Free particle

Discussion in 'Physics' started by logearav, Oct 22, 2011.

  1. logearav

    Thread Starter Member

    Aug 19, 2011
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    Revered members,
    A free particle is one which has no forces acting on it and since there is no forces acting on it, so there is no potential energy and E is entirely kinetic.
    My doubt is, when no forces act on a particle, it should remain at rest and possess potential energy only and Kinetic energy should be zero. How the converse happens here?
     
  2. #12

    Expert

    Nov 30, 2010
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    Back to Newton. The particle will remain at the same velocity and direction unless acted on by an external force. Kinetic energy does not have to be zero.
     
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  3. steveb

    Senior Member

    Jul 3, 2008
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    Yes, exactly.

    The reason is that you are free to chose any inertial frame of reference. An inertial frame of reference needs only to not be accelerating, which means it can be moving with any fixed velocity relative to the particle. So, kinetic energy is relative. If you chose a rest frame with zero velocity relative to the particle, then kinetic energy is zero. If you chose a moving reference frame then the kinetic energy is 0.5*m*V^2, where m is the mass and V is the velocity of the particle relative to the frame.

    Potential energy does require a force and since there are no forces, there can be no potential energy. In everyday life, where we apply Newton's laws, there are only two forces to think about: gravitational and electromagnetic. Even a spring force is fundamentally an electromagnetic force at the molecular level. In a universe with only one particle, there is no way to have other objects with mass or charge that can generate forces, hence there is no potential energy.
     
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  4. Wendy

    Moderator

    Mar 24, 2008
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    There are several problems here. A particle is in quantum foam, it is never a "free" particle. I assume this is a subatomic particle, which is in the world of quantum mechanics. You can not predict such particles as if they are in the macro world. Macro is large, our world. The two worlds, quantum and macro, might as well be different dimensions.

    If we are talking a particle of dust then yes, Newton's Laws apply.
     
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  5. someonesdad

    Senior Member

    Jul 7, 2009
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    The free particle is studied in both classical and quantum contexts. It is up to the OP to state which it is; although my guess is he was talking about classical stuff...
     
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  6. steveb

    Senior Member

    Jul 3, 2008
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    That's a good point Bill. We kind of assumed the OP is learning classical mechanics in the traditional sense. We no doubt made this interpretation based on the context, but the context is very vague when you really look at it.

    If we go beyond Newton to simple Quantum Mechanics, or simple relativity theory, our above answers are still valid because the concepts of kinetic energy and potential energy are general enough to hold up. Actually, they hold up even in quantum field theory. However, you are pointing out that the idea of a subatomic particle removed from the the properties of vacuum is not physical. I agree with you on that point, but this is invoking very high level physics principles that are meaningful at the Quantum Field theory level and beyond. We should stress to the OP not to get too hung up on your point when learning his basic physics principles, but it's a good point nonetheless.
     
    Last edited: Oct 22, 2011
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  7. Wendy

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    Mar 24, 2008
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    The main reason I pointed this out is the OP has tried to make the connection between macro and quantum in the past. He hasn't figured out there really are two worlds out there that haven't totally been reconciled.

    The experiments involving observers and no observers to create interference patterns or not is down right spooky in my book. This works for both electrons and photons. I have never heard a logical explanation, so I take it on face value. I am not the sharpest pencil in the cup, but I'm not the dullest either. I do like to read, there is no substitute for it, and the rules do change enough that it makes it even more interesting.

    Another thing in quantum mechanics. In many experiments a photon is a particle. It seems to unambiguous, but with other experiments it can split and take several different paths. This has been offered as potential proof for the many worlds theory, where our world overlaps subtly with other worlds like ours and we are seeing other paths in other worlds at the same time. I have no opinion, but it does make for interesting reading.

    When we get to the world of the very small some of the answers are just weird.
     
    Last edited: Oct 22, 2011
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  8. logearav

    Thread Starter Member

    Aug 19, 2011
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    Thanks to all, for your kind replies, revered members. I started reading about Schrodinger equation for a free particle in one dimension and found the term free particle and the explanation for it. Since, i can't understand, i opened a thread so as to elicit the views of learned members of this forum
     
  9. Wendy

    Moderator

    Mar 24, 2008
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    I know English is not your primary language (and respect you more for it). Particle is one of those words that have multiple meanings, in this case all related. A particle can be a subatomic particle or a particle of dust. Basically it means a small part of something.
     
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  10. russ_hensel

    Well-Known Member

    Jan 11, 2009
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  11. studiot

    AAC Fanatic!

    Nov 9, 2007
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    Now we know that you are discussing the Schrodinger equation, a better answer can be made.

    However there are two versions of this equation and it is important to know which one you mean.

    The two versions are time dependent and time independent.

    The time dependent version, which is implied by your requirement for a free particle, is applicable for instance to an individual electron travelling as part of a β ray.

    The time independent version is appropriate to an electron in an atom, restricted to its orbit or orbital.

    go well
     
  12. logearav

    Thread Starter Member

    Aug 19, 2011
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    iℏ∂ψ(x,t)/∂t = (-ℏ^2/2m)(∂^2ψ(x,t)/∂x^2)
    This is the equation i am referring . I think this is time dependent.
     
  13. studiot

    AAC Fanatic!

    Nov 9, 2007
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    OK consider a beam of freely moving electrons otherwise known as a beta ray.

    Quantum mechanics suggests each particle can be considered a wave so let us call the quantity that varies psi (\Psi) then

    \Psi  = {\Psi _0}\sin \left[ {2\pi \left( {\frac{x}{\lambda } - \frac{t}{T}} \right)} \right]

    Now De Broglies equation relates the momentum of a particle travelling at less than 20% of the speed of light to its quantum wavelength

    p = \frac{h}{\lambda }

    And Planks equation relates the total energy of an oscillator to its frequency

    E = h\nu

    Substituting these in yields

    \Psi  = {\Psi _0}\sin \left[ {2\pi \left( {\frac{{xp}}{h} - \frac{{Et}}{h}} \right)} \right]

    This may be also written in exponential form for ease of further calculation

    \Psi  = {\Psi _0}\exp \left[ {i\left( {xp - Et} \right)/\hbar } \right]

    Differentiate twice with respect to x and once with respect to t

    \frac{{{\partial ^2}\Psi }}{{\partial {x^2}}} =  - {\Psi ^2}\frac{{{p^2}}}{{{\hbar ^2}}}

    and

    i\hbar \frac{{\partial \Psi }}{{\partial t}} = E\Psi

    The kinetic energy of a particle with momentum p and mass m is

    {p^2}/2m

    So if V is the potential energy and E is the total energy we have

    \frac{{{p^2}}}{{2m}} + V = E

    substituting the partial equations from above and multiplying by \Psi

     - \frac{{{\partial ^2}\Psi }}{{\partial {x^2}}}\frac{{{\hbar ^2}}}{{2m}} + V\Psi  = i\hbar \frac{{\partial \Psi }}{{\partial t}}

    Which is the (one dimensional) time dependent Schrodinger equation

    Setting V = 0 (your boundary condition) yields your equation.
     
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  14. logearav

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    Aug 19, 2011
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    @studiot,
    What else I can say, except a big Thanks to you. In fact, thanks is too frugal a word to appreciate your efforts to make things clearer for me. Can i know why you multiplied by ψ in the last step?
     
  15. studiot

    AAC Fanatic!

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    If you perform the substitutions you will see that it is simply multiplying through by ψ to simplify.

    Note also the change from h to hbar in Planks constant hbar = h/2∏.

    Now that we have arrived at your equation you need to know something more.

    An oscillation or vibration is a variation of 'something' that stores energy the ψ field (strictly ψ squared).

    The actual wave shows an perpetual interchange between kinetic energy and potential energy held in the medium.

    A travelling wave (=time dependent) (such as yours) transfers energy from one point in space to another.

    A time independent ( =stationary)ot transfer or propagate energy in this way.

    Can you see the effect your particular boundary conditions have vis a vis potential energy in the medium?
     
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  16. logearav

    Thread Starter Member

    Aug 19, 2011
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    Thanks studiot.
    But how you changed the first expression into exponential form. I dont understand that step. Since i dont know how to type those expressions with this editor i attached them as images. Because e^iθ is cosθ+ isinθ. So, how you used that exponential function. Thanks again.
     
  17. studiot

    AAC Fanatic!

    Nov 9, 2007
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    Well there are two ways to go about this.

    Firstly you have quoted only one of the two Euler's formulae.

    If you take both, along with the formula for the sin of a sum and a great deal of algebra you can go from the sin to the exponential.

    \begin{array}{l}<br />
 {e^{i\theta }} = \cos \theta  + i\sin \theta  \\ <br />
 {e^{ - i\theta }} = \cos \theta  - i\sin \theta  \\ <br />
 \sin (A + B) = \sin A\cos B + \cos A\sin B \\ <br />
 \end{array}

    Alternatively we can switch to complex numbers as follows

    let

    z = i\theta

    Then
    {e^z} = \cos \theta  + i\sin \theta  =  Re (z) + Im(z)

    If we use exp(z) = exp(iθ) and take only the imaginary or real part we can select the sin or cosine

    \begin{array}{l}<br />
 \cos \theta  = Re(z) \\ <br />
 \sin \theta  = Im(z) \\ <br />
 \end{array}
     
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  18. logearav

    Thread Starter Member

    Aug 19, 2011
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    @studiot,
    if we take
    ψ=ψ0cos(2∏(xp/h -Et/h) , can we still use
    ψ=ψ0e^[i(xp-Et/\hbar)
     
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