New answers tagged

-1

The spacetime interval $ds^2 = -dt^2+dx^2+dy^2+dz^2$ is invariant (Set $c=1$). Since proper time $\tau$ is defined as the time measured in a co-moving frame, $ds^2 = -d\tau^2$. You can find the value of $ds^2$ by the parametric curve you have.


1

Yes it is correct. You can aply the covariant derivatives in $\phi^2$: $$\nabla_{\mu}\nabla_{\nu}\phi^2 = \nabla_{\mu}(2\phi \nabla_{\nu} \phi) = 2 \nabla_{\mu}\phi\nabla_{\nu} \phi + 2\phi\nabla_{\mu}\nabla_{\nu} \phi$$ and the same for the box term, but your calculation is correct!


1

Q1) The derivative is a linear operator hence \begin{equation} \partial_\alpha (x+ \xi)^\mu = \partial_\alpha x^\mu + \partial_\alpha \xi^\mu = \delta^\mu_\alpha + \partial_\alpha \xi^\mu \end{equation} The statement that those 2 factors get Taylor expanded is actually wrong, only the first factor gets Taylor expanded. Q2) To first order in the derivatives ...


5

With respect to the first question, to get the Einstein field equation, you must begin with an action of the form \begin{equation} S=S_{\text{EH}}+S_{\text{M}}, \end{equation} where \begin{equation} S_{\text{EH}}=\frac{c^{4}}{16\pi G}\int\mathrm{d}^{4}x\sqrt{-g}\,R \end{equation} is the Einstein-Hilbert action, whose variation with respect to $g_{\mu\nu}$ ...


2

Some preliminary consideration: In a uniform expanding universe the following equation holds $$\frac{GM}{R} \propto 1$$ with $c=1$ and $R$ is the reciprocal Hubble parameter. Now the assumption of Brans was that not only the geometry defines gravitation, but also the mass. So the gravitational constant should be depending on the mass. When you write this ...


0

I may as well convert my comment to an answer. The $\mu=\nu$ calculation is actually a contraction over all cases, not just the case not satisfying $\mu\ne\nu$. Since the latter obtains $0$, $R^\mu_\nu-G^\mu_\nu=A\delta^\mu_\nu$ in general iff $R^\mu_\mu-G^\mu_\mu=4A$.


0

In fact an Professor gave me a excellent paper which answers this question of mine. The paper is: Generating rotating regular black hole solutions without complexification by Mustapha Azreg-Ainou https://arxiv.org/abs/1405.2569


0

Check Appendix E of Burgess & Moore's The Standard Model: A Primer. There is a pretty good collection of results like that there. PS. Sorry to lose another soldier to the dark side that is the $(+---)$ metric (just kidding)


2

TL;DR: The main rule is that kinetic terms should be positive, cf. above comment by G. Smith. Examples: The Lagrangian density for a scalar field is $${\cal L}~=~\pm \frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi -{\cal V}(\phi), $$ with EL equations $$ \mp\Box\phi~=~{\cal V}^{\prime}(\phi),$$ if the signature Minkowski metric is $(\pm,\mp,\mp,\mp)$, ...


1

No. The basic invariant is $ds^2-cdt^2$ is a difference so you need an “i” somewhere if you’re gonna take a “usual” scalar product $(ds,icdt)\cdot (ds,icdt)$. It is more convenient to include it with $ct$ because $ds^2=dx^2+dy^2+dz^2$, or $d\vec s=(dx,dy,dz)$ so that’s fewer minuses. Note that the more “modern” approach define a metric $\eta_{\mu,\nu}=\...


2

The $\delta(\phi R)$ term will be: $$\delta(\phi R) = \delta(\phi g^{\mu\nu}R_{\mu\nu}) = \phi\delta g^{\mu\nu}R_{\mu\nu} +\phi\delta R_{\mu\nu}g^{\mu\nu} $$ The term: $\phi\delta g^{\mu\nu}R_{\mu\nu}$ is ready, here the variation of the inverse metric tensor is already a multiplying factor. Now the second term is: $$\phi\delta R_{\mu\nu}g^{\mu\nu} = \phi (...


2

The inner expression reads: $$ g_{ij} \frac{dx^i}{dt} \frac{dx^j}{dt} = \sum_{\text{all}(i,j)} g_{ij} \frac{dx^i}{dt} \frac{dx^j}{dt} $$ So you'd have to sum and then square root as the sum and square root operations are not interchangeable


0

Your question is a little confusing, so I'm going to explain what I think it's asking. Please let me know if I misunderstood it. Let us first define the object ${\tilde \epsilon}_{a_1\cdots a_n}$ as follows $$ {\tilde \epsilon}_{a_1 \cdots a_i \cdots a_j \cdots a_n} = - {\tilde \epsilon}_{a_1 \cdots a_j \cdots a_i \cdots a_n} , \qquad {\tilde \epsilon}_{12\...


0

The second equation does not directly follow from the first. You are correct in that \begin{equation} \tilde{\epsilon}_{\bar{\mu}_{0}..\bar{\mu}_{n}}| M|=\tilde{\epsilon}_{\mu_{0}..\mu_{n}}M^{\mu_{0}}_{\ \ \ \ \bar{\mu}_{0}}...M^{\mu_{n}}_{\ \ \ \ \bar{\mu}_{n}}, \end{equation} which is the full algebraic expression for the determinant of a matrix. Your ...


0

The Mandelstam variables $s, t,$ and $u$ are mostly used in $2 \to 2$ scattering and $1 \to 3$ decays. As for $2 \to 2$ scattering with initial momenta $p_1$ and $p_2$ and final momenta $p_3$ and $p_4$, they are defined as $s = (p_1 + p_2)^2 = (p_3 + p_4)^2$ $t = (p_1 - p_3)^2 = (p_2 - p_4)^2$ $u = (p_1 - p_4)^2 = (p_2 - p_3)^2$ They satisfy $s + t + u = \...


2

General Relativity is a mathematical framework within which we can construct Lorentzian manifold models of reality. In general, structures (e.g. the spacetime manifold) of a given model taken to represent some aspect of observable reality need not be physically real aspects of nature, whatever that might mean. Indeed, they're almost certainly not-- what are ...


4

Mathematical conditions such as Hausdorf or paracompactness apply to mathematical models of reality. They are introduced to prove theorems that apply to these models. Do not confuse mathematical models of reality with reality itself. Whether the universe is everywhere Hausdorf or paracompact is something to be decided by experiment. No amount of studying ...


6

The set is not predetermined but arises from physical/mathematical requirements of the given solution. GR is local theory and sufficinetly small region of spacetime is assumed to be isomorphic to open region of $\mathbb{R}^4.$ Globally the set is given by "gluing" these regions together until you arive at global solution you are satisfied with. GR ...


8

A manifold is a set - you don't need to put the manifold structure onto anything. Take a look at the first line of the wikipedia page for a manifold: a manifold is defined as a topological space which satisfies certain properties (and a topological space is a set of points). Intuitively: a manifold is a set which looks flat if you zoom in close enough on ...


1

The answer to your first question is yes, you just simply convert $p_{;\mu}$ to $p_{,\mu}$. The reason for that is that $p$ is a scalar function: see equation (1) in your question - you can take it as a definition of pressure and energy density: $$p = \frac{1}{3}(g_{\mu\nu} + u_\mu u_\nu) T^{\mu\nu}.$$ Here we used the signature of the metric in which $u^2 = ...


1

The light cone consists of all vectors $\mathbf V$ such that $\boldsymbol \eta(\mathbf V,\mathbf V) = 0$ (where $\boldsymbol \eta$ is the Minkowski metric). In Cartesian coordinates $(t,x,y,z)$, this means that $$\boldsymbol \eta(\mathbf V,\mathbf V) = -(V^t)^2 + (V^x)^2+(V^y)^2+(V^z)^2 = 0$$ or $$V^t = \pm \sqrt{(V^x)^2+(V^y)^2+(V^z)^2}$$ As mentioned by ...


1

In Special Relativity (so a bit of a reduction from GR), you use: $$\partial_{\mu}T^{\mu\nu} + u_{\mu}u^{\nu}\partial_{\alpha}T^{\alpha\mu} = 0$$ to obtain the SR Euler Equation. Similarly, try using $$\nabla_{\mu}T^{\mu\nu} + u_{\mu}u^{\nu}\nabla_{\alpha}T^{\alpha\mu} = 0$$ This equality follows from the fact that $u_{\mu}u^{\mu} = -1$ (just multiply each ...


1

After some more thought: ${\Lambda_\beta}^\rho$ is the inverse Lorentz transformation matrix and also satisfy the orthogonal matrix requirement: $${(\Lambda^T)_\beta}^\rho{\Lambda_\rho}^\mu={\Lambda^\rho}_\beta{\Lambda_\rho}^\mu=1$$ Taking the transpose of both sides, $${\Lambda^\mu}_\rho{\Lambda_\beta}^\rho=1=\delta^\mu_\beta$$ which is what I needed to ...


4

The relation $\Lambda^{\mu}_{\ \ \sigma}\,\Lambda^{\nu\sigma}=\eta^{\mu\nu}$, or more properly $$ \Lambda^{\mu}_{\ \ \sigma}\,\eta^{\sigma\tau}\,\Lambda^{\nu}_{\ \ \tau}=\eta^{\mu\nu}\ , $$ actually is the definition of a Lorentz transformation $\Lambda^{\mu}_{\nu}$. As such it cannot be proven true, unless you rest on some alternative definition. In case ...


1

You are trying to prove eq (18) in 2009.11827. That has nothing to do with the identity you are asking about in the question (which is completely incorrect, btw). To derive (18) in the ref. we simply need to vary the action w.r.t. the metric ($G_4$ is a function only of $\varphi$) $$ \delta \int \sqrt{-g} G_4 R = \int \left[ \delta \sqrt{-g} G_4R + \sqrt{-g}...


0

Yes, there is such a thing in GR. These are the rotations of tetrads. At each point of a space define an orthonormal basis: $$ (\vec e_{(a)}, \vec e_{(b)}) = \eta_{ab} $$ Where $a, b$ - denote the indices, corresponding to the local frame, in constract to the Greek spacetime indices $\mu, \nu$. The coordinate basis is related to the local basis, by some ...


3

This expression: $$(g^{\mu\nu} \nabla_\mu \nabla_\nu f) g_{\mu\nu} = \square f g_{\mu\nu} = (\nabla_\mu \nabla_\nu f)g^{\mu\nu}g_{\mu\nu} =\nabla_\mu \nabla_\nu f,$$ is wrong. This does not hold. The Box operator is a scalar quantity. It is defined as: $$\Box = g^{\mu\nu}\nabla_{\mu}\nabla_{\nu}$$ and for a four dimensional diagonal metric this is: $$\Box = ...


-1

Your expression is correct if you're using a derivative that is compatible with the metric, so if $\nabla_\mu g_{\rho\sigma}=0$, otherwise the d'Alembertian would bring other terms too. Edit: The reason why I think this is correct is $\Box f g_{\rho\sigma}=g^{\mu\nu}\nabla_\mu\nabla_\nu f g_{\rho\sigma}=\nabla^\nu\nabla_\nu f g_{\rho\sigma}$ $f_{,\rho\sigma}=...


3

There are two possibilities to look at this: Either the exact form of $V(B^2)$ is known and you can simply take the variation with respect to the $B^2$-dependency in that potential. More general, you can use use the chain rule as follows \begin{equation} \frac{\delta V(B^2)}{\delta B^2} \frac{\delta B^2}{\delta g^{\mu \nu}} \end{equation} The same ...


3

Yes, your action is of the form \begin{equation} S=\int\text{d}^{4}x\sqrt{-g}(\mathcal{L}_{\text{EH}}+\mathcal{L}_{\text{M}}), \end{equation} where \begin{equation} \mathcal{L}_{\text{EH}}=\frac{R}{2k^{2}} \end{equation} is the part whose variation with respect to $g_{\mu\nu}$ gives you the Einstein tensor in the equations of motion, and \begin{equation} \...


1

Yes. The field strength tensor is $F_{\mu \nu}$. The only two fundamental fields in your action are $g_{\mu \nu}$ and $A_\mu$ (unless you use Palatini variation and treat the connection as independent of the metric). So before preforming the variation, you should write the action in terms of these fundamental fields, and vary w.r.t each of them. The only ...


2

Yes. You have to vary $g^{ab}$ everywhere it appears.


2

You forgot to lower the $\mu$ index in ${(\mathcal{J}^{\rho\sigma})^\mu}_\nu$ before using its definition. Doing it makes $g$ appear as you need.


0

This is a good question, and there are probably technicalities that I'm not quite qualified to address. I'll give a non-rigorous answer and try to avoid any absolute statements. Geodesics are defined abstractly on some manifold, but they are a direct generalization of the idea of a straight line. Note that in a flat space a straight line is the "...


0

By parameterising a worldline from event $A$ to $B$ by the variable $\sigma$, then, denote the action $$S=\int ds,$$ which should then be written as an integral over the parameter $\sigma$. Next, recall the invariant measure $$ds^2 = g_{\alpha\beta}\,dx^\alpha dx^\beta .$$ Using the rules of total derivatives, we then get $$S = \int \sqrt{-g_{\alpha\beta} \...


1

The corresponding action $S=\int_{\sigma_i}^{\sigma_f} \mathrm{d}\sigma~L$ is the arc length between 2 spacetime events. The principle of stationary action (with Dirichlet boundary conditions) therefore leads to geodesics. See also e.g. this related Phys.SE post.


0

Hartle's Lagrangian is designed so that the corresponding Euler-Lagrange equation coincides with the geodesic equation for the metric $g_{\alpha\beta}$. That this is so is a standard exercise with Christoffel symbols that appears in most GR books.


3

As mentioned in ApolloRa's answer, in 2+1 dimensions there exist no asympototically flat black hole solutions. However, you can still solve the Einstein Field Equations to find the metric of a non-spinning point mass $M$. The answer is given by $$ ds^2 = -dt^2 +\frac{1}{(1-4GM)^2}dr^2 + r^2 d\phi^2$$ As you can easily check this metric is flat for all $r>...


10

No it is not. The Weyl tensor vanishes by definition in three dimensions, Einstein's equations (in the absence of matter) impose: $$R_{\mu\nu} = 0 \rightarrow R=0$$ and since $Riemann = Weyl + Ricci$ no geometry can be formed. The solution in three dimensional spacetime is the BTZ black hole (https://arxiv.org/abs/hep-th/9204099v3) which includes a ...


1

I've eventually found the answer. So we have : $$ \mathcal{L} = \frac{1}{2} m g_{ij}(\vec{x}(t))\dot{q}^i\dot{q}^j = - U(q,t) $$ Computing the different terms in Euler-Lagrange equation for a particle $k$: $$ \frac{\partial \mathcal{L}}{\partial \dot{q}^k} = \frac{1}{2}mg_{ij}\dot{q}^i\delta^j_k + \frac{1}{2}mg_{ij}\delta^i_k\dot{q}^j = \frac{1}{2} m g_{ik} \...


0

The link gives examples such as the union of two spheres, and in an example like this, any Riemannian metric defined on the spheres gets extended to one that is singular where the spheres intersect. This seems to me to be a setup that is guaranteed not to be of interest in classical relativity. The singularity guarantees that observers in a particular region ...


2

Force, in its Newtonian sense, is an integral part of general relativity. General relativity attributes the effects of gravitation and inertia to a dynamic 3D metric and a "dynamic" time lapse. But all other forces remain as they were. In fact we can say that it's force, momentum, and energy – not mass – that cause curvature and are affected by ...


1

Let me review an argument given in the most textbooks. First let's start by formulating postulates of special relativity. Postulate 1 "The laws of nature and the results of all experiments performed in a given frame of reference are independent of the translational motion of the system as a whole. More precisely, there exists a triply infinite set of ...


2

The divergence of the energy-momentum-stress tensor is zero. The time component of this expresses energy conservation, and the spatial components express momentum conservation. This means that the time-time component is energy density, the time-space components are energy flow / momentum density, and the space-space components are momentum flow. The ...


0

I get the impression that when people ask about the concept of force they assume that there a single universal definition of force. However, that is not the case. In the history of physics it is quite common for a word to be radically redefined. Example: physics used to have 'Caloric theory', but over time it became clear that Caloric theory had to be ...


0

A tip: use the Kronecker Delta in the derivative, and remember that the x coordinates are related to the generalized coordinates q, {q^dot}. The delta will basically kill that 1/2 factor, and the Christoffel symbol arises from the corrdinate transformation (parallel transport)


1

Having one temporal dimension and three spatial dimensions means that the metric tensor has signature $(1,3)$ or $(3,1)$ depending on your convention. Two temporal dimensions and two spatial dimensions would mean that the metric has signature $(2,2)$. Signature is a concrete property of a metric, so there is no sense in which a dimension could be both ...


0

I think that it is the opposite. In the framework of GR the concept of force keeps necessary and it is clearer than in Newtonian mechanics. It is necessary to postulate a gravitational force in Newtonian mechanics, otherwise accelerated movements of planets and apples would violate the second law. And a typical device to measure force as a spring, would ...


0

In Newtonian physics, every force can be described by a (three-)vector, including fictitious forces in accelerating coordinates. In special relativity, electromagnetic and normal forces can be described by (four-)vectors, but fictitious forces can't be. If you write special relativity in a manifestly covariant form with a metric tensor, then fictitious ...


0

This is a comment: Searching through the literature there are several derivations of Newtonian gravity from the General Relativity expressions. For small masses and velocities (flat space) Newtons equations are recovered. This is an example of the mathematics. Where General Relativity has to be used forces have no specific meaning. The GR equations have to ...


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