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2

Let us look at this prolem from a (relativistic) field theory perspective. The Hamiltonian for $\psi$ must contain a term of the form $$\nabla\psi^\dagger\cdot\nabla\psi$$ due to Lorentz invariance. Assuming $\psi$ to transform in a representation of U(1). Resulting in the following simultanious transformations: \begin{aligned}\psi\rightarrow U(x)\psi ... 2 Given that the regular (non-covariant) derivative of the adjoint satisfies \nabla \psi^\dagger = \left(\nabla \psi \right)^\dagger $$one likewise expects and defines$$ \tilde{\nabla}\psi^\dagger = \left( \tilde{\nabla}\psi \right)^\dagger = \left( \nabla \psi +ie\bf{A}\psi \right)^\dagger = \nabla \psi^\dagger -ie\bf{A} \psi^\dagger $$1 I'm not sure what might be confusing you. Assume, as in most cases with the Golden rule, that the transition rate is constant, \Gamma. So, for small times, the cumulative transition probability is W=\Gamma t. Think of the transition as leakage from a vessel. At t=0, no water has been lost, but with a constant rate of leakage, \Gamma, the longer ... 0 So you started with L' = L(|v|^2 + 2v\bullet \epsilon + \epsilon^2)  If you treat  2v\bullet \epsilon + \epsilon^2  as a variation of |v|^2, you may use the taylor's theorem treating |v|^2 as a variable, and you get the following expression:  L(|v'|^2) = L(|v|^2) + \frac{\partial L}{\partial |v|^2} (2v \bullet\epsilon + \epsilon^2) + higher order ... 3 Well, a good example is thinking in term of components. In several areas of physics, the math gets more intuitive when you think in terms of components of the vectors. So, instead of writing the vector \mathbf r for the position of a particle, you write x^i as the i-th component of a vector. The i in the top is to indicate a contravariant vector, ... 2 I think your last sentence shows you're on the right track. A one form is a linear functional that maps vectors to real numbers. You give it a vector as an input, and, as you say, it returns the rate of change in the implied direction. Let's say we have a path whose tangent at a point is defined by the vector v^j\,\partial_j - the differential operator ... 1 Answer to the OP can be obtained easily using the Savitzky-Golay (SG) smoothing-differentiation filter. Suppose we have noisy n-point data such as the temperature (T) vs. time (t) as in the OP. As per the OP we want to smooth the data, find the time rate of change, and the uncertainty that the SG procedure would introduce in the time rate of change. ... 7 I know how , in the physical sense -$$\frac {dv}{dt} = a$$But, even after thinking a lot, I am not able to see the fault in this -$$\frac {dv}{dt} = \frac {d(st^{-1})}{dt} = \frac {sd(t^{-1})}{dt} = s*(-1)*t^{-2} = \frac {-s}{t^2} = \frac > {-v}{t} = -a$$I know something is terribly wrong here but I'm just not able to ... 1 First, you should be careful with your choice of indices. What you have written in Eq. 1 implies a summation over \mu that I don't think you actually want. It is true that $$\frac{\partial F_{\mu\nu}}{\partial A_{\sigma}}=0,$$ but that is just because F_{\mu\nu} depends on the derivatives of A_{\mu} and not A_{\mu} ... 0 (v) I then differentiate this by v: t = d'(v) =\frac{mv}{k\sqrt{A^2-\frac{mv^2}{k}}} (am I crazy?) You made a mistake, \frac{d}{dv} d(v) doesn't equal time. For instance, \frac{d}{dv} d(v) can be the same at two different times. Consider the simplest case, a particle is at rest. Then v=0 and d(t)=const  so the derivative either ... 1 We know the relationship between enthalpy, energy, pressure and volume$$H = U + P V \, . \tag{1}$$Differentiation of this expression yields$$ dH = T dS + V dP \, .$$However,$$TdS = dQ$$where dQ is the amount of heat system received or gave away. Take the derivative of equation (1) with respect to T at constant P to get$$\left( \frac{dH}{dT} ...

-1

Yes, that's right. It's hard to be more explicit than that, since it's just the chain rule. As a product, $$\frac{d}{dt} \dot{\theta}^2 = \frac{d}{dt}(\dot{\theta}\dot{\theta}) = \dot{\theta}\ddot{\theta} + \dot{\theta}\ddot{\theta} = 2\dot{\theta}\ddot{\theta}.$$

0

${d \over dt} \theta = \dot \theta$ So what you are asking is basically ${d^2 \over dt^2} \theta^2$ First, ${d \over dt} \theta^2 = \dot{\theta^2} = 2 \theta \dot \theta$ Second, ${d \over dt}(\dot{\theta^2}) = {d \over dt} (2 \theta \dot \theta) = 2 \dot \theta^2 + 2 \theta \ddot \theta$

3

$\frac{DA^\mu}{D\lambda}=\frac{DA^\mu}{d\lambda}$ are two notations for the same object.

4

This is a covariant derivative along a world line (if you would not consider a world line the proper time $\tau$ would not make any sense). So you consider a curve in space time parametrized in dependence of the proper time $x^\mu(\tau)$. Then you have: \frac{DA^\mu}{d\tau} = \frac{\partial A^{\mu}\big(x(\tau)\big)}{\partial \tau} + ...

-1

According as you say, I can assume the function $\phi$=k $\frac{\mu}{\sigma}$ where k has the units of ms. Hence by partial differentiation, we get $\frac{\partial \phi}{\partial \mu}$ = $\frac{k}{\sigma}$ with units $ms/mV$. and for $\frac{\partial \phi}{\partial \sigma}$ = -k$\frac{\mu}{\sigma^2}$ with units $ms*mV/(mV)^2$ = $ms/mV$ Hence the units of ...

2

The derivative is the slope of the tangent line to the function at each point. For a function $y(x)$, the slope is in units of $y / x$. Hence, the units of the derivative of the function in your question are ms / mV.

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