# Tag Info

1

The conservative answer is: It depends on context. Different authors mean different things. See also this Phys.SE post.

3

It is definitely an ambiguous notation, but one that is quite conventional. You should interpret it as: $(\partial_a X_\mu)(\partial^a X^\mu)$. For instance, often the Klein-Gordon Lagrangian is written as: $$\mathcal{L} = \frac{1}{2} \partial_\mu \phi \partial^\mu \phi + \cdots$$ which should be interpreted as: $$\mathcal{L} = \frac{1}{2} (\partial_\mu ... 1 In my experience, the correct interpretation is$$ \partial_\mu X \partial^\mu X = (\partial_\mu X)( \partial^\mu X). $$Total derivatives are usually written clearly as$$ \partial_\mu(\ldots). $$2 The bounds of the integral have no dependence on any of the variables, and hence we may move the differential operator into the integrand,$$\frac{\delta}{\delta \eta (z)} \int \mathrm{d}^4 y \, S_F (z-y) \eta(y) = \int \mathrm{d}^4 y \, S_F (z-y) \delta^{(4)}(z-y)$$Evaluating the integral using the standard delta distribution identity, we obtain your ... 0 The question is somewhat ill-defined. This answer is based on what I gather the OP is asking. No. Generically, the covariant derivative of a tensor is zero only if the entire tensor is zero. Note how the covariant derivative is defined (taking example of a 1-form)$$ (\nabla_\mu A)_\nu = \partial_\mu A_\nu - \Gamma^\lambda_{\mu\nu} A_\lambda $$Now, ... 7 The result is sometimes called Flanders' lemma. The remarkable point is that it does not need that f is analytic, but just that it is C^\infty. So it does not relies upon the Taylor series as it could seem at first glance, since that series may not converge. It works in any open star-shaped neighborhood of points in \mathbb R^n. A set A\subset ... 0 I'll share a one dimensional example and see if this helps to shed any light on the matter. Suppose f(x) is an analytic function (i.e. the taylor series of f converges to f). Then we can write,$$ f(x) = f(a) + f'(a)(x-a) + f''(a)(x-a)^2/2 + \cdots  = f(a) + (x-a) \left[ f'(a) + f''(a)(x-a)/2 + \cdots \right]  = f(a) + (x-a) H(x) $$... 1 Let us star with the fundamental theorem of calculus for n=1: F(x) - F(a) = \int_{a}^{x} F'(s)ds Now use the substitution s=t(x-a)+a. This is because s rescales the interval [a,x] to [0,1]. Then ds = dt(x-a), then we get: F(x) - F(a) = (x-a)\int_{0}^{1} F'(t(x-a)+a)dt, and this proofs n =1. Now define a vector y^{\mu} = t(x^{\mu} ... 2 It is the Taylor expansion around a point a of a multivariable function where H_{\mu} is similar to the Hessian matrix but for order 1. You can find it in Vector Calculus of Marsden, Tomba, or just google it. This representation is written with the Einstein summation 1 I think your equation is slightly incorrect, should be:$$\frac{dv}{dk} = \frac{1}{k}\frac{d\omega}{dk} - \frac{\omega}{k^2} This is because $\omega = \omega(k)$, $\omega$ is a function of $k$, so you need to apply the chain rule to evaluate the derivative. The dispersion relation may also be a topic of interest.

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First recall that we define a co-vector field $\eta$ from a vector field $v$ via the flat map $\eta~=~v^{\flat}.$ In components we have $\eta_k = g_{ki} v^i.$ Equivalently, we can reconstruct the vector field $v$ via the sharp map $v ~=~\eta^{\sharp}.$ In components we have $v^i = (g^{-1})^{ik} \eta_k.$ Now let us return to OP's question. As Danu writes ...

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Try differentiating twice just as you said. Then to combine them, you can do a few different things: Visually compare the two expressions, and note that they're identical except for a multiplicative constant, or divide the two expressions and see if you actually do get $v^2$ or $1/v^2$, or multiply your expression for $\partial^2 y/\partial t^2$ by ...

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