# Tag Info

## New answers tagged general-relativity

0

For the first question, you do not need to interchange $\nabla_{\alpha}$ and $\nabla^{\alpha}$. We simply have $$\nabla_{\alpha}\varphi\nabla^{\alpha}\nabla_{\beta}\varphi \\ = \eta^{\alpha\gamma}\nabla_{\alpha}\varphi\nabla_{\gamma}\nabla_{\beta}\varphi \\ = \nabla^{\gamma}\varphi\nabla_{\gamma}\nabla_{\beta}\varphi \\ = ... 0 @RandyWelt Note that, first, theta cannot be pi/2 for a this specific problem. There exist no theta, J, M, r that solve the time dilation equation for theta = \pi/2. Even in my calculations, I was careful to take for example \pi/2 < \theta \leq \pi or 0 \leq \theta < \pi/2.Also, as an aside, for a Kerr metric, it is singular at points ... 1 It can be shown that \omega_{ab} = 0 \Leftrightarrow \omega^a \equiv \epsilon^{abcd}u_b \nabla_c u_d = 0. The latter quantity is known as the twist (or vorticity). In a local inertial frame it is easy to see that \vec{\omega} \sim \vec{\nabla}\times \vec{v} where \vec{v} is the 3-velocity field of the flow. This lends to the following interpretation ... 0 I think there is some confusion on your part here. a = 0.998 represents the "speed" at which the black hole is rotating, which is taken to be by Kip to be 99.8% the speed of light, i.e., it is a = 0.998*c. The "a" in the calculations for the Kerr metric, is very different. It is a length scale. This can be seen for example by considering a spherical shell ... 1 No that is not what you must prove. It is not true that if u^a is hypersurface orthogonal then \nabla_a u_b = \nabla_{(a}u_{b)}. In fact this is only true if u^a is geodesic. If u^a is hypersurface orthogonal then, by definition, u_{[c}\nabla_b u_{a]} = 0. Writing this out we have$$ u_c \nabla_b u_a - u_b \nabla_c u_a + u_a \nabla_c u_b -u_c ...

0

$M=10^8M_\odot,$ so $r_s=$ $2GM/c^2=$ $2.95\times 10^{11}m.$ And $\alpha = J/Mc,$ has units of distance, and we can set it to $\alpha =0.998 r_s/2,$ which is very close to the critical value of $r_s/2.$ Now you can put $d\tau=$ 1 hour, $dt=$ 7 years, and $\theta=90°$ into $$\left(\frac{d\tau}{d\ t}\right)^2=1-{\frac{2GMr ... 4 The idea behind that quote is that you can't really separate space and time in General Relativity, which is the most complete scientific theory concerning the geometry of space and time. Instead, it works best to consider them as one integrated thing, called spacetime. To go into a little more detail, first consider galilean spacetime. Here you can think ... 1 I only had a brief look at the paper, but to me it looks like that even after choosing A^{0}, in such a way that \nabla_{l}l=0 and l^{\mu}=(0,1,0,0) then you still have some freedom in what they call \theta^0 which will affect m^{\mu} and \overline{m}^{\mu} but not the first two. I will agree that there is no more freedom in A^{0}, only for ... 2 As far as I understood from my so far cursory look into a living review article by Poisson, Pound and Vega on The Motion of Point Particles in Curved Spacetime, it's a bit messy. But I think if you manage to go through GR, this should be manageable, as well. It will probably help if you've dealt with Green's functions before and even better if you've seen ... 0 General relativity and quantum fields evolution in curved space There's a problem with this I'm afraid. Gravity is all to do with curved spacetime, but that isn't curved space and curved time, it's a curvature of "the metric", or more simply, a curvature in your plot of measurements. See Baez: "Note: not the curvature of space, but of spacetime. The ... 4 To know what a closed timelike curve looks like, you just do like every spacetime metric. You compute geodesics and field equations and all of that. Unfortunately, things start getting complicated. Closed timelike curves have a lot of weird behaviours, especially when it comes to matter fields upon them. They may not have a properly defined Cauchy problem, ... 1 I got the answer reading a book from E. Poisson, what I was doing was indeed wrong, you have to start with the induced metric given by$$ h_{ab}= g_{\mu\nu}e^{\mu}_a e^{\nu}_b $$where$$e^{\mu}_a=\frac{\partial x^{\mu}}{\partial y^a}$$are the tangent vectors to curves of the hypersurface. Then, you just replace g by h in the usual relation ... 1 \require{cancel}I) OP's is considering Dirac fermions in a curved spacetime. OP's action has various shortcomings. The correct action reads^1$$ S~=~\int\!d^nx~ {\cal L}, \qquad {\cal L} ~=~e L, \qquad L~=~T-V,\qquad e~:=~\det(e^a{}_{\mu})~=~\sqrt{|g|},  T~=~\frac{i}{2} \bar{\psi} \stackrel{\leftrightarrow}{\cancel{\nabla}} \psi, \qquad V~=~ ...

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A closed timelike curve wouldn't actually "look" like anything because it's an abstract thing. You can't actually see any lightcones or worldlines. A metric is an abstract thing too, to do with your measurements of distance and time, typically made using the motion of light. And the crucial point is this: you don't travel along your worldline. You move ...

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I'm having difficulties understanding why a gravitational acceleration can be guaranteed to be locally equivalent to an accelerating frame. Actually, it can't. See section 20 of Relativity: the Special and General Theory where Einstein said this: “We might also think that, regardless of the kind of gravitational field which may be present, we could always ...

2

You never have to make energy come from nowhere, the fact that the Einstein tensor has zero divergence means thavyiu can write any spacetime and the corresponding stress-energy tensor will have zero divergence. And zero divergence means that changes in energy (or momentum) are effected only by energy (or momentum respectively) net flowing in or out of a ...

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It took me a couple of days to figure it out, but I can now answer my own question: The problem with directly substituting the expression for $\delta g_{\mu\nu}$ into that for ${\delta \Gamma^\alpha}_{\beta\mu}$ is that there is no obvious source for the term $\xi^\sigma \partial_\sigma { \Gamma^\alpha}_{\beta\mu}$ that has to be the leading term in the ...

1

Due to relativity, the clocks on the GPS satellites move fast by about 38 µs per day. Which would be a problem, but not that big a problem because they all move fast by the same amount. Still you'd need to synchronize the clocks from time to time, because the satellite's position in space also depends on the clock. HOWEVER, if you do that once a week, ...

-4

The speed of light wouldn't stop time but make it extremely slow. After u have traveled the distance light travels in a sec then you will pass 1 second in space time assuming you are still whole and have not turned into a photon which would break so many rules.

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If the floor of the elevator is exerting a force on me (due to some external force accelerating it) then this would be very different from a gravitational acceleration that would accelerate each part of my body equally. No, it wouldn't. The two situations are experimentally indistinguishable. That's one of the points of this thought experiment. An even ...

0

The answer is very simple and you may forget everything about lifts, elevators and so on and so forth. The gravitational force (which does exist, I wonder the comments above) is the only interaction in the universe where the dynamics does not depend on the mass of the particle (the so called statement that inertial mass is equivalent to gravitational mass), ...

-1

Say, if you have a SNIa standard candle, with the help of its luminosity curve, you can recreate its absolute luminosity. By knowing the absolute luminosity $L$ and the visible flux $F$, one can calculate the luminosity distance $d_{\rm L}=\sqrt{L/4\pi F}$. What is a physical distance? There is a Hubble distance, luminosity distance, angular diameter ...

1

It appears that all you have to do is integrate (6) to obtain: $$\alpha = \int dx \frac{2GM}{c^2}\frac{y}{(x^2+y^2)^\frac{3}{2}} = \frac{2GM}{c^2}\frac{xy}{y^2\sqrt{x^2+y^2}}=\frac{2GM}{c^2}\frac{x}{y\sqrt{x^2+y^2}} \equiv \frac{4GM}{c^2R}$$ So we obtain: $$R=\frac{2y\sqrt{x^2+y^2}}{x}$$

4

When MTW say the universe is isotropic, they mean it is isotropic everywhere i.e. at all points in the universe. It's easy to construct universes that are isotropic at a single point and not homogeneous, for example CuriousOne's suggestion of a ball with density that is a function of distance from the centre. However this ball is only isotropic if you are ...

-1

The fundamental concept that is often overlooked is that in General Relativity (i. e. whenever you have accelerated motion or gravitational fields) the definition of time interval itself depends on the path you follow in the space-time manifold; in particular, it is the integral of the $g_{00}$ component of the metric along the curve describing your ...

1

There are a few misconceptions in your scenario that cause the misunderstanding. First of all, by definition, causality means that if the time interval between two events is positive in one reference frame, then it is positive in any other reference frame of your choice and viceversa, provided the velocity the events propagate to be smaller than $c$. If, on ...

1

I agree with everything John Rennie said but let me just take a slightly different direction. Note that Schwarzschild space-time has a time-like Killing field $\xi^{\mu}$. In a stationary space-time such as this, one can define the Newtonian analog of gravitational potential by $\phi \equiv \frac{1}{2}\log(-\xi_{\mu}\xi^{\mu})$. One can then easily show ...

0

Your question doesn't have an answer because it isn't possible to split energy into a potential part and a kinetic part, or at least not in an observer independant way. The Schwarzschild metric is time independant and spherically symmetric, and these symmetries mean there are two conserved quantities that you can think of as total energy, $E$, and angular ...

2

You ask: If the clock is running slowly compared to a distant clock is this equivalent to the clock having a lower energy compared to a distant clock? but you have to very careful what you mean by energy in general relativity. As it stands your question too vague to be usefully answered. However in the weak field limit there is a sense in which time ...

-2

I don't believe there are any plausible experiments which have proved the theory of relativity and I do not believe that time can pass slower on one body than it can on another. Imagine a train moving at 100km per hour relative to someone standing by the train tracks. A person on the train turns on a torch. The torch is on the train and would have the same ...

0

Indeed as was commented before usually in physics these derivatives are 'derived' by postulating how the object transforms. Obviously there are more rigorous definitions, as is usually the case. In this particular case you could try to find the structure which gives you these 'covariant derivatives' in the first place. In books they are often referred to ...

2

Really, to answer this carefully, we have to really think through what a horizon is. And for a general spacetime, there are several different notions of horizon, and "event horizon" is probably the most difficult of them to work with. The formal definition of "event horizon" says "Let's go to the distant future, take every freely-falling path that ...

3

As an example, consider an exponential curve, $e^x$. This is not linear, right? If I plot it on a graph with the y-axis scaled logarithmically, it can be plotted as a straight line. In the case of your P2, you can use a linear y-axis, but locally contract and expand the x-axis as necessary to produce a straight line. Would anybody in their right mind do ...

2

Given that your reference is from a philosophy department and not a scientific dep't, I would take their positions with a grain of pseudosalt. It looks as though someone heard of nonEuclidean geometry and went hogwild on the premise. It's sort of like they're claiming they can construct a coordinate system which bends with that particle $P_2$ , which is ...

1

There is already a good answer from John Rennie. Here we will just mention that if the spherically symmetric metric (2) is supposed to be a vacuum solution to Einstein field equations with $\Lambda=0$, then Birkhoff's theorem shows that the metric (2) [after a possible reparametrization of the time coordinate $t$] is exactly the Schwarzschild metric. See ...

11

Let's start with what we mean by a horizon: The event horizon of an asymptotically-flat spacetime is the boundary between those events from which a future-pointing null geodesic can reach future null infinity and those events from which no such geodesic exists. A null geodesic is the path followed by a light ray, so the horizon marks the surface at ...

-2

You are standing on surface of a planet of mass $M$ and radius $R$. With which velocity $v$ you need to throw away from the planet an object of mass $m$ that it will not return? The gravitational force is $F=G\frac{mM}{r^2}$. The work to be done to move the object in the gravitational field of the planet from distance R to infinity is $A = ... -4 There is an amusing "trick" (so is not meant as an answer to your question) in Newtonian gravity (plus a bit of special relativity) to derive that the Schwarzschild radius should be "special" Since self-gravitational energy of massive bodies is always negative (at least in Newtonian gravity) $$- \frac{G M^2}{R}$$ And massive bodies have positive rest ... 5 The geometry of spacetime is described by an equation called the metric. This is analogous to Pythagoras' theorem but with some key differences. Start with a 2d plane, where we identify positions of points by their$(x, y)$coordinates. Suppose you move a distance$dx$then a distance$dy$, then the distance from your starting point,$ds$, is given by ... 4 This equation suggests a path integral, of a Lagrangian that contains terms for both general relativity and the Standard Model in highly abbreviated form. Compressing it all into one line is a stunt, of course, rather than an actually useful equation. :) It glosses over many technical issues (for instance that we don't actually know how to do quantum ... 0 "Invariant" in this case means "invariant under Lorentz transformations". This means that it is observed to be the same by different observers passing through one particular spacetime point; two different observers with differing velocities will have their observable quantities related by a Lorentz transformation. However, this doesn't imply that the ... 0 Your questions are indeed related. First let's describe spacetime as like a stack of photos, the photos being different times. Real things are in some region of one of the photos and in a possibly different region of a different photo but don't jump discontinuously from one region to another in between two individual photos. Note this is an analogy and lots ... 0 You can't just acquire velocity (and hence momentum) without an equal and opposite momentum going to something else. So you are wrong that there are things you can do to acquire velocity. Unless you want to leave parts of yourself behind or are capable of stealing momentum from, say, an electromagnetic field. However, if you are an extended body in a ... 1 What the author means is as follows. Consider the (un-normalized) vector field$\partial_r$where$r$is the radial coordinate;$\partial_r$is thus just the vector field orthogonal to the level sets$r = \text{const.}$or, equivalently, it is the vector field foliating said level sets. As an aside, note that Schwarzschild coordinates are perfectly valid ... -1 He says that there is nothing funny going around for r < rs because both dt² and dr² term changes sign. I'm perfectly fine with this You shouldn't be. When you aren't happy with some conclusion you should go back over everything with a fine tooth comb. Examine your assumptions, check your postulates, look closely at the things you've taken for granted. ... 2 The Schwarzschild metric as you've written it is only one particular coordinate system and the fact that$r$and$t$switch roles at the evnt horizon is an artifact of that coordinate system. There are other coordinate systems which make certain properties of the metric more intuitive. The ones that might be most useful for you are ones which can be drawn as ... 1 If you plot a continuous curve in spacetime, it could be a path of a body if the tangent to the curve exists and has positive squared interval. General relativity is a geometrical theory, so everything is written in a geometrical way and the geometrical generalization is the predictions the theory makes. So outside the event horizon your curve has to have ... 2 You cannot change your linear or angular momentum in open space at all. You need something to transmit it to. if you swing your legs your body will rotate in the opposite direction while you swing, and stop when you stop swinging. If you are out of fuel there is no way to accelerate. Only by releasing mass you could change momentum, as Bender well shows you, ... 0 The way you pose the question it seems you have in mind a solution with full translational symmetry in space, and rotational symmetry about the magnetic field direction at each point. I don't know if such a solution exists; if it does must be time-dependent. (That symmetry would imply that the spatial sections are flat, and the energy density is constant. If ... 1 Well, this is linked to what the cotetrad$e_\mu^I\$ is. It is customary to present the cotetrad as a diagonalization of the metric and indeed, we have: $$g_{\mu\nu} = e_\mu^I e_\nu^J \eta_{IJ}$$ Note here that the cotetrad has two kind of indices. The greek type corresponds to spacetime coordinates but the latin type indices are indices in the tangent ...

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