63

That's because in thermodynamics we sometimes use the same letter to represent different functions. For example, one can write the volume of a system as $V=f_1(P,T)$ (a function of the pressure and the temperature) or as $V=f_2(P,S)$ (a function of the pressure and the entropy). The functions $f_1$ and $f_2$ are distinct in the mathematical sense, since they ...


10

Your presumption that partial derivatives can be uniquely specified by a single coordinate is false. Consider the function $f : \mathbb{R}^2 \to \mathbb{R}$, $$ f : (x, y) \mapsto x + y \qquad \left(\frac{\partial f}{\partial x}\right)_y = 1 \, . $$ But I can replace the second coordinate by $z = x+y$. Now we have $$ f : (x, z) \mapsto z \qquad \left(\frac{\...


5

If a point is at a unit distance from the origin, and it makes an angle $p$ with the x-axis, then $\cos p$ is defined as the x co-ordinate of that point. $\sin p$ is defined as the y co-ordinate of that point. If a point is a distance $r$ from the origin and makes an angle $p$ wih the x-axis, then its x and y co-ordinates are $r \cos p$ and $r\sin p$ ...


4

While the OP's answer describes a perfectly valid interpretation, I would like to suggest another possible way of looking at it. In this approach, when one writes things like $\left(\frac{\partial V}{\partial P}\right)_T$ and $\left(\frac{\partial V}{\partial P}\right)_S$ in thermodynamics, the symbol $V$ stands for the same function in both cases; however, ...


4

Active transformation: the vectors and other geometric quantities change. Passive transformation: the vectors (with the exception of basis vectors) and other geometric quantities do not change, but the basis does (e.g. a coordinate basis), so the components of a vector change even though the vector itself does not. Local Lorentz transformation: the ...


3

The short answer is no, the Hawking radiation is not a coordinate dependent phenomenon. These (Kruskal-Szekeres and Tortoise) coordinates simplify the problem significantly when you are dealing with QFT in curved spacetime background (here Schwarzschild black hole background). Physically, any static observer at infinity would detect the same Hawking ...


3

Perhaps it is helpful to take a step back and review the definitions: In this answer, we will assume that the Lagrangian $L=T-U$ is the difference between kinetic and (possibly velocity-dependent) potential energy. Consider the (Lagrangian) energy function $$ h(q,\dot{q},t)~=~\left(\sum_{j=1}^n\dot{q}^j\frac{\partial }{\partial \dot{q}^j}-1 \right)L(q,\dot{...


3

This is a common misconception. A dipole is NOT a set of two opposite charges, so you do not have a dipole in your configuration. To be precise, you have a "physical dipole", two equal (in absolute value) and opposite ( sign) charges near each other. This is a usual configuration; but, as you have a finite numebr of point charges, you find the ...


3

Here are some concrete examples which may shed some additional light on the issue. Let's consider the manifold $\mathcal M = \mathbb R \times \mathrm S^1$, i.e. the cylinder. Points $p\in \mathcal M$ can be labeled by a triple $(z,a,b)$, where $z\in \mathbb R$ and $a^2+b^2=1$. It's critical to note that $(z,a,b)$ are not the coordinates of $p$ in some ...


3

TL;DR - I suspect your confusion lies in the Physics 101 example that e.g. the ordered pair ("temperature","pressure") does not define a vector because when we change our coordinates, temperature and pressure don't transform. However, if we are working in cartesian coordinates, the object (temperature)$\hat x$ + (pressure)$\hat y$ is a ...


3

The correct expansions are $$ \nabla_X \nabla_Y Z^{\ell} \enspace = X^i \, \nabla_i \Big(Y^j \, \nabla_j Z^{\ell}\Big) = X^i \, \nabla_i \Big(Y^j ( \partial_j Z^{\ell}+\Gamma_{jm}^\ell Z^m)\Big)=\\=X^i\partial_i\Big(Y^j( \partial_j Z^{\ell}+\Gamma_{jm}^\ell Z^m)\Big)+X^iY^j\,\Gamma_{in}^\ell( \partial_j Z^n+\Gamma_{jm}^n Z^m)=\\=X^i\partial_iY^j( \...


2

The spacetime interval between two events is always the same in any two inertial reference frames- that is, as long as the frames themselves are not accelerated. It doesn't matter if these events are points on the worldline of an accelerating particle, a particle of constant velocity, or just two events picked from spacetime at random.


2

Let’s ignore the kinematic time dilation and estimate the gravitational time dilation. The dilation factor, approximately $1+\frac{GM}{rc^2}$ at large $r$, varies between $1+\frac{1}{6.5}\approx 1.15$ to $1+\frac{1}{15}\approx 1.07$. That’s about a 4% variation around the average, and it seems consistent with the slight visible wobble around the plot’s ...


2

To provide a secondary view which may be more accessible, I have elected to write a second answer. Let $\mathbf V$ be a vector. Given some choice of basis $\{\hat e_1,\hat e_2\}$, we can expand $\mathbf V$ in component form as $$\mathbf V = \sum_i V^i \hat e_i = V^1 \hat e_1 + V^2 \hat e_2$$ Now we consider a rotation matrix $$R = \pmatrix{\cos(\theta) &...


2

UPDATE: Since the other answers are providing more of the answer, let me reveal my interactive visualization. As I mentioned in the comment to the OP, the "physical dipole" (composed of two equal-magnitude oppositely-signed point charges) is not the same as the "pure [or point] dipole". For the same dipole moment $\vec p$, they disagree, ...


2

The right expression should be $$\int d\mathbf{v}\rightarrow 4\pi \int_0^\infty v^2dv$$ This could be conclude from $$\int d\mathbf{v}\rightarrow \int_0^{2\pi}d\phi \int_0^\pi d\theta \sin\theta \int_0^\infty v^2dv=4\pi \int_0^\infty v^2dv$$ Where in the last step we have to perform angular integral.


2

You're in the right ball-park. One thing that is not explained well in my opinion is the distinction between covariant and contravariant components, and as you point out, how this relies upon the metric. If we fix a basis $e=(e_\alpha)$ in a vector space $V$, then for every vector $v \in V$, we have its components, $v^\alpha$. So far so standard. What's not ...


2

It is true, as the answer of benrg says, that the sphere in question does not necessarily have to be embedded in a space of higher dimension in which its centre would be a point in that higher-dimensional space. But one could in principle imagine that it was so embedded, so there does exist a coordinate system in which the sphere's centre in that higher-...


2

For mechanical system you can use this: Euler Lagrange \begin{align*} &\mathcal{L} =T-U\\ &\frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{\boldsymbol{q}}}\right)- \frac{\partial \mathcal{L}}{\partial \boldsymbol{q}} =\left[\frac{\partial \boldsymbol{R}}{\partial \boldsymbol{q}}\right]^T\boldsymbol{f}_a\qquad\qquad (1) \end{...


2

Consider the example that Altland suggests: let $\theta,\phi,\psi$ be the Euler angles specifying a rotation. Then $x,y,z \to x(\theta,\phi,\psi),y(\theta,\phi,\psi), z(\theta,\phi,\psi)$ and the derivatives $\partial x/\partial \theta$ etc. give the infinitesimal form of the rotation.


2

Three things I learned from studying these Relativity paradoxes are: The events always happen. If one observer sees "pole hit the back of the barn", so does the other. Most of the "paradox" stems from our intuition of absolute time and / or simultaneity. Two things that are simultaneous in one reference frame need not be simultaneous in ...


2

So to clarify, you're just trying to plot the vector using it’s x-y components, or or are you asking how to get the $X$ or $Y$ components of a vector? If it’s # 1 it seems like you already did it since there’s no $X$ component and the $Y$ component is in the negative direction. If it’s # 2 then you're probably looking for $x=R\cos\theta$ and $y = R\sin\...


2

Recall the triangle law of vectors. To find the x and y components, construct a triangle with the vector $F_e$ as the hypotenuse. This is where trigonometry comes in- since $F_{ex}$ is the base of the triangle and $F_e$ is the hypotenuse, you can now figure out the trigonometric relation between $\theta$, $F_e$ and $F_{ex}$.


1

You can trivially obtain the solutions to the shifted problem by replacing $x$ with $x-x_0$ in the original solutions. From there, it's a matter of elementary trigonometry (i.e. the angle addition formula) to express the result in terms of $\sin([\ldots]x)$ and $\cos([\ldots]x)$.


1

I see two problems in your question. The first is that the force laws in both of your examples make certain assumptions: the quantities ($\mathbf{r}$ in the first example and $x$ in the second example) are not absolutes, they are distances measured from some origin which represents something physical. In the case of the gravitational force, the centre ...


1

As stated in my comment, you can do this without making reference to normal coordinates. E.g. see Confusion about Lie derivative on metric. Here's a quick proof in local coordinates which I haven't seen answered here though. Take the standard form of Killing's equation, $$ \begin{align} \tag{1} \mathcal{L}_{\xi} g_{\mu \nu} &= 0 \\ &= \xi^{c} \...


1

For clarification, write $\vec{u} = u_x \hat{x} + u_y \hat{y} + u_z\hat{z}$ in components, and $d\vec{S} = \hat{r} R^2 \sin\theta d\theta d\phi$ $$ \vec{I} \equiv \oint \oint \vec{\nabla} \vec{u} \cdot d\vec{S} $$ \begin{align} I_\alpha = & \oint \oint \vec{\nabla} u_\alpha(\vec{r})\cdot d\vec{S}\\ =& R^2 \int_0^\pi \sin\theta d\theta \int_0^{...


1

That's a direct consequence of the transformation in spherical coordinates. In fact given a general vector in $\mathbb{R}^3$ with components $(x,y,z)$, we can go to spherical coordinates $(r,\theta,\phi)$ and the infinitesimal volume becomes $$d^3x \equiv dxdydz = r^2\sin\theta\, dr\,d\theta \,d\phi$$ where $r^2\sin\theta$ is the Jacobian of the ...


1

The potential near the two charges would be found by adding two terms. The formula you show is a good approximation if, R, is much larger than, a.


1

I was also blocking for a time on this point in the book, I came accross your question which was helpfull for me. In fact, I think the intuitive way is using the derivative of the product of functions: In 3.16, let assume $f$ to be the first term $ \frac{\partial x^{'a}}{\partial x^{d}} $ and $g'$ to be the second $ \frac{{\partial}^2 x^{d}}{\partial x^{'c}\...


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