New answers tagged

2

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 $r\sin p$ ...


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 $F_{ex}$.


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\...


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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}\...


1

First lets clear up some confusion. You say: To be conformal means that the physic is unchanged This is not what "conformal" means. Conformal means that the angles are unchanged. In the context of GR, this in particular means that the causal structure remains unchanged. The phyisics in the meantime can be quite different. In particular, the $uv$-...


1

There are two different ways to do differential geometry: the 'extrinsic' view, and the 'intrinsic' view. The extrinsic view is what you just described: you set up an embedding of your manifold inside a copy of $\mathbb R^n$, and you refer your manifold's geometry to that ambient space. The intrinsic view is what you're asking about: studying the geometry ...


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.


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The answer is that the pole can either fit or not fit depending on who is observing it. The observer standing to one side of the barn will consider that it fits if the trailing end of the pole enters the barn before the leading end leaves it. But that assumes he can view both ends of the barn simultaneously. Crucially, two simultaneous events for that ...


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 ...


0

So is this loss of simultaneity real or just because light takes more time to travel from left side? Yes, it is real, and yes, it is because light takes more time to travel from the left side. To understand why that is the case, we need to dive into what we actually mean when we say that two events are simultaneous. A common definition for two events to be &...


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 $\cos([\ldots]x)$.


0

In Spherical coordinates, the shift in $\pi/2$ will correspond to a charge located at the plane $z=0$. Since the polar angle $\theta$ is measured from "north" to "south" and goes from $0$ to $\pi$, a shift of $\pi/2$ means it exactly in the middle of the 2 hemispheres. you can also see that "mathematically": $$\delta(\theta-\...


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-...


0

The sphere has no poles really. It's fully symmetric with no distinguished points. The author of these notes has chosen to call the arbitrary origin point (normally taken to be our location) the "north pole", possibly to clarify that it's a point on the surface of the sphere and not its center. The origin can't be the center because there is no ...


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{...


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{...


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 ...


2

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 ...


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

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} \...


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This is actually more general than just diffeomorphisms, but it is a but more intuitive there since we are so used to coordinate transformations. For any quantity, one can simply decide to measure it in different units, different scale, or shift the scale. This doesn't change physical properties in any way, and in this sense corresponds to a gauge ...


0

Perhaps this defeats the point of the exercise, but it seems to me that if you want to know what the snowball does then why not just calculate the motion of the snowball, which has nothing at all to do with the motion of the roundabout. It just flies in a parabola, whose horizontal part is a straight line relative to the ground. If the initial velocity is ...


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 ...


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 ...


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 ...


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.


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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, ...


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 ...


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 ...


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) &...


0

If the relationship holds for perpendicular axes, it cannot be true for your sketch (where one or the other angle is smaller).


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 ...


1

Firstly, when we speak of contravariance or covariance we are talking about how various quantities transform under passive changes of coordinates and basis. Tensorial quantities are defined at the manifold level and don't depend on coordinates, so by definition they don't change at all. Consider a vector $\mathbf V$. If we choose a coordinate system $x\...


-1

What you have is the one form $d\phi$ which in coordinate $(x,y)$ is given by $$\frac{\partial\phi}{\partial x}dx+ \frac{\partial\phi}{\partial y}dy$$ $dx$ and $dy$ are one forms so they transform covariantly. Finally $\frac{\partial\phi}{\partial x}$ and $\frac{\partial\phi}{\partial y}$ are scalar function so they do not transform. One last word about this ...


0

Events do not "occur" in frames. They are measured or described relative to frames, but the events themselves occur in spacetime. For example the emission of a particle and absorption of that same particle later are events. Different inertial frames would assign different coordinates to these events, but the spacetime interval between them is the ...


0

It is not a paradox. Consider the case of two poles 100m tall that stand on Earth a thousand km apart. You stand next to one and I stand next to the other. Owing to the curvature of the earth, the vertical direction in your frame of reference is not parallel to mine- the two are angled somewhat. If you measure the height of my pole in your frame of reference,...


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 ...


0

Try to avoid the awkward variation due to the existence of a term of order $dt$ in the Lagrangian, I reformulate the problem using dissipation function and gneralized force. $$\tag{1} G(r, \dot{\phi}) = \frac{1}{2} \frac{dm}{dt} \left\{(r+R_E) \dot\phi - u\right\}^2 $$ where $ \dot m = \frac{dm}{dt}$ is the mass changed between $t$ and $t+dt$, negative ...


1

I think to answer your question, the point of the $V(r) = kr^2$ is not to distract, but because in one of the integrals you will need to make a change of variables $R = ar$ and you will get an $a^{-2}$ factor from the $r^2$. For the broader picture, this technique is often used to show the virial theorem from the variational method in quantum mechanics (and ...


0

Try writing out the S.E. \begin{equation} \frac{-\hbar^2}{2m} \Delta\psi+kr^2\psi=E\psi \end{equation} although cumbersome, it may be helpful to just solve the equation outright. In general it is not a good idea to assume the wavefunction is linear (this would usually make the wavefunction impossible to normalize over).


0

If we allow Hawking radiation, we can see that this time-journey cannot happen way at the boundary of a (hypothetical) developing Schwarzschild hole: the particle pairs of Hawking radiation are developed outside the event horizon; in this reference frame the particles are still just outside any possible horizon. Hawking radiation can therefore interact with ...


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, ...


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( \...


0

Instead of thinking about inertia it is clearer to think about momentum. When you throw a ball upwards on a moving train, it has two components of momentum: horizontal (due to the train's motion) and vertical (because you threw it). These two components are independent (ie motion in one direction has no impact on the motion in a perpendicular direction), the ...


0

I think actually that the light-clock experiment will work if light were confined and restricted to travel in a tube or cable stretched perpendicularly from the light to the detector. Otherwise, the light will miss the detector, for the reasons stated above. The problem might be in the way the experiment is usually described or illustrated.


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{\...


0

First you need to decide whether you want to use sum symbols or whether you want Einstein summation convention (implicit sums). The vector decomposition you have written down is actually Einstein: $$\vec V = V^i \hat e_i := \sum_i V^i \hat e_i$$ Second, you should keep upper indices upper, and lower indices lower, on both sides. Rather than writing (wrongly, ...


1

First of all, the last sign of the long expression should be a $+$, see: Since $\delta_F\phi$ accounts for the change of the field at the same point, $[\delta_F,\partial_\mu]=0$, then $$\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\delta_F(\partial_\mu\phi)=\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\partial_\mu(\delta_F\phi)=\partial_\mu\...


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