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2

First, I want to say that different people use different notation and I welcome any comments. I also feel as if I am about to enter a minefield. Here the answer is made up with examples of use of $d$, $\partial$ and $\delta$. I would say for $d$ that $dV \over dx$ would be the total derivative in one dimension for $V(x)$ where the potential $V$ is a ...


19

Typically: $\rm d$ denotes the total derivative (sometimes called the exact differential):$$\frac{{\rm d}}{{\rm d}t}f(x,t)=\frac{\partial f}{\partial t}+\frac{\partial f}{\partial x}\frac{{\rm d}x}{{\rm d}t}$$This is also sometimes denoted via $$\frac{Df}{Dt},\,D_tf$$ $\partial$ represents the partial derivative (derivative of $f(x,y)$ with respect to $x$ ...


1

There is no problem in treat Cristoffel symbols as tensors, because in some definitions they actually are tensors. If one defines abstractly a covariant derivative as an operator over tensors with the following properties: Linearity: $$ \nabla_c \left( \alpha A^{a_1,\dots,a_k}_{b_1,\dots,b_l} + \beta B^{a_1,\dots,a_k}_{b_1,\dots,b_l} \right)= \alpha ...


3

It doesn't. The covariant derivative is a map from $(k,l)$ tensors to $(k,l+1)$ tensors that satisfies certain basic properties. As such it cannot act on anything except tensors. The collection of components $\Gamma^a_{bc}$ does not constitute a tensor. If you got to this expression via something like $$ \nabla_d(\nabla_b A^a) = \nabla_d(\partial_b A^a) + ...


1

When I square $v_x$ and $v_y$, I get \begin{align} v(t)&=\sqrt{\left(\omega-\omega\cos\omega t\right)^2+\omega\sin^2\omega t}\\ &=\left[\omega^2+\omega^2\cos^2\omega t-2\omega^2\cos\omega t +\omega\sin^2\omega t\right]^{1/2}\\ &=\omega\sqrt{2-2\cos\omega t} \end{align} which, due to the square root term, is slightly different than what you have. ...


3

You should always work with \begin{equation} \nabla_\rho K_\sigma=\partial_\rho K_\sigma-\Gamma^\mu_{\rho\sigma} K_\mu \end{equation} even if $K_\mu$ is a constant vector. Here $\partial_t K_t=0$ but you must consider the Christoffel symbols. For example: \begin{equation} \nabla_t K_t=\partial_t K_t-\Gamma^\mu_{tt} K_\mu=-\Gamma^t_{tt} K_t \end{equation} ...


4

This is just the chain rule: $\frac{\partial L}{\partial u}=\frac{\partial L}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial L}{\partial y}\frac{\partial y}{\partial u}$. And similar thing for $\frac{\partial L}{\partial v}$. We have the derivative of $x$ and $y$ with respect to $u$ and $v$.


2

The notation $\nabla_1$ refers to the gradient with respect to the first coordinate $\mathbf{r}_1$. I think the most transparent way to do the derivation is to switch to the notation $\partial/\partial\mathbf{r}_1$, then expand the derivative using the multivariable chain rule, and then switch back to the nabla notation: $$\begin{align}\nabla_1 &\equiv ...


1

Other sources do not point out that term I have problems with. Other sources explicitly assume a constant angular velocity and thus ignore that component. The wikipedia article you cited is correct. In any case, I want to know how you evaluate that derivative. Given any vector quantity $\mathbf q$ that is the same (other than component ...


1

If you just follow your nose, then... $$\left(\frac{d}{dt}\right)_{rotating} {\bf{\Omega}} =\frac{d\bf{\Omega}}{dt}+\bf{\Omega}\times {\bf{\Omega}}$$ Do you know what the second term is equal to? Hopefully this clears up the problem you have.


-1

$$ \mathbf V = \mathbf v + \boldsymbol \Omega \times \mathbf r $$ The derivative of $\mathbf V$ in the inertial frame is indeed, $$ \mathbf A = \frac{\mathrm D \mathbf v}{\mathrm Dt} + \frac{\mathrm D \boldsymbol \Omega}{\mathrm D t} \times \mathbf r + \boldsymbol \Omega \times \frac{\mathrm D \mathbf r}{\mathrm Dt}.$$ You are right, both $\mathbf v$ and ...


3

I see now your problem and I believe that I can help. Let's begin from the velocity formula $$v_i = v_r + Ω \times r .\tag{1}$$ Let's take the derivative of $v_i$ IN THE INERTIAL frame, $$a_i = \left(\frac{dv_r}{dt}\right)_i + \left(\frac{dΩ}{dt}\right)_i \times r + Ω \times v_i .$$ Here we use as much as we can our formula $(dF/dt)_i = (dF/dt)_r + Ω ...


2

Ahh, Richard Fitzpatrick. Great guy. Ok, If you start with the second set of expressions, use the appropriate double-angle-formula and then assume the "angle" $2\Omega \sin\lambda t$ is small (note that the $t$ is not within the sin function!), you get the first expressions, e.g. $$\cos(\theta+\phi) = \cos\theta\cos\phi - \sin\theta\sin\phi,$$ and then ...


3

Note that Fitzpatrick states towards the beginning, The following solution method exploits the fact that the Coriolis force is much smaller in magnitude that the force of gravity: hence, $\Omega$ can be treated as a small parameter Generally, when statements like that are made, powers (greater than 1) of the term in question are considered to be zero: ...


2

What you are missing is that you are calculating the Jacobian, not simply multiplying $d\psi$ by $d\bar\psi$. The determinant also goes downstairs instead of upstairs, because that's how Grassmann numbers roll. See http://en.m.wikipedia.org/wiki/Berezin_integral for details.


4

It's pretty much what Javier Badia said in the comments: Grassmann numbers anticommute. $$\chi_1 \chi_2 = -\chi_2 \chi_1\tag{1}$$ or in this case, $\chi\bar\chi = -\bar\chi\chi$. Note that this implies the square of any Grassman number is zero, if you set $\chi_1 = \chi_2 = \chi$ in equation (1). Using these properties and some very careful algebra, you ...


2

It's the chain rule: $$ \partial_x f(y(x)) = \partial_y f(y(x)) \cdot \partial_x y(x)$$ Your vector field $A^\mu$ depends on the worldsheet coordinates only through the worldsheet coordinates $x^\mu$. Thus, when how $A^\mu$ behaves under an infinitesimal shift on the world-sheet you need to take into account how $A^\mu$ depends on $z$ - that's $\partial_z ...


1

Question 1: $$\frac{dS}{dV}=\left(\frac{\partial S}{\partial T}\right)_V\frac{dT}{dV}+\left(\frac{\partial S}{\partial V}\right)_T\frac{dV}{dV}$$ This doesn't make much sense, because is not a well defined expression. The differential $$ \tag{A} dS=\left(\frac{\partial S}{\partial T}\right)_VdT+\left(\frac{\partial S}{\partial V}\right)_TdV$$ is just ...


1

This proof from Griffiths book introduction to electrodynamics Consider the vector function $$\vec{a}=\frac{1}{r^2}\hat{r}$$ At every location $\vec{a}$ is directed radially outward ; if ever there was a function that ought to have a large positive divergence, this is it. and yet, when you actually calculate the divergence, you will get ...


4

Indeed the answer is not zero but $-4\pi\delta(r)$ (Dirac delta function). The formula of divergence can be found in any standard textbook on mathematical physics, for example chapter 2 of Mathematical methods for physicists by Arfken. But since this function is singular at $r=0$ we must be careful. At any other points is easy to calculate it. It is ...


1

The covariant derivative for a general tensor of the form $T^{a_1\dots a_n}_{b_1 \dots b_n}$ is given by, $$\nabla_c T^{a_1\dots a_n}_{b_1 \dots b_n} = \partial_c T^{a_1\dots a_n}_{b_1 \dots b_n} + \Gamma^{a_1}_{cd}T^{d\dots a_n}_{b_1 \dots b_n} + \dots - \Gamma^d_{c b_1}T^{a_1\dots a_n}_{d \dots b_n} - \dots$$ Taking the covariant derivative of a ...


1

A covariant derivative of a tensor is itself a tensor. Actually, when we say something is covariant (or invariant under coordinate transformation), we mean that thing is a tensor. So, in this case $\nabla_\mu V^\nu\equiv T_\mu{}^\nu$. Now calculate $\nabla_\alpha T_\mu{}^\nu$ easily. \begin{equation} \nabla_\alpha T_\mu{}^\nu=\partial_\alpha ...


0

As it looks like another question I've supplied an answer to might be duplicated here (and hence closed), I am going to provide a similar but not identical answer here. In words - divergence is the flux of something into or out of a closed volume, per unit volume. The best visual picture I have of this is a fluid flow. Imagine water spewing out of a tap - ...


0

Divergence can be thought of as the flux of a vector field per unit volume. It is positive if there is a net flux out of a small volume and negative if there is a net flux inwards. When you say "its diagram" - of course there are different ways of plotting vector fields. Perhaps the most common way is using field lines. In which case it can be ...



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