# Tag Info

4

Here is an algebraic approach to understand the edge state. Let us start from a generic Dirac Hamiltonian for the bulk fermions in the $d$-dimensional space. $$H=\sum_{i=1:d}\mathrm{i}\partial_i\alpha^i+m(x_i)\beta,$$ where $\alpha^i$ and $\beta$ are anti-commuting gamma matrices ($\{\alpha^i,\alpha^j\}=2\delta^{ij}$, $\{\alpha^i,\beta\}=0$, $\beta\beta=1$), ...

2

Certainly, it would be ideal if we could solve most differential equations analytically, without resorting to approximation. However, for most systems, it is simply not the case, and this often manifests due to the inherent complexity of the problem. Though, there are exceptions, such as the equation of motion for a pendulum without the small angle ...

3

This is a quite general question. Whether or not one should use an approximation depends on several things. Disclaimer: my answer is not restricted to differential equations and contains examples from perturbation theory, but the general idea still applies. The most important question is whether an approximation makes sense from a physical point of view. ...

0

If no approximations were used or all orders were kept, the outcome of the theory would presumably match perfectly the corresponding experiment that is being modeled. The approximate equations used to model such experiments are good enough if the predictions match the results within the desired uncertainty. As for how much some approximation is off as ...

0

The admittance looking into the network "$Y_{net}$" can be expresed as: $$Y_{net} = \dfrac{1}{R + \dfrac{1}{Y_{net} + Y_{diode}}}$$ Rearranging as a quadratic and solving: $$Y_{net}^2R + Y_{net}Y_{diode} - Y_{diode} = 0$$ $$Y_{net} = \dfrac{\sqrt{Y_{diode}}\sqrt{4R +Y_{diode}}}{2R}$$ Where $$Y_{diode} = \frac{nV_t}{I_s}e^{-\frac{V_f - I_fR}{V_t}}$$ ...

0

To 'derive' conservation of Energy from $\vec{F} = m \vec{a}$, we take a dot product ($\hat{i} \cdot \hat{i} = 1$) which means that we have one (scalar multi-variable non-linear differential) equation with potentially many unknowns. Energy is nice because it provides a common language with all the physical sciences, but in classical mechanics, it's mostly ...

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This question has been marinating in my own mind for some time since asking here, and seeing how the respected community of this site has not yet answered this head on in the awesome manner I have seen other answers, I will share my "answer" as far as i have been able to find it on my own. [EDIT: this answer has been edited to more directly answer the ...

3

These solutions are preferred because they directly embody the scale invariance of the equation. In general, when a physical problem has some sort of symmetry - like the parabolic dilation invariance of the heat equation - then this establishes a corresponding action of the symmetry group on the solutions. The canonical forms based on dimensionless ...

0

Intuitively a lot of equations we use to describe physical phenomena such as $exp(x)$ take in only dimensionless variables. Moreover in physics we're always taking the derivative or the log of things, and it works out to be much simpler if you're not always having to deal with a dimensioned constant factor that drops out every time you take the derivative.

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The key concept is that the units of measurements you chose to use make no difference to the physical behaviour. I.e you expect the solution behaviour to be independent of the units of measurement used. This means you should get the same solution if we chose to measure $x$ is in meters, and $t$ is in seconds and $D$ is in $m^2/s$ as if we chose to measure ...

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Let's say your goal is to describe the shape of some object, such as a box. You could create a completely arbitrary ruler and measure the three axes of the box, coming out for example with lengths of 11.72, 23.44, and 35.16 of your arbitrary ruler units. Or you might look at your results more closely and think hmm, something is going on here, since the ...

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Dimensionless equations have the advantage that they work for any value of the parameters. They are scale invariant. So the solution in terms of a single dimensionless variable applies to all values of $D$ and $t$. It also allows the definition of characteristic values for the dynamic variables. In your example, one could say $u_0$ = ...

1

Because it is easier for dimensionless quantities to be combined in arbitrary polynomial terms (or other terms e.g exponential) with no loss (or extra) factors. Think like "characteristic times" used in exponential factors. Especially quantities appearing in solutions of the form $e^{a} = \sum_0^\infty \frac{a^n}{n!}$, one can see why a dimensionless ...

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