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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 ...

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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 $x$ ...

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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) + ...

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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. ...

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

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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 $v$.

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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 ...

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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. \nabla_\alpha T_\mu{}^\nu=\partial_\alpha ...

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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 - ...

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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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