New answers tagged

-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,...
user avatar
0 votes

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 ...
user avatar
  • 1,119
0 votes

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,...
user avatar
  • 5,553
0 votes

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: $$x= t^2$$ $$dm = [\lambda 2t dt]$$ ...
user avatar
  • 4,376
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 ...
user avatar
  • 75
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 ...
user avatar
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 ...
user avatar
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 ...
user avatar
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 ...
user avatar
  • 2,667
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 ...
user avatar
  • 546
1 vote

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-\...
user avatar
  • 13.4k
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 ...
user avatar
  • 170k
1 vote

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 $\vec {\nabla T}$ ...
user avatar
  • 2,499
0 votes

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 ...
user avatar
  • 551
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{\...
user avatar
  • 7,614
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{\...
user avatar
  • 51.8k
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 ...
user avatar
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 ...
user avatar
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, $$\...
user avatar
  • 27.9k
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(\...
user avatar
  • 2,256
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 \...
user avatar
  • 36
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 ...
user avatar
  • 8,955
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 $x$ ...
user avatar
  • 6,873
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 ...
user avatar
  • 2,499
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.
user avatar
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 ...
user avatar
  • 4,062
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 ...
user avatar
  • 2,256
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 ...
user avatar
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} \...
user avatar
  • 34.3k

Top 50 recent answers are included