energy conservation - Virtual Photon in Electron Scattering Feynman diagram

If we know that the virtual photon emitted in an electron scattering Feynman diagram violates the energy and momentum conservation laws (though temporarily), why do we accept it as a feasible diagram representing another perspective of reality?

homework and exercises - Time dilation in special relativity

Suppose a star ship is moving with some velocity. Two light pulses one in direction similar to star ship another opposite to it is shot towards the space ship. Then how time inside space ship adjust to make velocity of light constant?

Is quantum field theory defined by its lattice regularization?

One statement I've heard many times is that QFT is "defined" by the lattice, or that the "only" definition of QFT is on the lattice (when such definition exists, e.g in pure Yang-Mills theory). I've heard that from many people I respect, but I have my doubts. Specifically - the lattice is an algorithmic definition of those quantities that can be calculated in Euclidean spacetime. As I said in my answer to this question, this is a proper subset of all physical quantities you may be interested in calculating in general, and many physical situations do not have Euclidean formulation. Is there some response to this objection I am missing, or am I being naive somehow?

particle physics - Could the universe be accurately simulated with an infinitely powerful computer?

This would mean that every event happens because of what has hapened before it and there is no randomness factor. At a microscopic level, the motion of atoms is a result of the motion of other atoms around it. When an atom or molecules moves, it may hit another atom or molecule which will acquire part of its force. This means that every motion is related to each other. At a macroscopic level the same rule applies. If I behave a certain way, it is because of the education I have received, my manners, my way of thinking, which I have acquired from my family, friends, and other people surrounding me. In short, everything has a reason to happen and nothing pops out of the blue. For example, me asking that question comes from me and a friend discussing this topic. This happened because I am interested in science and this thought crossed my mind when I was bored.

Is this true, or is there a randomness factor?

If it was true, then how comes our world exists as it is? Since the Big Bang, everything must have been uniform, and there is no reason for there to be irregularities, like planets or suns.

Do quantum mechanics add the randomness factor?

TL;DR: Would it be possible to simulate the Universe with an extremely powerful computer? (One that could simulate every particle at the same time.)

NB: some may think that this question is a duplicate of the question: "Is the Universe deterministic?" In my opinion it is not since that question focuses on quantum mechanics, while this one on all aspects of the universe.

buoyancy - Can the buoyant force on an object be seen as reduction in weight on a scale?

Consider a container with some fluid of density $\rho_l$ and volume $V_l$. This is kept on a measuring device and has weight $\rho_l V_lg$. Now, consider a block of density $\rho_b$ and volume $V_b$. This block is put into the fluid and here, its apparent weight equals $\rho_b V_bg - F_b$ , where $F_b$ equals the buoyant force which equals $V_b\rho_lg$.

Therefore, the apparent weight of block equals $gV_b(\rho_b - \rho_l)$.

What happens if you take the weight of this whole apparatus? Will it equal $$gV_b(\rho_b - \rho_l) + \rho_lV_lg,$$ or $$\rho_bV_bg + \rho_lV_lg?$$

Answer

It will actually weigh $F_b+\rho_lV_lg$ which is $\rho_lV_bg+\rho_lV_lg$, and not one of the two options you say.

Consider the liquid as the system and the block as an external body. Now we know that the liquid applies a buoyant force on the block. According to Newton's third law, the block will apply a reaction force on the liquid, equal in magnitude and opposite in direction. Thus total force on liquid is $F_b+F_g$ which gives $\rho_lV_bg+\rho_lV_lg$.

Note that this is the weight when the block is attached to the spring balance, and suspended in the liquid. If the block is kept on the floor not attached to the spring balance, the reading weight will be different.

Edit:

In the case when the block is resting on the floor, the weight will simply be $\rho_lV_bg+\rho_lV_lg$, because when you consider the block and liquid as a system, the buoyant force will become an internal force and cancel out on the whole system. So the only force responsible for the weight will be gravity.

quantum mechanics - Heisenberg's uncertainty principle for mean deviation?

The Heisenberg uncertainty principle states that

$$\sigma_x \sigma_p \ge \frac{\hbar}{2}$$

However, this is only for the standard deviation. What is the inequality if the mean deviation, defined as

$$\bar \sigma_x=\int_{-\infty}^{\infty} \lvert x-\bar x\rvert\,\rho(x) \ \mathrm{d}x = \int_{-\infty}^{\infty} \lvert x-\bar x\rvert\ \lvert\Psi(x)\rvert^2\ \mathrm{d}x$$

is used as the measure of dispersion? This measure of dispersion generally gives values less than the standard deviation.

Is there a positive number $\lambda$ such that

$$\bar \sigma_x \bar \sigma_p \ge \lambda$$

holds in general?

Answer

We can assume WLOG that $\bar x=\bar p=0$ and $\hbar =1$. We don't assume that the wave-functions are normalised.

Let $$ \sigma_x\equiv \frac{\displaystyle\int_{\mathbb R} |x|\;|\psi(x)|^2\,\mathrm dx}{\displaystyle\int_{\mathbb R}|\psi(x)|^2\,\mathrm dx} $$ and $$ \sigma_p\equiv \frac{\displaystyle\int_{\mathbb R} |p|\;|\tilde\psi(p)|^2\,\mathrm dp}{\displaystyle\int_{\mathbb R}|\tilde\psi(p)|^2\,\mathrm dp} $$

Using $$ \int_{\mathbb R} |p|\;\mathrm e^{ipx}\;\mathrm dp=\frac{-2}{x^2} $$ we can prove that¹ $$ \sigma_x\sigma_p=\frac{1}{\pi}\frac{-\displaystyle\int_{\mathbb R^3} |\psi(z)|^2\psi^*(x)\psi(y)\frac{|z|}{(x-y)^2}\,\mathrm dx\,\mathrm dy\,\mathrm dz}{\displaystyle\left[\int_{\mathbb R}|\psi(x)|^2\,\mathrm dx\right]^2}\equiv \frac{1}{\pi} F[\psi] $$

In the case of Gaussian wave packets it is easy to check that $F=1$, that is, $\sigma_x\sigma_p=\frac{1}{\pi}$. We know that Gaussian wave-functions have the minimum possible spread, so we might conjecture that $\lambda=1/\pi$. I haven't been able to prove that $F[\psi]\ge 1$ for all $\psi$, but it seems reasonable to expect that $F$ is minimised for Gaussian functions. The reader could try to prove this claim by using the Euler-Langrange equations for $F[\psi]$ because after all, $F$ is just a functional of $\psi$.

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Testing the conjecture

I evaluated $F[\psi]$ for some random $\psi$: $$ \begin{aligned} F\left[\exp\left(-ax^2\right)\right]&=1\\ F\left[\Pi\left(\frac{x}{a}\right)\cos\left(\frac{\pi x}{a}\right)\right]&=\frac{\pi^2-4}{2\pi^2}(\pi\,\text{Si}(\pi)-2)\approx1.13532\\ F\left[\Pi\left(\frac{x}{a}\right)\cos^2\left(\frac{\pi x}{a}\right)\right]&=\frac{3\pi^2-16}{9\pi^2}(\pi\,\text{Si}(2\pi)+\log(2\pi)+\gamma-\text{Ci}(2\pi))\approx1.05604\\ F\left[\Lambda\left(\frac{x}{a}\right)\right]&=\frac{3\log2}{2}\approx1.03972\\ F\left[\frac{J_1(ax)}{x}\right]&=\frac{9\pi^2}{64}\approx1.38791\\ F\left[\frac{J_2(ax)}{x}\right]&=\frac{75\pi^2}{128}\approx5.78297 \end{aligned} $$

As pointed out by knzhou, any function that depends on a single dimensionful parameter $a$ has an $F$ that is independent of that parameter (as the examples above confirm). If we take instead functions that depend on a dimensionless parameter $n$, then $F$ will depend on it, and we may try to minimise $F$ with respect to that parameter. For example, if we take $$ \psi_{n}(x)=\Pi\left(x\right)\cos^n\left(\pi x\right) $$ then we get $$ 1< F\left[\psi\right]<1+\frac{1}{12n} $$ so that $F[\psi_n]$ is minimised for $n\to\infty$ where we get $F[\psi_{\infty}]=1$.

Similarly, if we take $$ \psi_{n}(x)=\frac{J_{2n+1}(x)}{x} $$ we get $$ F[\psi]=\frac{(4n+1)^2(4n+2)^2\pi^2}{64(2n+1)^3}\ge \frac{9\pi^2}{64} \approx1.38791 $$ which is, again, consistent with our conjecture.

The function $$ \psi_n(x)=\frac{1}{(x^2+1)^n} $$ has $$ F[\psi]=\frac{\Gamma (2 n)^2 \Gamma \left(n+\frac{1}{2}\right)^2}{(2 n-1) n! \Gamma (n) \Gamma \left(2 n-\frac{1}{2}\right)^2}\ge 1 $$ which satisfies our conjecture.

As a final example, note that $$ \psi_{n}(x)=x^n\mathrm e^{-x^2} $$ has $$ F[\psi]=\frac{2^n n! \Gamma \left(\frac{n+1}{2}\right)^2}{\Gamma \left(n+\frac{1}{2}\right)^2}\ge 1 $$ as required.

We could do the same for other families of functions so as to be more confident about the conjecture.

Conjecture's wrong! (2018-03-04)

User Frédéric Grosshans has found a counter-example to the conjecture. Here we extend their analysis a bit.

We note that the set of functions $$ \psi_n(x)=H_n(x)\mathrm e^{-\frac12 x^2} $$ with $H_n$ the Hermite polynomials are a basis for $L^2(\mathbb R)$. We may therefore write any function as $$ \psi(x)=\sum_{j=0}^\infty a_jH_j(x)\mathrm e^{-\frac12 x^2} $$

Truncating the sum to $j\le N$ and minimising with respect to $\{a_j\}_{j\in[1,N]}$ yields the minimum of $F$ when restricted to that subspace: $$ \min_{\psi\in\operatorname{span}(\psi_{n\le N})} F[\psi]=\min_{a_1,\dots,a_N}F\left[\sum_{j=0}^N a_jH_j(x)\mathrm e^{-\frac12 x^2}\right] $$

Taking the limit $N\to\infty$ yields the infimum of $F$ over $L^2(\mathbb R)$. I don't know how to calculate $F[\psi]$ analytically but it is rather simple to do so numerically:

The upper and lower dashed lines represent the conjectured $F\ge 1$ and Frédéric's $F\ge \pi^2/4e$. The solid line is the fit of the numerical results to a model $a+b/N^2$, which yields as an asymptotic estimate $F\ge 0.9574$, which is represented by the middle dashed line.

If these numerical results are reliable, then we would conclude that the true bound is around $$ F[\psi]\ge 0.9574 $$ which is close to the gaussian result, and above Frédéric's result. This seems to confirm their analysis. A rigorous proof is lacking, but the numerics are indeed very suggestive. I guess at this point we should ask our friends the mathematicians to come and help us. The problem seems interesting in and of itself, so I'm sure they'd be happy to help.

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Other moments

If we use $$ \sigma_x(\nu)=\int\mathrm dx\ |x|^\nu\; |\psi(x)|^2\qquad \nu\in\mathbb N $$ to measure the dispersion, we find that, for Gaussian functions, $$ \sigma_x(\nu)\sigma_p(\nu)=\frac{1}{\pi}\Gamma\left(\frac{1+\nu}{2}\right)^2 $$

In this case we get $\sigma_x\sigma_p=1/\pi$ for $\nu=1$ and $\sigma_x\sigma_p=1/4$ for $\nu=2$, as expected. Its interesting to note that $\sigma_x(\nu)\sigma_p(\nu)$ is minimised for $\nu=2$, that is, the usual HUR.

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$^1$ we might need to introduce a small imaginary part to the denominator $x-y-i\epsilon$ to make the integrals converge.

newtonian gravity - Why can't we feel the Earth turning?

The Earth turns with a very high velocity, around its own axis and around the Sun. So why can't we feel that it's turning, but we can still feel earthquake.