New answers tagged differentiation
1
vote
From material derivatives to partial derivatives in the wave equation
It happens that the convective term $\mathbf{\dot x}\cdot \nabla \mathbf u$ describes the transport of the field $\mathbf u$ over the flow, but in waves there is no transport of matter.
Another way to ...
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0
votes
Why must the total time derivative only be a linear function of velocity?
It comes down to how the differentiation works. The function I mentioned, $f(q, t)=q^2$ would indeed be a linear function once differentiated:
$$\frac{d}{dt}(q^2)=\frac{d}{dt}(q\cdot q)$$
Which, when ...
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Why must the total time derivative only be a linear function of velocity?
The Taylor expansion is:
$f(x+\delta)=f(x)+f'(x)\delta+...$
Here $x=v^2$, and you know that
$v^2+\delta=(\vec{v}+\vec{\epsilon})^2=v^2+2\vec{v}.\vec{\epsilon}+\epsilon^2$
thus,
$\delta=2\vec{v}.\vec{\...
- 1,809
2
votes
Accepted
Covariant derivative acting on Dirac delta function
Let's be a bit systematic. Let $M$ be a smooth $m$-dimensional manifold. Suppose that $M$ is orientable and oriented (so that we can use $m$-forms as densities). Let $\xi:E\rightarrow M$ be a smooth ...
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Clarification of Ehrenfest theorem
The Newton's equation of motion in potential $V(x)$ are:
$$
\dot{x}=\frac{p}{m},\dot{p}=-\frac{dV(x)}{dx}
$$
or simply
$$m\ddot{x}=-\frac{dV(x)}{dx}=F(x),$$
where $F(x)= -\frac{dV(x)}{dx}$ is called (...
- 65.1k
0
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Arguing that the time derivative of $\exp(-iHt)$ is $-iH\exp(-iHt)$ without taylor expansion
But I cannot see how I can ignore the eigenvectors as if they do not exist and use chain rule
Do you want to prove it without using the spectrum or Taylor expansion?
Let $U(t) = e^{-iHt}$. By ...
- 3,379
0
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Arguing that the time derivative of $\exp(-iHt)$ is $-iH\exp(-iHt)$ without taylor expansion
If $H$ is a hermitian matrix with eigenvectors $$H|n\rangle=\epsilon_n|n\rangle,$$
(note that the eigenvectors and eigenvalues do not depend on time)
then we could write
$$
H= \sum_n|n\rangle\langle n|...
- 65.1k
0
votes
Arguing that the time derivative of $\exp(-iHt)$ is $-iH\exp(-iHt)$ without taylor expansion
Since the Hamiltonian is a self-adjoint operator, you can use the spectral theorem to write
$$H = \int_\mathbb{R} \lambda \, dP_\lambda.$$
Then, as the exponential is a measurable function,
$$
\exp(-...
- 1,314
4
votes
Accepted
Understanding the definition of the covariant derivative
In the case of a scalar field, the covariant derivative equals the partial derivative.
In the case of a tensor field of higher rank, there are further terms involving connection coefficients: for rank ...
- 61.8k
3
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Understanding the definition of the covariant derivative
The covariant derivative of a covector field $V_\mu$ is:
$$ V_{\mu;\nu} = V_{\mu,\nu} - \Gamma^\lambda{}_{\mu\nu} V_{\lambda} $$
The covariant derivative of a scalar field is just its partial ...
- 3,348
3
votes
Accepted
Need help in understanding Tangential Acceleration
Derivatives speak to the instantaneous behavior at a point. It is possible to have a 1st derivative that is non-zero and a 2nd derivative that is 0 at a point. They're simply measuring two different ...
- 51.7k
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"Deriving" the covariant derivative
I think you just need to find an operator $D_{\mu}$ such that
$$
D'_{\mu}\phi' = e^{i\beta}D_{\mu}\phi.
$$
For some arbitrary phase $\beta$. The normal derivative is a starting point.
$$
\partial_{\mu}...
- 3,379
1
vote
Derivative for the Maxwell field
$$\frac{\partial(\partial_{\mu}A^{\sigma})}{\partial(\partial^{\nu}A_{\lambda})}$$
I can't understand whether I must raise the lower index of the partial derivative, and lower the one of the vector ...
- 23.3k
3
votes
The definition of the Lie Derivative
It's just because you want to compare two objects at the same point. In differential geometry you can NOT compare objects at different points since they live in different spaces. The pull back allows ...
- 541
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Why are Weyl's Equations composed of only first-order derivatives?
Equations 2nd-order in time (like the Klein-Gordon Equation) yield negative energy solutions which are not consistent if interpreted as wave functions of physical particles (as Schroedinger and Dirac ...
- 1,897
3
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Is it ever possible that the object is moving with a velocity such that its rate of change of speed is not constant but acceleration is constant?
In general if $v$ denotes the velocity, the rate of change of speed is
\begin{align*}\frac{\text{d}|v|}{\text{d}t} &= \frac{\text{d}}{\text{d}t} \sqrt{ \left< v, v \right> } \\&= \frac{1}...
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6
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Is it ever possible that the object is moving with a velocity such that its rate of change of speed is not constant but acceleration is constant?
Yes, this happens all the time. Fire a gun, or throw a ball, or do just about anything that involves making something move. And ignore things like air resistance, curvature of the earth and so on.
...
9
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Is it ever possible that the object is moving with a velocity such that its rate of change of speed is not constant but acceleration is constant?
Prelude - a (hopefully) fun but counterintuitive geometrical fact
A nice fact which may be a bit counterintuitive, is that if you have a square with diagonal length $\ell$ and this length varies in ...
- 3,348
26
votes
Is it ever possible that the object is moving with a velocity such that its rate of change of speed is not constant but acceleration is constant?
Hint: In the projectile motion (without drag) the acceleration $\vec{a}=\frac{d\vec{v}}{dt}$ is constant. However $\frac{d|\vec{v}|}{dt}$ is not constant, since it is negative when the projectile is ...
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