The  is one of the measures of spacetime curvature. For flat (Minkowski) spacetime it is zero. The dimensions of the Kretschmann scalar are $[L]^{-4}$. What does that physically signify about the geometry of spacetime?
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How do you interpret the  (in general relativity)? What can you tell from it?    The Kretschmann scalar is defined as   $$K = R_{abcd} R^{abcd} $$  where $R_{abcd}$ is the Riemann curvature tensor.
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I am following Carroll's GR book. He explain that it is convention to parameterize geodesics of photons by a parameter $\lambda$ such that  $$p^\mu ~=~ \frac{d x^{\mu}}{d \lambda}.\tag{3.62}$$  But this is the definition of 4-velocity for a massive particle in the case of $\lambda=\tau$ equal to proper time. My question then, is $$u^\mu = p^\mu$$ for all massless particles? I ask because it's well-known that proper time $\tau$ freezes for massless particles.
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Consider a scientist who is accelerating at a constant rate along the x direction, in flat minkovski space. The relevant coordinate transformations are  $$\begin{cases}t'=t \\ x'=x+\frac{1}{2}at^2 &amp;&amp; a=\text{const}  \\ y'=y \\ z'=z\end{cases}$$  The question is, do you expect the curvature tensor to vanish in the primed coordinates?   One argument is that the curvature tensor is indeed a tensor. Since the space is flat, the curvature tensor will vanish for a stationary observer and hence will vanish in all coordinate systems. This makes sense because the curvature of space-time is an intrinsic property of the manifold and doesn't  care about which reference frame (RF) we measure it in.  But when i try to apply the equivalence principle (EP) to answer the same question i wind up contradicting myself. Which is worrying because it likely means that my basic understanding of the concept is flawed.   The EP says that the scientist cannot decide whether his laboratory is uniformly accelerating in flat space or if his laboratory is fixed on the surface of his home planet, who's surface conveniently results in an acceleration of 'a' outward from the planet (recalling that a freely falling observer is now the definition of unaccelerated). So actually, the primed coordinates aren't those of someone who is accelerating uniformly in flat space, they're  equivalently the coordinates of someone fixed in a gravitational field, the space of which is curved and so has a non zero curvature tensor.   Please correct where my logic fails, it's something quite basic and fundamental so i would really like to straighten it out. Thanks.
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What is the exact shape of the universe? I know of the balloon analogy, and the bread with raisins in it. These clarify some points, like how the universe can have no centre, and how it can expand equally everywhere in all directions. But they also raise some questions, like if you are on the surface of a balloon and travel in 1 direction, you will eventually return to your starting point, is it possible our universe have this feature?  If it has, or had, would this be a symmetry of sorts ($\psi(x)=\psi(x+R)$), and as such have a conserved quantity associated with it (by Noether)? Assuming "small curled up dimensions" wouldn't these dimensions have this type of symmetry, what are the associated conserved quantities? Is it known exactly what the geometrical shape of the universe is? (on a large scale) (I am not talking about only the observable universe). How does one define the "size" of a dimension, is this scale only applicable to curled up ones? Is it possible to describe to a layman the shape of the universe without resorting to inept analogies?
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Any rank 2 tensor would do the same trick, right?  Then, which is the motivation for choosing the metric one?    Also, if you help me to prove that $g^{kp}g_{ip}=\delta^k_i$, I would be thankful too ^^
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How should I prove the following in general relativity?   A flat spacetime can be covered by Minkowski coordinate neighborhoods. A flat spacetime with the trivial topology can be covered by a global Minkowski coordinate chart.
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The question is as follows.  Imagine we have two closed geodesics on a surface, where these two geodesics bound a cylinder together and the geodesics are diffeomorphic to circles.  Can the cylinder have $K&gt;0$ everywhere or $K&lt;0$ everywhere?  Here $K$ is Gaussian Curvature.   What's confusing me about this question is that a cylinder has Gaussian Curvature of $K=0$ everywhere, so wouldn't the answer to both questions be no?  But then I don't get the point of the question with the geodesics.  Thank you in advance
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What is the physical significance of  $$\nabla\times\boldsymbol{V}~?$$  I mean I read 'curl V represents the rotation of the vector $V$. My question what is it about the term $\nabla\times\boldsymbol{V}$ that it represents the rotation of the vector?
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It is usually said that existence of discrete spacetime violates Lorentz symmetry. What quantity is used to quantify such violation? I mean could someone points a reference for a derivation that shows such analysis.  My other question is: if Lorentz symmetry is violated, does that imply space-time is discrete? or not necessarily?)
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Let us consider the Euclidean two-sphere, defined by the embedding in the three dimensional Euclidean space as $$ \mathbf n \cdot \mathbf n = 1\,, $$ where $\cdot$ denotes the standard scalar product. The metric on the sphere, in some coordinates $x^i$, is expressed as  $$ \gamma_{ij}=\mathbf e_i \cdot \mathbf e_j\,, $$ where $\mathbf e_i=\partial_i\mathbf n$. For instance, in the standard spherical coordinates  $$ \mathbf n=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta) $$ and $$ \gamma_{\theta\theta}=1\,,\qquad \gamma_{\phi\phi}=\sin^2\theta\,,\qquad \gamma_{\theta\phi}=0\,. $$ We define the Laplace-Beltrami operator on the sphere by $ \Delta = \gamma^{ij} D_iD_i $, where $\gamma^{ij}$ is its inverse and $D_i$ is the associated Levi-Civita connection.  I would like to prove that $$\Delta \mathbf n = -2 \mathbf n$$ and that in higher dimensions the same holds with $2$ replaced by the dimension of the sphere. I came to believe that this is true by an explicit check in spherical coordinates in dimensions $3$, $4$ and $6$.  Considering that parallel transport of a given tangent vector $\mathbf v$ defined at the point $x+dx$ to the point $x$ is defined by keeping constant its embedding components and then projecting it on the sphere at the point $x$, we have $$ \mathbf v_{\parallel}(x+dx,x)=\mathbf v(x+dx)-\mathbf v(x+dx)\cdot \mathbf n (x)\, \mathbf n(x) $$ hence $$ D_i\mathbf v\, dx^i = \mathbf v_{\parallel}(x+dx,x)- \mathbf v (x)= (\partial_i\mathbf v+\partial_i\mathbf n \cdot \mathbf v\, \mathbf n)dx^i $$ where we have used $\mathbf n \cdot \partial_i\mathbf v+\partial_i\mathbf n \cdot \mathbf v=0$ and $$ D_i\mathbf v = \partial_i\mathbf v+\partial_i\mathbf n \cdot \mathbf v\, \mathbf n\,. $$ Applying this to the basis vectors $\mathbf e_j =\partial_j\mathbf n$ affords $$ D_i\mathbf e_j = \partial_i \mathbf e_j+\gamma_{ij}\mathbf n\,. $$ But unfortunataly I am not able to go further.
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I'm having some trouble understanding the counting procedure for the number of independent components of Riemann curvature tensor $R_{iklm}$ in 4D spacetime. (The answer is supposed to be 20, but I'm trying to understand how so!)  Let's do it sequentially:   Since we are looking at a rank 4 tensor in 4D, $R_{iklm}$ could bear $4^4 = 256$ components in general. (I follow this argument.)  Since the tensor is anti-symmetric w.r.t. first and second pair of indices, i.e. $$R_{iklm} = - R_{ikml} = - R_{kilm}$$ all 256 are not independent. e.g. $R_{1234} = - R_{2134}$, and those where these antisymmetric indices get repeated, vanish. e.g. $R_{1134} = - R_{1134}$, and hence $=0$. Thus, the number of independent components is far lesser. (I follow this qualitative argument.)   Trouble begins when we have to count how many. An attempt towards the answer can be found at , from which I quote verbatim:     The two conditions show that all components where either the first and second indices, or the third and fourth indices are equal must be zero. In four dimensional spacetime, this means that at most $\left[\ {}^4C_2\ \right]^2 = 36$ components can be non-zero, since we can choose two distinct values for both the first and last pairs of indices.   I don't get the above argument cleanly: of course with repeated indices, we get a vanishing answer, so the indices have to be distinct, but how does number of independent components amount to "choosing two distinct values for both the first and last pairs of indices"?  i.e. what is the origin of the latter half of the sentence?     An alternative reasoning is adopted in a comment made below the post in the aforementioned link:      It was stated,…. “The two conditions [(1) &amp; (2)] show that all components where either the first and second indices, or the third and fourth indices are equal must be zero.”      …This is correct as stated, since an interchange of two indices that are equal refers to the same tensor component, and the only number that can be equal to the negative of itself, is zero. However this only accounts for 112 components being zero; with 256 – 112 = 144 still left that are not proven to be zero.      A 4 X 4 anti-symmetric matrix is one where the 6 bottom-left triangle of components are equal to the negative of the 6 upper-right triangle of components, and so dependent, and that the diagonal components are zero (already counted).      So, for the nj-matrix, get rid of the 6 dependent, for each of the lm combinations,.. that’s 6*4*4=96. For the lm-matrix, again get rid of the 6 dependent, for each of the nj combinations,.. that’s 4*4*6=96. That’s 96nj + 96lm – 84 (that are counted twice) = 108. Now, 256 – 112 (above) – 108 = 36.   The spirit underlying this answer looks cleaner, but there are some missing steps:  How do we count that 112 vanish? How do we count that 84 were counted twice as per the former argument?  I'm sure that if I am able to understand this part, I would be able to figure out the rest of the counting myself too (i.e. slashing 36 to 21 because of the symmetry argument, and reducing one more due to the Bianchi identity constraint.)  Any help in clearing the haze would be greatly appreciated.
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Studying the scalar field and Klein-Gordon equation in quantum field theory I came across this definition for the inner product in the space of the solutions of the K.G. equation:  $$\langle \Phi_1 | \Phi_2 \rangle = i\int \mathrm{d}\vec{x}(\Phi_1 ^* \overleftrightarrow{\partial_0}\Phi_2) = i\int \mathrm{d}\vec{x} (\Phi_1 ^* \partial_0\Phi_2 -  \Phi_2 \partial_0\Phi_1^*).  $$  I see that this definition should be invariant under Poincaré transformations, but I couldn't prove it.    Moreover I couldn't find the reason why such a scalar product is introduced. Aren't there other possible scalar products? Why choose this one?
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General relativity says that spacetime is a Lorentzian 4-manifold $M$ whose metric satisfies Einstein's field equations.  I have two questions:   What topological restrictions do Einstein's equations put on the manifold?  For instance, the existence of a Lorentz metric implies some topological things, like the Euler characteristic vanishing. Are there any experiments being done or even any hypothetical experiments that can give information on the topology?  E.g. is there a group of graduate students out there trying to contract loops to discover the fundamental group of the universe?
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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.
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This is something that has been bothering me for a little while. The usual procedure that I've seen is to write the proper time as the line integral $$\tau=\int_\gamma d\tau$$ along some curve $\gamma$. That curve that minimizes this is a geodesic (assuming Levi-Civita connection here). The definition $d\tau^2=-g_{\mu\nu}dx^\mu dx^\nu$ leads to $$\frac{d\tau}{d\lambda}=\sqrt{-g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^\mu}{d\lambda}}.$$ So, using , we have $$\int_\gamma d\tau=\int_{\lambda_i}^{\lambda_f}\sqrt{-g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^\mu}{d\lambda}}d\lambda,$$ where $\gamma(\lambda_i),\gamma(\lambda_f)$ are the beginning and end points of the curve. It can be easily verified that this is parameterization invariant. It is standard to normalize $\gamma$  $$g_{\mu\nu}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}=-1.$$ Thus it would seem that $$\int_\gamma d\tau=\int_{\lambda_i}^{\lambda_f}\sqrt{-g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^\mu}{d\lambda}}d\lambda=\int_{\tau_i}^{\tau_f}d\tau=\tau_f-\tau_i.$$ Thus $\gamma$ does not even play a role it seems. Thus when we introduce a one-parameter family $\{\gamma_\varepsilon\}$ and take the variational derivative, we get 0 arbitrarily. What gives? How can we vary this integral without changing the bounds?
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This was supposed to be a long question but something went wrong and everything I typed was lost. Here goes.   Why is  $k = \dfrac{1}{4\pi\epsilon_0}$ in ?  Is this an experimental fact?  If not, what is the significance of this definition?
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Given an $n$-dimensional geodesically complete connected Riemannian manifold $M$, we want to prove that the dimension of its isometry group is $$\dim {\rm ISO}(M) \leq \frac{n(n+1)}2.$$  Does it suffice to say that, since Euclidean space $\mathbb{R}^n$ is expected to be maximally symmetric, and the number above is the dimension of its isometry group, namely translations plus rotations, then the bound must be true for any other manifold? Do you know a more rigorous proof?