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Infinitesimal volume elements do not have to be cubes. Some familiar examples come from typical solids of revolution problems from calculus 1/2. Typically one discusses using either the "disk/washer" or "cylindrical shells" methods for finding the volume of the solid. As you can guess, the former method uses infinitesimally thin disks/...


17

Your comments (and to a lesser extent, your Question) indicate a severe confusion about ever having an infinitesimal volume. You never construct an infinitesimal volume. Infinitesimal volumes appear at the end of a limiting process. Where do the infinitesmial rectangular parallelepipeds you are discussing appear? They appear in the limit of an iterated ...


16

$\gamma:\mathbb R\rightarrow M$ is a curve whose image lies in the spacetime $M$, so $\gamma(t)$ is the event at parameter value $t$ along the curve. $\gamma'(t) \in T_{\gamma(t)}M$ is the tangent vector to the curve at parameter value $t$, and $\gamma''(t)\in T_{\gamma(t)} M$ is the rate of change of this tangent vector with respect to the parameter $t$. ...


7

Different coordinate systems have different kinds of volume elements; The volume elements are a consequence of how the grid lines of the coordinate system are set. Volume element can be generated by nudging the parameters which describe points in the space by infinitesimal amounts and figuring out the volume of the region generated as a consequence. This is ...


6

What we get out of this is the ability to define vectors where there is no obvious way to choose a vector from $\mathbb R^n$. In particular, it allows us to define the tangent space of a manifold without first embedding the manifold in a real vector space. If we have a $k$-dimensional submanifold $M\subseteq\mathbb R^n$, then we can literally take the ...


6

Consider the local positional coordinate vector $\mathbf{r}$ which can be used to define the location of any arbitrary point in space, but here in our case, since it only depends on a single time parameter, it therefore only describes a curve within a submanifold of said space. It's acceleration vector is given by the second order derivative with respect to ...


5

An infinitesimal is by definition a length that is really, really small. I think that your question arises due to a misunderstanding of what infinitesimals are. Infinitesimals are not easy to understand, they can be understood either as a limit as a quantity goes to zero or in terms of the hyperreal numbers. As the hyperreal concept is relatively new ...


5

Measurement instruments are not infinitely precise, however It is possibile to distinguish objects using them. This is possible when the precision of them permits it. The precision of an instrument around measured values is the physical corresponding of a neighborhood of a point. The fact that two measures can be distinguished by means of sufficiently ...


5

It’s not so much a question of what is theoretically correct, more a question of which shape of region allows us most easily to pass to the limit and derive a differential equation or an integral (which is usually the goal of this step). The choice of region often depends on the symmetry of the problem. In problems with cylindrical symmetry it is common to ...


4

You are using two different sets of basis vectors. $\frac{\partial\mathbf e_r}{\partial \theta} = \mathbf e_\theta$ and $\frac{\partial \mathbf e_\theta}{\partial \theta} = -\mathbf e_r$ hold for the orthonormal polar basis, in which the metric takes the form $$g_{ij} = \pmatrix{1 & 0 \\ 0 & 1}$$ The connection coefficients you quote arise from the ...


4

It’s not $\nabla$ which is behaving strangely, it’s $\frac{d}{d x}$. You note that $A\cdot\nabla\ne\nabla\cdot A$, by which you mean $A\cdot\nabla f\ne\nabla\cdot(A f)$ in general for a test function $f$. But it’s also true that $g\frac{d}{d x}\ne\frac{d}{d x}g$, in the same sense that $g\frac{d f}{d x}\ne\frac{d}{d x}(g f)$ in general. That second ...


4

To take an entirely different approach to the various integration-related approaches of other answers .... You appear to be perfectly comfortable defining "an infinitesimal cube", as a cube with sides of infinitesimal length. Let's go one step further ... lets say: "the infinitesimal cube with a vertex at the origin, and lying within the ...


3

You are correct in the interpretation. I will try adding a few comments. If you compare to usual vector calculus in flat space, when one has a curve $\vec{r}(t)$, the acceleration corresponds exactly to the second derivate of the trajectory or alternatively the time derivative of the velocity, here all being a curve of vectors in 3D (trajectory, velocity, ...


3

You asked: "Does it make sense to take an infinitesimal volume of shape other than a cube?" Yes, and no. If you have an integral $$\int\int\int \text{something} \mathrm d\alpha \mathrm d\beta\mathrm d\gamma$$ then you do have cubes in $\alpha\times\beta\times\gamma$. But those cubes in $(\alpha, \beta, \gamma)$ will be distorted into other shapes ...


3

Am I correct in understanding that there is no mathematical distinction between vectors and one-forms on a (pseudo-)Riemannian manifold endowed with a metric? No. Vectors and one-forms are distinct mathematical objects with distinct transformation behavior under changes of coordinate chart. There are also subtleties when defining pushforwards and ...


2

Let me do it for once and for all. Though the question has been answered by spiridon, I would like to give a formal derivation as spiridon's answer involves guess work. We have a situation where we need to calculate the inverse of a partitioned matrix. So let us first derive a general formula for the inverse of partitioned matrices and then we shall apply it ...


2

It's best to think relativisticly. We are solving the Maxwell equation $d\star (d A)=\star J$ where $$ A=-\phi dt+ A_x dx +A_ydy+A_zdz\\ J= -\rho dt + j_x dx+j_ydy+j_z dz $$ are 1-forms in four dimensions and $\star$ is the Hodge dual. So $\star J$ is a three form. The $j_x$ appearsin $\star J$ with $dy\wedge dz \wedge dt$ so it's only a 2-form if you ...


2

Remember that $\mathbf Y_x$ is the vector (not vector field) from $\mathbf Y$ at the point $x$. A vector eats a function and spits out a real number. As both $\mathbf Y_{\phi_t x}(f)$ and $\mathbf Y_x(f)$ are numbers, you can subtract them without issue.


1

Spacetime topology certainly has importance in physics. In condensed matter physics, topological phases of matter are systems which are sensitive to spacetime topology. Many of the interesting ones are two dimensional, so they can in principle be fabricated into interesting surfaces in a lab. Alternatively, through clever tuning of non-local interactions, ...


1

The two most important properties of an infinitesimal volume used for integration are a) its side is shorter than that of any other volume you care to specify b) the value of the property that is a function of its dimensions, this is the thing you're integrating over a volume, is the same on any edge or vertex of the volume. That means it doesn't matter ...


1

Yes, it absolutely does sometimes make sense to have infinitesimals be shapes other than cubes. Particularly when the metric space being used is not necessarily Euclidean. In Walter Rudin's Principles of Mathematical Analysis, in the early part of his formulation of the general form of Stoke's Theorem (i.e., for arbitrary finite dimensional metric spaces), ...


1

There are (at least) two occasions when it is useful to distinguish contravariant and covariant tensors from each other. Most objects are either naturally covariant or contravariant, and if metric variations are considered, which is pretty often, it may happen that the covariant or contravariant forms of objects do not get varied or get varied differently. ...


1

The second one. The definition is the double covariant derivative. The first covariant derivative acts on a scalar field, and so it is just a derivative. The second is a covariant derivative on the vector field $\partial_i \phi$, which can be shown to equal the second expression you gave.


1

It's not quite so easy to prove the general case $\phi \in (0, \alpha)$ (the singularity can't be removed in general), but fortunately, for the case $\alpha = 2\pi$, it is fairly easy. First let's write our metric in a more appropriate form. This spacetime can be described with two coordinate patches, $(U_1, f_1)$ and $(U_2, f_2)$, with \begin{eqnarray} f(...


1

The question is too broad to get an answer, which is expected and desired. If the question is about how to grasp the understanding the notations and mathematics behind the general relativity - the good and rather easy introduction is Nakahara's book - "Geometry, Topology and Physics" https://www.amazon.com/Geometry-Topology-Physics-Graduate-Student/...


1

It's because the $\{dx^{'b}\}_{b=1}^n$ are linearly independent differential forms; in fact they form a basis for the cotangent space of the manifold, moreover they are the dual basis to $\left\{\frac{\partial}{\partial x^{'b}}\right\}_{n=1}^n$. So, if you have a linear combination of them which vanishes \begin{align} \sum_{b=1}^nA_b\, dx^{'b} &= 0 \end{...


1

Maybe it becomes clear for the example of polar coordinates in the plane: $$\mathbf e_r = \left(\frac{\partial x}{\partial r} , \frac{\partial y}{\partial r}\right) = (\cos(\theta) , \sin(\theta))$$ $$\mathbf e_{\theta} = \left(\frac{\partial x}{\partial \theta} , \frac{\partial y}{\partial \theta}\right) = (-r\sin(\theta) , r\cos(\theta))$$ The derivatives ...


1

I don't think the setting of the configuration space is of any consequence. In that sense I will say there is no geometrical reason. That said, key to Hamilton's stationary action is a property that lends itself very well to visual/geometrical demonstration. I will demonstrate that Hamilton's stationary action capitalizes on the following property of ...


1

The OP's calculation seems alright. If we proceed along that line the required expression can be achieved rather easily. First, I note that, $$D_\alpha N_\beta=\partial_\alpha N_\beta-{}^{(3)}\Gamma^\gamma_{\alpha\beta}N_\gamma\neq \partial_\alpha N_\beta-{}^{(4)}\Gamma^\gamma_{\alpha\beta}N_\gamma=\partial_\alpha N_\beta-\Gamma^\gamma_{\alpha\beta}N_\gamma,$...


1

First of all, note that the covariant derivative of a vector is a tensor. This means that if we have two vectors $A_\mu$ and $A^\nu$, then their covariant derivatives $\nabla_\rho A_\mu$ and $\nabla_\rho A^\nu$ are tensors. These two tensors satisfy the transformation law: $$(\nabla_\rho A_\mu)=g_{\mu\nu}(\nabla_\rho A^\nu)$$ The same transformation can be ...


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