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Rockazella

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Rockazella

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russ_watters

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F=MA is equally valid at all times. But there is a catch - at high relativistic speeds, the variation of M is enough to notice.Originally posted by Rockazella

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FZ+

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Hurkyl

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Originally posted by Rockazella

This depends on what kind of accuracy you're interested in. If the particle is moving at relativistic speeds, i.e. close to the speed of light, then the force is not related to the acceleration as F = ma. It's given by F = dP/dt where P = mv = m_o*v/sqrt[1 - (v/c)^2] (m_o = electron rest mass). If the speed is not relativistic, i.e. v << c, then F = ma will work fine and there's no distinction between m and m_o. These are classical concepts. For example: In particle accelerators where particles are travel at almost the speed of light, then the particles leave tracks in bubble chambers and you can actually see the trail of bubbles - there are of course different ways of detecting the paths of particles.

If you're talking about the trajectory on the quantum level then a trajectory has little meaning. One then speaks in terms of quantum states or what many people call a "wave function." The "wave" has meaning only on a statistical basis though.

Pete

- #7

Originally posted by russ_watters

F=MA is equally valid at all times. But there is a catch - at high relativistic speeds, the variation of M is enough to notice.

F = ma is wrong. It'd equal to dp/dt in many cases. In those cases F = ma is an equality.

At the quantum level F = ma is meaingless. If a particle is loosing mass then F = ma is wrong. The correct equation is F = dp/dt

Pete

- #8

Originally posted by christench

The force is used in quantum mechanics in the sense of a potential energy function. i.e. the F in F = -grad V is not used by the V is.

And an electron is not a wave. An electron has a wave function associated with it and then it has meaning only in a statistical sense.

Pete

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But, if you must take in consideration the wave properties, you should use a quantum mechanics expression involving the wave function of the particle. Example: Calculation of bands in a semiconductor.

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No. The electron is not a wave. It's not a particle. It's not either. It's something else. Some people use the term "wavicle" but not I.Originally posted by christench

And no. The wave function has only a statistical meaning. For example: If the wave function of a particle is Psi(x,t) then the meaning of this is defined as follows. The quantity

P = |Psi(x,t)|^2 dx

is the probability that, if the postition of the particle is measured then the probability that the particle will be in the interval of width dx center at x at the time t is P. The quantum state is a quantity which contains all the information about the particle. All this information is of a statistical nature.

I created this page

http://www.geocities.com/physics_world/qm/probability.htm

to describe the meaning of a quantum state using die as an example. It's not a perfect analogy since the die are not quantum particles - but it gives you the idea of what it means

Pete

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FZ+

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I thought the commonest description is of "both, but not at the same time".No. The electron is not a wave. It's not a particle. It's not either. It's something else. Some people use the term "wavicle" but not I.

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russ_watters

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My QM isn't very good - can you give me some more? Why would "m" being a variable make f=ma wrong? If the mass goes up due to high relativistic speed, can't you just plug that new mass back into the equation? What do you mean by losing mass? And what are "p" and "t": position and time?Originally posted by pmb

F = ma is wrong. It'd equal to dp/dt in many cases. In those cases F = ma is an equality.

At the quantum level F = ma is meaingless. If a particle is loosing mass then F = ma is wrong. The correct equation is F = dp/dt

Pete

Sure, you wouldn't want to integrate the acceleration function into a velocity or position function, but at an instantaneous time, I don't see why you can't still use it to find instantaneous acceleration. Or why couldn't you just insert the equation for relativistic mass in place of "m" and end up with valid acceleration, velocity, and position functions?

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My QM isn't very good - can you give me some more? Why would "m" being a variable make f=ma wrong? If the mass goes up due to high relativistic speed, can't you just plug that new mass back into the equation? What do you mean by losing mass? And what are "p" and "t": position and time?

The variation of mass makes F=ma inexact at the classical level. It’s not necessary to take in consideration relativistic scenarios or QM ones.

One of the typical examples is a rocket. The propulsion force is acting over a system which is losing mass (the propellant burned).

The Force acting over a particle is:

F= dp/dt where p is the momentum and t is the time.

p = mv by definition of momentum, where m is the mass and v the velocity.

That is: F = vdm/dt + mdv/dt

Usually the first term is null because the system doesn’t change their mass (we obtain F=mdv/dt=ma). But in the rocket example, you must take in consideration the two terms.

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That's not what I meant. For f = ma to be meaningful the "a" has to be meaningful. Acceleration is defined in terms of the particle's position as a function of time. However in quantum mechanics one cannot do this. A particle does not have a classical trajectory. It's not possible to determine the particle's position as a function of time because that would mean continuous measurements of the particle's precise location. But the position can't be measured precisely without the momentum being affected. Classically you can picuture this as everytime you measure the position the velocity changes. The end-all is that it's not meaningful to speak of a trajectory and therefore it's not meaningful to speak of acceleration. It's only meaningful to speak of averagges. For example: If <p> is the average momemtum and the particle is moving in a potential V then the equation of motion is found using Ehrenfest's theorem which isOriginally posted by russ_watters

My QM isn't very good - can you give me some more? Why would "m" being a variable make f=ma wrong? If the mass goes up due to high relativistic speed, can't you just plug that new mass back into the equation? What do you mean by losing mass? And what are "p" and "t": position and time?

d<p>/dt = <-grad V>

As far as loosing mass I was refering to the use of f = ma in general in mechanics. You shouldn't think of this as the definition of force - For example: Suppose you wanted to cqalculate the position of a rocket as a function of time given a certain inittial mass and thrust etc. If you try to use f = ma you'll get the wrong answer. The mass of the rocket is constantly decreasing. The correct formula is f = dp/dt.

How would you measure the acceleration?Sure, you wouldn't want to integrate the acceleration function into a velocity or position function, but at an instantaneous time, I don't see why you can't still use it to find instantaneous acceleration.

Or why couldn't you just insert the equation for relativistic mass in place of "m" and end up with valid acceleration, velocity, and position functions? [/B]

You'd get the wrong result. The expression F = ma is not relativistically correct. If you want to express the force as a function of accleration then you'd have to express it in terms of longitudinal mass M(L) and transverse mass M(T) (Transverse mass happens to be equal to relativistic mass().

M(L) = m_o[1-(v/c)^2]^(3/2)

M(T) = m_o[1-(v/c)^2]^(1/2)

Let A(L) = component of acceleration vector parallel to the direction of motion

Let A(T) = component of acceleration vector transverse to the direction of motion

F =M(L)A(L) + M(T)A(T)

Pete

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Originally posted by christench

The electron is not a wave. It's not a particle. It's neither. Do you have the Feynman Lectures? From V-III pagbe 1-1

Every time an electron is detected its a single entity, not part of something like one gets with a wave. So when an electron is detected its behaves like a particle. However if you run a large number of experiments then what you'll find is that the measurements of the postion of the electron, in general, have a wave pattern to them - and that includes interference. And the interference pattern is there even when the electron can't interact with another electron in the experiment.Newton thought that light was made of particles, but then it was discovered that it behaves like a wave. Later, however (in the beginning of the twentieth century), it was found that light did indeed sometimes behave like a particle. Historically, the electron, for example, was thought to behave like a particle, and then it was found to behave like a wave. So really it behaves like neither. Now we have given up. We say: "It is like 'neither'"

There is one lucky break, however - electrons behave just like light. The quantum behavior of atomic objects (electrons, protons, neutrons, photons, and so on) is the same for all, they are "particle waves," or whatever you want to call them. So what we learn about the properties of electrons (which we shall use for our examples) will apply also to all "particles," including photons of light.

So Feynman is quite right when he says that an electron is neither a particle nor a wave. And most physicists agree with that. I know of now quantum mechancis text which states differently. (there is Bohm's theory though but it's largely rejected)

Pete

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Actually the opposite is true. An object 'is' what it behaves like. The precise description is to say that it's neither a particle nor a wave.Originally posted by FZ+

I thought the commonest description is of "both, but not at the same time".

How it behaves depends on the experiment and how what you're measuring. See my post where I quote Feynman - he's about the most accurate that I can find on this topic and I agree with him 100%. Every single quantum mechanics text I have or have ever used says the same thing.

Pete

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jackle

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Tunnelling in quantum mechanics makes a mockery of the idea of a force.

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russ_watters

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I don't want to measure it, I just want to mathematically model it.Originally posted by pmbWell thats just it - the mass of the rocket decreases at a known rate (kt) that you can apply to the formula f=ma. Substitution gives you f=(m-kt)a. You can still integrate that into an accurate position function. I'm sure I did exactly this problem in high school physics (or maybe calculus).As far as loosing mass I was refering to the use of f = ma in general in mechanics. You shouldn't think of this as the definition of force - For example: Suppose you wanted to cqalculate the position of a rocket as a function of time given a certain inittial mass and thrust etc. If you try to use f = ma you'll get the wrong answer. The mass of the rocket is constantly decreasing.

How would you measure the acceleration?

Ok, that makes a lot more sense - I thought an electron was big enough to model using rough Newtonian motion. I realize when around an atom, there is no definable position function (hence electron "cloud"). But what about when free and say fired from an electron gun? Does this affect the resolution of a TV?That's not what I meant. For f = ma to be meaningful the "a" has to be meaningful. Acceleration is defined in terms of the particle's position as a function of time. However in quantum mechanics one cannot do this. A particle does not have a classical trajectory.

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That's not quite correct. A single electron does not display wwave properties. And the wave cunction is not the shape of an elecron by any means whatsoever.Originally posted by christench

You can see the wave nature of an electron with just one electron, you dont need many.

No offense but I'm afraid that you have some serious misconceptions about quantum mechanics.

Pete

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See web.media.mit.edu/~sibyl/projects/cognition/science/rocket.htmlOriginally posted by russ_watters

Well thats just it - the mass of the rocket decreases at a known rate (kt) that you can apply to the formula f=ma. Substitution gives you f=(m-kt)a.

I don't want to measure it, I just want to mathematically model it.[/B]

Its not that simple in quantum mechanics. In fact its a crucial point. If you can't state exactly how to measure it in principle then it's not meaningful - that's the lesson of quantum mechanics. This is very different from Newtonian mechanics.

More later

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And single electrons do exhibit wave properties. A single electron will diffract, or undergo Bloch oscilation, or tunnel....

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jcsd

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Originally posted by pmb

That's not quite correct. A single electron does not display wwave properties. And the wave cunction is not the shape of an elecron by any means whatsoever.

No offense but I'm afraid that you have some serious misconceptions about quantum mechanics.

Pete

Yes, in the standard (Copenhagen) intepretation, the wavefunction has no physical analogue, but in the De Broglie-Bohm theorum you get 'matter waves'.

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Originally posted by christench

And single electrons do exhibit wave properties. A single electron will diffract, or undergo Bloch oscilation, or tunnel....

A single electron does not exhibit wave properties per se. Only a large number of identical experiments each with a single electron, or a large number of electrons in a single experiment will wave characteristics manifest themselves. Tunneling is a quantum effect with no mechanical analogy. But it can be envisioned as single particle violated energy conservation but consistent with the time-energy uncertainty principle.

As far as difraction goes - Let me use Young's double slit experiment. A source of light is directed towards the double slit. The source is very weak. On the screen behind it the location of the photon is registered. The source is adjusted so that only one photon is registered at a time. But only one photon is detected and as such it is not a wave. In fact that's about what one would expect for particles. I.e. if a beam of particles impinges on a single slit then there is a spread of particles on the screen. But only *one* particle does not show any wave property. However if the experiment is done a large number of times and each 'photon detection' location recorded then in the analysis the data will show a diffraction pattern. Or you can do the experiment once but have a large number of experiments done with a single photon and the results compared - then there's a diffraction pattern. The pattern is clearly defined by the wave-function and only then as a probability distribution

But not a single particle - that's why they call it the wave-particle duality

Pete

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