juanrga

Quantum correlations are no part of a hidden variable theory, but part of standard quantum mechanics. As stated in #6 entanglement is just a quantum correlation. Moreover, the Aspect experiment has not ruled out hidden variable theories. MigL is correct. What Aspect found experimentally is that we are saying to you [http://en.wikipedia.org/wiki/Alain_Aspect]:

Science does not deal with the "absolute truth" this is a task best left for religions, each one of the which has its own "absolute truth" . Science deals with evidence: observations and experiments. Scientific knowledge is accumulative. Scientific theories are not shown to be wrong and abandoned, but older scientific theories are incorporated as special cases of newest theories.

Before 1930 Einstein raised his doubts on the consistency of quantum mechanics. But after 1930 he maintained the view that quantum mechanics is logically consistent but incomplete. See Einstein and the quantum theory for a review. Einstein was also a critic of the old Bohr-Heisenberg interpretation and was one of the pioneers of a modern and rigorous interpretation of quantum mechanics which is named the statistical interpretation

An image is "a reproduction of the form of a person or object". A quantum particle such as the electron is smaller than light and has no form. You cannot imagine the image of something that has no image. As said above a quantum particle such as the electron is not, neither looks as, a round ball. In fact, an electron is pointlike, it has no volume. The concept of quantum particle is an abstract concept, it cannot be visualized in terms of ordinary daily experience, a quantum particle does not look as anything that you have seen in your life. Formally, the concept of a quantum particle such as the electron is rather close to the concept of material point (another abstract concept) used in classical mechanics, except that the properties and the behaviour are very different. The correct understanding of a quantum particle, a physical object, is obtained from the representation of that system using the formalism of quantum mechanics (a theory of physics). For instance, the Schrödinger equation says, in an exact and unambiguous way, how a quantum particle moves in space. No image or picture can substitute the Schrödinger equation.

No. This is standard material found in any textbook on QM. To complement a bit the answer already given by MigL. Textbooks explain how to derive the uncertainty inequality for any pair of non-commuting observables A and B. If A and B do not commute then a quantum particle cannot be in state with values for both. Momentum and position are only a special case of this. A and B do not need to be two different quantities, they can represent components of the same vector, for instance, the components of the angular momentum. No quantum particle can be in a state [math]\Psi_\mathbf{L}[/math] with angular momentum [math]\mathbf{L}[/math], because the particle cannot have [math]L_x[/math], [math]L_y[/math] and [math]L_z[/math] all at once. You cannot measure what does not exist.

As shown by Wheeler and Feynman all the effects usually attributed to electric fields, magnetic fields, electromagnetic fields can be explained using a theory of only particles. Quantum fields can be also eliminated http://rmp.aps.org/abstract/RMP/v21/i3/p425_1 http://rmp.aps.org/abstract/RMP/v67/i1/p113_1 We do not need to introduce a medium to understand what is happening. One of the first advances of modern physics was to eliminate the unobservable aether as medium for the electromagnetic waves (this was the born of the special theory of relativity). We would not gain anything by introducing some other unobservable medium. Yes, the CERN link has an artistic header image. But in no part of the text the webpage says that particles are round balls with colorful labels. No, a quantum particle does not look as "a spinning probability wavefunction". A wavefunction is an abstract function which is not even defined in the ordinary space. The state of a particle is given by a wavefunction only in some special cases (free particle in a pure state) and only in some formulations of QM. Finally the concept of spin in quantum mechanics is not the classical concept of spinning around an axis. Quantum mechanics is very clear at this point. It says that a quantum particle cannot be in a state [math]|\Psi\rangle[/math] with a well-defined value (eigenvalue) of both position and momentum. Either the particle is in a position eigenstate [math]|x\rangle[/math] and has a well-defined position, or in a momentum eigenstate [math]|p\rangle[/math] and has a well-defined momentum, but not both because position and momentum are non-commuting observables in QM. You cannot measure what does not exist: a corollary of the above QM restriction is that you cannot measure the position and momentum of a quantum particle.

Who said you that the concept of particle implies having both a position and a momentum? A Newtonian particle must have both, but a quantum particle does not. In quantum mechanics, and in particle physics, a particle is not defined as a little sphere with both a position and a momentum. If you do not like the term "particle" you can change the name and use "XAXFDBGTRJHGN" or anything of your invention, but the physics remain and maintain in mind that we scientist use the term particle:

It means that the quantum particle cannot be in a quantum state [math]\Psi[/math] with both a given position and a given momentum. If the particle cannot be in that state, then you cannot measure/observe the particle in that state. Next is the best funny explanation of what happens when someone violates the principle

Friedmann spacetime is not static, but depends on time. The scale factor is not completely determined by density alone, pressure and cosmological constant also play a role. Friedmann cosmologies are not valid at early times and probably will be not valid at long times. Yes your guess is correct. However, those relativistic wavefunction equations are not even aceptable for one-particle systems and were abandoned when quantum field theory was developed. In quantum field theory the Klein Gordon equation is no more a wavefunction equation but a mere identity for a field operator. Moreover, quantum field theory is defined in a dummy spacetime without physical meaning. Wave-functions do not persist for macroscopic objects. E.g. even if we accept that the state of a cat is given by a wavefunctiion at a given instant, the wave-function is destroyed by decoherence almost instantaneously.

As stated at CERN website:

Notice that I answered to the same post that you in that thread and my answer is outside of the quote. Yes it is annoying.

Stationary spatial wavefunctions are [math]\psi_r=\psi_r®[/math] only for one-particle objects. Adding time-dependence [math]\Psi_r=\Psi_r(r,t)[/math]. Spatial many-body wavefunctions are described in an extended [math]3N+1[/math] space. If you add spin then there is more variables in the wavefunctions [math]\Psi=\Psi(r_1,r_2,...r_N,s_1,s_2,...s_N,t)[/math].

It says, in words, that the energy of a free particle at rest (zero velocity) is the product of its mass and of the speed of light squared. Momentum has units of mass x velocity. Energy has units of mass x velocity squared. Energy would not be imaginary but infinity for a massive body moving at c. Or said in another way: you need apply a infinite energy to accelerate the object up to c. To "make sense" according to your usual experience in real life? Maybe, but at high velocities the behaviour of objects is very different from the ordinary low-velocity behaviour over which you base your experience. In reality it makes sense why a massless object has to move at c. A massless object has not the traditional concept of inertia and cannot be accelerated as when you accelerate a rock. Massless objects are forced to move always to the same speed, and it seems natural that this speed is c, which is an universal constant. A more detailed and rigorous explanation of why massless objects are forced to travel at c requires the use of the relativistic equations.

As stated in #10 the uncertainty principle of quantum mechanics holds even in absence of any detector. The principle is really about the quantum state of the particle not about measurements.

Space volume expansion in non-bounded regions: this includes intergalactic regions.

Imagine a large quote with two paragraphs. E.g. If you put the cursor in the middle of both paragraphs and click twice the enter key the quote is automatically splinted into two parts and you can reply to each one of the paragraphs by separate.

I agree on that there are fundamental issues with the current cosmological models, but the recession velocity is not an ordinary velocity. Galaxies are not "travelling" at speeds faster than light, but the space between galaxies is expanding with a recession velocity faster than that of light. I do not find any problem with such recession speeds. No known law is violated. The Big bang is not an ordinary explosion. It is often presented as such in pictorial or popular presentations, but it is not. In an explosion matter moves in a fixed space, and there is a centre of the explosion. This is not how the Big Bang works. The Big Bang deals with the expansion of the own space and there is not a true centre.

The expansion of the universe is unrelated to dark energy or dark matter and cannot be explained as a repulsive force, because in general relativity neither gravitation nor space expansions are described by any force. The origin of the expansion is on the Big Bang, but there is not universally accepted theory about what happened at the first instant of time. Our confirmed theories cease to work much before.

The physics content is almost all wrong.

Quantum mechanics says that the particle cannot have both a position and a momentum (speed), with independence of our measurements. A quantum object has a given property when its wavefunction is an eigenfunction of the quantum operator corresponding to that property. Quantum mechanics says that there is not wavefunction that can be eigenfunction of both position and momentum operators and therefore no quantum object can have both a position and a momentum (speed). This is the reason why wavefunctions are either functions of position [math]\Psi(x,t)[/math] or functions of momentum [math]\Psi(p,t)[/math] but not of both. Note: Nonrelativistically the momentum p is given by p=mv where v is the velocity and m the mass.

Unless I am completely wrong, it seems that you are trying to avoid the uncertainty principle of quantum mechanics, but this is not possible. Heisenberg believed that the uncertainty was the result of our measurements. In fact still some popular presentations of the principle says that we cannot know position and momentum because any measurement introduces a perturbation and changes, for instance, the speed of the quantum object. This is not the real reason. We know today that the principle is unrelated to our measurements. Quantum mechanics says that the quantum object cannot be in some wavefunction with a given position and momentum. Therefore it does not matter if you imagine a detector that does not physically interact with the object. The quantum object continues without having a given position and momentum because the laws of quantum mechanics say that there is not any wavefunction that was at the same time eigenfunction of the position operator and eigenfunction of the momentum operator.

Yes, for instance the quantum confinement effect. There is not a sharp boundary but almost all the effects on atoms and molecules are quantum.

Bending in Newtonian theory uses tricks, because Newtonian theory is only valid for low velocities and does not really apply to relativistic particles such as photons. Start with the Newtonian potential energy [math]V = - \frac{GMm}{r}[/math] and use the trick [math]m = p/c[/math] by substituting the speed of light on the Newtonian momentum [math]p=mv[/math] (it is a trick because this expression is only valid for speeds much smaller than c and because for a photon m=0) [math]V = - \frac{GMp}{rc}[/math] This potential can be now used in the Newtonian equation of motion [math]\frac{dp}{dt} = \frac{\partial}{\partial r} \frac{GMp}{rc} = - \frac{GMp}{r^2c}[/math] but gives one-half of the observed bending. Then the second trick consists on multiplying the Newtonian bending by a factor 2. Using a relativistic theory one obtains the correct bending and without any trick.

I don't know if this is happening to everyone else, but with the upgrade my notifications settings were deleted and I was not being notified of replies to forums I am following or where I am posting. I have changed the settings manually again. Check your profile settings!

As said before [math]h_{\mu\nu}[/math] is the gravitational potential. It is obtained from solving the field equations for specific problems. It is rather simple to check that the gravitational interaction term [math]T^{\mu\nu}h_{\mu\nu}[/math] has units of energy. One way is to see that the term is found in the Lagrangian for gravitation and Lagrangians have units of energy. The second way starts from noticing that [math]h_{\mu\nu}[/math] is dimensionless (as any textbook explains) whereas the [math]T^{\mu\nu}[/math] has units of energy: [energy x dimensionless] = [energy].