# Tag Info

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Where there is matter there is energy and vs versa.

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Because if your manifold is not differentiable (and even then, at least $C^3$), you end up doing non-linear distribution theory and having to use Colombeau algebras, and trust me, you do not want that.

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Do clocks measure conformal time? No. I would venture to say that clocks don't literally measure time at all. A clock isn't some cosmic gas meter with time flowing through it. What clocks do is "clock up" some kind of regular cyclical local motion and show you a cumulative result that you call the time. A pendulum clock clocks up the swings of a pendulum. A ...

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However it makes sense that gravity can't travel faster than light because of the force-carrying photons Whilst it makes sense that gravity can't travel faster than light, we don't actually know this for sure. What we do however know is that the force of gravity is not conveyed by photons. Even electromagnetic force is not conveyed by photons - hydrogen ...

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There is a thing called separation, it physically meaningful, and exists between events. There is a thing called coordinate difference, it is only meaningful in that you can use the metric to get actual separations from it. I mention the above because it might be the root causes of your misunderstandings. There is nothing wrong with your formula for the ...

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Depends on what do you mean by $\epsilon^{\tau\lambda\mu\nu}$. When it represents the Levi-Civita symbol, while you can technically perform this operation component-wise, it would be quite meaningless, since the Levi-Civita symbol is not the representation of a tensor field. However, in coordinate index-based formulations of general relativity and ...

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I'll answer to 1 Maxwell equations are already relativistic, but - in a flat spacetime. You can write Maxwell equations for a general metric $g_{\mu \nu}$ (The original Maxwell equations are formulated for a flat spacetime - $g_{\mu \nu} = \eta _{\mu \nu}$ ). One way this can be done is by the following algorithm: Transform coordinates to a local ...

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Firstly, notice that the Frequency of light used to measure time will not remain constant. I wont use this but do take note. Secondly notice that If the object is rigid, then and One end is fixed at x = 0, then the other end in the co-ordinate systems used will be at $\Delta x = \frac{\Delta L}{a(t)}$ So Now light using equation 1 from your question we ...

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Good question. I think the event horizon has to be absolute, because as you suggested, light either gets out or it doesn't. I venture to suggest that isn't in accord with what most here would say is current teaching, but here's a couple of interesting facts: 1) Light is not redshifted when it ascends, and nor is it blueshifted when it descends. You can work ...

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The definition of the event horizon is `the boundary of the past of future null infinity', so it is the surface beyond which nothing can escape to infinity. It isn't defined with reference to any observer. A consequence of the definition is that an observer can never really determine where the event horizon is, since its location depends on all future ...

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You might be interested to have a look at the the site The Universe in Problems. This is a community maintained web site, so the problems are very variable in style and difficulty. The downside of this is that many of the problems will not suit your current level of expertise, but on the other hand the upside is that there is bound to be some fraction of ...

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Potential energy has absolutely nothing to do with stress-energy or pressure. The following reference is a good source about the origin of the pressure term in the stress-energy tensor: "Momentum due to pressure: A simple model" by Kannan Jagannathan in American Journal of Physics 77, 432 (2009);  http://dx.doi.org/10.1119/1.3081105 Potential energy ...

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Bad question- the speed of light should be rephrased as "Speed of electromagnetic radiation" and then specify what wavelength of radiation we're talking about and which part of the neutron star is being discussed. Nobody has determined the EM properties of the inside of neutron stars so the question is not on target. Different parts of a neutron star have ...

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Any metric is locally equivalent to a symmetric matrix, which is always diagonalizable. Just taking a reference system co-rotating with the black-hole and you will get a diagonal form to the Kerr metric.

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Since $\rho$ is already a summed-over dummy index in the first equation, we can't introduce it again. Instead, multiply both sides of the first equation by $g_{\sigma\mu}$: \begin{align} g_{\sigma\mu} \Gamma^\mu_{\nu\lambda} & = -\frac{1}{2} g_{\sigma\mu} g^{\mu\rho} (\partial_\nu g_{\lambda\rho} + \partial_\lambda g_{\rho\nu} - \partial_\rho ...

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Your professor is telling you something that is absolutely fundamental to a proper understanding of relativity. Suppose we draw out the trajectory of some object on a space time graph, we may get something like this: The path traced out by the object(the blue curve) is called the world line. The length of the world line, $s$, is equal to $c\tau$, where ...

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The shortest world line (geodesic path) is given by the GR coordinate system. GR has $x_0$ (ct), $x_1$ (x), $x_2$ (y), $x_3$ (z). Often in GR, c is set to 1 (c = 1). Distance squared ($ds^2$) is given by $$ds^2 = -dt^2 + dx^2 + dy^2 + dz^2$$. If one second passes, the shortest distance is to stay in the same spatial location, that is $$ds^2 = -dt^2$$ ...

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That discussion is about a spherical shell and does not apply to the Earth nor to the sun. In the latter situations assuming uniform mass distribution we know that time is dilated by $t(\infty)/t(r)=\sqrt{g_{00}}=\sqrt{1 - \frac{2GM}{rc^2}}$, where $r$ is the distance from the center. For a spherical SHELL however we have to go back and forth between the GR ...

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Space-time is founded on the idea that mass/energy do not change, but that space, that which contains all things, changes. Consequently the concept of time is reflective of this change in space. In other words, the speed of C is constant, and consequently so is mass. The only thing that is relative is the changing of space and our idea of time. "Seeing" ...

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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 ...

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In order to be able to even define a metric, you need tangent vectors, since these are the arguments to the metric, and to have tangent vectors you need differentiability.

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A very short answer might be that in general relativity, spacetime can be curved. To estimate how much it's curved, you need to be able to calculate the rate of change, that is done by differentiating the co-ordinate system you are using to map each region of spacetime you are dealing with.

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How efficient would be these energy extraction processes when applied to micro black holes? I fear the efficiency would be zero percent. See the Blandford–Znajek process on Wikipedia where you can read that the power can be estimated as the energy density at the speed of light cylinder times area: $$P=B^2\left(\frac{r}{r_c}\right)^4 r_c c=\frac{B^2 r^4 ... 1 You set \rho equal to one for no reason. In detail the expression for \Gamma is:$$\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].$$And since g_{01} equals zero (since your metric is diagonal), all four partial ... 3 As you travel through the warped spacetime, you would not notice much difference. This is because any spacetime regions you are likely to ever encounter look exactly the same as flat spacetime locally. This is great news for us because it means we'd always be able to assume that propelling ourselves forward will actually make us go forward and not to the ... 0 In light of the comments above, regarding the time energy relation, I would need to read more about the time energy relation and later rephrase this question. Normally I would delete the question, but it may be of value to other newbies like myself, making the same wrong assumption that the time energy relation is equivalent to the position momentum ... 3 The theorem does not apply, as we do not have spherical symmetry. All we have is rotational symmetry about a preferred axis. In fact, the gravitational field outside a rotating object will be Kerr, which only reduces to Schwarzschild in the case of no rotation. Otherwise, there will be time-space terms in the metric, making it not static. Still, Kerr is ... 0 It is quite easy to make a solution with unequal masses. First start with a single Schwarzschild solution of mass m and extend it however you like for areal coordinate r<3m. Then take a Schwarzschild solution for a mass M in the range m3m and the solution for mass m in the region r<3m and sew them together at the region r=3m. Since M>m this requires ... 1 You can always transform to the coordinates of the traveler which experiences no motion in space but only in time (with proper time!). That way  d\tau=dt  and  dx=dy=dz=0  .  |u|^2 = \eta_{\alpha \beta} u^{\alpha} u^{\beta}=-1  - which is a scalar. Scalars are invariant under any coordinate transformation, so even after you return to your original ... 5 Use:$$ u_{\alpha}= g_{\alpha\beta}u^{\beta} $$where g_{\alpha\beta} is the metric tensor. 0 The way the equations are presented seems unnecessarily obscure, as there are only two equations that matter:$$ \frac{d^2r}{dt^2} = \frac{-4M^2+2Mr+(r-5M)r^3}{r^3}\,\dot{\phi}^2  \frac{d^2\phi}{dt^2} = \frac{2(-3M+r)}{(2M-r)r} \, \dot{r}\dot{\phi} $$These come from the geodesic equation expressed using coordinate time. So you start at some ... 2 is there any requirement to determine (and possibly correct for) the perturbation, or "shift", of any and all primary frequency standards, besides the described "shift due to ambient radiation"? Yes. These are called "systematic errors" and they are the order of business pretty much all day, every day, at the metrology labs that implement frequency ... 2 First, as has been said in the comments, this has nothing to do with General Relativity per se and can be perfectly explained within Newtonian gravity. The answer is yes, depending on what you mean by weight, since, after all, the building will pull you to the side. Weight is a force and forces are vectors; in this case, your weight will be longer and ... 1 Most G.R.textbooks introduce time as an extra dimension... So although I can not mentally imagine this, I think of it as an extra line, "orthogonal" to the 3 normal space dimensions. Don't. It's a dimension in the sense of measure, not in the sense of freedom of motion: I can hop forward a metre but you can't hop forward a second. Think about what a clock ... 5 Everyone who has been interested in modern science has heard explanations (certainly simplifications) of general relativity, mostly that space is curved. I'm afraid that those explanations that say space is curved are misleading. See Baez: "Similarly, in general relativity gravity is not really a 'force', but just a manifestation of the curvature of ... 0 You can arrive at equations (2) and (3) by plugging the Robertson-Walker metric into the Einstein field equations. The universe is modelled as a perfect fluid so the energy-momentum tensor is (T_{ab})=\text{Diag}(\rho,P,P,P) and the first equation comes from the T_{00} (time) part of the field equations, the second from the spacial part once you have ... 0 "Straight lines" in curved space are geodesics. But the geodesics that define particle paths are in the pseudo-Riemannian manifold of space-time, they are not geodesics in space! As a planets path is not closed in space-time (but some kind of spiral), massive bodies do not induce loops in the topology of the space-time. (Actually the loop is not even closed ... 17 Calling orbits loops is a dangerous line of thinking. Objects that are not under the influence of other forces follow geodesics, which are the curved space equivalent of straight lines. And, while it's tempting to say that the orbit of a planet is effectively a loop in spacetime, let me try to convince you why such a simplification should be avoided. Yes, ... 14 No. A loop has to start and end at the same point. In GR that means it has to start and end at the same spacetime point i.e. the same point in time as well as the same point in space. Such loops are called closed timelike curves, and with the exception of some obviously non-physical geometries they do not exist. 1 Your understanding of time is roughly correct - it is another direction, "orthogonal" to our usual 3 space directions. But keep in mind that you can think of it purely mathematically as well. The position of an object as a function of time can be written as x(t), y(t), and z(t), but why not write that exact same thing in four dimensions?$$(t,x,y,z)$$... 0 If a theory contains divergences, there is no need to assume it is plain wrong. It probably won't be the full theory of everything, but that's the only criticism that can be leveled at it. It can still be in perfect agreement with experiments below very-high-energy. The issue with QED is that QED is not a complete theory. In the real world, QED is just a ... 1 In a classical context, LRL vector is conserved only for potentials behaving like \frac{k}{r}, indeed we can see the general construction of LRL vector : \begin{eqnarray} \frac{d\vec{p}}{dt}\times \vec{L} &=& -\partial_r v(r) \frac{\vec{r}}{r} \times \vec{L}, \nonumber\\ \mu r^3 \frac{d\hat{r}}{dt} &=& ... 1 TL;DR: It may be helpful to think of the above-diagonal orange components as "energy flux" rather than "momentum density"; if you do this, the interpretations in terms of shears and pressures become more natural. Here's another way to think of the stress-energy tensor. First, you're hopefully familiar with the notion of the energy-momentum four-vector: ... 3 Constraint equations, in the sense of the "Einstein constraint equations", arise when you try to write out an initial value formulation of GR. The idea behind an initial value formulation, in general, is to show that if I hand you a set of data about the fields everywhere in space at some initial moment in time, that you can always find a solution to the ... 2 First off, traveling at constant velocity in flat spacetime is not the same as traveling g in a uniform circular motion. Quite the contrary, free falling towards the gravitational source is actually equivalent to moving with constant velocity in flat spacetime. This is so because the objects are following a geodesic path defined by the geodesic equation. I ... 1 This shows the situation as viewed by the Schwarzschild observer i.e. an observer far from the black hole: (Note that the angle \theta is not connected to the Schwarzschild \theta coordinate.) The angle \theta is (obviously) given by:$$ \tan\theta = \frac{b}{a} = \frac{b}{r_2 - r_1} $$But we've calculated the angle using the Schwarzschild r ... 2 The Schwarzschild metric describes the geometry of the spacetime containing a time independant spherically symmetric mass and nothing else. In other words the spacetime has to have existed unchanged for an infinite time and continue to exist unchanged for an infinite time, and there must be nothing else in the universe. Obviously there is no object in the ... 1 By "derived the Schwarzschild metric" I assume you mean calculating the exterior solution of the form$$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 ...

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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 ...

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Page 25 is in the section on special relativity. In section 2.1 the $\xi$ is basically a label based on the initial position. If you think of it as a function of $x^\mu$ then on page 23 there are nonzero $\xi^i_{,\mu}$. I'd you are referring to the equation after eq. 2.7 the author is tracking the motion if a particular label, and that is why there is no ...

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