Tag Info

New answers tagged

4

Given your question, it seems likely that your misunderstanding comes from a limited sense of vectors, fields, and partial derivatives. So there's a lot of education that we have to cover in a very short time. Multivariate functions When we transition from a function $f(x)$ to a field, which is a function of many variables $f(x, y, z)$, we suddenly have ...


1

The work-energy theorem leads us to the following result; \begin{equation} \oint \vec F\cdot d\vec s=0 \end{equation} \begin{equation} \oint \vec F\cdot d\vec s=\underbrace{\int \int }_{\text{surface}}(\nabla \times \vec F)\cdot d\vec n \end{equation} Using the rules of vector calculus there must exist some scalar function such that; \begin{equation} \vec ...


1

Let $F$ be a force field. Assuming that the force field is a conservative vector field, then it follows that the line integral of the force field is zero $$\oint_{O} F \cdot dr = 0$$ The del operator $\nabla$ is defined in 3 dimensions as $$\nabla = \langle\frac{\partial}{\partial{x}}, \frac{\partial}{\partial{y}}, \frac{\partial}{\partial{z}}\rangle$$ ...


6

On spherical coordinates, the gradient of a general function $V$ is: $$ \nabla V = \frac{\partial V}{\partial r}\mathbf e_r + \frac{1}{r}\frac{\partial V}{\partial\theta}\mathbf e_\theta + \frac{1}{r\sin\theta}\frac{\partial V}{\partial\phi}\mathbf e_\phi $$ If $V(r, \theta, \phi)$ only depends on $r$, that is $V = V(r)$, which is exactly the case of the ...


3

I am not sure what is the path $C$ you are integrating over? In your definition you evaluate $U(C)$ which in the present case of force is independent on the explicit path you choose but still depends on initial and final point, i.e. $U(p_1,p_2)$. In your final result it seems you are actually 'walking' three times the path $p_1=(-\infty,y,z)$ to ...


2

"The answer is that since we are proud physicists and not nitpicking mathematicians we will just wing it when the need arises" This quote is taken from A. Zee's Quantum Field Theory in a Nutshell, and it summarizes the attitude of physicists to mathematics. (At least in an undergraduate level) Since we are physicists, most of our mathematics isn't rigorous. ...


1

Divergence at a point measures the flux flowing out from an infinitesimal volume around that point. Suppose you have a volume $v$ in vector field $\mathbf{h}$. Divide it into two parts: $$v = v_1 +v _2$$. Now the flux out of volume $v$ is given by : $\int \mathbf{h}\cdot da_v = \int \mathbf{h} \cdot da_{v_1} + \int \mathbf{h}\cdot da_{v_2}$. Start dividing ...


1

Think about it one more time. If $\vec{F}$ has continuous partial derivatives, then $$\vec\nabla\cdot\vec{F}=\sum_i \frac{\partial F_i}{\partial x_i}$$ is also continuous. If a function is continuous, it's approximately constant on sufficiently small volumes: that's pretty much the definition of continuity! So your original understanding was just fine. ...


0

In your last equation $U$ is a function or $n$ variables. Which of these does your $x$, $y$, and $z$ in that equation represent? To find the contribution to the potential energy due to the action of forces on a particular particle, one has to take the partial of the potential with respect to the variables representing the position of that particle. I ...


2

It's not taking partial derivatives with respect to an observed particle's position, but rather the space of all possible positions of that particle. Think of the potential energy as being defined prior to the particle having an actual path. Really, at heart, these things are defined on a phase space not on ordinary physical space.


0

The purpose of all this is to calculate and derive potential equation of the gravitation field. Let us assume that we have symmetrical sphere object that "generates" gravitational field. We want to know if the field depends on the objects "homogeneity" (correct me if this is not quite the right term). So from Gauss theorem (it has different names, but they ...


6

It depends on how the quantity in question transforms. Almost always, densities in the form of "stuff per unit volume" and generally the "stuff" (like a charge) is a scalar (a number of things - number of elementary charges), but the volume it is contained in is observer dependent, owing to the Lorentz contraction. Therefore the density is ...


1

Your sign is wrong when computing $$\frac{\partial{(B^2)}}{\partial{(\partial_{y} A_x)}}.$$ The only term in $B^2$ that contains $\partial_{y} A_x $ is $(B_z)^2 = (\partial_{x}A_y - \partial_{y} A_x)^2 ,$ and clearly by the chain rule, $$ \frac{\partial{(B_z)^2}}{\partial(\partial_{y}A_x)} = -2B_z$$ which disagrees with what you have by a sign. Fixing ...


1

Comments to the question (v3): The vielbein $e^A{}_{\mu}$ in the Cartan formalism is an intertwiner $$\tag{1} g_{\mu\nu}~=~e^A{}_{\mu} ~\eta_{AB} ~e^B{}_{\nu} $$ between the curved (pseudo) Riemannian metric $g_{\mu\nu}$ and the corresponding flat metric $\eta_{AB}$. Here Greek indices $\mu,\nu,\lambda, \ldots,$ are so-called curved indices, while ...


0

Consider the metric in comoving conformal coordinates: $ds^2 = a^2(\eta)[-d\eta^2 + \delta_{ij}dx^i dx^j]$. It is easy to show that $\partial_{\eta}$ solves $\mathcal{L}_{\partial_{\eta}}g_{\mu\nu} = \frac{2}{a(\eta)}g_{\mu\nu}$; furthermore $\partial_{\eta}$ is clearly time-like.


1

A normal Killing vector is obtanied by solving the usual Killing equation $\nabla_{(\mu} \xi_{\nu)} = 0$ A conformal Killing vector is obtanied by solving a slightly different equation, the conformal Killing equation: $\nabla_\mu \xi_\nu + \nabla_\nu \xi_\mu = 2 \alpha g_{\mu \nu}$ where $\alpha$ is obtanied by taking the trace of the equation above. ...


4

Nice question. One way to think about it is that given a metric $g$, the statement $\mathcal L_Xg = 0$ says something about the metric, whereas $\nabla_Xg = 0$ says something about the connection. Now what $\mathcal L_Xg = 0$ says, is that the flow of $X$, where defined, is an isometry for the metric, while $\nabla_Xg = 0$ says that $\nabla$ transports a ...



Top 50 recent answers are included