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Thread: Acceleration of A Classical Particle without A Force Acting on The Particle

  1. #1 Acceleration of A Classical Particle without A Force Acting on The Particle 
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    Excuse me ... .

    In sight, it is impossible that a classical particle can be accelerated without a force acting on it ... . In other word ... , the velocity of the particle cannot change if the force acting on it is zero, sightly ... .

    But ... , in fact, this opinion is not always true generally ... .

    Really, the Newton’s second law of motion is a differential equation


    where the and the , respectively, are the force and the momentum of the particle as function of the time ... .

    If the velocity and mass of the particle, respectively, are and as function of time , then ... . Thus, in general,


    where is rate of the mass, and is its acceleration ... . The last expression doesn’t only obey in non-relativistic cases, but also in relativistic cases ... .

    If we set , then ... . In this case, alias alias alias ... . Thus (by subtituting or by differentiating), this yields ... . [Here, the and the denote the initial mass and the initial velocity, respectively ... .]

    In another case, if , then ... .

    Therefore, really, the force merely changes the momentum rather than the velocity ... .


    Last edited by trfrm; April 14th, 2013 at 10:18 PM.
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    Quote Originally Posted by trfrm View Post
    Excuse me ... .

    In sight, it is impossible that a classical particle can be accelerated without a force acting on it ... . In other word ... , the velocity of the particle cannot change if the force acting on it is zero, sightly ... .

    But ... , in fact, this opinion is not always true generally ... .

    Really, the Newton’s second law of motion is a differential equation


    where the and the , respectively, are the force and the momentum of the particle as function of the time ... .

    If the velocity and mass of the particle, respectively, are and as function of time , then ... . Thus, in general,


    where is rate of the mass, and is its acceleration ... . The last expression doesn’t only obey in non-relativistic cases, but also in relativistic cases ... .

    If we set , then ... . In this case, alias alias alias ... . Thus (by subtituting or by differentiating), this yields ... . [Here, the and the denote the initial mass and the initial velocity, respectively ... .]

    In another case, if , then ... .

    Therefore, really, the force merely changes the momentum rather than the velocity ... .
    This stuff is really bad. Care to try again, given the information that . May I suggest that you forget about the nonsense . This doesn't exist.
    If you want to do "classical domain", just set


    Last edited by Howard Roark; April 14th, 2013 at 11:20 PM.
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  4. #3  
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    Quote Originally Posted by xyzt View Post
    This stuff is really bad. Care to try again, given the information that . May I suggest that you forget about the nonsense . This doesn't exist.
    If you want to do "classical domain", just set
    O.K. ... . Thanks ... .

    We have known that the relativistic momentum is , where and ... . [Here, the is not the initial mass, but is the rest mass ... .] Then,



    where and ... , so

    ... .

    Thus ... (by subtituting), this yields


    which be known as the relativistic force ... .

    [Note that the special relativity (without quantum aspect) is sometimes classified into the classical mechanics subject ... . I saw it in the Classical Mechanics textbook by Hebert Goldstein, 1980 ... .]
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    Quote Originally Posted by trfrm View Post
    Quote Originally Posted by xyzt View Post
    This stuff is really bad. Care to try again, given the information that . May I suggest that you forget about the nonsense . This doesn't exist.
    If you want to do "classical domain", just set
    O.K. ... . Thanks ... .

    We have known that the relativistic momentum is , where and ... . [Here, the is not the initial mass, but is the rest mass ... .] Then,



    where and ... , so

    ... .

    Thus ... (by subtituting), this yields



    which be known as the relativistic force ... .

    [Note that the special relativity (without quantum aspect) is sometimes classified into the classical mechanics subject ... . I saw it in the Classical Mechanics textbook by Hebert Goldstein, 1980 ... .]
    Better but still wrong, you should get:

    Last edited by Howard Roark; April 15th, 2013 at 02:06 AM.
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  6. #5  
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    Hence, if the force was zero, then the acceleration should not be zero ... . In other word (in the relativistic case without presence of force), , so alias alias

    ... .

    N.B. : The and the , respectively, are the initial gamma and the initial velocity (at ) ... .

    Because the final velocity has same direction with the initial velocity, then


    Thus, in this case, if ... .
    Last edited by trfrm; April 15th, 2013 at 01:00 AM.
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    Quote Originally Posted by trfrm View Post
    Hence, if the force was zero, then the acceleration should not be zero ... . In other word (in the relativistic case without presence of force), , so alias alias

    ... .

    N.B. : The and the , respectively, are the initial gamma and the initial velocity (at ) ... .

    Because the final velocity has same direction with the initial velocity, then


    Thus, in this case, if ... .
    I don't know where you are trying to get, you simply have:




    meaning that you get

    A little further algebraic manipulation shows that the above is impossible unless .

    Last edited by Howard Roark; April 15th, 2013 at 12:09 PM.
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    Oh O.K. ... . From the identity of vectors , then alias ... .

    Thus,

    alias

    ... .

    Since , then

    ... .

    In the last equation, we can see that there are transversal and longitudinal components of the force ... .
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    Quote Originally Posted by trfrm View Post
    Oh O.K. ... . From the identity of vectors , then alias ... .

    Thus,

    alias

    ... .

    Since , then

    ... .

    In the last equation, we can see that there are transversal and longitudinal components of the force ... .
    It's getting worse, you should really arrive to :

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    Quote Originally Posted by xyzt View Post
    I don't know where you are trying to get, you simply have:




    meaning that you get

    A little further algebraic manipulation shows that the above is impossible.
    Here ... , incidentally, ... .

    Quote Originally Posted by xyzt View Post
    It's getting worse, you should really arrive to :

    It turns out that the equation is not different from the equation ... .
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    Quote Originally Posted by trfrm View Post
    Quote Originally Posted by xyzt View Post
    I don't know where you are trying to get, you simply have:




    meaning that you get

    A little further algebraic manipulation shows that the above is impossible.


    It turns out that the equation is not different from the equation ... .


    If this means const, i.e. (i.e ) so . This is the correct chain of inference, what you posted so far, isn't.
    Last edited by Howard Roark; April 15th, 2013 at 11:12 AM.
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    I think it is important to distinguish pure mathematics from physics in this context. Mathematically it is certainly possible to come up with all sorts of relations affecting speed and acceleration even in the absence of a force, but that does not necessarily mean those relations are physically meaningful. In the context of physics, changing the state of motion of an object requires changing its momentum, and, at least classically, that will require a force.
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    Quote Originally Posted by xyzt View Post
    If this means const, i.e. (i.e ) so . This is the correct chain of inference, what you posted so far, isn't.
    In this context (this relativistic case) ... , it is right ... .

    But, in general case, if , then (where the index denotes the initial state) ... , then ... , alias ... .

    Because in general, the mass is not constant, then


    which must not to be zero ... .

    But, however, in the relativistic case, , so the last equation becomes

    ... .

    The possible solution of the last equation is ... .


    Quote Originally Posted by Markus Hanke View Post
    I think it is important to distinguish pure mathematics from physics in this context. Mathematically it is certainly possible to come up with all sorts of relations affecting speed and acceleration even in the absence of a force, but that does not necessarily mean those relations are physically meaningful. In the context of physics, changing the state of motion of an object requires changing its momentum, and, at least classically, that will require a force.
    This seemly comes from the origin definition of the force as derivative of the linear momentum respect to time, that is the Newton’s second law of motion as a differential equation [Goldstein, Classical Mechanics, 1980, Chapter 1] ... .
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    Quote Originally Posted by trfrm View Post
    Quote Originally Posted by xyzt View Post
    If this means const, i.e. (i.e ) so . This is the correct chain of inference, what you posted so far, isn't.
    In this context (this relativistic case) ... , it is right ... .

    But, in general case, if , then (where the index denotes the initial state) ... , then ... , alias ... .

    Because in general, the mass is not constant, then


    which must not to be zero ... .
    This is nonsense.



    But, however, in the relativistic case, , so the last equation becomes

    ... .

    The possible solution of the last equation is ... .


    Quote Originally Posted by Markus Hanke View Post
    I think it is important to distinguish pure mathematics from physics in this context. Mathematically it is certainly possible to come up with all sorts of relations affecting speed and acceleration even in the absence of a force, but that does not necessarily mean those relations are physically meaningful. In the context of physics, changing the state of motion of an object requires changing its momentum, and, at least classically, that will require a force.
    This seemly comes from the origin definition of the force as derivative of the linear momentum, that is the Newton’s second law of motion as a differential equation [Goldstein, Classical Mechanics, 1980, Chapter 1] ... .
    I don't know why you insist on using the antiquate notion of relativistic mass, this is the source of confusion. Using the modern notion of invariant mass makes things a lot more straightforward. And yes, is the only acceptable solution for the equation:

    Last edited by Howard Roark; April 15th, 2013 at 12:16 PM.
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    Quote Originally Posted by xyzt View Post
    I don't know why you insist on using the antiquate notion of relativistic mass, this is the source of confusion. Using the modern notion of invariant mass makes things a lot more straightforward. And yes, is the only acceptable solution for the equation:

    I’m sorry ... . There, and are the initial and final relativistic mass, respectively ... .
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    Quote Originally Posted by trfrm View Post
    Quote Originally Posted by xyzt View Post
    I don't know why you insist on using the antiquate notion of relativistic mass, this is the source of confusion. Using the modern notion of invariant mass makes things a lot more straightforward. And yes, is the only acceptable solution for the equation:

    I’m sorry ... . Hence, and ... .
    I know, no one is using "relativistic mass" anymore, it is messy and a never ending source of errors and misunderstandings.
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