# Tag Info

6

The curvature of the universe can be derived from the temperature fluctuations in the Cosmic Microwave Background. For a given amount of radiation, baryons, dark matter and dark energy in the universe, these temperature fluctuations can be calculated theoretically, and compared with observations, and so one searches for the values that yield the ...

6

I believe you've already spotted the answer to your question with this sentence: And a black hole will shift the trajectory entirely. This is all dependent on the proximity to the source of gravity. You can "shift light" (bend its trajectory) as much as you want with as little mass as you want using a black hole. Just let the light get arbitrarily ...

4

You're quite correct that the metaphor is misleading, and indeed you'll find professional relativists tend to be rather scornful of it. There are a number of problems with it, of which the problem you mention is just one. For example the diagram implies only space is bent, while the bending is of spacetime so time is bent as well. The diagram also implies ...

3

$g^{\alpha\beta}$ is symmetric in $\alpha$ and $\beta$, while $R_{\alpha\beta\gamma\mu}$ is anti-symmetric in $\alpha$ and $\beta$, so the contraction $g^{\alpha\beta}R_{\alpha\beta\gamma\mu}$ is necessarily $0$, and cannot be $R_{\gamma\mu}$. Moreover, it is not correct to say, that if the contraction of $2$ tensors with another tensor (here the metric ...

3

It seems that OP is pondering the following. What happens in a field theory [in OP's case: GR] if spacetime $M$ has a non-empty boundary $\partial M\neq \emptyset$, and we don't impose pertinent (e.g. Dirichlet) boundary conditions (BC) on the fields $\phi^{\alpha}(x)$ [in OP's case: the metric tensor $g_{\mu\nu}(x)$]? I) Firstly, it should stressed ...

2

I really don’t like that diagram. No, REALLY. I think it conveys a bad intuition that may confuse you. I don’t like what it does with the connexion coefficients (Christoffel symbols). Here’s why. In a general manifold, tangent spaces of course are not comparable at different points. This is in contrast with the situation in Euclidean space, which can be ...

2

For reference, Weinberg p. 378: A metric space is said to be homogeneous if there exist infinitesimal isometries (13.1.3) that carry any given point $X$ into any other point in its immediate neighborhood. Equation (13.1.3) defines an infinitesimal transformation and (13.1.5) concludes the Killing equation $\xi_{\sigma;\rho} + \xi_{\rho;\sigma} = 0$ ...

2

In principle, yes, you can specify the stress tensor and solve the resulting equations, but in practice, this is hard to do because the field equations are non-linear PDEs...darn. The simplest possible example is the case in which the stress tensor vanishes; $T_{\mu\nu} = 0$ namely the vacuum equations. The field equations with vanishing cosmological ...

2

At the level of understanding the data and observations we have up to now, General Relativity describes well what we perceive of the Cosmos and Quantum Field Theory what we observe in the microcosm of elementary particles and their interactions. The two have not been joined up to now, i.e. there is no accepted unified theory that joins smoothly these two ...

1

Not only the position in the gravitational field is important, but also the velocity. Consider the Schwarzschild metric $$\text{d}\tau^2 = \left(1 - \frac{2GM}{rc^2}\right)\text{d}t^2 - \frac{1}{c^2}\left(1 - \frac{2GM}{rc^2}\right)^{-1}\left(\text{d}x^2 + \text{d}y^2 +\text{d}z^2\right),$$ where $\text{d}\tau$ is the time measured by a moving clock at ...

1

There is a standard way to find out if the spacetime around you is curved. Surround yourself with a sphere of small test masses and wait and see what happens. If the sphere stays exactly the same shape you're in flat spacetime but if it changes shape or volume you're in a curved spacetime. In the case of the ISS the test masses nearer the Earth than you ...

1

Firstly, it should be made clear that being isotropic is a very special and rare property. (A spacetime can never be truly isotropic because no isometry can map spacelike vectors to timelike vectors, for example, so I'll talk about "space" being isotropic). There are very few spaces isotropic around every point, only very few spaces will even be isotropic ...

1

Partly this answer is just gathering together the comments above, though there are a couple of points that haven't been mentioned. Firstly, as mentioned in the comments electromagnetic waves do gravitate and the links in the comments cover this well. In the early universe (for the first 47,000 years after the Big Bang) EM radiation was the dominant ...

1

the emitter and the receiver move with constant velocities relative to an inertial frame and $v$ is the constant velocity of the receiver relative to the emitter and away from it. No, both the emitter and the receiver are accelerating, and the receiver has gained an extra velocity $v$ between the time the photon was emitted and the time it was received. ...

1

I'll try to explain parallel transport first: This image shows how a vector generally does not remain parallel to itself under parallel transport. The vector here is moved along the path A-N-B and is constrained to remain in the tangential plane to the spherical surface at all points of the path because the geometry does not know of the "third" dimension ...

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