F=ma on small masses?

  • Thread starter Rockazella
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I know that QM must be used to describe atomic sized systems, but is classical still applicable? Take an electron for example. I dont the the mass of an electron so I'll just say its m=x. If an electron were to have a force of 10N applied to it, would its acceleration be 10/x, or do these equations have no use at this size?
 

russ_watters

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Originally posted by Rockazella
I know that QM must be used to describe atomic sized systems, but is classical still applicable? Take an electron for example. I dont the the mass of an electron so I'll just say its m=x. If an electron were to have a force of 10N applied to it, would its acceleration be 10/x, or do these equations have no use at this size?
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.
 

FZ+

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F = dp/dt = rate of change of momentum. Theoretically, this should apply to all sorts of things - including impacts with photons.
 
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christench

Actualy you cant really think of applying a force like that to an electron. It doesnt make much sense because its a wave. The action of the electron to a froce very much depends on the state of the electron.
 

Hurkyl

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Blah, F=ma is a bad thing to try to apply at relativistic speeds, since transverse and longitudinal inertia are different at relativistic speeds. F=dp/dτ is the way to go!
 
P

pmb

Originally posted by Rockazella
I know that QM must be used to describe atomic sized systems, but is classical still applicable? Take an electron for example. I dont the the mass of an electron so I'll just say its m=x. If an electron were to have a force of 10N applied to it, would its acceleration be 10/x, or do these equations have no use at this size?
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
 
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pmb

Re: Re: F=ma on small masses?

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
 
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pmb

Originally posted by christench
Actualy you cant really think of applying a force like that to an electron. It doesnt make much sense because its a wave. The action of the electron to a froce very much depends on the state of the electron.
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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curro_jimenez

You can use classical equations if the wave properties of the particle (i.e the electron) are not important for the problem. Example: Calculation of the electron mobility in a semiconductor. Habitually this a first approach, semi-classical.

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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christench

An electron is a 'wave', and the 'wave function' certainly has more meaning than statistical. The reason it isnt actualy a wave is because we dont solve a wave equation, but a different equation.
 
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pmb

Originally posted by christench
An electron is a 'wave', and the 'wave function' certainly has more meaning than statistical. The reason it isnt actualy a wave is because we dont solve a wave equation, but a different equation.
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.

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
 

FZ+

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

russ_watters

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Re: Re: Re: F=ma on small masses?

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
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?

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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christench

An electron is ALWAYS a 'wave'. It is only modeled as a particle at the classical limit, rather like Newtons laws are used at non-relativistic limits. The only reason for making such an approximation is to make the math easier.
 
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curro_jimenez

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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pmb

Re: Re: Re: Re: F=ma on small masses?

Originally 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?
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 is

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.


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.
How would you measure the 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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pmb

Originally posted by christench
An electron is ALWAYS a 'wave'. It is only modeled as a particle at the classical limit, rather like Newtons laws are used at non-relativistic limits. The only reason for making such an approximation is to make the math easier.
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
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.
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.

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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pmb

Originally posted by FZ+
I thought the commonest description is of "both, but not at the same time".
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.

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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Forces and electrons

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

You can see the wave nature of an electron with just one electron, you dont need many. In physics a thing is what we model it as. An electron can always be modeled as a wave, and sometimes as a particle if certain approximations are good. Therefore the electron is a wave according to the accepted fundamental(rather than approximate) model. Its particulate nature only comes about when we ask classical questions about it. The wave function is the shape of the electron, it does not just give us the probability of finding the electron. You can interpret the whole thing differently like Bohm, so then the electron is a particle and the wave function some quantum interaction term. This gives even more meaning to the wave function. This interpretation is usually ignored because it requires the quantum potential, which comes from nowhere. Feynman knew the electron was a wave, he even expands wave functions in terms of trajectories which are sometimes classical and sometimes not. Part of his work describes how the wave function gets to a classical limit, but it is still a wave function.
 

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