I'm learning general relativity from the book " ". The field equations are derived by the Hilbert - Einstein action and are written in the form  $$ R_{\mu\nu}-\frac{1}{2}R\,g_{\mu\nu}+\Lambda\,g_{\mu\nu}=kT_{\mu\nu}$$ where $k$ is acostant that should be determined. The authors find this $k=\frac{8\pi G}{c^4}$ by imposing the well known conditions of "Newtonian Limit". The first requirement is the following      The particles in free fall moving along geodesics induced by $g$.   This is a postulate? By the Einstein equivalence principle we know that locally we can eliminate the effects of gravity and so living in a (flat) Minkowskian space-time but there is not mention of the above statement. I agree on the fact that once we have a curved space-time, in presence of gravitational fields, it is natural to require that the free falling particles move along the geodesics, but this can be proved or it is postulated?
119309
Trying to teach myself general relativity. I sort of understand the derivation of the geodesic equation $$\frac{d^{2}x^{\alpha}}{d\tau^{2}}+\Gamma_{\gamma\beta}^{\alpha}\frac{dx^{\beta}}{d\tau}\frac{dx^{\gamma}}{d\tau}=0.$$ which describes  "how" objects move through spacetime. But I've no idea "why" they move along geodesics.   Is this similar to asking why Newton's first law works? I seem to remember reading Richard Feynman saying no one knows why this is, so maybe that's the answer to my geodesic question?
21634
Can someone suggest a textbook that treats general relativity from a rigorous mathematical perspective? Ideally, such a book would   Prove all theorems used. Use modern "mathematical notation" as opposed to "physics notation", especially with respect to linear algebra and differential geometry. Have examples that illustrate both computational and theoretical aspects. Have a range of exercises with varying degrees of difficulty, with answers.   An ideal text would read a lot more like a math book than a physics book and would demand few prerequisites in physics. Bottom line is that I would like a book that provides an axiomatic development of general relativity clearly and with mathematical precision works out the details of the theory.  Addendum (1): I did not intend to start a war over notation. As I said in one of the comments below, I think indicial notation together with the summation convention is very useful. The coordinate-free approach has its uses as well and I see no reason why the two can't peacefully coexist. What I meant by "mathematics notation" vs. "physics notation" is the following: Consider, as an example, one of the leading texts on smooth manifolds, John Lee's Introduction to Smooth Manifolds. I am very accustomed to this notation and it very similar to the notation used by Tu's Introduction to Manifolds, for instance, and other popular texts on differential geometry. On the other hand, take Frankel's Geometry of Physics. Now, this is a nice book but it is very difficult for me to follow it because 1) Lack of proofs and 2)the notation does not agree with other math texts that I'm accustomed to. Of course, there are commonalities but enough is different that I find it really annoying to try to translate between the two...  Addendum (2):  For the benefit of future readers, In addition to suggestions below, I have found another text that also closely-aligns with the criteria I stated above. It is, Spacetime: Foundations of General Relativity and Differential Geometry by Marcus Kriele. The author begins by discussing affine geometry, analysis on manifolds, multilinear algebra and other underpinnings and leads into general relativity at roughly the midpoint of the text. The notation is also fairly consistent with the books on differential geometry I mentioned above.
9129
In General Relativity Einstein's equation implies that stress-energy tensor on its RHS is conserved (has vanishing divergence), due to the Bianchi identity. Considering variational principles leading to Einstein's equation leads to conclusion that this stress tensor is equal to the variational derivative of full action with respect to the metric tensor. However, on several occasions I heard people stating that quite generally one can define stress tensor for a field theory in this way and it is automatically conserved. In flat spacetime and without any coupling to gravity! I wonder if this is true. I don't see a reason why it should be.
2538
I'm looking for a rather intuitive explanation (or some references) of the difference between the metric of a curved space-time and the metric of non-inertial frames.  Consider an inertial reference frame (RF) with coordinates $\bar x^\mu$, in flat spacetime $\eta_{\mu \nu}$ (Minkowski metric).   If I have well understood, on one hand, I can go to an accelerated RF by change of coordinates $x^\mu(\bar x)$. The metric is given by: $$\tag{1}g_{\mu \nu}(x) = \frac{\partial \bar x^{\alpha}}{\partial x^{\mu}} \frac{\partial \bar x^{\beta}}{\partial x^{\nu}} \eta_{\alpha \beta}$$ On the other hand, I know that a curved space-time with metric $q_{\mu \nu}$ cannot be transformed to Minkowski $\eta_{\mu \nu}$ by coordinate transformation. In other words there does NOT exist any coordinate $x^\mu(\bar x)$ such that (in the whole coordinate patch): $$\tag{2}q_{\mu \nu}(x) = \frac{\partial \bar x^{\alpha}}{\partial x^{\mu}} \frac{\partial \bar x^{\beta}}{\partial x^{\nu}} \eta_{\alpha \beta}\qquad \leftarrow \text{(does not exists in curved space)}$$   So far, everything is more or less ok... But my question is:    What is the difference between $q_{\mu \nu}$ and $g_{\mu \nu}$? I mean, in both cases a particle would "feel" some fictitious forces (in which I include the weight force due to the equivalence principle).  What physical situation can $q_{\mu \nu}$ describe and $g_{\mu \nu}$ cannot?   I additionally know that by change of coordinates $q_{\mu \nu}$ is locally Minkowski. But still, I can't see clearly the difference.
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The principle of equivalence - that, locally, you can't distinguish between a uniform gravitational field and a non-inertial frame accelerating in the sense opposite to the gravitational field - is dependent on the equality of gravitational and inertial mass. Is there any deeper reason for why this equality of "charge corresponding to gravitation" (that is, the gravitational mass) and the inertial mass (that, in Newtonian mechanics, enters the equation $F=ma$) should hold? While it has been observed to be true to a very high precision, is there any theoretical backing or justification for this? You could, for example (I wonder what physics would look like then, though), have the "charge corresponding to electromagnetic theory" equal to the the inertial mass, but that isn't seen to be the case.
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Firstly I think shades of this question have appeared elsewhere (like , or ).  Hopefully mine is a slightly different take on it.  If I'm just being thick please correct me.  We always hear about the force of gravity being the odd-one-out of the four forces.  And this argument, whenever it's presented in popular science at least, always hinges on the relative strength of the forces.  Or for a more in depth picture .  But, having had a single, brief semester studying general relativity, I'm struggling to see how it is viewed as a force at all.  A force, as I understand it, involves the interaction of matter particles with each other via a field.  An energy quantisation of the field is the force carrying particle of the field.  In the case of gravity though, particles don't interact with one another in this way.  General relativity describes how space-time is distorted by energy.  So what looked to everyone before Einstein like two orbiting celestial bodies, bound by some long distance force was actually two lumps of energy distorting space-time enough to make their paths through 3D space elliptical.   Yet theorists are still very concerned with "uniting the 4 forces".  Even though that pesky 4th force has been well described by distortions in space time.  Is there a reason for this that is understandable to a recent physics graduate like myself?     My main points of confusion:   Why is gravity still viewed as a force? Is the interaction of particles with space time the force-like interaction? Is space-time the force field? If particles not experiencing EM/weak/strong forces merely follow straight lines in higher-dimensional space (what I understand geodesics to be) then how can there be a 4th force acting on them?     Thanks to anyone who can help shed some light on this for me!
84967
I have often heard that the gravitational field has spin $2$. How can I read the spin of the field from the Einstein-Hilbert action  $$S=\int \! \mathrm{d}^4x \,\sqrt{|g|} \, \mathcal{R} \, \, \, ?$$
18173
Does such a set of equations exist or does our universe have an intrinsic curvature that can't be explained by an embedding in a flat Minkowski space of 1 higher dimension? Even if general relativity can be explained by such equations, it still doesn't prove our universe is embedded in a flat Minkowski space of higher dimension.
12176
Whenever I read about the curvature of spacetime as an explanation for gravity, I see pictures of a sheet (spacetime) with various masses indenting the sheet to form &quot;gravity wells.&quot; Objects which are gravitationally attracted are said to roll down the curved sheet of spacetime into the gravity well. This is troubling to me, because, in order for objects on the locally slanted spacetime sheet to accelerate, gravity must be assumed. Therefore I ask; does the explanation of gravity as the curvature of spacetime assume gravity? If yes, what is the point of the theory? If No, what am I missing?
84806
I understand how to derive the spacetime interval being invariant for Minkowski space, but I've never seen any derivation of it in general curved spacetime. Is the invariance just derived for Minkowski space and then postulated that it holds for all metric tensors in general relativity, or is there a proof to show it is invariant in general relativity?
61266
We know that the equation of motion of particle can be derived from the respective action. But in the book I am reading, the author is saying:     ... timelike worldline of a massive particle is parametrised by proper time, $x^a=x^a(\tau)$. The velocity (tangent) vector is $u^a \equiv \dot{x}^a(\tau)$. This is normalised as $u^au_a =-c^2$.   How can one derive e.o.m?
111399
I'm writing this question since I haven't find exhaustive enough answers searching on the Net. I'm writing an application for simulating General Relativity orbits calculating the acceleration from Christoffel's symbols, which are already derived from the Schwarzschild metric in isotropic coordinates:  $$\frac{d^2x^\alpha}{d\tau^2}=-\Gamma{^\alpha}{_\mu}{_\nu} \frac{dx^\nu}{d\tau} \frac{dx^\mu}{d\tau} \hspace{7.1em} (1)$$  My problem is: how should $\frac{dx^0}{d\tau}$ be defined from arbitrary spacial components of the four-velocity?  $$u^i=\frac{dx^i}{d\tau}=\text{arbitrary quantity} \hspace{4.4em} (2)$$  In special relativity, the time component of four-velocity, in terms of the spacial components, is:  $$u^0=\frac{dx^0}{d\tau}=\frac{cdt}{d\tau}=\sqrt{c^2+|u^i|^2} \hspace{4em} (3)$$  where $|u^i|^2 = (u^1)^2+(u^2)^2+(u^3)^2$ is the magnitude squared of the spacial components of the four-velocity. This comes from the Minkowskian metric. Then I derived a similar equation from the Schwarzschild metric in isotropic coordinates:  $$u^0=\frac{cdt}{d\tau}=\sqrt{\frac{B}{A}(c^2+B^2|u^i|^2)} \hspace{4em} (4)$$  where $A=(1-\frac{GM}{2R})^2$ and $B=(1+\frac{GM}{2R})^2$. If this is correct, it's quite straightforward to find that the latter definition gives a greater value than the one from special relativity, since $B &gt; 1$ and $B &gt; A$. Does this mean that the four-velocity magnitude is not constant in General Relativity, as it depends also on radial distance? (I knew that the four-velocity always equals the speed of light)
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My question is, can the (classical) Lagrangian be thought of as a metric? That is, is there a meaningful sense in which we can think of the least-action path from the initial to the final configuration as being the shortest one? Then the equations of motion would be geodesics - not in physical space, but in phase space, with the metric defined by the Lagrangian. Below is a slightly expanded outline of this idea, outlining why I think it might make sense.  (Classical) Lagrangian mechanics is derived by assuming the system takes the path of minimum action from its initial state to its final one. In the formalism of classical mechanics we normally write time $t$ as a special type of coordinate, quite distinct from the other (generalised) coordinates $\mathbf{q}$.   However, if we want we can lump these together into a single set of "generalised space-time coordinates" $\mathbf{r} = (t, q_0, q_1, \dots)$, to be considered a vector space of dimension $1+n$, where $n$ is the number of degrees of freedom. Then the action is just a function of curves in $\mathbf{r}$-space, and the equations of motion are determined by choosing the least-action path between two points, $\mathbf{r_0}$ and $\mathbf{r_1}$. In this picture, $L$ depends upon the position and direction of the curve at each point along it, and the action is obtained by a line integral along the curve.  This has a pleasingly geometrical feel, and so my question is, is there a meaningful sense in which I can think of the least-action path as being the shortest path, with the Lagrangian playing the role of a metric on $\mathbf{r}$-space? Or is there some technical reason why $L$ can't be thought of as a metric in this sense?  Note that this isn't in general the same as a metric defined on space-time. For an $m$-particle system in $3+1$ space-time, $\mathbf{r}$ will have $3m+1$ dimensions, rather than 4. I'm interested in the notion of a metric in $\mathbf{r}$-space defined by $L$, rather than the more usual picture of $L$ being defined in terms of a metric on space-time.
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Einstein's equation is  $$8 \pi T_{ab} = G_{ab},$$  where the left side contains the stress-energy tensor and the right side contains the Einstein tensor.  Is there exactly one unique stress-energy tensor corresponding to a given spacetime curvature? Or is it possible for one curvature (i.e. one spacetime metric) to be produced by multiple different mass configurations?
116893
In science documentaries that touch on general relativity, it is often said that gravitational pull isn't an actual a pull (as described by classical physics), but rather one body travelling in a straight line on the curved space around another massive body.  I don't know if I misunderstood that or if it's just an oversimplification meant to help the uninitiated understand the concept better, but something doesn't seem to follow. If the Earth travelling around the Sun is just moving along a straight line in the curved space, shouldn't light also be trapped in orbit around the Sun?  I expect the actual equations in the theory also take into account a body's velocity, not just the curvature.
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I have read the following statement:      "The Einstein-Hilbert action can be seen as an effective field theory: setting $g_{\mu\nu}= \eta_{\mu\nu}+h_{\mu\nu}$, one gets a free Lagrangian for the field $h_{\mu\nu}$ plus interactions."   Now, how is it possible to define all the interaction terms?  EDIT: I ask apologize if my statement was appeared unclear: I have created a  which is more precise; please don't continue to consider seriously this one.
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I want to delve into Cartan's formulation of General Relativity but it don't really know where to look. Does anyone have any book suggestions?