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1

Mathematically, that's due to superposition. Both masses produce some gravitational field, which add together to give the "net field". (The same goes for electromagnetism, where one may add electric/magnetic field strengths,electric potentials etcetera for every point in space.) Ever so slightly changing the field strength at the well. As gravity gets weaker ...


1

Your question is many questions. Let me see if I can be of some help for the first two: It appears so. These equations should suffice to trace the projection of the orbit of the ray of light in the equatorial plane. I think this problem also appears in the $a=0$ case (Schwarzschild) and it is related to the fact that you cannot observe anything, light ...


2

Yes, that is quite right. A 'null' separation between two events is also called 'lightlike' since it implies that the points are on each other's light cone (the property is symmetric; it is in fact an equivalence relation now that I think about it)


1

Spacetime curvature and gravity are not two distinct concepts, they are one and the same. You can say that one "translates into" (can be represented and looked at as) the other, they do not have a cause-and-effect type of relationship. Gravity simply is the curvature of spacetime. To do a quick analogy - it's like someone saying either "Au revoir" or ...


-1

Space-time is affected by energy forms, like gravity, electro-magnetic flux-but in yet unknown ways. Space-time shifts under influence of these fields, but you will agree that the mass and electro-magnetic field too shifts mass and electro-magnetic fields! Hence, we may notice immeasurable change in mass under the influence of the energy-for mass and ...


1

As in your post, let $\gamma : [a,b] \rightarrow \mathcal{M}$ be a smooth curve onto a Lorentzian manifold. You might be interested in this math.SE post, indicating that all points in more than two Lorentzian dimensions can be joined by a spacelike curve, so space-like chord curves are utterly uninteresting, as they can be constructed from taking a small ...


0

Every single comment and answer was very useful. I think i have found an answer in an old paper by Clark Glymour (Minnesota studies in philosophy of science, volume III, pp.50-60): "It has recently been noted (Ellis, 1971; Dautcourt, 1971; Ellis and Sciama, 1972; Glymour, 1972; Trautman, 1965) that in some general relativistic cosmologies various global ...


0

You have better than baez, look http://settheory.net/cosmology http://settheory.net/general-relativity It is directly applied to an important example (universal expansion) The expression is simpler (relating 1 component of the energy tensor to 3 components of the Riemann tensor) The relation between energy and curvature is not only expressed but also ...


-1

I found many explanations for this type of questions http://settheory.net/cosmology http://settheory.net/general-relativity It's better than "The Meaning of Einstein's Equation" (John Baez). In particular - It is directly applied to an important example (universal expansion) - The expression is simpler (relating 1 component of the energy tensor to 3 ...


0

As far as I have understood from this paper, they have given some observational limits to the value of $\Omega_{k,0}$, but this article concludes asserting that "there is no evidence from Planck for any departure from a spatially flat geometry". Taking $\Omega_{k,0}=0$ and the value for $\Omega_{r,0}$ given at this post, one can compute the above integral ...


9

Actually you're quite correct, though possibly not in the way you expected. Ordinary velocity isn't an invariant because obviously different observers moving at different speeds will measure different velocities. However there is an invariant form of velocity called the four velocity that is an invariant under special relativistic (i.e. Lorentz) ...


2

As has been discussed in many questions around here (e.g. here), relativity tells us only about local properties and behavior of a space-time. There are some exceptions when we make global assumptions - if we have a space of globally and strictly constant positive curvature, non-trivial topology is imminent because the space has to be the 3-sphere ...


9

I'm not going to provide a full answer here, because I don't know the answer, but I want to give some statements that illustrate quite nicely the kind of problems one would face when determining topology of anything: We know spacetime is a manifold. That means, locally, it looks just like $\mathbb{R}^4$. That's already a bummer. We can't do jack at one ...


0

Gravity can be seen as a gauge theory of the Lorentz group (which acts on the tangent space). These was pointed out by Kibble and Sciama during the 50s and 60s. As John said before, it's better seen in terms of differential forms. Another reference you might find interesting is the Lecture notes on Chern-Simons gravity by Jorge Zanelli (available in ...


3

Just to add to John Rennie's answer, the objects where we expect to see the largest frame dragging effects are spinning black holes. There, there is actually a surface called the ergosphere (outside of the event horizon), where it is impossible for observers to stay stationary with respect to observers far from the black hole. In a sense, their reference ...


5

I disagree with your premise that fields and curvature are different at all. The gravitational field strength tensor is (or, can be seen as, but usually isn't) the Riemann curvature tensor of spacetime. Likewise, the electromagnetic field strength tensor is the curvature tensor of a gauge principal bundle. The fields and the curvatures are not distinct ...


1

I am assuming that zero (rest) mass particles don't interact gravitationally with each other and other particles. That's not a valid assumption in general relativity. Particles with zero invariant mass have energy and momentum and, thus, gravitate. Essentially, in general relativity, the density and flux of energy and momentum are the sources for ...


1

The trajectory of an object, whether massless or not, in a curved spacetime is given by the geodesic equation. $$ {d^2 x^\mu \over d\tau^2} + \Gamma^\mu_{\alpha\beta} {dx^\alpha \over d\tau} {dx^\beta \over d\tau} = 0 $$ Solving this is a formidable problem unless there is some helpful symmetry that simplifies the working, but in the particular case of ...


3

The spacetime outside a spinning mass is described by the Kerr metric. To explain how the Kerr metric produces frame dragging is hard, because it's not something for which there's an easy intuitive model. Frame dragging arises because the spacetime geometry links the angle measured around the spinning object to time, and this means the angle changes with ...


1

No, a massive body is able to bend light around it, which is called gravitational lensing. This has been observed multiple times. EDIT Photons are massless. Otherwise, they would not travel at the maximum speed, which is called speed of light. Keep in mind, that gravitational lensing is not a part of Newtonian mechanics. You need general relativity for ...


1

This is my first problem, as the modulus of a vector shouldn't be negative. First, while there are many useful properties of introductory linear algebra you should keep in mind with GR, thinking in Cartesian terms with positive definite matrices simply has to go. Vectors in relativity can very much have negative norm. Even though it's not often done in ...


5

There is theory that light cone shape does not depend on the reference frame in which it is viewed. So why we draw light cones near black hole differently? In general relativity, frames of reference are local, not global. Each of the light cones in your diagram corresponds to a certain local frame of reference. An observer using that frame of reference ...


1

Light travels along paths with a metric interval of zero. In flat spacetime this would be drawn as a light cone with a 45 degree opening angle in a standard Minkowski space time diagram.Things get a bit weirder in GR when spacetime is curved by mass/energy. In GR, the concept of an invariant speed of light only applies locally in non-accelerating frames of ...


5

Let's start at the beginning: The setting for relativity - be it special or general - is that spacetime is a manifold $\mathcal{M}$, i.e. something that is locally homeomorphic to Cartesian space $\mathbb{R}^n$ ($n = 4$ in the case of relativity), but not globally. Such manifolds possess a tangent space $T_p\mathcal{M}$ at every point, which is where the ...


0

There are, in fact, much better procedures than the background subtraction used by Gibbons and Hawking. For their procedure to work one must be able to embed the relevant regulating surface into an appropriate "background" spacetime. In some cases you can do this, but in general it is not possible for spacetime dimension greater than 3. To complicate ...


1

Now g and g′ are two different metrics on the manifold and so should predict different physics, but somehow they are the same? The field equations require four coordinate conditions for a unique solution. For the Schwarzschild line element, a coordinate condition is that the surface area of each sphere is $4\pi r^2$. For the line element of the 2nd ...


1

When you change coordinates, you just choose a different atlas for $\mathcal{M}$. This does not change the metric $g$, as it is defined independent of coordinates. Now, your Schwarzschild example has little to do with this, since there you are talking about a diffeomorphism $t : \mathcal{M} \rightarrow \mathcal{M'}$. Yet, if $\mathcal{M'}$ is not already ...


1

Changing the coordinates of a metric does not change the underlying physics, nor the solution. A metric $g$ describing a manifold $M$, if under diffeomorphisms is transformed to a new metric $g'$ it will still describe the original $M$.


3

This is really a comment, but it got a bit long for the comment field. I'd guess that, like me, your experience in physics is from an area where solving differential equations is a routine part of the job. We're used to analysing a problem, writing down a differential equation that encapsulates the physics and solving it, analytically if we're lucky or in ...


1

As i can remember, for the case of classical electrodynamic the d.o.f counting start after the assumption of the Bianchi Identity, and at the end the desired result came all from the gauge freedom. In fact $\partial_{[\mu}F_{\nu\alpha]}=0$ only choose a suitable form for $F_{\mu\nu}$, for example $$F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu.$$ After ...


1

You can think of diff as bianchi id. The additional 4 dof is killed by the fact that 4 of the 4 of the EFE are constraints.


1

The continuity equation (without sources) is usually written as follows $$\partial_t \rho + \nabla \cdot \mathbf{j} = 0$$ If you identify $\rho$ as the mass density, integrate over some volume $V$ and use the divergence theorem you get the result that you mention in your question. Namely, the change in mass in $V$ equals the amount of mass flowing through ...


4

The Ricci tensor is built adding up some of the components of the Riemann tensor, as the definition specifies: $$ R_{\mu\nu} = R^{\lambda}_{\ \mu\lambda\nu} $$ The repetition of two indices above and below means that all these components must be summed: $$ R^{\lambda}_{\ \mu\lambda\nu} = R^{0}_{\ \mu 0\nu} + \dots + R^{3}_{\ \mu 3\nu} $$ This operation ...


3

I've said it before and I will say it again: There are no frames travelling at the speed of light As David Z says in the very link you give, it is meaningless to ask what you would perceive travelling at the speed of light. You cannot. And even though there are particles that can, there are no frames associated with them. Have a look at the Lorentz boost. ...


2

The time dilation due to motion in a circle, relative to an observer at the centre, is just the usual Lorentz time dilation due to the velocity of the motion. If you're interested, in my answer to Is gravitational time dilation fundamentally different than other forms of time dilation? I showed how this is derived from the metric. Anyhow, as you say, the ...


6

Your second method is correct. To compare, say, the magnetic field with what you find in Jackson, you really need to realize that there's an assumption that you have unit basis vectors there, and that the cross product is actually a hodge dual (which will invoke factors of the square root of the determinant of the metric). These will make direct ...


0

The second form is the way in which the metric was written in the age of Kaluza and Klein. Why? out of embarrassment. If you keep the scalar field when considering the action you get an scalar. That was an undesired feature those times and that's why they hid it making it constant (actually they made it equal to 1). Now, is there a reason why the first is ...


0

In "The Scalar-Tensor Theory of Gravitation", of Yasunori Fujii and Kei-ichi Maeda you can find explicitly the solution, in Appendix C (pag. 195). Personally, I really didn't like this book and even this demonstration it's very difficult to follow. So I did it in another way. Use the usual theory for the GR part, and isolate this term: $\int d^4 x ...


1

I'm not entirely sure that I understand your question, but what leaps out at me is it's wrong to suggest that the energy of the light rays are decreased by an amount proportional to the difference in gravitational potential between the place where the light source is located and the potential of the place where an observer is detecting the light. ...


2

This is a partial answer for the first part of your question. You are not completely right. In the Schwarzschild metrics, for instance, the ratio of the energies (or frequencies) for the photons, is given by (in $c=1$ units) (here no Doppler effect is included, we only check gravitational effect): $$\frac{E(r_1)}{E(r_2)} = \sqrt{\frac{1 - 2 \Phi(r_2)}{1 - ...


1

It's not an uncommon interpretation that the Alcubierre drive acts as a multiplier of an existing subluminal velocity. The trouble is that the Alcubierre metric describes the drive in constant motion that doesn't change with time. It tells us nothing about how the drive accelerated to that speed or decelerated from it. The drive is moving in whatever ...


3

Let $\Delta_S$ and $\Delta_G$ be the time dilation effects due to General Relativity (gravity) and Special Relativity (motion) respectively (i.e. the clock rate on the satellite due to SR and GR is $1 - \Delta_S + \Delta_G$, signs chosen for simplicity). If these are small, they can be approximated as : \begin{eqnarray} \Delta_S &=& 1 - \sqrt{1 - ...


3

You know that in GR you need a locally Minkioski spacetime. This, in each point of your manifold you can change the coordinates so that the metric is diagonal, and the square of the infinitesimal displacement is $ds^2=\left(ct\right)^2-x^2-y^2-z^2.$ So here is where the $c$ come from. Then, when you want to compute the coupling constant $k=\frac{8\pi ...


1

Notation: I will use overdot for differentiation with respect to $\tau$, overtilde for partial differentiation with respect to $x^0 = t$, and prime for partial differentiation with respect to $x^1 = r$. (Edit: removed overloading of $\lambda$, sorry.) I assumed a general $\nu = \nu(t,r)$; reading the question more carefully, they're functions of $r$ only, ...


0

I can only think of three ways that you could get back to your starting point by travelling in a straight line: the universe is closed in the Friedmann sense the universe is not simply connected i.e. closed in a topological sense the universe as a while is rotating i.e. it's a Gödel universe In case 1 there is no special speed to get back to where you ...


11

Why is there no curvature outside this spherically symmetric, non-rotating, uncharged body that still has mass? I suspect you're getting confused by the fact that the Ricci tensor $R_{\mu\nu} = 0$ and therefore the scalar curvature $g^{\mu\nu}R_{\mu\nu} = 0$. This is always the case in regions of space where the stress-energy tensor is zero. The ...


1

Your answer is dependent on many assumption. Your idea of coming back to the starting point was mostly "plausible" before the discovery of dark energy. If the current size and dynamics of spacetime is as we believe as of today, then you will never return back because there will be more space created ahead of you as you travel that what you cover even at the ...


3

The geodesic equation (GE) $$\tag{1} {d^2 x^\mu \over d\lambda^2} + \Gamma^\mu_{\alpha\beta} {dx^\alpha \over d\lambda} {dx^\beta \over d\lambda} ~=~ 0$$ depends on the parametrization: The GE (1) holds when the parameter $\lambda$ is affinely related to the arc length $s=a\lambda+b$ of the geodesic. This can e.g. be deduced from the fact that eq. (1) is ...


0

I agree with the previous answers that the cited blog article is naive and superficial. But then all great ideas have humble beginnings. Along similar lines of thought a respected scientist called Walther Ritz (http://en.wikipedia.org/wiki/Walther_Ritz) developed a sophisticated model of electric forces in 1908 which explains gravity as a small net effect ...


1

If space-time of a black hole is infinitely curved, how can new volume be created for these particles to occupy? The spacetime of a black hole isn't infinitely curved. Only at the spacetime singularity within a black hole is the curvature infinite. The spacetime near, at, and within the horizon is highly curved but not infinitely so. I recommend ...



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