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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/...


22

This can only be answered by pointing out the difference between special and general relativity. There is the historically motivated definition, which is still in widespread use in popular and semi-popular accounts, according to which special relativity only deals with inertial frames and coordinates (1a), while general relativity deals with accelerated ...


19

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 ...


14

Why do we have to revise the definitions of momentum and force in special relativity? I am going to downplay the theoretical reasons, because they aren’t what’s important. What’s important is that the Newtonian definition of momentum is useless because experimentally in relativistic particle collisions it isn’t conserved. When you change the definition to $\...


8

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

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 ...


6

Let us propose a 3-vector quantity of the form $$ {\bf p} = m f(v) {\bf v} $$ where $\bf v$ is the velocity of a particle, $m$ is an invariant scalar (a property of the particle) and $f(v)$ is some function of speed, to be discovered. Let us then propose that this quantity $\bf p$ is conserved in a two-body collision: $$ {\bf p}_{A,i} + {\bf p}_{B,i} = {\bf ...


5

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 ... let's say: "the infinitesimal cube with a vertex at the origin, and lying within the ...


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

I liked Batiatus' answer where we distinguish between flat spacetimes and curved ones, I feel like that gives you a good answer of questions (1) and (2). To answer question (3) I would question this premise that special relativity doesn't “do” acceleration, because I believe that that's the only thing it does. So a remarkable thing to my mind is that the ...


3

This is a problem I've struggled with before. The issue is that most textbooks are not clear enough in their definitions of the fixed and rotating reference frames. I've tried to be as comprehensive as possible in my answer, I hope it helps! To describe the motion of a rigid body through space we must make use of two separate reference frames: Firstly, a ...


2

The word frame is intended to refer to physical matter. For example we talk of the Earth frame, and if you are travelling by car you will naturally use the car as your reference frame when considering objects in the car. A reference frame is the matter relative to which coordinates are defined. I have defined these in The Large and the Small: A reference ...


2

First things first: Software questions are off-topic at this site. So I'll focus my answer from the perspective of the title question, with a hint toward poliastro. Is ECI a pseudo-inertial frame? This question is being asked from a Newtonian perspective. (Poliastro is fully Newtonian.) The answer is of course it is a pseudo-inertial frame. The Earth ...


2

For starters, your question can be reduced to "why do we need to revise momentum" only, because force can be defined in terms of momentum via $$\mathbf{F} = \frac{d\mathbf{p}}{dt}$$ The reason we are interested in force as a measure of the changes in momentum is because, seen from this perspective, it is interaction that is the more basic concept, ...


2

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), ...


2

True vectors are geometric invariants. They have different representations in different co-ordinate systems but the actual geometric object does not change. In classical physics, the force vector for the most part is a 'true' vector in the sense that you need 0-coordinates to describe it. I could say that a force of 10N is acting on your centre of mass and I ...


2

From the linked article it shows exactly what the vector $\hat{\theta}$ means There it is, a unit vector pointing around the "hoop" direction, tangent to the surface and perpendicular to the axial direction $\hat{z}$.


2

Is this a valid frame? I mean valid in the sense that in Newtonian mechanics acceleration frames are not valid because they experience ghost forces. You do experience a "ghost" force in Rindler coordinates; there will appear to be a force opposite the direction of acceleration (much like when a car accelerates, objects in the car appear to have an ...


2

The note that this is a single scalar equation is probably intended to clarify that the $0$ on the right hand side is a scalar, and not, for example, the zero vector $\vec 0$. So the function $g$ on the left hand side is a scalar function of $\vec r$, $\dot {\vec r}$ and $t$. A simple example would be a particle that is constrained to move in a circle ...


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

Think of a vector $\vec{g}_r$ resting on a rotating frame, which without loss of generality has its axis of rotation out of the plane as seen below: The rotation angle is $\theta$ at any instant and the vector $\vec{g}_r$ is expressed in terms of the local coordinate vectors $\hat{i}_r$ and $\hat{j}_r$ as $$ \vec{g}_r = x_r \hat{i}_r + y_r \hat{j}_r \tag{1}$...


1

I'll answer myself as this could be helpful for someone else, but the credit belongs to @Frobenius who has pointed me to his useful answer in the following link: https://physics.stackexchange.com/a/252265/190100 It is now clear to me that it was notation that got me confused. All vectors here ($\vec {u}, \frac {d\vec {u}}{dt}$, unit rotating vectors) are ...


1

We did take the sign problem into account by using $|g|$ when multiplying with a tensor density. If this sounds like cheating: there is something called orientation and if the manifold we're working is orientable (which is usually what we encounter Schwarzschild, Kerr, RN) then while defining new coordinates $x^{\mu'}=x^{\mu'}(x^{\mu})$ we go for orientation ...


1

You can measure torque about any point you like, so you can pick an arbitrary origin and calculate torque about that origin. The torque of a single force depends on where you measure it from. So if there is not an obvious axis, like a hinge or a pivot, how do you decide where to measure torque from ? Well, if the net force on an object is zero then the ...


1

We can generalize the concept of integration thusly: given a space $S$ with a measure $m$ and given a function $S \rightarrow \mathbb R$, for each natural number $n$, separate $S$ into disjoint subsets, none of which has measure greater than $2^{-n}$. For each subset, take the supremum of $f$ over that subset, multiply it by the measure of the subset, and ...


1

Some important points I have gathered from discussions, that might help someone to completely dispel my doubts. Thanks to all! One does not set out to construct a volume element, rather, it occurs as a result of the grid system we are using, through a limiting process. Naturally, the shape of the volume element will depend on the system. A question on this: ...


1

Yes, one can rotate coordinate system in linearized gravity (and perform Lorentz boosts and finite translations). Just these transformations would not be infinitesimal diffeomorphisms. Remember, that linearized gravity is Poincare invariant field theory in addition to the gauge symmetries of infinitesimal diffeomorphisms.


1

Kronecker deltas are simply constants, being either $1$ or $0$ depending on the values of $i$ and $j$. An expression like $\{q_{i} + \delta q_{i}, p_{j} + \delta p_{j}\} = \delta_{ij} - \epsilon\{\delta_{ij}, g\}$ should really be thought of as $9$ different expressions for each possible value of $i$ and $j$, and in each of those expressions the Kronecker ...


1

That's the thing: the displacement is so small that "it has no space to vary". We're drawing it big, but we are considering so small dispalcements that curves can be approximated by circle arcs. The real curve might start diverging from that point aftewrwards, but in such small interval, teh tangent point can be extended to a tiny environment ...


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{...


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