New answers tagged

0

I’m working on the equation but I don’t know what does the letters in the equation mean. Can you tell me what the letters in the equation mean? (Words or numbers) This is the equation: 48M^2/r^6 If the letters have a exact number, I hope you can write it down ,too.


2

Let me say some words about how we determine symmetries in physics. If we were to imagine closing our eyes and forgetting we ever saw the world, we would have no idea what symmetry groups might be found when we open our eyes. The flip side of this is that, before we open our eyes, any symmetry group is possible. As soon as we open our eyes and look at things,...


3

Hint: $$ \epsilon^{\rm symbol}_{a_1\ldots a_d} e^{a_1}_{\mu_1}\ldots e^{a_d}_{\mu_d} ~=~ \det(e^a_{\mu})\epsilon^{\rm symbol}_{\mu_1\ldots \mu_d} ~=~ \sqrt{|\det(g_{\mu\nu})|}\epsilon^{\rm symbol}_{\mu_1\ldots \mu_d} ~=~ \epsilon^{\rm tensor}_{\mu_1\ldots \mu_d}. $$


2

Observe that the definition of the determinant and its change of sign under interchanging rows or columns tells us that $$ \epsilon^{a_1\ldots a_d} e^{\mu_1}_{a_1}\ldots e^{\mu_d}_{a_d}= \epsilon^{\mu_1\ldots\mu_d} {\rm det}[e^\mu_a], $$ and remember that there is a connection between the $e^\mu_a$ and the metric.


2

I will try to rephrase a bit how I understand what Weinberg is saying. A priori we only know what for two different inertial reference frames we have $ds^2 = ds'^2$ when $ds^2 = 0$. We then try to find the most general coordinate transformations of spacetime which possess this property. This way we discover conformal transformations. However, when we try to ...


1

This is a complementary answer to Qmechanic's, which provides somewhat more detailed calculations. The physical quantity identical from the viewpoint of both spacetimes $\mathscr{V}^+$ and $\mathscr{V}^-$ is the respective induced metrics. Intuitively, it is the tangential projections of the spacetime metrics $g_{\mu\nu}^\pm$ onto the surface. This is the ...


1

First lets clear up some confusion. You say: To be conformal means that the physic is unchanged This is not what "conformal" means. Conformal means that the angles are unchanged. In the context of GR, this in particular means that the causal structure remains unchanged. The phyisics in the meantime can be quite different. In particular, the $uv$-...


2

No, there is not such general rule that you are looking for. But you can count the maximum number of independent Christoffel symbols in a general space. In fact, for each independent component of the metric tensor, there are, at most, $N$ distinct Christoffel symbols. Let me first start with an example. If you consider a two-dimensional Cartesian coordinate ...


0

There seems to be some confusion here, at least in terminology. If $\mathrm d\mathbf s=(\mathrm dx^0,\mathrm dx^1,\mathrm dx^2,\mathrm dx^3)$ is an infinitesimal displacement along a null worldline, then $\mathrm ds^2 \equiv \mathbf g(\mathrm d\mathbf s,\mathrm d\mathbf s) = g_{\mu\nu} \mathrm dx^\mu \mathrm dx^\nu = 0$. This is not really a statement ...


3

This is a classic misuse of the Einstein summation convention. The correct expression in your final equation is $$ \gamma^{\mu} \partial_{\mu} = \sum_{\mu = 0}^3 \gamma^{\mu} \partial_{\mu} = \gamma^0 \partial_0 + \vec{\gamma} \cdot \vec{\nabla} $$ The reason for your confusion is because you're used to taking the Minkowski inner product between two ...


0

Lorentz indices are indeed raised and lowered with the metric tensor. But note that in QFT there are also other indices than Lorentz indices and these are raised an lowered with the appropriate tensor.


3

I believe it is just a matter of notation inconsistency, common in physics. Manifolds are by definition locally Euclidean (i.e. locally look like $\mathbb{R}^n$, the $n$ dimensional Euclidean space). There is no metric given a priori. In Special Relativity, we also define a metric of Lorentzian signature, and call this structure Minkowski spacetime. This ...


2

The value of $\kappa$ different from the value of $1$ (which is what we have in the usual form of Schwarzschild solution) corresponds to the deficit (for $\kappa<1$) or excess of solid angle. If we consider a sphere of constant radius $r=R$ when $R\gg M$, the distance from the central object (black hole) is approximately $\kappa^{-\frac12} R$, while the ...


0

Let me try to add some comments to the already given explanation, which focuses on there being more dimensions in actual Schwarzschild spacetime. So we first have to polish the question a bit. The statement claiming that the Schwarzschild space-time (independent from the coordinates used to describe it) is isomorphic to the complex-split plane is false in ...


2

If this were a 1+1 dimensional spacetime then you'd be correct, but there are two other spatial dimensions not shown on the chart. Each point on the chart is really a sphere whose radius is a function of $X^2-T^2$. The event horizon is a 3-cylinder (direct product of a 2-sphere and a line), but with a degenerate metric along the length of the cylinder. If ...


2

Those two expressions are equivalent. When contracting tensor indices, it doesn’t matter which one is up and which is down. So it doesn’t matter which one you contact with the (inverse) metric first.


3

Yes. An example is a FLRW spacetime with equation of state $p=- ρ/3$. The second Friedmann equation for this equation of state ensures that $\ddot a ≡ 0$, and this means that Ricci tensor component $R_{tt}\sim \ddot a ≡0$. So, with 4-velocity $V^i$ of the comoving frame $R_{ij}V^i V^j≡0$, while spatial components of the Ricci tensor remain nonzero.


1

Since $P_{muon}$ is the 4-momentum, its square magnitude is the mass-squared times $c^2$, which can written in terms of an observer's decomposition of $P_{muon}$ into a temporal part, the relativistic energy $E_{muon}/c$ (which includes the kinetic energy) and the spatial part, the relativistic momentum $\vec p_{muon}$, where $$P_{muon}^2=m_{muon}^2c^2=(E_{...


2

In general, a $(p,q)$-tensor eats $p$ covectors and $q$ vectors and returns a real number; such an object has $p$ indices upstairs and $q$ indices downstairs. The components of the metric tensor are written $g_{ij}$ because the metric $\mathbf g$ is a $(0,2)$-tensor which eats two vectors (and no covectors) and returns their inner product. If you see the ...


1

$s_0$ is just the unit matrix. It is not explicitly written in the Hamiltonian - every entry in the $4\times4$ matrix is a $2\times2$ unit matrix itself, but since it's just a unit matrix, a simplified notation is used.


1

You have to rearrange and factor terms in a particular way and then use properties of derivatives. For full derivation with explanations on each step, see the blog


1

The good thing about 4-vectors is that the inner product between two of them using the proper metric is always preserved, no matter from which reference frame you take it! I don't know how familiar you are with the concept of metric, but an easy intuition of it is that it is the set of rules that define the mathematical structure of the inner product. It's ...


1

As stated in my comment, you can do this without making reference to normal coordinates. E.g. see Confusion about Lie derivative on metric. Here's a quick proof in local coordinates which I haven't seen answered here though. Take the standard form of Killing's equation, $$ \begin{align} \tag{1} \mathcal{L}_{\xi} g_{\mu \nu} &= 0 \\ &= \xi^{c} \...


1

$$\nabla^{a}u=g^{a\alpha }\nabla_{\alpha }u\overset{u\space is\space scale\space,\Gamma=0}{=}g^{a\alpha }\partial_{\alpha }u= g^{at }\partial_{t }u+g^{ar }\partial_{r }u=\frac{a^{2}}{a^2-r^2}(\frac{\partial}{\partial t})^{a}+(\frac{\partial}{\partial r})^{a}$$


0

This is actually more general than just diffeomorphisms, but it is a but more intuitive there since we are so used to coordinate transformations. For any quantity, one can simply decide to measure it in different units, different scale, or shift the scale. This doesn't change physical properties in any way, and in this sense corresponds to a gauge ...


1

The trick is the difference between intrinsic and extrinsic curvature. Intrinsically curved spacetimes are curved relative to their own coordinate system. On intrinsically curved space(times), Euclid's laws, in particular, the parallel postulate, are not valid, and the geometry is necessarily different. Mathematically, there is an object called the Riemann ...


2

It might be helpful for you to read up on Riemannian and flat manifolds. But, yes, it is possible for some flat N-D flat manifold to appear curved in some (N+1)-D hyperspace. At least, that's my recollection. I'll go through my old notes soon and edit this comment as / if appropriate. EDIT: There a difference between intrinsic and extrinsic geometry. You ...


-1

There is a Bianchi classification of 3d Lie algebras upto isomorphism. Hence this also classifies all 3d Lie groups. Now, in the ADM formalism of General Relativity, we consider the evolution of a 3d spatial slice. The symmetry group of this slice is a Lie group. When this symmetry group is 3d, we can use the Bianchi classification. This gives us a ...


2

You're in the right ball-park. One thing that is not explained well in my opinion is the distinction between covariant and contravariant components, and as you point out, how this relies upon the metric. If we fix a basis $e=(e_\alpha)$ in a vector space $V$, then for every vector $v \in V$, we have its components, $v^\alpha$. So far so standard. What's not ...


0

The proper time is the time experienced by the particle itself. When you thus choose your reference frame as that of the particle itself, you fix your spatial coordinates at the particle's location. So $dx^i=0$.


-1

The function $t$ is simply time coordinate. So if I call schwarzschild time coordinate $t'$ we have $t=t'$. Then $$X^b\equiv g(X,.)=g(\partial_{t'},.)=g_{t'j}dx^j=g_{t't'}dt'$$ since metric has no off-diagonal elements and $$g(X,X)dt=g(\partial_t',\partial_t')dt=g_{t't'}dt=g_{t't'}dt'$$ since function $t$ is our $t'$ coordinate. The two expressions are the ...


0

A good reference for such things is Wald's General Relativity. In that book, you'll find the following result: Theorem. Let $(M,g)$ (a manifold with a metric) be globally hyperbolic. Then There exists a global time function, that is a map $t:M\to \mathbb{R}$ such that $-dt$ is future-directed and timelike; Surfaces of constant $t$ are Cauchy surfaces, and ...


0

consider Killing's equation for schwarzschild: $$\nabla_{(a}v_{b)} = 0$$ consider the ansatz $v_{b} = a(r)dt$ Then, This gives us one nontrival equation (t,r): $$\begin{align} 0 &= \partial_{r}a - \Gamma_{rt}{}^{t}a\\ &= \partial_{r}a - a\frac{M/r^{2}}{1-2M/r} \\ ln(a) &= ln(C) + (1/2)ln\left(1-2M/r^{2}\right)\\ a &= C\left(\sqrt{1-2M/r}\...


1

I remember that I have also struggled with the sign of the minimal coupling term several times in the past. Therefore, I feel compassionate with your attempts to derive it. Would you accept the non-relativistic Schrödinger equation for the bare electric field of the nucleus (i.e. without magnetic potential) as "some principled way to understand what's ...


0

Events do not "occur" in frames. They are measured or described relative to frames, but the events themselves occur in spacetime. For example the emission of a particle and absorption of that same particle later are events. Different inertial frames would assign different coordinates to these events, but the spacetime interval between them is the ...


1

Note that your attempt is not technically correct, because you can't repeat a dummy index, so you should use (for example) $g_{\alpha\nu}$ instead of $g_{\mu\nu}$, since $\mu$ is already in use. Apart from that, since $g^{\mu\nu}g_{\mu\nu}=\delta^\nu_\mu$ is not correct either, it would be $g^{\mu\nu}g_{\mu\nu}=\delta^\mu_\mu=\delta^0_0+\delta^1_1+\delta^...


2

The second-to-last paragraph in the question starts with $$ p^\mu\to i\partial^\mu. $$ This implies $$ p_\mu\to i\partial_\mu, $$ because the index on both sides is lowered using the same metric (regardless of the sign convention for the metric). This equation can be separated into \begin{align*} p_0 &\to i\partial_0 \\ p_j &\to i\partial_j. \...


0

The point is that, for the Levi-Civita connection, we can find Riemann normal coordinates around any point P. In those coordinates $\partial_\alpha g_{\mu\nu}=0$ and the ${\Gamma^\alpha}_{\mu\nu}=0$ at P. Now the formula for the Lie derivative $$ [{\mathcal L}_Xg]_{\mu\nu}= X^\alpha \partial_\alpha g_{\mu\nu} + g_{\mu\alpha}\partial_\nu X^\alpha + g_{\alpha\...


0

Apart from the fact that your first equation is missing some $i$s in the exponential, the relative minus difference is just due to convention and it's not really a problem. I could have defined the transformation of the field operator (as some authors do) not as $$U^\dagger(a) \phi(x)U(a)$$ but as $$U(a)\phi(x)U^\dagger(a)$$ and this would have solved the ...


2

The spacetime interval between two events is always the same in any two inertial reference frames- that is, as long as the frames themselves are not accelerated. It doesn't matter if these events are points on the worldline of an accelerating particle, a particle of constant velocity, or just two events picked from spacetime at random.


0

Hamiltonian characterizes a probability distribution function. And so does Fisher information. The two are connected obviously. Renormalization offers ways to view the data at different scales. Hence with renormalization, Hamiltonian or Fisher information would change accordingly.


0

First there is a bijection between Minkowski space, $M$ and Hermetian 2x2 matrices, $Hm[2]$, this is given by $X: M \rightarrow Hm[2]$ where $x=(x^\mu) \rightarrow x^\mu.\sigma_\mu$. We then see that $det[X(x)] = |x|^2$, where the latter is the square of the Minkowski norm. Next, we define an action of $SL(2,C)$ on $Hm[2]$ by $S.A:=SAS^\dagger$ where $S \in ...


1

OP particular question boils down to the use of the polarization identity, cf. e.g. this related Phys.SE post. More generally for the group homomorphism, see e.g. this related Phys.SE post.


-1

The "Expansion of Space" is a pedagogical device which makes no sense in language except as a local effect in descriptions of the movement of objects (as in, "they expanded the space in the living room by moving the couch into the dining room, thereby decreasing the space available there"), and is, consequently, even less likely to make ...


0

Let $C$ be a field such that one had a covariant derivative $\partial_\mu+C_\mu$ satisfying: \begin{align*} &(D_\mu \Psi)'=e^{i\alpha(x)}D_\mu \Psi \\ &\Longrightarrow (\partial_\mu+C'_\mu)[e^{i\alpha(x)}\Psi]=e^{i\alpha(x)}(\partial_\mu+C_\mu)\Psi \\ &\Longrightarrow C'_\mu\Psi=C_\mu\Psi-i\partial_\mu \alpha(x)\Psi \end{align*} So one can see ...


3

The quick answer is that yes, it does make sense, if we interpret symmetric to mean self-adjoint. Symmetry of a bilinear form $g$ (or higher rank tensor) means $g(X,Y)=g(Y,X)$, but symmetry of a linear operator $A$ means $\langle X,A(Y)\rangle = \langle A(X),Y\rangle $ where the brackets denote some chosen inner product. In that sense, symmetry of a linear ...


6

There can be different notions of "symmetric". Also, keep in mind that $\delta$ is just the index form of the identity operator. The first one pertains to the transpose of an operator, which is just the adjoint for a real case. Generally, for the transpose of a tensor, we have $$\left<v,M w\right>=\left<M^T v,w\right>\,,$$ or in index ...


1

Since $V$ is a Killing vector field, $\nabla_iV_j=-\nabla_jV_i$. Using this in equation 2, the desired relation follows.


0

The covariant derivative is independent of the chart used, if the value at a selected point $p$ is zero in one chart, it is zero in every chart. Using geodetic curves from $p$ allows to project the tangent space at $p$ to the manifold, this is called the exponential chart. Parallel transport along these geodetic curves gives constant coefficients in the ...


Top 50 recent answers are included