# Tag Info

## Hot answers tagged differentiation

6

Ah you got caught by something subtle! Planck's law gives a probability density function in terms of the frequency $B(\nu,T)$. To convert to the probability density function in terms of $\lambda$ (let's call it $\tilde{B}(\lambda,T)$), you can't just set $\nu=c/\lambda$. You need to include a Jacobian factor. In more detail: What remains invariant under a ...

5

You are comparing the spectral radiance in frequency $B(f,T)$ with the spectral radiance in wavelength $B(\lambda,T)$. Because these are differential distributions, you cannot simply replace $f=c/\lambda$ to go from one to another. You must take into account the Jacobian of transformation. $$d\Omega=B(f,T)df = B(\lambda,T)d\lambda$$ \begin{align} \implies ... 5 I think you're being somewhat misled by your terminology and by the ambiguity of the standard, compact notation. Note that some texts use only the symbol "\mathrm{d}/\mathrm{d}\tau", or only "\mathrm{D}_\tau" (or similar symbols), for both operations you mention. Because its meaning is determined by what it is applied to. A general ... 4 This sort of derivation is done by differentiating the coupled equations and then substituting the results back into the original, undifferentiated equations to eliminate the undesired variables. In this case, if I were asked to reproduce the last equation, I would: Isolate q from the first equation. Differentiate this equation twice to find expressions ... 3 The short answer is yes, it does. A covariant derivative describes the change of a vector field under parallel transport along a given vector. It does not single out changes in direction from changes in magnitude, but rather returns a vector that describes the change of both. To extract one or the other, you can take the dot product and orthogonal projection ... 3 Because position is not a vector. 2 I want to try to lift the confusion. First, I want to introduce two important aspects. Physics should not change if you change your coordinate system. What does the real life care about the coordinate system you use in you calculations? Further, there are two facts to realize, because we generally choose an easy path in calculations: we usually align our ... 2 The gradient \nabla can be thought of as the "components" of the differential operator d. More concretely, if \phi is a scalar function, then\text{d}\phi = \frac{\partial\phi}{\partial x}\text{d}x + \frac{\partial\phi}{\partial y}\text{d}y + \frac{\partial \phi}{\partial z}\text{d}z = \nabla \phi\cdot (-),$$where (-) is a placeholder for ... 2 Let us combine the terms containing a and b respectively:$$\dfrac{1}{\sqrt{x^2+a^2}} \Bigg( 1-\dfrac{x^2}{x^2+a^2}\Bigg)=\dfrac{1}{\sqrt{x^2+b^2}}\Bigg(1-\dfrac{x^2}{x^2+b^2} \Bigg)$$Simplifying this you get-$$\dfrac{a^2}{(x^2+a^2)^{3/2}}=\dfrac{b^2}{(x^2+b^2)^{3/2}}$$And you can proceed further. 2 Your idea is correct: you have to exploit Stokes' formula for a decreasing sequence of surfaces \Sigma_n tending to a point p. Here the surfaces of the sequence should be viewed as restrictions of an initially given surface \Sigma containing p. When the surfaces tend to p, the averaged value with respect to the area of the line integral of \vec{a}... 2 If you're looking for a concrete example, perhaps the best one comes from Maxwell's equations. The first of Maxwell's equations is$$\nabla \boldsymbol\cdot \mathbf{E}=\frac{\rho}{\varepsilon_0} \tag{1}$$Here \mathbf{E} is the electric field, which is vector valued, \nabla\boldsymbol\cdot is the divergence operator, which essentially measures how much \... 2 The right hand side of your equation is an explicit function of r_{i} and the r_{k}s. One just needs to apply the chain and product rules from introductory calculus. The expression will be super-ugly, and will have explicit \dot r's in it, but I don't see what difficulty you're having with the computation. Perhaps the only complication is that your ... 1 It is somewhat problematic to rigorously define \partial_i\partial_j\frac{1}{r} in 3D distribution theory, cf. e.g. this Math.SE post. Nevertheless, due to the identity$$ \nabla^2\frac{1}{r}~=~-4\pi\delta^3(\vec{ r}),\tag{A}$$it makes heuristic/physical sense to assign$$ \partial_i\partial_j\frac{1}{r}~=~-\frac{4\pi}{3}\delta_{ij}\delta^3(\vec{ r}) ~+~ ...

1

It doesn't make a lot of sense to talk about the "time derivatives of the unit vectors of the basis themselves", because the unit vectors themselves are constant with respect to time. Instead, in curvilinear coordinates, the basis vectors are vector fields, with different values from point to point in space. The unexpected terms that arise in the ...

1

No, there isn't any difference in the physics. However, in this paper are distinguished in an attempt to make things clearer, reserving upper indices for time, and lower indices for spatial dimentions. Still, this is just a choice and you could write everything in subscript as long as you don't confuse time and spatial dependence. The difference in upper or ...

1

There's no way to know for sure with just these images posted, but most likely this is just a difference in convention between two sources or a difference in convention from a single source for how indices for time are printed vs. indices for spatial dimensions. It looks very much like the second group is written with a convention in which time indices are ...

1

The missing conceptual step to close the reasoning after your last equations is that $$0=\left(T-\frac{\partial U}{\partial S}\right)dS+ \left(-p-\frac{\partial U}{\partial V}\right) dV$$ must be valid for every $dS$ and for every $dV$. Therefore, if you take the special case $dS=0$, $dV \neq 0$, we must have $p=-\left.\frac{\partial U}{\partial V}\right|_S$....

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