} ^{2} = {x}^{2} + {y}^{2} + {z}^{2}$$ And set it equal to the distance the light would ... 1 The true meaning of all the trouble is that coordinates are just tools for calculations and don't have any intrinsic meaning in GR. You should never interpret the coordinates alone. These can be very misleading and in particular become singular in some regions of spacetime (because coordinates are local). And never trust the labels of coordinates: Fact that ... 1 Invariants are useful, in general, because they represent something that all observers can agree upon. Relativity showed us that the concept of time-intervals, spatial-distances, and even sequences of events can be drastically different from different observers. So how can one observer 'relate' to another? I.e. how could I, standing still, figure out what ... 1 The spacetime interval invariance property allows us to, for example, compare the rate of time passing for two observers moving at relative velocities to each other. Although no observer in the universe is at complete rest, the interval is a benchmark for comparison of the physical effects of differences in velocity, or indeed location. Say one observer is ... 1 The reason to consider such metrics was not a particular interpretation, but that for this ansatz one could hope to find some exact solutions. It is known as the Kerr-Schild metric, and its role in the process of finding exact solutions has been described by Kerr in Wiltshire, Visser, Scott (eds.), The Kerr Spacetime. One can, of course, try to develop ... 1 $$\nabla_\mu A^\nu = g^{\nu\alpha} \nabla_\mu A_\alpha = g^{\nu\alpha} ( \partial_\mu A_\alpha - \Gamma^\lambda_{\mu\alpha} A_\lambda ) = g^{\nu\alpha} \partial_\mu ( g_{\alpha\beta} A^\beta ) - g^{\nu\alpha} \Gamma^\lambda_{\mu\alpha} g_{\lambda\beta} A^\beta$$ This simplifies to$$\nabla_\mu A^\nu = \partial_\mu A^\nu + ( g^{\nu\alpha} \partial_\mu ... 1 Explicitly working with the components of$g$and$\Gamma$can be messy. Here is a nicer way to derive the coordinate expression for the covariant derivative of a vector (or any tensor) without raising or lowering indices. Consider vector fields$A^\mu, B_\mu$. The covariant derivative of a scalar is the same as the coordinate derivative (in any coordinate ... 1 In special relativity there are two major assumptions: -the laws of physics are the same in all inertial frames -the speed of light that you observe is always the same, (thus independent of the relative motion between the light source and the observer). From this two assumptions follows the famous Lorentz transformations. In these Lorentz transformations ... 1 The set of transformations that leaves the speed of light unchanged is the Lorentz group. Representation theory enables us to investigate the irreducible representations of the Lorentz group. The lowest-dimensional representations act on scalars four-vectors However, take note that usually we consider representations of the corresponding Lie algebra ... 1 The concept of 'straight' is a bit ill defined in GR and has no real definition. In fact in a sense the geodesics themselves be seen as 'straight' lines; they are the shortest paths connecting 2 points (this is what in normal Euclidean space would be a 'straight line') In the LC connection they are the integral curves of some vector field$V$with$ ...

1

I do not buy into the view that spacelike geodesics are not physical entities or, at best, can only be understood in tachyon terms. Mathematically spacelike geodesic paths are well understood in 3+1 spacetime. They can be computed in principle given a metric. Given 2 events in a local (but not infinitesimal) part of manifold they can be joined by a unique ...

Only top voted, non community-wiki answers of a minimum length are eligible