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How to obtain orthonormal tetrad basis for an infalling observer?

Let me give you the general idea. When constructing an orthonormal frame of an observer, you want to start with the zeroth leg, which is the four-velocity of the observer. In this case you are looking ...
Void's user avatar
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What is the difference between a local and global coordinate transformation?

$x \rightarrow x' = (1+\sigma)x$, acts identically on each point. Similarly, if $\epsilon(x) = kx$, then scaling transformation is global as well.
Shark's user avatar
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Does the singularity travel along the time axis?

In a curved spacetime there is no unique coordinate chart that covers all of spacetime, so the whole notion of "the time axis" is ill-defined. Moreover, the singularity itself is not part of ...
Eric Smith's user avatar
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Does the singularity travel along the time axis?

We can't observe the singularity but we can observe the event horizon since we are attracted towards it. This comment reflects a Newtonian approach and neglects GR where the interpretation is that ...
Wookie's user avatar
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The angular momentum of zero mass limit of Kerr metric

The Kerr metric describes a rotating body. $a = J/Mc$ is the angular momentum per unit mass of the body. $a$ characterizes the rate of rotation. In the limit $M \rightarrow 0$, the metric becomes the ...
mmesser314's user avatar
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Why if the metric tensor components are constant then SR applies?

Let me summarize some ideas in the comments and the other answer and add an important point regarding frames that I think is interesting to keep in mind. In Lorentzian signature, if you go to an ...
Gravitino's user avatar
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Why if the metric tensor components are constant then SR applies?

The metric is diagonalizable, yes, and then with further coordinate transformations it can be converted to Minkowski; see, for example, Schutz exercise 6.3 where he guides you through the steps.
Lewis Kirby's user avatar
3 votes

Partial derivatives vs Covariant derivatives in polar coordinates

As OP correctly points out connections introduce a concept of differentiation of tensor fields or more in general of sections of vector bundles that takes into account how the bases of the fibers ...
Mr. Feynman's user avatar
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5 votes

Partial derivatives vs Covariant derivatives in polar coordinates

Covariant derivatives take into account for both component and basis changes, thereby applicable for curved spaces - where partial derivatives only take component changes into account - is this ...
Lenard Kasselmann's user avatar
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Is my understanding of canonical transformations flawed?

Hint: In symplectic notation $$z^I~=~(q^i,p_i)\qquad\text{and}\qquad Z^I~=~(Q^i,P_i),$$ and assuming no explicit time dependence in the transformation $Z^I=f^I(z)$, OP has shown that $$\begin{align}\...
Qmechanic's user avatar
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Equating 2 sides of EFE

First, note that for scalar functions, the covariant derivative reduces to the partial derivative. So for scalar functions, it is true that if the covariant derivative is zero at a point, then the ...
Andrew's user avatar
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2 votes

The center of the Schwarzschild black hole

Equation for $r$ in terms of $r'$ (in the question as originally posed) has them the wrong way round. In notation $r$=isotropic, $r'$=Schwarzschild coordinate it should be $$r' = r (1 + a/r)^2 .$$ ...
Andrew Steane's user avatar
1 vote

Difference and meaning of index the derivative operator

Here is a brief summary: In (relativistic) physics, it is standard to adorn a local coordinate $x^{\mu}$ with a superindex. We define a shorthand notation for the partial derivative $\partial_{\mu}:=...
Qmechanic's user avatar
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3 votes

Difference and meaning of index the derivative operator

You can eventually (if you need to) learn a more rigorous treatment later, so let me instead provide a cookbook approach: An object with an open index means that its value changes when the observer ...
SonerAlbayrak's user avatar
3 votes

The center of the Schwarzschild black hole

The metric in isotropic coordinates (1) and (2) has a (coordinate) singularity at $r=a$. Consequently, there is no a priori relationship between the metric for $r>a$ and $r<a$. They are both ...
TimRias's user avatar
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3 votes
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Why can the dot product of two vectors be expressed as a differential?

I would imagine that the simplest way to show this is to note that the position vector $\mathbf x$ can be expressed in either basis: $$x'^j \hat e_j' = \mathbf x = x^i \hat e_i$$ A given set of ...
J. Murray's user avatar
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1 vote

Why can the dot product of two vectors be expressed as a differential?

First observe, that for any matrix, we can pick up its $i,j$ entry by applying it first to a column vector that is zero apart from $1$ at its $j$'th position and then dotting it into a similar vector ...
Amit's user avatar
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3 votes
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How does the wavefunction transform under an arbitrary change of variables?

TL;DR: As the overall phase of the wavefunction is not physical, OP's question has a non-unique answer that ultimately comes down to a choice of convention. Within a given class of situations we often ...
Qmechanic's user avatar
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7 votes
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Vector addition for differentials in the context of electric potential

I think that your professor is showing the differential vector for infinestimal change in each coordinate component. The diagrams correspond to cartesian, cylindrical and spherical coordinate systems ...
Muhammed Çağlar TUFAN's user avatar
3 votes

Vector addition for differentials in the context of electric potential

The infinitesimal displacement $d\vec{s}$ is derived from $\vec{s}$. In cartesian coordinates : $\vec{s}=x\vec{i}+y\vec{j}+z\vec{k}$ $$d\vec{s}=dx\vec{i}+xd\vec{i}+dy\vec{j}+yd\vec{j}+dz\vec{k}+zd\vec{...
M06-2x's user avatar
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1 vote

Vector addition for differentials in the context of electric potential

The diagrams are slightly misleading, because the infinitesimal changes in angular quantities are shown as being quite large. It may be more helpful to draw separate diagrams showing how $\vec{s}$ ...
FTT's user avatar
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0 votes
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Direction of Area Vector of a simple 2D figure

For a open surface, it is usually in that direction which make acute angle with electric field.
Alv's user avatar
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1 vote

Time dilation and understanding which is $\Delta t$ and which is proper time $\tau$

superman is not at rest There is no preferential frame in relativity. The problem supposes both Louis frame and Superman frame as inertial, with a relative velocity of $0.7$c. The two events (say ...
Claudio Saspinski's user avatar
0 votes

Time dilation and understanding which is $\Delta t$ and which is proper time $\tau$

Superman's wrist watch is at rest with respect to superman. So the time range measured by superman's wrist watch is smaller (i.e. the time runs slower for superman and for his watch) than the ...
Frederic Thomas's user avatar
1 vote

Time dilation and understanding which is $\Delta t$ and which is proper time $\tau$

I've always found talk of an observer confusing in Special Relativity. Much clearer, in my opinion, to use the idea of an inertial frame of reference (frame, for short). The proper time between two ...
Philip Wood's user avatar
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Time dilation and understanding which is $\Delta t$ and which is proper time $\tau$

Whatever Superman does, he does right where Superman is. Superman kisses Lois goodbye right where Superman is. Twenty hours later in Superman's proper time $\tau$, Superman fixes the space probe right ...
g s's user avatar
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Time dilation and understanding which is $\Delta t$ and which is proper time $\tau$

Superman is at rest, in Superman's frame ($S'$). It's good to assign a frame name to each observer, btw. It increases clarity. In $S'$, Lois's frame ($S$) is moving at $\beta=-0.7$. Use events: $$ E_i ...
JEB's user avatar
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Time dilation and understanding which is $\Delta t$ and which is proper time $\tau$

According to superman's wrist watch, superman is not moving, and instead Earth and space probe are moving.
naturallyInconsistent's user avatar
0 votes

Inertia tensor but expressed after undergoing rotation

Yes, the interpretation behind $$ {\rm I}_{\rm world} = R \, {\rm I}_{\rm local} R^\intercal \tag{1}$$ where ${\rm I}_{\rm local}$ is the diagonal matrix representing the MMOI tensor on the body frame,...
John Alexiou's user avatar
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1 vote

Direct conversion of cartesian velocity to spherical velocity and vice-versa

Notice that $\frac{d\theta}{dt}$ and $\frac{d\phi}{dt}$ do not have units of velocity (m/s), so right there you have a problem. Velocity in the $\hat \theta$ direction is "how much distance $ds$ ...
RC_23's user avatar
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0 votes

Direct conversion of cartesian velocity to spherical velocity and vice-versa

Spherical coordinates describe the position of a point in space in terms of a radius and two angles. Depending on the convention this looks like $$ \vec{p} = \pmatrix{x \\ y \\ z} = \pmatrix{ r \cos \...
John Alexiou's user avatar
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1 vote

General Relativity manipulating tensors, tensor indices meaning

About the first part, you nearly answered your own question: it is indeed exactly because the definition of tensors arises naturally out of multilinear maps such as $t(e^a,e^b)$ that the order of ...
Amit's user avatar
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1 vote
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Direct conversion of cartesian velocity to spherical velocity and vice-versa

This is one of those things that will, upon closer inspection, turn out to be horrible and confusing. Let us compute some stuff, with suggestive notation chosen to make the understanding easier. ...
naturallyInconsistent's user avatar
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Completeness of radial null geodesics in spacetime of Schwarzschild interior solution

What I was missing is the notion of analytical continuation. While the coordinate is $r$ is always positive geodesics should be continued to the antipodal direction $\theta \rightarrow \pi-\theta,~~~\...
JanG's user avatar
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0 votes

Find canonical transformation for $P$ given canonical transformation for $Q$

TL;DR: The answer depends on the context. Examples: If we assume that a canonical transformation here means a symplectomorphism, then indeed OP's method is correct: the determinant of the Jacobian is ...
Qmechanic's user avatar
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Components of velocity 3-vector of light in spherical coordinates

$$v_r = c\\v_\vartheta=\vartheta_0\\v_\varphi=\varphi_0$$ Basically by definition, or visual inspection of how the relations are meant to work.
naturallyInconsistent's user avatar
2 votes
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Kinetic equation material derivative

Using the coordinate system in Fig 5.1, the electric field can be written as: \begin{equation} \mathbf{E} = E\left(\cos \vartheta \hat{e}_v - \sin \vartheta \hat{e}_\vartheta \right). \end{equation} ...
ZinnTheJackal's user avatar
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ADM decomposition incompatible with black hole?

In general, one needs to use a set of horizon-penetrating coordinates. These can be illustrated using a Penrose diagram of a Schwarzschild black hole: (If you do not know what a Penrose diagram is, ...
Void's user avatar
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-1 votes

ADM decomposition incompatible with black hole?

I feel like it should be improbable to decompose a black hole in these constraints. Given that that the space like hypersurfaces cannot circumvent the singularity in the black hole metric, it would ...
Tasnuva Jahan's user avatar

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