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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 ...


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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.


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(1) V = v + Ω x r . The derivative of V in the inertial frame is indeed, A = (Dv/Dt) + (DΩ/Dt) x r + Ω x Dr/Dt. You are right, both v and r are described according to the rotating axes. Though, for the observer in the inertial frame they rotate together with those axes. So, for doing the derivative according to the inertial frame, you derivate again ...


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I see now your problem and I believe that I can help. (I apologize that I never had time to learn how to write with LaTeX underscipts - for a symbol F with underscript b I write F_b). Let's begin from the velocity formula (1) v_i = v_r + Ω x r . Let's take the derivative of v_i IN THE INERTIAL frame, a_i = (dv_r/dt)_i + (dΩ/dt)_i x r + Ω x v_i . Here we ...


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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 ...


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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: ...


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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.


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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 ...


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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 ...


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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 ...


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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 ...


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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 ...


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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. \begin{equation} \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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The identity in question is given as, $$(\nabla_\beta u^\alpha) u_\alpha=(\nabla_\beta u_\alpha)u^\alpha$$ Expanding the left-hand side, we find, $$(\nabla_{\beta}u^\alpha)u_\alpha = (\nabla_\beta g^{\alpha \delta}u_\delta)u_\alpha = (u_\delta \nabla_\beta g^{\alpha\delta} + g^{\alpha\delta}\nabla_\beta u_\delta)u_\alpha$$ The Levi-Civita connection is ...


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$$T_{\mu\nu} = \partial_\mu\phi \partial_\nu \phi + g _{\mu\nu} (-1/2) (\partial_\nu\phi \partial^\nu \phi +m^2\phi^2) $$ $$\partial^\mu T_{\mu\nu}=\partial_\mu\partial^\mu\phi \partial_\nu \phi+\partial_\mu\phi\partial_\nu\partial^\mu \phi-\frac{1}{2}\partial_\nu(\partial_\tau\phi \partial^\tau \phi +m^2\phi^2)$$ We have equation of motion ...


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Consider a superball falling toward a hard tile floor. The ball will bounce rather nicely. How to model this collision? If you look very closely at the bounce, a normal force from the floor starts acting on the ball when the bottom of the ball hits the floor. At this point in time, the top of the ball doesn't "know" that the bottom of the ball has hit the ...


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but I doubt how can there be change of momentum at an instant? It isn't clear to me what you mean here. Do you mean you doubt that momentum can be discontinuous? If so, then your doubt is justified. Classically, momentum cannot actually be discontinuous at some time $t_0$ since that would imply an actually infinite force acting at $t_0$. If you're ...


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Maybe this graph helps to illustrate the point. Consider a constant force that is applied for a finite interval $\Delta t$. The slope in the $p$ graph at any given time is $\frac{dp}{dt}$. Now, in order to have a step in $p$ (an instantaneous impulse), it is necessary to get an infinite slope, hence an infinite force is required which is physically not ...


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All objects interact via a force that has a finite range. Ultimately this would be one of the four forces electromagnetic, strong, weak and gravity. Since these forces change smoothly with distance the rate of change of momentum can never become infinite.


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When you write this: $$D_{t_2} S=-x_2^2\csc^2t_2.\tag6$$ You obviously mean total time derivative by $D_{t2}$. Which means: $D_{t2}S=dS/dt_2=\frac{\partial S}{\partial q}\frac{\partial q}{\partial t_2}+\frac{\partial S}{\partial t_2}\frac{\partial t_2}{\partial t_2}$ In our case: $q=x_2$ and you're missing this part: $\frac{\partial S}{\partial ...


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Firstly, I'd like to recommend Tristan Needham's book Visual Complex Analysis which is an excellent text and very accessible to physicists. In his own words (p. 515): The Laplacian of $\Phi$ at $p$ measures the amount by which the average value of $\Phi$ on an infinitesimal circle centered at $p$ exceeds the value of $\Phi$ at $p$ itself. More ...


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Imagine an elastic fabric streched over a flat frame. If the frame is then bent out of the plane, the shape of the fabric will follow. The equation $ \nabla^2 u = 0 $ can be used to model such, where $u$ represents the displacement of the frabric with respect to the initial configuration. As explained here the laplacian $\nabla^2 u(x_0)$ is proportial to ...


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The job of calculus is to handle quantities that vary over the domain of the problem at hand. Often, and particularly in introductory physics, we care about quantities that vary in time. We cannot put them into our equations as constants. We also often care about the interrelationship of these quantities. So for example now velocity is given by ...


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How else can you recover displacement from a non-linear velocity without integrating? How else would you calculate acceleration of a non-linear velocity without differentiating? ADDED: If your question is how to solve $\displaystyle v(t) = \frac{dx}{dt} = b\sqrt{x}$, let us know.



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