wavelength - Visible light spectrum to color space

I need to be able to convert an arbitrary emission spectrum in the visible spectrum range (i.e. for every wavelength between 380 and 780, I have a number between 0 and 1 that represents the "intensity" or dominance of that wavelength), and I need to be able to map any given spectrum into a particular color space (for now I need RGB or CIE-XYZ). Is it possible?

For the spectrum say I have the emission spectrum of a white light, then every wavelength in the spectrum will have an intensity of 1, whereas for a green-bluish light I'd have most of the wavelengths between 500 and 550 with an intensity close to 1, with other wavelengths gradually dropping in intensity. So the first spectrum should be converted to pure white whereas the other one would be converted to a green-bluish color in any color space.

Is there a way to do this?

Answer

Human eye has three types of color receptors which respond differently to different parts of the spectrum. See this chart.

One way to tackle your challenge is to basically simulate what the eye does: you take the spectrum as input, calculate how much it would excite each of the three color receptors based on their sensitivity to different parts of the spectrum and then use the three resulting numbers as RGB corresponding to the spectrum.

In order to compute the excitation level, you can integrate the product of the sensitivity S_C(λ) of each of the three color receptors with your spectral power distribution P(λ) to obtain the three RGB numbers:

\begin{equation} R = \int_{0}^{+\infty} S_R(\lambda) P(\lambda) d\lambda \end{equation} \begin{equation} G = \int_{0}^{+\infty} S_G(\lambda) P(\lambda) d\lambda \end{equation} \begin{equation} B = \int_{0}^{+\infty} S_B(\lambda) P(\lambda) d\lambda \end{equation}

For prototyping you can probably just assume the sensitivity S_C(λ) functions to be appropriately scaled and translated Gaussian functions of the wavelength. As you refine your model you should seek better sensitivity functions for each of the three types of color receptors.

classical mechanics - Intuition - why does the period not depend on the amplitude in a pendulum?

I'm looking for an intuition on the relationship between time period and amplitude (for a small pertubation) of pendulums. Why does the period not depend on the amplitude? I know the math of the problem. I am looking for physical intuition.

Answer

The higher you are, the greater the maximum velocity and maximum potential energy.

Consider one pendulum lifted higher than a second both released at the same time. When the higher pendulum reaches the starting point of the second, it already has a velocity greater than 0.

This higher velocity allows the higher pendulum to complete its swing in the same amount of time as the lower, even though it has a longer path.

Since I'm at a computer now I will address a majority of what is said in the comments. For starters, this does not exactly apply to pendulum; only approximately, and the approximation gets worse as $\theta$ increases.

A good way to visualize this is through the Tautochrone curve which is a frictionless curve where for all heights, the time to fall is the same (this is the equivalent of a pendulum period if you ignore the backswing, or have 2 of these curves mirrored; which will be a perfect mirror of the front swing if energy is conserved in the system).

In this scenario, the accelerations work out perfectly (under the same gravity) so that they all arrive at the same time. This is unlike the circular motion, which is only approximately correct for small angles. The interesting thing to note is that looking at a small displacement of a tautochrone curve; it looks approximately circular if you only look at a small section near the bottom. This is an intuitive way to explain why a circular pendulum approximately has this behaviour with small angles.

(Henning mentioned a tautochrone curve in his answer as well. It seemed to be an appropriate way to add more intuition to this)

electromagnetism - Have we directly observed the electric component to EM waves?

For example, has anyone has directly observed charges oscillating due to standing EM waves? I am particularly interested because it'd demonstrate that radiation has a transverse electric component to it. Anything else (historical or modern) that shows that light has a transverse electric component would also be gladly invited.

Answer

Yes, we have. As other answers have explained, this is easy to do in the radio regime, but over the past fifteen years or so we've been able to do it for light too.

The landmark publication here is

Direct measurement of light waves. E. Goulielmakis et al. Science 305, 1267 (2004); author eprint.

which broke new ground on a method called attosecond streaking that lets us see things like this:

$\qquad$

On the left you've got the (mildly processed) raw data, and on the right you've got the reconstruction of the electric field of an infrared pulse that lasts about four cycles.

To measure this, you start with a gas of neon atoms, and you ionize them with a single ultrashort burst of UV radiation that lasts about a tenth of the period of the infrared. (For comparison, the pulse length, $250\:\mathrm{as}$, is to one second as one second is to $125$ million years.) This releases the electron out of the atom, and it does so at some precisely controlled point within the infrared pulse. The electric field of the infrared can then have a strong influence on the motion of the electron: it will be forced up and down as the field oscillates, but depending on when the electron is released this will accumulate to a different impulse, and therefore a different final energy. The final measurement of the electron's energy, as a function of the relative delay between the two pulses (top left) clearly shows the traces of the electric field of the infrared pulse.

universe - What are the most realistic ways of high speed space propulsion?

Liquid and solid chemical fuels in rockets are very expensive and inefficient. I have heard of solar sails but what are the most realistic space travel fuels that will be used in the future to get close to the speed of light?

Answer

In a lot of ways this is a technology---rather than physics---question, but lets look at some limits imposed by physics.

• For rockets there are two numbers that matter: the velocity relative the spacecraft with which the fuel can be expelled (called the specific impulse) and the fraction of the original mass that is fuel. For very high mass fractions the spacecraft can have delta V equal to several times the specific impulse. That means that if you plan to stop at the other end your maximum velocity perhaps twice is the specific impulse. If you need to come home without refueling it is around the specific impulse.

That's the killer for fast burn rockets. We still use them for launch from the surface because they can develop the necessary thrust (our high efficiency engines (ion drives and the like) are all low thrust at the present time).

That suggests one of two strategies:

1. Leave your engine behind. This is the mechanism used by laser sails, star wisps, etc.


2. Pick up your fuel as you go. The strategy of ram-scoops.

Sail technology is--at least in principle--within reach of our current competence, but building and running one capable of efficient interstellar travel is beyond the current Gross Planetary Product. Star wisps are a subset of sails, but require considerable expertise in nanotechnology before they are useful even for exploration.

A second disadvantage of sails for traveling to other start in person, is that you're counting on politicians and policy makers you left behind to continue funding your trip...

Ram-scoops are a big unknown, as we don't know how to build them as yet. A BOTE calculation for a minimal, naive hydrogen fusion ram-scoop limits their top speed to about 14% of c (under the maximum drag assumption that the fuel must be brought to rest relative the craft).

torque - Component of angular momentum perpendicular to the rotation axis in rigid body rotation

I have difficulties in understanding, in the rotation of a rigid body, the properties of the component of the angular momentum vector $ \vec {L} $ which is perpendicular to the fixed axis of rotation $ z $. I will call this component $ \vec {L_n } $. Suppose that the angular velocity $ \Omega $ is constant in direction but can vary in magnitude.

$ | \vec {L_ {n, i}} | = m_i r_i R_i \Omega cos \theta_i \implies | \vec {L_n} | = \Omega \sum m_i r_i R_i cos \theta_i$

Can I therefore say that $ | \vec {L_n} | \propto | \vec {\Omega} | $ (1)?

If so, suppose to apply a torque perpendicular to the $z$ axis and parallel to $ \vec {L_n} $, so that the magnitude of this vector increases. Follows from (1) that there should be an angular acceleration $ \vec {\alpha} $, although we are in the absence of a torque with an axial component.

This would go against the fact that $ I_z \vec {\alpha} = \vec { M_z}$ (Where $I_z $ is the moment of inertia with respect to the $z$ axis and $ M_z $ is the axial component of the exerted torque).

gravity - Does the potential energy for a given photon increase or decrease in quanta?

As a photon leaves a strong gravitational field, it loses energy and redshifts. Is the exchange in potential energy of a photon characterized by energy quanta?

Answer

No. A photon of a given frequency $f$ is exactly one quantum of energy. An electromagnetic wave has a total energy given by $E_\text{total} = \langle N\rangle hf$, where $\langle N\rangle$ is the number of photons in the wave.

When an EM wave escapes from a gravitational potential well (or falls into one), it's the energy of each individual photon that changes; in other words, the unit in which energy is quantized gets smaller or larger. The number of units (number of photons) stays the same.

quantum mechanics - Density of states of 3D harmonic oscillator

Consider the following passage, via this image:

5.3.1 Density of states

Almost all of the spin-polarized fermionic atoms that have been cooled to ultralow temperatures have been trapped by magnetic fields or focused laser beams. The confining potentials are generally 3D harmonic traps. So let's consider this case in more detail. You might be interested to note that Fermi's original paper on fermionic particles considered this case, not the 3D box case above. As we saw previously, ignoring the zero-point energy in each dimension the eigenvalues (accessible energy states) are given by $\epsilon(n_x, n_y, n_z)=n_x\hbar\omega_x + n_y\hbar\omega_y + n_z\hbar\omega_z$. In order to evaluate the various integrals, we first need to obtain the density of states per unit energy. A rough way to do this is to simply set $k_i=n_i$, so that $$\epsilon^2 = k_x^2(\hbar\omega_x)^2 + k_y^2(\hbar\omega_y)^2 + k_z(\hbar\omega_z)^2 \equiv k^2(\hbar \overline\omega)^2,$$ where $\overline \omega = (\omega_x\omega_y\omega_z)^{1/3}$ is the mean frequency, and $dk_i/\epsilon_i=1/\hbar\overline \omega$. Because $k_i=n_i$ now rather than $k_i=\pi n_i/L$, th 3D density of states is given by $$g(\epsilon) = \frac{k^2}{2} \frac{dk}{d\epsilon} = \frac{\epsilon^2}{2(\hbar\overline\omega)^3}.$$

for the first displayed equation,

shouldn't be $\epsilon^2 =\epsilon_{n_x}^2 +\epsilon_{n_y}^2 + \epsilon_{n_z}^2 + 2\epsilon_{n_x}\epsilon_{n_y} + 2 \epsilon_{n_x}\epsilon_{n_z} + 2\epsilon_{n_y}\epsilon_{n_z}$..?

if I assume $\omega_i=\omega$ for $i=x,y,z$

by

$\epsilon_{n_x}=\hbar \omega n_x $

$\epsilon_{n_y}=\hbar \omega n_y $

$\epsilon_{n_z}=\hbar \omega n_z $

$\epsilon_{n_x,n_y,n_z}=\hbar \omega(n_x +n_y +n_z)$

let $\vec{k}=(k_x,k_y,k_z)$ where $k_i=n_i$

$$\epsilon_{n_x}^2 +\epsilon_{n_y}^2 + \epsilon_{n_z}^2 = \hbar^2 \omega^2 (k_x^2 + k_y^2 +k_z^2 ) = \hbar^2 \omega^2 k^2 \not=\epsilon^2~?$$

And for second displayed equation, why it's not $$\frac{\pi k^2}{2} = \frac{1}{8}4\pi k^2~?$$

Answer

The author is working by analogy with the 3D box case. The 3D box worked out easily because $\epsilon\propto k^2$, so that the number of states up to each energy could use the volume of an ellipsoid, which is well known. He/she makes the volume of an ellipsoid work in a "rough way" for the harmonic oscillator. He/she puts a state where each 3D box state is (with $k_i=n_i$), and calls this $\epsilon^2$ to get the energies right along each axis.

Your suggestion of $4\pi k^2/8$ was correct for the 3D box, but here he/she is using the 3D box result and correcting using $k_i=\pi n_i/L \to k_i=n_i$.

One way of calculating the 3D density of states directly is to use integral 4.634 of Gradshteyn and Ryzhik (7th). I can supply more details if interested. Also see http://arxiv.org/abs/cond-mat/9608032.

electrostatics - Conductors and Uniqueness Theorem

I'm working with Griffiths Electrodynamics, and he introduces a uniqueness theorem:

First Uniqueness Theorem: The potential $V$ in a volume $\Omega$ is uniquely determined if (a) the charge density throughout the region, and (b) the value of $V$ on the boundary $\partial\Omega$, are specified.

I'm a little confused how Griffiths uses this theorem in examples: In the classic image problem (finding the potential due to a point charge $q$ a distance $d$ above an infinite, grounded conducting plane), the trick is to forge the original problem and the configuration of $q$ and its mirror image of the point charge $q$ through the plane (and this new charge is $-q$). The boundary conditions are given ($V=0$ on the plane, and $V\rightarrow 0$ far from the point charge), but how do we know that the charge distribution $\rho$ is the same in this second scenario as it is in the first?

Answer

This is a typical case of a problem which is clear enough physically speaking, but mathematically messy. Where rigorous results are folkloristically employed to achieve some result which, actually, would need much more care in deriving it... But presumably, mathematical details would not change the physical picture. Here the difference between theoretical physics and mathematical physics evidently shows up.

Actually, that uniqueness theorem is not properly used in the example you mention. As it stands in your initial statement, the uniqueness property holds when $\Omega$ is an open and bounded subset of $\mathbb R^n$ and $\varphi$ is continuous in $\overline{\Omega}= \Omega \cup \partial \Omega$ and it is $C^2(\Omega)$ satisfying Poisson's equation $\Delta \varphi = \rho$ in $\Omega$ itself.

(The proof of uniqueness is a trivial consequence of a celebrated theorem on harmonic functions $\phi$ in $\Omega$, i.e., functions verifying $\Delta \phi =0$ in $\Omega$, which are continuous in $\overline{\Omega}$. That theorem establishes that, if $\Omega$ is open and it closure is compact, $\max_{\overline{\Omega}} |\phi|$ is reached in a point of $\partial \Omega$. Thinking of $\phi$ as the difference of two solutions of Poisson's equation filling the same boundary conditions, the uniqueness property easily arises.)

When $\Omega$ is not bounded, as in the mentioned example, where $\Omega = \{(x,y,z) \in {\mathbb R}^3\:|\: z>0\}$, one must add further requirements on the behaviour of $\varphi$ for $||(x,y, z)|| \to +\infty$, and there are a number of possibilities.

However the mentioned example suffers for another problem. In the above mentioned uniqueness result, $\rho$ is continuous because $\Delta \varphi$ is. In the considered example instead $\rho$ is singular, properly speaking is a Dirac delta. There are several possibilities to deal with this problem. The simplest one is replacing the point charge with a given spherically symmetric distribution - with total charge $q$ and confined in a bounded spherical small region - and continuously vanishing at the boundary of that region. In the rest of my answer I assume it. Another possibility, technically more complicated is to remove the point occupied by the charge from $\Omega$. In this case the uniqueness result cannot be exploited as it stands because $\Omega$ acquires another part of boundary, where the potential diverges. Other approaches based on Green's identities instead of the maximum principle could be implemented in that case.

In this case in $\Omega$ the charge is $q$ (that is the $\rho$ distribution you mention) with distance $d$ from $\partial \Omega$ and $\varphi=0$ on $\partial \Omega$ because $\partial \Omega$ is a grounded conducting plane. The value of $\varphi$ on $\partial \Omega$ is constant, we are free to assume that it is $0$. The true boundary condition here is that $\varphi$ attains on $\partial \Omega$ the same value it attains for $||(x,y, z)|| \to +\infty$.

Let us pass to consider the situation where two charges stay at the reciprocal distance $2d$ along the $z$ axis, focussing on what happens in $\overline{\Omega}$ (not outside it) regarding charge distributions and boundary conditions of $\varphi$.

The $\rho$ distribution in the half space $\Omega = \{(x,y,z) \in {\mathbb R}^3\:|\: z>0\}$ is the same as in the previous case: There is the charge $q$ at distance $d$ from the plane at $z=0$.

Also the boundary conditions of $\varphi$ on $\partial \Omega$ and for $||(x,y, z)|| \to +\infty$ are the same as for the other case: The plane at $z=0$ is equipotential in view of the symmetry of the problem and the value of $\varphi$ thereon is the same as the value of $\varphi$ for $||(x,y, z)|| \to +\infty$.

Therefore, applying the uniqueness property in $\overline{\Omega}$, we are committed to conclude that the potential $\varphi$ in the region $\Omega \cup \partial \Omega$ is the same in both cases.

ADDENDUM. Actually one can use an argument arising from the theory of elliptic regularity to deal with uniqueness in situations where in a bounded region are present point charges described by Dirac deltas. The idea relies upon the following elliptic regularity result.

If $\phi$ is a distribution verifying $\Delta \phi =f$ (in weak sense) for a smooth ($C^\infty$) function $f$, then $\phi$ is a $C^\infty$ function up to zero measure set.

(It is worth stressing the the above result immediately entails the fantastic fact that harmonic functions are always $C^\infty$ and not only $C^2$, actually it is possible to prove that they are real analytic.) This result leads to the following uniqueness theorem which can be improved making weaker some hypotheses on the behaviour of the function on the "regular" boundary.

Theorem. Suppose $\Omega \subset \mathbb R^n$ is non-empty open and $\overline{\Omega}$ is compact. Let $p \in \Omega$ and consider the problem: $$\Delta \varphi(x) =\rho \quad x \in \Omega \setminus \{p\}$$ with boundary conditions $$\varphi|_{\partial \Omega} = f$$ where $$\varphi \in C^2(\Omega \setminus \{p\}) \cap C^0(\partial \Omega \cup \Omega \setminus \{p\} )$$ and $f \in C^0(\partial \Omega)$ and $\rho \in C^0(\Omega \setminus \{p\})$ are assigned. If both $\varphi_1$ and $\varphi_2$ are solutions of the problem and $$ \lim_{x\to p} (\varphi_1(x)- \varphi_2(x))=0\:,$$ (where the limits can diverge or not exist, if considered separately in order to embody the case of a point charge at $q$) then $$\varphi_1 = \varphi_2\:.$$

PROOF. With the given hypotheses, evidently $\phi:= \varphi_1-\varphi_2$ is continuous on $\Omega$, therefore it is a distribution for test functions, $h\in C_0^\infty (\Omega)$. If $B_\epsilon$ is a small ball around $p$ with radius $\epsilon$, using continuity of $\phi$ in particular, integrating by parts and defining $\Omega_\epsilon := \Omega \setminus B_\epsilon$, we have $$\int_\Omega \phi \Delta h d^nx = \lim_{\epsilon \to 0^+}\int_{\Omega_\epsilon} \phi \Delta h d^nx = \lim_{\epsilon \to 0^+} \int_{\Omega_\epsilon} (\Delta \phi) h d^nx = \lim_{\epsilon \to 0^+} \int_{\Omega_\epsilon}(\rho-\rho) h d^nx$$ $$= \lim_{\epsilon \to 0^+} 0 = 0\:.$$ All that means that $\phi$ is a distribution solving $\Delta \phi =0$ in distributional sense. Therefore, in view of the mentioned elliptic regularity property, it is a smooth function up to a zero measure set. Since $\phi$ is continuous on $\Omega \setminus \{p\}$ and extends to a continuous function at $p$ (which has zero measure), $\phi = \varphi_1-\varphi_2$ is a smooth function everywhere in $\Omega$. In particular, by continuity of second derivatives the smoothly extended function $\phi$ verifies $\Delta \phi =0$ in the whole set $\Omega$ in the proper sense. By construction, we end up with a function $\phi$ which is $C^\infty(\Omega) \cup C^0(\overline{\Omega})$, satisfying $\Delta \phi =0$ in $\Omega$ and $\phi =0$ on $\partial \Omega$. In view of the standard uniqueness result, $\phi=0$ in $\overline{\Omega}$, i.e. $\varphi_1= \varphi_2$ in $\overline{\Omega}$. QED

general relativity - Why is light described by a null geodesic?

I'm trying to wrap my head around how geodesics describe trajectories at the moment.

I get that for events to be causally connected, they must be connected by a timelike curve, so free objects must move along a timelike geodesic. And a timelike geodesic can be defined as a geodesic that lies within the light cone.

I want to know why exactly null geodesics define the light cone. Or, why null geodesics define the path of light.

Also, if there's a better explanation why matter follows timelike geodesics, that would also be welcome.

Answer

Even in curved spacetime, you can perform a coordinate transformation at any location ("move to a freely falling frame") such that your metric is locally flat , and takes the form \begin{equation} ds^2 = -c^2 dt^2 + dx^2 + dy^2 + dz^2\end{equation}

If you consider a null trajectory where $ds^2 = 0$, then the above equation takes the form

\begin{equation} cdt = \sqrt{dx^2 + dy^2 + dz^2}. \end{equation}

This is the statement that "the speed of light times the differential time interval, as measured by an observer in a freely falling frame at the location in consideration, is equal to the differential physical distance traveled along the trajectory, measured by that same observer." From Einstein's equivalence principle, this is precisely the way that light must behave.

quantum information - Bell polytopes with nontrivial symmetries

Take $N$ parties, each of which receives an input $s_i \in {1, \dots, m_i}$ and produces an output $r_i \in {1, \dots, v_i}$, possibly in a nondeterministic manner. We are interested in joint conditional probabilities of the form $p(r_1r_2\dots r_N|s_1s_2\dots s_N)$. Bell polytope is the polytope spanned by the probability distributions of the form $p(r_1r_2\dots r_N|s_1s_2\dots s_N) = \delta_{r_1, r_{1, s_1}}\dots\delta_{r_N, r_{N, s_N}}$ for all possible choices of numbers $r_{i,s_i}$ (in other words, each input $s_i$ produces a result $r_{i,s_i}$ either with probability 0 or 1, regardless of other players' inputs). Polytopes of this kind are of interest in quantum information theory.

Every Bell polytope has a certain amount of trivial symmetries, like permutation of parties or relabelling of inputs or outputs. Is it possible to give an explicit Bell polytope with nontrivial symmetries? (e.g. transformations of the polytope into itself that takes faces to faces and is not trivial in the above sense) In other words, I'm interested whether a specific Bell scenario can possess any "hidden" symmetries

Bell polytopes in literature are usually characterized by their faces, given by sets of inequalities (Bell inequalities), which, however, usually do not have any manifest symmetry group.

quantum mechanics - Why is the dimension of the set separable states $dimmathcal H_1+dimmathcal H_2$?

Please can you help me to understand how the dimension of the set of separable states is $\dim \cal H_1 + \dim \cal H_2$?

This is the relevant passage:

So far, we have assumed implicitly that the system is made of a single component. Suppose a system is made of two components; one lives in a Hilbert space $\cal H_1$ and the other in another Hilbert space $\cal H_2$. A system composed of two separate components is called bipartite. Then the system as a whole lives in a Hilbert space $\cal H = \cal H_1 \otimes \cal H_2$, whose general vector is written as $$\left|\, \psi \right\rangle = \sum_{i,j} c_{ij} \left|\,e_{1,i}\right \rangle \otimes \left|\,e_{2,j}\right\rangle, \tag{2.29}$$ where $\{|\,e_{a,i}\rangle\}$ ($a=1,2$) is an orthonormal basis in $\cal H_a$ and $\sum_{i,j} |c_{ij}|^2 = 1$.

A state $|\,\psi \rangle \in \cal H$ written as a tensor product of two vectors as $|\,\psi \rangle = |\,\psi_1 \rangle \otimes |\,\psi_2\rangle$, ($|\,\psi_a\rangle \in \cal H_a$) is called a separable state or a tensor product state. A separable state admits a classical interpretation such as “The first system is in the state $|\,\psi_1\rangle$, while the second system is in $|\,\psi_2\rangle$.” It is clear that the set of separable states has dimension $\dim \cal H_1 + \dim \cal H_2$.

Answer

Note that the space of separable states is not a vector space, and in particular not a subspace of the total Hilbert space: the sum of two separable states is unlikely to be separable. So dimension here means something more general than vector space dimension.

Having said that, I would disagree with the author on his dimension! I would say that the space of (nonzero) separable states has dimension $\dim \mathcal{H_1}+\dim\mathcal{H_2}-1$.

To specify a separable state, we can supply an element of each of $\mathcal{H_1}$ and $\mathcal{H_2}$, which means $\dim \mathcal{H_1}+\dim\mathcal{H_2}$ complex numbers. However, there is a redundancy here, because we can change each by an overall scaling ($|\psi_1\rangle\mapsto\lambda|\psi_1\rangle, |\psi_2\rangle\mapsto\lambda^{-1}|\psi_2\rangle$) without changing the product state, which reduces the dimension by 1.

A couple of simple examples:

1) If $\mathcal{H_1}$ is 1-dimensional (completely trivial!), then all states are separable, and $\mathcal{H_1}\otimes\mathcal{H_2}\simeq\mathcal{H_2}$.

2) If both $\mathcal{H_1}$ and $\mathcal{H_2}$ are two-dimensional, we can write a state of $\mathcal{H_1}\otimes\mathcal{H_2}$ as a 2x2 matrix. The separable states have proportional columns/rows, so are exactly the same as matrices of determinant zero. If we exclude 0, this is a 3-dimensional submanifold.

electricity - Can we generate electric current by injecting electrons into a copper panel

Suppose I take a vacuum tube and accelerate electrons in it by electric fields then collide it on a copper plate.

Will the electrons then go inside it and generate an electric current by putting a positively charged panel on the opposite end to create a potential difference?

If yes how can I extract the electrons from the positive plate back into the vacuum tube?

Thanks! Also you can see a diagram I made for what I am trying to say,

Answer

It will help if you study this diagram of what a vacuum tube is

If a cathode is heated, it is found that electrons from the cathode become increasingly active and as the temperature increases they can actually leave the cathode and enter the surrounding space.

When an electron leaves the cathode it leaves behind a positive charge, equal but opposite to that of the electron. In fact there are many millions of electrons leaving the cathode. As unlike charges attract, this means that there is a force pulling the electrons back to the cathode. Unless there are any further influences the electrons would stay in the vicinity of the cathode, leaving the cathode as a result of the energy given to them as a result of the temperature, but being pulled back by the positive charge on the cathode.

The electrons are produce by heating a filament. Electrons are bound in atoms and even facing a vacuum stay there, unless kicked out, which is what heating a filament does. If you place the tube in an electric circuit, if the filament is not heated, no current will flow through because there is nothing to conduct electricity in the vacuum. Extracting electrons from the filament allows a current to form and close the circuit when they hit the other plate. In this form to get electrons again out of the second plate one would have to somehow heat the second plate.

Better study a bit the link given above.

Planetary-sized pure quantum states

Picture a planet wandering intergalactic space. Such a planet would only couple to vacuum flucuations and the cosmic microwave background. (Ignore stray Hydrogen atoms.)

If this planet started as a pure quantum state, how fast would that state lose its coherence?

In such a system, clearly there are many more degrees of freedom that are isolated from the environment compared with those coupled to the outside. So I want to know if those isolated DOF somehow protect the purity of the quantum state.

Answer

I am just going to quote Schlosshauer as being pertinent to this question and discussion in comments.

Reference: Decoherence and the Quantum-to-Classical transition (page 84):

To summarize, we have distinguished three different cases for the type of preferred pointer states emerging from interactions with the environment:

1. The quantum-measurement limit. When the evolution of the system is dominated by $H_{int}$, i.e. by the interaction with the environment, the preferred states will be eigenstates of $H_{int}$ (and thus often eigenstates of position).


2. The quantum limit of decoherence. When the environment is slow and the self-Hamiltonian $H_S$ dominates the evolution of the system, a case frequently encountered in the microscopic domain, the preferred states will be energy eigenstates, i.e., eigenstates of $H_S$


3. The intermediary regime. When the evolution of the system is governed by $H_{int}$ and $H_S$ in roughly equal strengths, the resulting preferred states will represent a compromise between the first two cases. For instance in quantum Brownian motion the interaction Hamiltonian $H_{int}$ describes monitoring of the position of the system. However, through the intrinsic dynamics induced by $H_S$ this monitoring also leads to indirect decoherence in momentum. This combined influence of $H_{int}$ and $H_S$ results in the emergence of preferred states localized in phase space, i.e. in both position and momentum.

electricity - Dependent and Independent Variables in an Electrical Experiment

So, for my science fair I had to test the gauge of a copper wire vs electrical resistance. What I did was I set up a series circuit involving a 6v battery, a light bulb, a multimeter, and a voltmeter. I would change out three different wires with three different gauges.

• For the 4-gauge wire, I got 0.004V and 0.04A.


• For the 8-gauge wire, I got 0.003V and 0.04A


• For the 12-gauge wire, I got 0.001V and 0.04A.

In this experiment, what would the independent and dependent variables?

We also have to fill out a table like this:

Answer

A dependent variable is one that changes as a result of another change in your experiment. It is a variable you measure in order to find out what happened when something in your experiment changed. The measured value depends on other parts of the experiment.

An independent variable is one that you are free to modify. It is set before the experiment begins and doesn't change until the next experiment. This value is chosen before the experiment and so is independent of other parts of the experiment.

In your experiment, you chose the gauge of the wire before the experiment began and measured the current and voltage after turning on the circuit. The gauge of the wire does not change during a measurement, so it is the independent variable. The current and voltage you measure depends on the wire gauge, so these are the dependent variables.

vacuum - Why do scientists think Higgs fields is metastable?

I have recently learn and read some articles about the false vacuum in quantum fields theory.
I wanted to know why scientists think that Higgs field could be in a metastable state instead of vacuum state like the others fields.

Can anyone explain me why ?

Vacuum state are when fields or particles have their lowest energy level possible
Metastable state are when the fields or particles reach a lower level but not the lowest one.
Am i right ?
Thank in advance, Creekorful

Answer

I have recently learn and read some articles about the false vacuum in quantum fields theory.

Here is a definition:

In quantum field theory, a false vacuum is a metastable sector of space that appears to be a perturbative vacuum, but is unstable due to instanton effects that may tunnel to a lower energy state

A scalar field φ in a false vacuum. Note that the energy E is higher than that in the true vacuum or ground state, but there is a barrier preventing the field from classically rolling down to the true vacuum. Therefore, the transition to the true vacuum must be stimulated by the creation of high-energy particles or through quantum-mechanical tunneling.

The existence of a false vacuum is a hypothesis that some theories may use, but it is not the Higgs mechanism.

I wanted to know why scientists think that Higgs field could be in a metastable state instead of vacuum state.

The Higgs field is not metastable in the sense of a false vacuum. It is dependent on the energy available and is symmetric for high enegies of the interactions involved ( order of 100 GeV) , and broken for lower energies.

An example of symmetry breaking analogous to what happens with the Higgs field:

Spontaneous symmetry breaking simplified: – At high energy levels (left) the ball settles in the center, and the result is symmetrical. At lower energy levels (right), the overall "rules" remain symmetrical, but the "Mexican hat" potential comes into effect: "local" symmetry is inevitably broken since eventually the ball must roll one way (at random) and not another.

At each energy the vacuum is stable, it is not a false vacuum. The difference between the vacuum for the other fields , like the electron field and the photon field, is in the vacuum expectation value.

In quantum field theory the vacuum expectation value (also called condensate or simply VEV) of an operator is its average, expected value in the vacuum.

For all other particles the vacuum expectation value in the standard model is zero, for the Higgs it is 246GeV, at the broken state .

The conditions for symmetry existed at the beginning of the universe, according to the Big Bang model. The symmetry is broken at about 10^-10 seconds in the life of the universe (temperatures correspond to 100 GeV) and the Higgs field acquires the large vev.

cosmology - Does the universe have a center?

If the big bang was the birth of everything, and the big bang was an event in the sense that it had a location and a time (time 0), wouldn't that mean that our universe has a center?

Where was the big bang? What is there now? Are we moving away from the center still? Are these even valid questions?

Answer

The big bang was everywhere, because distance didn't exist before it, so from one perspective, everywhere may be the centre (especially as some theorists think, the universe doesn't have an edge)

The real issue is that the question shouldn't matter, as we can only gain information from distances within our visible radius and once we get to that limit, what we see gets closer to the big bang so it all looks closer to the centre.

Tricky eh

homework and exercises - Ascertain the height an object has fallen from given force exerted and mass

An object of a given mass falls from an unknown height. If the force exerted by the object on contact with the ground is known, how would you ascertain the height from which the object fell?

Answer

The answer is a definite maybe!

Ignoring air resistance the velocity a falling object hits the ground can be calculated using the appropriate SUVAT equation or by equating potential energy lost with kinetic energy gained. However this only tells you the speed the object hits the ground, and how hard it hits the ground depends on how fast it decelerates.

If you can measure the force as a function of time during the collision you can calculate the total impulse, and since this is equal to the momentum change you can calculate the original momentum and hence the original velocity. Just calculating the peak force doesn't help because it could be a hard slow moving object decelerating suddenly or a soft fast moving object decelerating slowly.

I haven't gone into specifics, since this looks rather like a homework question, but this should direct you to the physics you need to answer your problem.

Quantum Entanglement Simulation

Suppose there is a digital "quantum entanglement simulator" with a pair generator and two spin detectors, and it

1. Allows to create as many pairs as you want (100 pairs per single click)


2. Allows to measure spin at the detectors in any combination of angles


3. Allows you to set angle combinations randomly, at any time

My understanding is - it must demonstrate below correlations in order to qualify as a successful digital simulator for entanglement -

1. Anti-correlation, when particles of any pairs are measured in same angle, they always have opposite spin



2. Statistical correlation-1 - If you measure numerous particles at a detector, in a specific angle, the outcome must be ~50% up, ~50% down


3. Statistical correlation-2 - If you measure numerous pairs at two detector, in two angles A and B, the correlation (both up, or both down) should be SQUARE of sin((A-B)/2) times the total number of pairs measured at this angle combination.

My question is - Is there any other type of correlation that needs to be implemented?

If so, what is that correlation?

If no other correlation is required then

Follow-up question is - I have created an online simulator that is functional, and been tested for all three types of correlations at various angle combinations. May I ask people here to try to find flaws with the statistical correlation exhibited by the simulator.

I am not asking to test the simulator (which I have done myself), I mean to verify the correlations.

FYI - The simulator page is pretty self explanatory.

You can provide dummy information as it does not validate the email address.

Just remember the email address you entered, and password for login purpose.

Here is the simulator Does not work on mobile devices.

...

quantum mechanics - Does a photon instantaneously gain $c$ speed when emitted from an electron?

An excited electron looses energy in the form of radiations. The radiation constitutes photons which move at a speed $c$. But, is the process of conversion of the energy of the electron into the kinetic energy of the photon instantaneous. Is there a a simple way to visualize this process rather than math?

quantum mechanics - Triangle inequality Clebsch-Gordan coeffcients

The Clebsch-Gordan coefficients can only be non-zero if the triangle inequality holds: $$\vert j_1-j_2 \vert \le j \le j_1+j_2$$ In my syllabus they give the following proof: $$-j \le m \le j$$ $$-j_1 \le m_1 \le j_1$$ and $$-j_2 \le m_2 \le j_2$$

When $m$ takes its maximal value, $m = j$, $m_1 = j_1$ and $m_2 = j_2$, and we get:

1) $-j_1 \le j-j_2 \le j_1$ which implies $j_2-j_1 \le j \le j_1+j_2$

2) $-j_2 \le j-j_1 \le j_2$ which implies $j_1-j_2 \le j \le j_1+j_2$

which should prove the triangle inequality.

This proof looks really simple, but I don't completely understand it though. It seems that I'm missing some essential reasoning, and I can't find where. Why for instance do they take for $m_1$, $m_2$ and $m$ all maximal values? Can't I also take $m$ maximal and $m_1$ minimal? This would give bad results though. So I really don't understand it, and I hope that someone can clarify it.

mass - Massive Gauge Bosons without Higgs fields

In a possible theory like our Standard model but without a Higgs i.e.:

$$ \mathcal{L}=i\bar{\Psi}_f\gamma_\mu D^\mu\Psi_f-\text{Tr}[G^b_{\mu\nu}G^{b\,\mu\nu}] $$

where $b,f$ run over the typical species we have in the standard model (SM), and all fields are in the same representation as in the SM.

In this context it is sometimes stated that, although there is no Higgs, there would be a mass generation mechanism for the gauge bosons of $SU(2)$ because of QCD. This happens via the chiral quark condensate $\langle q_L q_R\rangle\neq 0$. (Or statements like "the gauge bosons eat up the pion")

My question is now, how can I see that this generates a mass for the $SU(2)$-gauge bosons? Usually using methods of spontaneous symmetry breaking, I would put a vacuum expectation value for some field and see that it results in a term that behaves like a mass term. But this won't work here because there is no term involving quarks and bilinear in gauge bosons.

quantum mechanics - Is Stephen Wolfram's NKS, an attempt to explain the universe with cellular automata, in conflict with Bell's Theorem?

Stephen Wolfram's A New Kind of Science (NKS) hit the bookstores in 2002 with maximum hype. His thesis is that the laws of physics can be generated by various cellular automata--simple programs producing complexity. Occasionally (meaning rarely) I look at the NKS blog and look for any new applications. I see nothing I consider meaningful. Is anyone aware of any advances in any physics theory resulting from NKS? While CA are both interesting and fun (John Conway, Game of Life), as a theory of everything, I see problems. The generator rules are deterministic, and they are local in that each cell state depends on its immediate neighbors. So NKS is a local deterministic model of reality. Bell has shown that this cannot be. Can anyone conversant with CA comment?

Answer

Wolfram's early work on cellular automata (CAs) has been useful in some didactical ways. The 1D CAs defined by Wolfram can be seen as minimalistic models for systems with many degrees of freedom and a thermodynamic limit. Insofar these CAs are based on a mixing discrete local dynamics, deterministic chaos results.

Apart from these didactical achievements, Wolfram's work on CAs has not resulted in anything tangible. This statement can be extended to a much broader group of CAs, and even holds for lattice gas automata (LGAs), dedicated CAs for hydrodynamic simulations. LGAs have never delivered on their initial promise of providing a method to simulate turbulence. A derivative system (Lattice Boltzmann - not a CA) has some applications in flow simulation.

It is against this background that NKS was released with much fanfare. Not surprisingly, reception by the scientific community has been negative. The book contains no new results (the result that the 'rule 110 CA' is Turing complete was proven years earlier by Wolfram's research assistant Matthew Cook), and has had zero impact on other fields of physics. I recently saw a pile of NKS copies for sale for less than $ 10 in my local Half Price Books store.

cosmological inflation - the nature of the big bang

If space-time expanded together with matter then why do physicist bother extrapolating backwards the expansion back to a point in time? I mean does that really tell us anything? I mean if the speed of the passage of time (measured by atomic vibrations) is dependant on the local space-time geometry, and the measurement of space is dependant on having 'stuff' (light, matter) in it to measure the distances, then it seems to me that the concept of space, time, its interacting fields, and the resulting quantum actualities are all dependant on each other for their survival as a concept.

What explanatory power have we really added to the discussion on the origins of reality if we talk about a finite time periods back to the infinite? How can a finite time period escape from the infinite? Therefore what are we really learning by saying the universe is 14 billion years old? In my ignorance, its like we are applying logic based on newtonian physics to post Einstein physics and walking away happy that we said something sensible when we just got ourselves (and me especially ;) really confused (I'm sure the reality is different, thats why I'm asking the question :)).

Furthermore, if everything (within the fabric of nothing) expanded together then how could we prove anything expanded at all? The red-shift must be showing that matter expanded FASTER than the space-time right? Otherwise there would be no measurable difference. How could we ever measure that if space-time is expanding with it?

I guess what I'm asking is 3 questions (that are really all the same question); why do we talk about space-time expanding when "expansion" is a term that needs space to make sense. Wouldn't the permission to use that term require our space-time existing in a greater spacial reality?

Why do we talk about matter expanding when if the space-time expanded with it then it really didn't expand at all.

Why do we talk about a finite time period when the maths shows the passage of time retreats to infinity and therefore we haven't really explained anything.

Answer

We started from observations: galaxies and clusters of galaxies have been found to be retreating from our galaxy and each other.

Hubble's law is the name for the observation in physical cosmology that: (1) objects observed in deep space (extragalactic space, ~10 megaparsecs or more) are found to have a Doppler shift interpretable as relative velocity away from the Earth; and (2) that this Doppler-shift-measured velocity, of various galaxies receding from the Earth, is approximately proportional to their distance from the Earth for galaxies up to a few hundred megaparsecs away.1 This is normally interpreted as a direct, physical observation of the expansion of the spatial volume of the observable universe.

Once we model the space around us expanding, the way a loaf of bread expands when rising in the oven, then the question of matter also expanding is answered by the same observation: if it were, we would not be able to detect the expansion because our units would be changing in tandem.

Since we do see expansion, lets stretch the bread analogue a bit further> Let us put half walnuts in the bread. What will happen is the walnuts will be expanding with the bread, but their size will not change. Why? Because they are solid, held together with molecular forces and while the dough is elastic and expands distancing one walnut from the other the intermolecular forces keep the walnut intact.

In a similar way the collective gravitational Newtonian force which holds for local areas of the universe is much stronger than the current expansion of space, and the electromagnetic and strong forces are even stronger. Thus matter holds together. Radiation emitted and traveling through expanding space is affected and that is why we can deduce the velocities of galaxies and clusters receding from us.

It was not always so . We have made a pictorial representation of this expansion with the model of the Big Bang:

We have had for some time data of the Cosmic Microwave Background radiation (CMB), a snapshot of what happened 380.000 years from the singularity. Recently the BICEP2 experiment claims to give us a snapshot of what happened at 10^-32 seconds from the singularity, the effect of quantization of gravity. Our model of the history of the colored region, before the CMB was released, depends on our knowledge of interactions from particle physics and the standard model that fits the particle data very well, and also data from nuclear physics.

This is a model that blends what we know of theory and experiment and observational astronomy, it is the state of the art at the moment.

Is there really a singularity at the beginning of time? We do not know. The general relativity solution says so, but it is not a quantized solution. Quantization is necessary during the inflationary period in the beginning, to explain the uniformity of distribution of matter in the observed universe. Most physicists expect that once quantization of general relativity is formulated correctly singularities may disappear, but we do not know. At the moment it looks as if there existed a singularity at the beginning of space time, one can fit all observations with such a model.

newtonian mechanics - What are the properties of an 'ideal' pendulum?

This question mentions a phrase ideal pendulum.

What are the properties of an ideal pendulum(if there is any such term)?

Internet search tells me that there are two types of pendulum:Simple and compound.

Answer

An ideal pendulum, as in the question you refer to, is one without friction, air resistance, a point-mass bob - i.e. the bob is not a real massive object but just a point, etc. It's defined that way to make the maths simpler. Real, actual pendulums only behave in approximately the same way.

The ideal pendulum can, however, have any length or bob-weight, so it does not have any special proportions.

Two-point function of massless scalar theory in 2d CFT

Following the derivation of the massless free-boson two-point function given in Di Francesco, Mathieu and Sènèchal, I had an apparently stupid doubt. Look at the attached picture.

Where does the contribution $\lim_{\rho \rightarrow 0} \rho K'(\rho)$to the integral in Eq. (2.100) go? Do they suppose it is zero? In this case Eq.(2.101) is not consistent with this requirement.

The only possible explanation I found is that the behaviour in $\rho\sim 0$ is not well-defined but I am not satified.

Any ideas??

Answer

This term is not there because you are integrating over a disk, and the disk has only one boundary at $r$, and no boundary at $0$. Here is how to see it more explicitly. Note that $$ \int_D -g\partial^2K(x,0)d^2x = -g\int_{\partial D}\nabla K(x,0) \centerdot dS = -2\pi g r K'(r). $$ Then we see that if $K(r)$ has a $\log$-singularity near $r=0$, then $\partial^2K$ reproduces the delta-function, while $m^2 K$ is integrable and thus does not contribute to the delta-function. Overall, we conclude $$ 1=\int_D g(-\partial^2+m^2)K(x,0)d^2x = -g\int_{\partial D}\nabla K(x,0) \centerdot dS + \int_D g m^2 K(x,0)d^2x= 2\pi g\left\{ -r K'(r)+m^2\int_0^r d\rho\,\rho K(\rho)\right\}. $$

fluid dynamics - How does the size of the hole affect the exit velocity?

I took water bottles and drilled several different sizes of holes on the bottom. During the experiment, I found that the water bottle with a smaller hole takes a lot longer time to leak than those with larger holes.

Does this mean that the exit velocity of the liquid (water) decrease as the size of hole decreases due to the viscosity of water? And should the exit velocity be directly proportional to the area of the hole?

newtonian gravity - Approximations in elliptical orbits

I am learning about orbits and getting very confused as to what is exactly true and what is an approximation.

Namely the following points:

• It is often said that the planet has an elliptical orbit with the sun at one focus. I think this is an approximation in assuming that the mass of the sun is so much greater than that of the planet that its motion is negligible compared with the planet, but in fact both have an elliptical orbit about their common center of mass (as in this SE post).

But the derivation we used in our classes for the elliptical motion of a planet is about the sun, and not for both bodies about a common center of mass. The other derivations I have seen also show this orbit about the sun. The derivations all initially assume the sun as a center, and use the gravitational potential around the sun. I am guessing that the approximation comes in because, if you assume the sun at one focus, then the kinetic energy is not simply $0.5mv^2$ because the sun is actually a non inertial reference frame. Is this correct?

Finally, is angular momentum really conserved if we place the sun at a new focus? Why/why not? Linear momentum is clearly not conserved. It would be if we considered the whole system (the planet and the sun). So is linear momentum conserved?

Answer

Ignoring that the derivation assumes a two body problem and Newtonian gravity, there's no approximation here. With these assumptions, a bound orbit of one body about another is an ellipse, with either body viewed as being fixed, and the fixed body being one of the foci of the ellipse.

This is of course a non-inertial perspective. The fixed body is accelerating toward the orbiting body. From the perspective of an inertial frame in which the center of mass is fixed rather than one of the bodies, both bodies are in elliptical orbits about the center of mass, with the center of mass being the common focus of the two ellipses.

To show that these are indeed equivalent, suppose we know that object B is orbiting the center of mass of a two-body system in an ellipse, with the center of mass at one of the two foci of the ellipse. This means that the polar coordinates of object B with the origin at the center of mass is $$\vec r_B = \frac {a_B (1-e^2)}{1+e \cos\theta} \,\,\hat r$$ where $a_B$ is the semi-major axis length of the orbit, $e$ is the eccentricity of the orbit, and $\theta$ is the angle on the orbital plane subtended by B's closest approach to the center of mass, the center of mass, and object B itself.

What about the other object? It's position in this center of mass system is constrained by $m_A \vec r_A = -m_B \vec r_B$. Thus it too moves in an ellipse with the center of mass as one of the foci: $$\vec r_A = \frac {a_A (1-e^2)}{1+e \cos\theta}(-\hat r)$$ where $a_A = \frac{m_B}{m_A} a_B$.

Finally, what about the displacement vector between the two? This is $$\vec r_B - \vec r_A = \left(1 + \frac {m_B}{m_A}\right) \frac {a_B(1-e^2)}{1+e\cos\theta} \,\, \hat r \equiv \frac {a(1-e^2)}{1+e\cos\theta}\,\,\hat r$$ where $a \equiv \frac{m_A+m_B}{m_A}a_B = \frac{m_A+m_B}{m_B}a_A$.

This is of course yet another ellipse. It can be looked at in two different ways: From the perspective of object $B$ orbiting a fixed object $A$, in which case object $A$ is at one of the foci of this ellipse, or from the perspective of object $A$ orbiting a fixed object $B$, in which case object $B$ is at one of the foci of this ellipse.

Why does bright pure red/green/blue light look white if bright enough?

I believe that white light is white because it contains every wavelength, as white by itself has no frequency by itself.

Can somebody explain why really bright pure red, green, or blue LEDs seem white to us and to cameras?

Also, can a bright light ever be pure(eg is it possible to emit a pure red/pure blue/pure green bright light only with wavelengths in the range of that light), or will the inherent brightness of a light force other wavelengths into the light?

Answer

The photoreceptors of the eye respond a little bit to light frequencies different from their peak sensitivity. If you look at the spectral sensitivity of the eye you will see that the response for the red and green receptors cover most of the visible spectrum (blue is a bit narrow). Camera sensors will be somewhat similar but with different curves and sensitivity.

Image from Wikipedia

This means that if you have a sufficiently strong light at some frequency, you will activate the receptors maximally. It might seem that you would get far more response from the most fitting receptor, but neuron firing rate has a ceiling: beyond a certain point they just fire maximally. Same thing for camera sensors, that saturate. At this point you just get the same output from all of them.

Note that this model predicts that sufficiently deep blue light might not look white to the human eye.

Super-intense spectrally pure light is essentially a description of a laser. There is nothing inherent in light that would force it to spread out spectrally. In a material such as air there will be effects due to light-matter interactions - it might scatter from atoms and particles producing a line broadening doppler effects, it might ionize atoms so they absorb and re-radiate light as spectral lines, or heat up matter to radiate broadband blackbody radiation.

condensed matter - Transition between 2D and 3D quantum systems

Quantum Hall effect and anyonic particles are examples that occur in a two-dimensional system. However, experiments for such systems can only be realized in a pseudo-2D environment, where the third spatial dimension is much smaller than the other two dimensions. How do we expect the results from such experiments to differ from a true 2D system? In particular, how/when does an anyon begin or cease to exist when we transit between a 2D and 3D system?

Answer

The results from the experiment does not differ significantly from the "true 2D" system, in fact, this is why experiments and theory agree so well!

Consider a semiconductor heterostructure GaAs/GaAsAl. At the interface, alignement of the Fermi level at both sides of the semiconductor crystals creates a triangular potential well at the GaAs side of the interface. This potential well is quite narrow so that quantization in one direction can be effectively assumed. In fact, putting numbers for GaAs, $m^{\ast}=0.067m_e$, $n_{2\text{D}} \simeq 10^{15}m^{-2}$, $\epsilon^\ast=13\epsilon_0$ you get that the typical quantization energy is $\Delta E \simeq 20 $meV.

Due to the triangular well, the $3\text{D}$ electron wavefunction [consider nearly-free electrons in the effective mass approximation] is modified as

$\Psi_{k_x,k_y,n\sigma}(\mathbf{r})=\dfrac{1}{A^{1/2}}{\rm e}^{i k_x x}{\rm e}^{i k_y y}\zeta_{n}(z)\chi_\sigma$

with $\zeta_n(z)$ the $n$-th eigenfunction of the triangular well with energy $\varepsilon_n^z$. THe total energy can be written as

$\varepsilon_{k_x,k_y,n} = \dfrac{\hbar^2}{2m^\ast}(k_x^2+k_y^2) + \varepsilon_n^z$

The Fermi energy can be obtained using that $k_F^2=2 \pi n$, $\varepsilon_F\simeq 10$ meV.

The difference between the highest occupied energy is $\Delta E -\varepsilon_F \simeq 10$ meV. This yields $T \simeq 100$ K.

Conclusion A quick conclusion of this calculation is the following: at temperatures $T \ll 100$K all the occupied electron states have the same orbital in the $z$ direction and promotion to other orbital requires an excitation energy of at least $10$ meV. If this is not provided, the system has indeed lost one degree of freedom and it is dynamically a true $2\text{D}$ system. Thus $2\text{D}$ systems can exist in Nature!

Concerning the second question, anyon statistics exists only in two-dimensional systems. It cannot exist in $3\text{D}$, and the reason is topological: in two dimensions the configuration space of $N$ particles is multiple connected and closed path of a particle which encloses another particle cannot be "shrinked" to a point [mathematically this is called "compatification"]. On the other hand, for higher dimensions, the configuration space is simply connected and we lose the possibility of distinguish between the interior and the exterior of a closed path.Hence, in the transition from $2\text{D}$ to $3\text{D}$ you simply lose the possibility of interpolate between Bose and Fermi statistics.

standard model - Can quarks be considered real and elementary?

In our current theories all hadrons are made up of quarks and gluons.

This view reduces considerably the big family of hadrons by providing a very logical structure in which all quantum properties values of the hadrons are originated from the valence quarks' quantum properties. But since they cannot be observed isolated, in the sense that electrons or positrons can be, the theory assumes color confinement is just part of the game.

However actual calculations of QCD use a view where they are quarks and gluons are distributed in a space-time volume, which is needed to describe much of the rich phenomena of high energy physics.

But all of this makes me wonder:

Are quarks real and elementary? I mean, is there more evidence supporting that they are more than just a model that works?

Since hadronic matter cannot by split into pieces smaller than the smallest hadron, couldn't reality be explained also by a model where all observable hadrons are the elementary ones, which follow rules for transforming into other ones, etc.?

By the way, this last idea was already presented by Hagedorn on his very famous known paper, in the end where he tries to give a phylosofical point of view to his model, so I wonder why was it discarded.

Answer

As mentioned in the OP quarks vastly simplify the theory of hadrons, like atoms did chemistry, and despite confinement Rutherford-like experiments were performed for them too, by Friedman, Kendall and Taylor who received the Nobel prize for it in 1990: "unexpectedly large numbers of electrons being scattered at large angles provided clear evidence for the pointlike constituents within nucleons. These constituents are now understood to be quarks."

But can quarks still be considered non-existent? Technically, yes. This is the answer Mach gave about atoms in the 19th century: they are just fictions, and the theory can be re-arranged in a way that eliminates them, e.g. by connecting only measurable quantities to each other, and purely mathematically (Mach and a leading chemist Ostwald even refused to mention atoms in their works after 1870). This remained true even after Rutherford's experiments, and it remains true today despite the technology that (ostensibly) "allows them to be imaged, split and smashed". Indeed, one can even eliminate everyday objects and reduce everything to sensations, as some positivists suggested. But doing so will make for a very unattractive theory.

Of course, it can also go the other way: at the end of 19th century the ether was a solid element of reality. Some even expected a theory of everything out of it, like Michelson in 1902, see Kragh's Quantum Generations, p. 4:

"The day seems not far distant when the converging lines from many apparently remote regions of thought will meet... Then the nature of the atoms, and the forces called into play in their chemical union... the explanation of cohesion, elasticity, and gravitation — all these will be marshaled into a single compact and consistent body of scientific knowledge... one of the grandest generalizations of modern science ... that all the phenomena of the physical universe are only different manifestations of the various modes of motion of one all-pervading substance — the ether."

And then the ether was no more. But that does not happen very often.

discrete - What is the smallest amount of energy?

It seems that we can get infinitely small but we will never reach a finite amount of energy that we could call the smallest amount. How can this be explained?

oscillators - Why dosn't amplitude increase when drive frequency is above resonance?

Why doesn't amplitude increase when the frequency of external periodic force increases above the natural frequency of the vibrating object?

Answer

In one sense this is a consequence of the meaning of resonance. Resonance is identified by a maximum in amplitude. If amplitude increased above (or below) the resonant frequency then we have not found a maximum in amplitude.

Perhaps what you are really asking is, why is there a maximum in amplitude when driving frequency $f$ equals the natural frequency $f_0$? This was answered by What is the qualitative cause for a driven oscillator to have a max. amplitude during resonance?

At the simplest level this is because at the natural frequency $(f=f_0)$ the driving force is always in phase with the natural motion of the system. The driving force is then always adding energy to the system, which will increase indefinitely unless there is some form of damping (eg friction) which removes energy from the system at a faster rate as amplitude increases.

At all other frequencies the driving force is sometimes in phase and sometimes out of phase with the natural motion of the system. Sometimes it adds energy, sometimes it takes energy away.

However, this assumes that both the natural motion and the driving force are both sinusoidal. If the driving force is applied as an intermittent impulse or "kick" then sub-harmonic frequencies of $f_0/n$, where $n$ is an integer, will also result in resonance. For example, if the system oscillates once every second, giving it a "kick" in the right direction once every 2 or 3 or 4 etc seconds will also increase the amplitude of the system, provided that the energy imparted with each kick is not outweighed by the loss of energy in between. See Non-resonant but efficient frequencies.

quantum mechanics - Expected value of $xp$ in harmonic oscillator

I wanna find out the expected value of the $xp$ operator for the $n$-th excited state of the harmonic oscillator, i.e. calculate the value $\langle n|xp|n \rangle$. I express the position and momentum operators in terms of the ladder operators:

$$x = \frac{x_0}{\sqrt{2}} \left( a+a^{\dagger}\right)$$ $$p = \frac{p_0}{i\sqrt{2}} \left( a-a^{\dagger}\right)$$

Where $x_0,p_0$ are the scales, $x_0p_0 = \hbar$. Since $$\langle n|a^m|n \rangle =0$$

For every $m$, and likewise for $a^\dagger$. So the only term that doesn't yield a zero expectation value after multiplying will be the commutator term, yielding $-1$. Which still leaves me with an imaginary answer of $\frac{i \hbar}{2}$. Where's the mistake?

Answer

As you have already worked out, $(xp)^\dagger = p^\dagger x^\dagger = px \neq xp$, so that the product of two self-adjoint operators is not needfully self-adjoint, and indeed from this equation we see that the product is self adjoint if and only if $p$ and $x$ commute.

To complete your thinking, how indeed do we calculate the product of the two measurements if we impart measurement $p$ first, then $x$? As you have already worked out for yourself, it can't be $\langle n|xp|n \rangle$.

Suppose our beginning quantum state is $\psi$ and impart $p$, getting measurement $\mu(p,\,\alpha)$ if the measurement forces our state into eigenvector $\psi_\alpha$, with probability $|\langle \psi_\alpha|\psi \rangle|^2$. Then, we impart $x$, and the mean value of this measurement is $\langle\psi_\alpha |x|\psi_\alpha\rangle$.

So the mean value of the product of the two measurements is:

$$\langle\psi|\left(\sum\limits_\alpha \langle\psi_\alpha |x|\psi_\alpha\rangle\,\mu(p,\,\alpha)\,|\psi_\alpha\rangle\langle\psi_\alpha|\right)|\psi\rangle$$

which we can write $\langle\psi|P_x|\psi\rangle$ where $P_x$ is the self adjoint observable:

$$P_x=\sum\limits_\alpha \langle\psi_\alpha |x|\psi_\alpha\rangle\,\mu(p,\,\alpha)\,|\psi_\alpha\rangle\langle\psi_\alpha|$$

This observable, interestingly, has the same eigenvectors as $p$ (being a superposition of projectors $|\psi_\alpha\rangle\langle\psi_\alpha|$ onto the eigenstates $\psi_\alpha$ of $p$), but with eigenvalues, i.e. measurements $\langle\psi_\alpha |x|\psi_\alpha\rangle\,\mu(p,\,\alpha)$ instead of $\mu(p,\,\alpha)$ (as is the case with the operator $p$). Naturally in the above $\alpha$ is an index variable ranging over the set of appropriate cardinality and integrals can replace the sums where appropriate.

Now, if we did measurement $x$ first, then $p$, the "observable" that would work out the appropriate mean value of the product of the measurements would be:

$$X_p=\sum\limits_\alpha \langle\tilde{\psi}_\alpha |p|\tilde{\psi}_\alpha\rangle\,\mu(x,\,\alpha)\,|\tilde{\psi}_\alpha\rangle\langle\tilde{\psi}_\alpha|$$

an operator with the same eigenvectors as $x$ (being a superposition of projectors $|\tilde{\psi}_\alpha\rangle\langle\tilde{\psi}_\alpha|$ onto the eigenstates $\tilde{\psi}_\alpha$ of $x$), but with eigenvalues, i.e. measurements $\langle\tilde{\psi}_\alpha |p|\tilde{\psi}_\alpha\rangle\,\mu(x,\,\alpha)$ instead of $\mu(x,\,\alpha)$. So in this case the mean of the product will be different from when we impart the measurements in the opposite order.

In your case, the application of these formulas is going to be tricky and needs to be handled with the appropriate theory of tempered distributions, since $x$ and $p$ do not have $\mathbf{L}^2(\mathbb{R})$ eigenfunctions. Unfortunately, this is not a calculation I have ever done, so I can't write up a full solution yet.

classical mechanics - Rigorous version of field Lagrangian

In Classical Mechanics the configuration of a system can be characterized by some point $s\in \mathbb{R}^n$ for some $n$. In particular, if it's a system of $k$ particles then $n = 3k$ and if there are holonomic constraints then in truth $s$ lies in some submanifold of $\mathbb{R}^n$. Even if the constraints are not holonomic, the configuration of a system can still be given by elements of some finite dimensional smooth manifold.

In that case, the Lagrangian becomes a smooth function $L: TM\to \mathbb{R}$ where $TM$ is the tangent bundle of the configuration manifold. Given coordinates $(q^1,\dots,q^n)$ on $M$ we can therefore make coordinates $(q^1,\dots,q^n,\dot{q}^1,\dots,\dot{q}^n)$ on $TM$ such that $q^i$ on $TM$ is really $q^i\circ \pi$ and $\dot{q}^i$ is characterized by the fact that if $v \in T_aM$ is

$$v = \sum_{i=1}^n v^i\dfrac{\partial}{\partial q^i}\bigg|_a$$

Then $\dot{q}^i(v) = v^i$. In that way, differentiating with respect to $q^i$ and $\dot{q}^i$ is perfectly well defined and Lagrange's Equation is totally meaningfull

$$\dfrac{d}{dt} \dfrac{\partial L}{\partial \dot{q}^i}(c(t),c'(t)) = \dfrac{\partial L}{\partial q^i}(c(t),c'(t))$$

When it comes then to studying fields like electromagnetic fields and so on, things get a little messy. Now, the system is the field and a configuration of the field is not anymore a certain list of numbers but a function like $\mathbf{E}: \mathbb{R}^3\to T\mathbb{R}^3$ or $\phi : \mathbb{R}^3\to \mathbb{R}$.

If we insist in building a configuration space $M$ it will be infinite dimensional and locally modeled on Banach Spaces. If we try to mimic Lagrangian formalism here, it'll end up in some infinite dimensional bundle, and this is not something nice to work with.

Now, most books work formally. For example, they let $\mathcal{L} = \dfrac{1}{2}g_{\mu\nu}(\partial^\nu \phi)(\partial^\mu\phi)- \dfrac{1}{2}m^2\phi^2$. Then they compute formally:

$$\dfrac{\partial \mathcal{L}}{\partial \phi} = -m^2\phi \\ \dfrac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)} = \partial^\mu \phi$$

And then Lagrange's Equations becomes

$$\dfrac{\partial \mathcal{L}}{\partial \phi} = \partial_\mu \dfrac{\partial \mathcal{L}}{\partial (\partial_\mu\phi)}\Longrightarrow \partial_\mu\partial^\mu\phi + m^2\phi = 0$$

Now this brings some questions:

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  First of them it is not clear on which space this $\mathcal{L}$ is defined and where it takes values. Some people say it is just a $3$-form on spacetime, but it doesn't seem like that, it looks like a scalar to me.

  
  


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  Second, we take derivatives of $\mathcal{L}$ with respect to functions. This is much confusing to me. It even conflicts the first point of view, if $\mathcal{L}$ is a $3$-form it can only be differentiated with respect to the coordinates of the manifold on which it is defined.

  
  

So how can we make all of this rigorous? I mean, in which space is $\mathcal{L}$ defined? What these derivatives really mean and why they make any sense at all? How to make a connection between this and the Classical Mechanics Lagrangian formalism?

Answer

Let us start from Minkowski spacetime $M$ and construct the trivial bundle $\Phi=\mathbb R \times M \to M$ whose sections $\phi : M \ni p \mapsto (p,\phi(p))$ are the scalar fields you want to discuss their dynamics.

Since you correctly wish to see the partial derivatives of $\phi$ as variables independent from $\phi$ itself (this is your second raised issue), the convenient space is the so called first jet bundle $j^1 \Phi$.

I will not enter here into the details of the mathematical notion of jet bundle, I will simply illustrate how it can be used to clarify your issues.

$j^1 \Phi$ is a fiber bundle over $M$ such that each fiber at $p\in M$ has the structure (is diffeomorphic to) $\mathbb R \times \mathbb R^4$. The first factor $\mathbb R$, on shell, embodies the information of $\phi(p)$ and the second $\mathbb R^4$ on shell refers to the derivatives $\partial_\mu \phi(p)$ at the same point of the basis $p$. However, in general these components must be viewed as independent variables: They are related just when the equations of motion are imposed, i.e., on shell.

Coming back to your first issue, in this picture, the Lagrangian is a map $${\cal L} : j^1\Phi \to \mathbb R$$ so that, ${\cal L}= {\cal L}(p, \phi(p), d_\mu(p))$. Euler-Lagrange equations determine sections $$M \ni p \mapsto (p, \phi(p), d_\mu(p)) \in j^1\Phi$$ and read $$\partial_\mu \left(\frac{\partial {\cal L}}{\partial d_\mu}\right) - \frac{\partial {\cal L}}{\partial \phi} = 0\:, \quad \partial_\mu \phi = d_\mu\:.$$

You see that the field equations themselves establish that $d_\mu = \partial_\mu \phi$, otherwise $\phi(p)$ and $d_\mu(p)$ would be independent variables.

Also in classical mechanics the convenient picture is that of a jet bundle (more natural than the one based on a tangent bundle). In that case, $M$ is replaced by the line of time $\mathbb R$ and each fiber $Q_t$ of the fiber bundle $\Phi \cong \mathbb R \times Q$ is the configuration space at time $t$ covered by coordinates $q^1,\ldots, q^n$. In this sense $\Phi$ is the spacetime of configurations. All Lagrangian mechanics is next constructed in $j^1\Phi$. Here the fiber $A_t$ at $t\in \mathbb R$ admits natural local coordinates $q^1,\ldots,q^n, \dot{q}^1,\ldots, \dot{q}^n$. The Lagrangian function is nothing but a map $$j^1\Phi \ni (t,q^1,\ldots,q^n, \dot{q}^1,\ldots, \dot{q}^n) \mapsto L(t, q^1,\ldots,q^n, \dot{q}^1,\ldots, \dot{q}^n)\in \mathbb R$$

and Euler-Lagrange equations now read $$\frac{d}{dt} \left(\frac{\partial L}{\partial \dot{q}^k}\right) - \frac{\partial L}{\partial q^k} = 0\:, \quad \frac{dq^k}{dt} = \dot{q}^k\:.$$

electromagnetism - Is it true that any system of accelerating charges will radiate?

I was recently told by a physics teacher that "any system of charges in which at least some of the charges are executing some sort of accelerated motion, will radiate and lose energy". This refers to classical electrodynamics (I'm not entering into quantum mechanics), that is, this statement should either be provable or disprovable from Maxwell's equations. I tried to think of a simple system of charges that would disproof this statement, since I intuitively think that it isn't true, but I haven't come up with any. The one example I could think of was a positive charge sitting on top of a negative charge, so that the net charge is zero. No matter how they move it is obvious that there will be no radiation. But my teachers did not accept this trivial example.

So my question is whether the statement is true or not, plus a proof or a counterexample of some charges executing accelerated motion without radiating.

Answer

The claim that accelerated charges must radiate is simply false. There are very many simple situations in which they do, but in general things should be examined on a case-by-case basis; there is not simple thumb rule like "acceleration yields radiation."

The simplest way to see this is to consider a wire carrying a constant current. This situation is magnetostatic, with well-understood electromagnetic field given by Biot-Savart law... which trivially lacks any radiation, as the field is constant. This is independent of the shape of the wire, and yet, if the wire is not straight, the charges within it must accelerate at some point.

A more exotic could be a uniformly rotating ring of charge. Both charge density and current density will be constant, and will produce constant electric and magnetic fields. This can be experimentally realized with toroidal superconductors.

general relativity - Are fully raised/lowered versions of Kronecker delta tensors?

I am confused. I have two textbooks contradicting each other, at least, it seems to me so. The first one – "Field theory" by Landau & Lifshitz says that by lowering or raising one index of Kronecker delta one gets covariant/contravariant metric tensor(in Minkowski space). The second one "Introducing Einstein’s Relativity (1992)" by Ray d’Inverno shows the opposite, namely, by lowering or raising an index of Kronecker delta one gets an object which apparently is not a tensor( the last sentence says that in this link). Is there a contradiction or the problem is with my understanding?

Answer

The metric tensor $g : TM\times TM \to \mathbb{R}$ is by definition a $(0,2)$-tensor and transforms like one.

You question is not about the metric tensor, but about the Kronecker delta, and for which index positions it defines a tensor.

To define a tensor by its components, we must fix a coordinate system $x$ on our manifold $M$ since the components of a $(0,2)$-tensor $T$, for instance, are defined as coefficients in the expansion of $T$ in the basic tensors $\mathrm{d}x^\mu$: $$ T= T_{\mu\nu}\mathrm{d}x^\mu\otimes\mathrm{d}x^\nu$$ and we now want to examine for which index positions on the $\delta$ the tensors $$ \delta^{ab}\partial_a\otimes\partial_b\quad\text{and}\quad\delta_{ab}\mathrm{d}x^a\otimes\mathrm{d}x^b\quad\text{and}\quad{\delta^a}_b\mathrm{d}x^a\otimes\partial_b$$ are defined independent of the chosen coordinate system.

To that end, we examine the transformation behaviour of $\delta^{ab},{\delta^a}_b,\delta_{ab}$. What we find is that under a coordinate transformation $x\mapsto y(x)$, \begin{align} \delta^{ab} & \mapsto \sum_i \frac{\partial y^a}{\partial x^i}\frac{\partial y^b}{\partial x^i}\\ \delta_{ab} & \mapsto \sum_i \frac{\partial x^i}{\partial y^a}\frac{\partial x^i}{\partial y^b}\\ {\delta^a}_b &\mapsto {\delta^a}_b \end{align}

Therefore, a tensor that has components $\delta^{ab}$ or $\delta_{ab}$ in one coordinate system does not have those components in another system, so just writing down $\delta^{ab}$ does not specify a tensor unless you also specify a coordinate system in which the tensor has these components.

On the other hand, a $(1,1)$-tensor that has components ${\delta^a}_b$ in one system has those in all, therefore, ${\delta^a}_b$ defines a tensor without need for a particular coordinate system.

Now, since it defines a $(1,1)$-tensor, we can raise and lower its indices. However, contrary to what one might expect, the fully raised and fully lowered versions are not $\delta^{ab}$ and $\delta_{ab}$, but instead $g^{ab}$ and $g_{ab}$.

So, lowering/raising the index on the $\delta$ with the metric tensor¹ gives proper tensors, "raising/lowering" the index by just writing both indices rasied on the $\delta$ does not give a well-defined tensor.

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¹Here, "raising/lowering" with the metric tensor refers to the musical isomorphism that a vector $v = v^\mu\partial_\mu$ has an associated covector $v^\flat = g_{\mu\nu}v^\nu\mathrm{d}x^\mu$ with components $(v^\flat)_\mu = g_{\mu\nu}v^\nu$, and similarily lowering an index of a tensor means contracting it with $g_{\mu\nu}$, and raising it contracting it with $g^{\mu\nu}$. So, lowering the index of ${\delta^a}_b$ means $(\delta^\flat)_{ab} = g_{ac}{\delta^c}_b = g_{ab}$ by definition of the Kronecker delta.

cosmology - How large is the universe?

We know that the age of the universe (or, at least the time since the Big Bang) is roughly 13.75 billion years. I have heard that the size of the universe is much larger than what we can see, in other words, much larger than the observable universe. Is this true? Why would this not conflict with what is given by Special Relativity (namely, that faster-than-light travel is prohibited)?

If it is true, then what is the estimated size of the universe, and how can we know?

Answer

It is indeed more useful to cite the age of the universe, because this defines the region in space which is observable, a 13.75 billion lightyear sphere (approximately).

Clearly, however, the entire universe could be more than 13.75 billion years across in diameter; that number is merely a radius. For example, let's suppose a naive view of the expansion of the universe which doesn't include inflation or dark energy. At the moment of the big bang, photons rush off in every direction at the speed of light (again, naive cosmology - ignore the fact that the universe is opaque for 300,000 years).

These photons are all moving out from one point at the speed of light. We imagine, then, an expanding sphere whose surface is defined by the furthest point which light has so far reached. This sphere is expanding in volume very quickly indeed - the radius is expanding at the speed of light.

So by now, 13.7 billion years later, the radius is 13.7 billion light years. The diameter of the sphere is twice that, 27.4 billion light years. The volume is volume of a sphere with radius r=13.7 GLy, which is $4\pi r^3 / 3 =57.4$ billion cubic lightyears.

This is shows that the universe can easily be much larger than 13.75 billion light years across. Also, note that, if the earth is formed on the expanding sphere, no one from earth will ever be able to "catch up" and see the other side of the sphere, since that side is still expanding. This is what people mean when they say that the universe is larger than we can see.

Now, this answer is wrong. Do not go quoting these numbers. It doesn't take into account inflation or the expansion of the universe. No one knows enough about either of those two effects to give a really good precise number for the size of the universe, but you can be certain that they only result in a bigger universe. One lower-bound for the radius of the universe is 39 billion light years, based on some analysis of the cosmic microwave background.

CPT symmetry as it applies to matter/antimatter

I've heard/read a couple different definitions for antimatter: it is charge-reversed matter (typical definition, for the lay person at least), or it's time-reversed matter (as described in QED).

Then there's CPT symmetry, which means that a universe with all charge and time direction reversed, and mirror imaged spatially, should evolve according to the same laws as our universe.

OK, my question(s). If, in our universe, you were able to take an electron and reverse its charge, parity, and time direction, is that CPT-reversed particle identical to the original electron? So for example, would it annihilate if it came into contact with an electron, or would the two interact just like two regular electrons?

Another way to phrase this is to ask, if antimatter is charge-reversed or it's time-reversed, if you take matter and both charge- and time-reverse it, is it back to the original state? And does P play any role here (could you also describe antimatter as parity-reversed matter)?

Aside from oversimplifying, I may be mixing two different concepts together here (confusing "follows the same laws" with "is the same thing", roughly speaking). Thanks in advance for any answers or clarifications!

special relativity - Accelerating particles to speeds infinitesimally close to the speed of light?

I'm in a freshmen level physics class now, so I don't know much, but something I heard today intrigued me. My TA was talking about how at the research facility he worked at, they were able to accelerate some certain particle to "99.99% the speed of light". I said why not 100%, and I didn't quite understand his explanation, but he said it wasn't possible. This confused me. Since the speed of light is a finite number, why can we go so close to its speed but not quite?

Edit: I read all the answers, and I think I'm sort of understanding it. Another silly question though: If we are getting this particle to 99.99% the speed of light by giving it some sort of finite acceleration, and increasing it more and more, why cant we increase it just a little more? Sorry I know this is a silly question. I totally accept the fact we cant reach 100%, but I'm just trying to break it down. If we've gotten so close by giving it larger and larger acceleration every time, why cant we just supply it with more acceleration? And how much of a difference is there between 99.99% the speed of light, and the speed of light? (I'm not really sure if "difference" is a good word to use, but hopefully you get what I'm asking).

Answer

By special relativity, the energy needed to accelerate a particle (with mass) grow super-quadratically when the speed is close to c, and is ∞ when it is c.

$$ E = \gamma mc^2 = \frac{mc^2}{\sqrt{1 - (\text{“percent of speed of light”})^2}} $$

Since you can't supply infinite energy to the particle, it is not possible to get to 100% c.

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Edit: Suppose you have got an electron (m = 9.1 × 10^-31 kg) to 99.99% of speed of light. This is equivalent to providing 36 MeV of kinetic energy. Now suppose you accelerate "a little more" by providing yet another 36 MeV of energy. You will find this this only boosts the electron to 99.9975% c. Say you accelerate "a lot more" by providing 36,000,000 MeV instead of 36 MeV. That will still make you reach 99.99999999999999% c instead of 100%. The energy increase explodes as you approach c, and your input will exhaust eventually no matter how large it is. The difference between 99.99% and 100% is infinite amount of energy.