Tag Info

Hot answers tagged

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.


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 ...


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

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 ...


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 ...


3

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 ...


2

(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 ...


2

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 ...


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 ...


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 ...


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 ...


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 ...


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

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.


1

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.


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 $r$ ...


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 ...


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: ...


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

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? $$(t,x,y,z)$$ ...


1

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 ...


1

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 ...



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