New answers tagged differentiation
-2
votes
Is every $dm$ piece unequal when using integration of a non-uniformly dense object?
Mass is seen as a scalar. In differential geometry, it's volume or density forms that are integrated. So even mass would be seen to have a directional character given by the density form. Nevertheless,...
0
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Is every $dm$ piece unequal when using integration of a non-uniformly dense object?
The correct answer is "it depends".
In calculus the area under a curve $y=f(x)$ is usually approximated using sections $\delta x$ that are the same size, and the limit is taken as the size ...
0
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Is every $dm$ piece unequal when using integration of a non-uniformly dense object?
There are several possible answers to this question depending on the framework you use.
The way you are thinking about "dm" suggests you are thinking about Riemann integral. In this integral,...
0
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Is every $dm$ piece unequal when using integration of a non-uniformly dense object?
the answer to your question: not necessarily. depends upon your choice of variables you wish to evaluate dm with
$$M= \int dm$$
$$M = \int \lambda dx$$
Let:
$$x= t^2$$
$$dm = [\lambda 2t dt]$$
...
2
votes
Is every $dm$ piece unequal when using integration of a non-uniformly dense object?
The $ dm = \rho(x) dx $ expression tells us that $ m $ and $ x $ do not have generally equal binnings. When you integrate according to $ dm $, it is uniform w.r.t. mass, but not w.r.t. length, and ...
5
votes
Is every $dm$ piece unequal when using integration of a non-uniformly dense object?
$\newcommand{\d}[1]{\mathrm{d}{#1}}$In general you can use a change of variables, to write
$$\d{m} = \frac{\d{m}}{\d{x}}\d{x}=:\rho(x) \d{x},$$
where $\rho(x)$, defined as $\frac{\d{m}}{\d{x}}$, is ...
1
vote
Accepted
Is every $dm$ piece unequal when using integration of a non-uniformly dense object?
I don't quite get the question, but yes, every piece of $dm$ has unequal mass, but equal length. And when you write $ dm = λdx $, the $λ$ you are writing is called the local linear mass density.
In ...
0
votes
Why is instantaneous velocity tangent to trajectory?
In a trajectory where $x = x(t)$ and $y = y(t)$, it is right that $v_x = x'(t)$ and $v_y = y'(t)$ are respectively tangent to the curves $x\times t$ and $y\times t$.
In a small neighborhood of a point ...
1
vote
Accepted
Why is instantaneous velocity tangent to trajectory?
The answer is in the definition of instantaneous velocity. While moving along a path a small change in position ${\rm d}\vec{r}$ over a small time frame ${\rm d}t$ the instantaneous velocity is ...
3
votes
Accepted
Commutator between covariant derivative and a field
It may help to evaluate the commutator on a function, i.e.
$$[\partial_\mu,\Phi]f=\partial_\mu(\Phi f)-\Phi\partial_\mu f=(\partial_\mu \Phi)f+\Phi \partial_\mu f-\Phi \partial_\mu f.$$
The last step ...
1
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Finding the Euler-Lagrange equation for a scalar field
The Lagrange density is ($c=1$)
\begin{equation}
\mathscr L\left(\phi\left(\boldsymbol x,t\right),\boldsymbol\nabla\phi(\boldsymbol x,t),\dot{\phi}(\boldsymbol x,t)\right)=-\frac12\left(\dot{\phi}^2-\...
4
votes
Accepted
Proof that the Euler-Lagrange equations hold in any set of coordinates if they hold in one
Yes, the generalized coordinates $(q^1,\ldots, q^N)$ are assumed to be independent, i.e. no constraints, and the cotangent vectors $(\mathrm{d}q^1_p,\ldots,\mathrm{d}q^N_p)$ at each point $p$ are ...
1
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Determine the meaning of a gradient of a graph
It really depends on the specific situation. The general meaning of gradient is the spatial rate of change in a quantity. For example in a temperature field $T(x,y)$, the gradient $\vec {\nabla T}$ ...
0
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Higher dimension derivatives
That will depend on how you contract the indices, which means, on how you are applying it, but neither of what you wrote makes much sense, some possibilities would be ($\mu_5$ is gonna be an index of ...
2
votes
Accepted
Taking the second time derivative of a scalar field
But I'm having a hard time simplifying the second derivative:...
$$
\frac{d^2\phi}{dt^2}=
\frac{\partial^2\phi}{\partial^2 t}
+
\vec{a}\cdot \vec{\nabla}\phi
+
\vec{v}\cdot\left( \frac{d}{dt}\vec{\...
1
vote
Question about Wald's example of a "derivative operator"
An explicit example may help. Consider the manifold $\mathbb R^2$ and a vector field $V$ which, in a Cartesian coordinate system $(x,y)$, takes the form
$$V = -y\frac{\partial}{\partial x} + x\frac{\...
1
vote
Question about Wald's example of a "derivative operator"
He doesn't get to just define the object in some coordinate system and call it a tensor. And he can't just declare that its components, by definition, transform however they need to, so that the ...
0
votes
Question about Wald's example of a "derivative operator"
Wald's definition of a tensor is not based on their coordinate transformation laws. Instead, it is akin to how mathematicians define a tensor. Earlier in Chapter 1 (I think), he defines a tensor to ...
1
vote
Why can we change $dt$ with $(dt/dp)_s dp$?
This is not how I would have solved this problem. I would first have written: $$ds=\left(\frac{\partial s}{\partial T}\right)_PdT+\left(\frac{\partial s}{\partial P}\right)_TdP=0$$In addition, $$\...
2
votes
Accepted
Why can we change $dt$ with $(dt/dp)_s dp$?
The first step is just the following model:
$$f(b)-f(a)=\int_a^b f'(x)\,dx=\int_a^b\frac{df}{dx}(x)\,dx$$
After that it's the triple product rule:
$$\left(\frac{\partial x}{\partial y}\right)\left(\...
2
votes
Accepted
Why does the material derivative and transport theorem look different?
The Reynolds transport theorem reads
$$\frac{\mathrm{d}}{\mathrm{d} t} \int_{\Omega} \phi\, \mathrm{d}V= \int_{\Omega} \frac{\partial \phi}{\partial t} \, \mathrm{d}V + \int_{\partial \Omega} \phi \...
0
votes
Standing Wave Equation: Why does assuming a small slope $du/dx$ not make $d^2u/dx^2$ negligible as well?
The two entities, du/dx and d/dx(du/dx), don't have the same units.
That means they cannot be compared. If one is 'small' in some sense,
the other need not be. That second derivative, with the ...
0
votes
Standing Wave Equation: Why does assuming a small slope $du/dx$ not make $d^2u/dx^2$ negligible as well?
From my own experience in the past, there is a common beginner's mistake in interpreting notation. It makes a big difference if $du/dx(x_0)=0$ for a single location $x_0$ or $du/dx(x)=0$ for all $x$ ...
0
votes
Standing Wave Equation: Why does assuming a small slope $du/dx$ not make $d^2u/dx^2$ negligible as well?
In the video, he is not assuming $\frac{du}{dx}$ is small. It has a definite nonzero value. He is taking the limit as $\Delta x$ gets small (goes to zero, or becomes an infinitesimal $dx$).
That ...
3
votes
Standing Wave Equation: Why does assuming a small slope $du/dx$ not make $d^2u/dx^2$ negligible as well?
Let's take an example signal: $y=\cos(x)$ about $x=0$.
$\frac{dy}{dx}=-\sin(0)=0$;
$\frac{d^2y}{dx^2}=-\cos(0)=-1$.
It is important to note that one is not simply the square of the other.
3
votes
Accepted
Simple difference between module of velocity and time derivative of module of position
The first describes the rate at which the distance between the object and the (often arbitrary) origin is changing, whereas the second is the actual speed of the object (the speed being the magnitude ...
3
votes
Accepted
Why are formulas in Physics represented in form of differentiation?
In physics, observables are usually modeled as numerical quantities, and laws are relations between these quantites.
A mathematical relation between numerical quantities, typically showing how one ...
0
votes
Why are formulas in Physics represented in form of differentiation?
Mostly formulas are represented in differential form.
As you mentioned yes, most of the formulas is written in differential form. It comes from one of the most important branch of Mathematics called ...
1
vote
Accepted
Variation of the Lagrangian
You can use the chain rule to handle any implicit dependence on $x$,
\begin{equation}
\partial_\mu \mathcal{L}\left(\phi(x), \partial_\nu \phi(x)\right) = \frac{\partial \mathcal{L}}{\partial \phi} \...
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