juanrga

Yeah, I've heard that. I just don't think even our best minds can know that for a fact. I feel that to many people take what someone else has said as the absolute truth. Is it true? Maybe. But what if it's wrong? Then we will continue to believe that and stop all effort to research any other thought that might not fit with the current thinking but which may in fact be right. The scientists that come up with any given theory have a vested interest in making sure that theory is maintained. But of course it won't really matter. If there's a Big Crunch or an Infinite Expansion we'll be long dead.

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.

He also didn't like quantum physics because he didn't think it made any sense.

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

When you say the word Quantum Particle What image comes into your mind then ? If not a round ball with the standard model Rows and columns of such particles.

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.

Do you see a number of laws and Math equations Say Schroenigers equation, Etc Etc.

If so what do you see them housed in ? what does it look like ? a nano something or other ? Some superstring topography, ? Where and how are the laws and maths formulae stored ? What do you see in your Head ?

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. I believe this your interpretation is wrong.

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.

Juanrga

 

 

Yes , that is all very well, but surely if we find that the descriptor is misleading and confusing, is there not a case for moving on slightly to ease the way for clarity of thought, particularly for those approaching physics and stumbling with all this mystery around quantum behaviour.

 

Perhaps you are saying that above all else the atomic particles are PARTICLES first and foremost. ? But with all the FIELD science,; electric, magnetic, Electro-magnetic, quantum field, Higgs Field, etc etc is there not a case to view things more in that field direction say, or some other image , than the particle direction.

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

Surely there must be some happy medium where we can understand what is happening at the quantum level

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 standard model is gaining beauty by the day, but in your link to Cern it still shows the particle as a round ball , surely we need to get off this image IF as you and I know is probably incorrect. ( I' ll probably be struck down ). It probably looks more like a spinning probability wave function.

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.

Are we not digressing from the truth with this nuance in the position and momentum of the EUP? I believe that it was talking about measurement. That one cannot measure to great precision simultaneously the position and momentum of a quantum particle. This, I believe, does not imply that they do not possess both at every instance, but that you cannot measure to a high degree of accuracy and simultaneously both quantities of a quantum particle.

Not having position and momentum, and measurement of position and momentum are two different phenomenon and misunderstanding this could lead to great doldrums. Like the above cartoon, if you think to focus your attention on momentum, you loose accuracy in measuring its position, and vice versa.

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.

Then that must tell us something about the reality, or nature of a " particle" at the quantum level.. If it cannot have BOTH momentum and position , then its not a particle as our language infers. We should perhaps try to generate a new description of these entities. !

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:

everything in the universe is found to be made from twelve basic building blocks called fundamental particles, governed by four fundamental forces.

Juanrga.

 

Does that mean, by what you say, that the particle actually does Not intrinsically posses both position and momentum, or that we can not measure, utilize or observe both position and momentum at the same time ?

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

So as not to be misleading...

 

as i understand GR, as applied to Friedmann cosmologies, space-time is a static entity.

Friedmann spacetime is not static, but depends on time.

At all times, that space-time is filled with matter-energy; and at each time, the density of matter-energy determines the scale factor (and rate of change thereof).

The scale factor is not completely determined by density alone, pressure and cosmological constant also play a role.

For the Friedmann equations, one assumes, first and foremost, the global topology of space-time, e.g. closed (k=+1). Thereafter, that global topology is immutable. So, if (say) you choose to model a closed cosmology, then, ipso facto, the density within that space-time fabric must always be greater than the corresponding critical density -- the equations force everything else to adjust, so as to maintain the chosen topology. So, assuming the accuracy of the Friedmann cosmologies; then if our cosmos is closed today, then it always has been, and always will be.

Friedmann cosmologies are not valid at early times and probably will be not valid at long times.

To be Relativistically invariant, i would guess that Relativistic QM equations, e.g. Klein-Gordon equation, "must" treat wave-functions as fully (3+1)D objects, which transform in Lorentz-invariant ways.

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 persist.

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.

So, everything is not made of energy?

As stated at CERN website:

Everything in the universe is found to be made from twelve basic building blocks called fundamental particles, governed by four fundamental forces.

guys i don't know what is happening, i have serious issues. working from another machine (.still IE8). the edit function of a post does not understand that I want to answer after the Quote. I do it again & again, it does not work. Look below. It's getting really worriyng...

 

http://www.scienceforums.net/topic/71281-universe-expansion/?p=718716

 

yesterday, from another machine (out of a number of 5), i was unable to see the whole "my profile" pulldown menu. Only the upper part with half my avatar.

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.

But aren't wave-functions [math]\Psi®[/math] 3D objects?

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].

I have question; What exactly does E=MC^2 mean?

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.

I thought I had a basic understanding, but the more I learn, well questions just seem to appear out of nowhere.

 

It seems to be nothing more than the formula for momentum, where energy takes the place for momentum.

Momentum has units of mass x velocity. Energy has units of mass x velocity squared.

Then there is the understanding that nothing with mass can move at C which would make E an imaginary value.

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.

And one more question; if anything with a mass of zero has to move at C, it would seem to make sense that anything with a mass greater than zero would have to move also. This is not exactly worded like a question, but it is a question.

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.

The Uncertainty principle holds as long as there is a detector, since the detector would have to interact with that particle to detect it.

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, or intergalactic distance increment ?

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

I don't understand that.

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.

20th century cosmological conjectures have reached a self-destructive stage due to:

1) Excessively far fetched implications (e.g., entire galaxies at the edge of the known universe are proffered to travelling faster than the speed of light in vacuum);

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.

2) Gross internal inconsistencies (e.g., per very definition of explosions, the big bang would require the universe to be empty for about 13.7 billion light years from Earth, with galaxies then decreasing in speed, in dramatic disagreement with astrophysical evidence, while the background radiation can be easily proved to have been absorbed by galaxies and intergalactic media billions of years ago); and

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.

It is pretty clear that the universe is expanding because of the galaxies moving away from one another. But how does that work? I read that is what dark matter or dark energy does. I'm curious how dark energy could use gravity as a repulsive force, or is there another idea of how this might be happening?

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.

Hi guys,

 

I was discussing physics with a friend of mine on another forum until someone showed up talking about Quantum Theory so I thought I might check with you guys if the information presented is accurate or not (I think it's not, but I'll let you decide).

 

What he said : "

 

Is he smoking something or does the above actually have some basis in truth?

The physics content is almost all wrong.

Should the [current] inability to "measure" position and momentum mean that a particle at a moment in time does not have a given position and speed?

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.

i heard that any object above absolute zero should radiate some form of energy. so, can't a particle's position be pointed at any instant if i place very sensitive radiation detectors all around its path? can't i pinpoint the paritcle's position and velocity at any given instant like this? the detectors are not physically interacting with the particle and therefore can't be responsible for any change in its velocity or position. can anyone please clear my doubt on this topic.

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.

Do nanomaterials exhibit quantum effects ?

Yes, for instance the quantum confinement effect.

At what distance do we start considering quantum effects, do normal atoms and molecules exhibit quantum effects.

There is not a sharp boundary but almost all the effects on atoms and molecules are quantum.

We know that light does bend around a mass under the General Theory of Relativity, but why does light also bend under Newton's Law? and how is it calculated?

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!

 

I must have overlooked that picture in your earlier posts - not to mention that I still don't see a definition of h or how your claim follows from the term. I'm not even convinced that your "gravitational interaction" term actually has units of energy as you claim - but that's just silly nitpicking and doesn't really add something.

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].

"The gravitational interaction is given by <some unspecified physical quantity with units of energy>" isn't exactly more helpful to me than "the gravitational interaction is given by <some number>".

The term [math]T^{\mu\nu}h_{\mu\nu}[/math] is not "unspecified" but a well-known term with precise properties. Next picture is from Feynman textbook on gravitation

Note that Feynman adds an extra lambda coupling constant, because this helps him in a perturbation expansion in orders of lambda, but usually lambda is absorbed into h and the expansion made in terms of orders of h.