Real equation of Force F = ma not dp/dt

Discussion in 'Alternative Theories' started by martillo, Aug 4, 2019.

  1. martillo Registered Senior Member

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    895
    This is an extract of "Appendix A" in my manuscript on a new theory which can be accessed through the link in my profile if someone would be interested.

    The real Equation of Force is F = ma

    Today's Physics is stating that the Equation of Force is F = dp/dt.
    We will analyze the equation of motion of rockets to see that the real Equation of Force is:
    F = ma

    A rocket has variable mass in its trajectory and it's important to see its motion equation.
    Let m be its variable mass at any instant in its movement composed by the mass of the rocket plus the mass of its contained fuel.
    I have made a search in the internet about rocket motion equations and all the sites agree in the equation:
    F = m(dv/dt) = –ve(dm/dt) where ve is the speed of the fuel expelled relative to a frame at rest.

    One web site:
    http://www.braeunig.us/space/propuls.htm

    They all agree that the force acting on the rocket is due to the expelled mass and is F = –ve(dm/dt) and that the equation of motion is F = m(dv/dt) = ma.
    I assume the equation have been completely verified experimentally with enough precision from a long time ago.
    It is evident that it is used the equation:
    F = ma
    for the force and not:
    F = dp/dt

    ... ... …

    This indicates that today's Physics is wrong stating the Equation of Force as F = dp/dt.

    The right equation for force is F = ma even when mass varies.

    Note that the natural derivation of the famous equation E = mc2 by Relativity Theory has no sense since it is based in the wrong relation F = dp/dt.
    Relativity Theory becomes a wrong theory since it is based on a wrong law.
     
    Last edited: Aug 5, 2019
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  3. origin In a democracy you deserve the leaders you elect. Valued Senior Member

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    No that is not right because p=mv so dp/dt =m dv/dt.
    Therefore F = ma = m dv/dt = dp\dt.
     
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  5. martillo Registered Senior Member

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    895
    No...
    p = mv
    dp/dt = d(mv)/dt = mdv/dt + vdm/dt
    While mdv/dt = ma is only the first term in the equation
    dp/dt = ma only if dm/dt = 0
    Not the same…
     
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  7. origin In a democracy you deserve the leaders you elect. Valued Senior Member

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    No, m is a constant. The derivative of a constant is 0!

    Wait, isn't this the same mistake you have been making for a couple of years now? Oh well, you get to believe whatever you want.
     
  8. Q-reeus Valued Senior Member

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    3,417
    Neglecting relativistic corrections where p = (gamma)mv, your formal derivation is trivially correct and also how it is physically. But it must be interpreted and applied correctly. Failure to do so can lead to concluding as you have that it's somehow wrong.
    Hint: in an inertial frame where initially the rocket is stationary, net momentum remains zero regardless of rocket speed over time.
     
  9. martillo Registered Senior Member

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    895
    No. m is not constant. Not for Rocket Dynamics, not for Relativity Theory.
     
  10. Q-reeus Valued Senior Member

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    3,417
    Hint in #5 evidently not taken. Continuing then....
    Expelled exhaust gases obviously provide the thrust force F = -vdm/dt (that second term in your derivation of dp/dt). Where m is the mass of exhaust gas expelled per second, and v is the velocity of expelled gas wrt instantaneous rocket frame.
    Thus F = -vdm/dt is the negative of rate of change of exhaust gas momentum.
    System net momentum is constant. Action-reaction. Hence the rocket itself picks up an opposite rate of change of momentum which is just F already given.
    The fact that the net rocket mass M-mt (M initial net rocket mass, t the duration of rocket burn) is time varying merely means the instantaneous acceleration, and accumulated velocity, is more complicated to evaluate than assuming a constant net M. It has no additional effect on rate of change of rocket momentum which is always just F = -vdm/dt (assuming a constant burn rate of course).
    Still maintaining formal dp/dt is somehow wrong?
     
    Last edited: Aug 6, 2019
  11. martillo Registered Senior Member

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    895
    Seems right for m excepting the last phrase.
    Everywhere in the rockets dynamics world it is stated that the force on the rocket is given by the "thrust equation":

    F = ma = mdv/dt = -ve(dm/dt)
    where ve would be the velocity of the expelled fuel relative to the rocket (excuse me the change in notation reserving v for the velocity of the rocket in a rest frame).

    It is clair that it is considered F = ma and not F = dp/dt which would give something different:

    "Formal dp/dt" is for me: dp/dt = mdv/dt + vdm/dt being given by the definition p = mv and simple derivation.

    What I consider wrong is to mantain stating that F = dp/dt is always valid, while it is actually valid only under the assumption dm/dt = 0 meaning a not varying constant mass.
    So the situation is the inverse as it is currently considered: F = dp/dt is actually valid when constant mass is present (dm/dt = 0) while F = ma is the general formulation even for a varying mass.
    As F = dp/dt is largelly applied in Relativity, even when mass varies, I conclude Relativity fails on this and so it becomes a wrong theory.
     
    Last edited: Aug 6, 2019
  12. paddoboy Valued Senior Member

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    Please Register or Log in to view the hidden image!

    You've concluded that here and also elsewhere, yet the last time I looked Relativity still matches our observational data and is the gravity theory of choice.
    Seems like you are pissing into the wind.

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  13. martillo Registered Senior Member

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    895
    You can think what you want. Is your opinion and your decision.
    I'm based on the largelly experimentally verified rockets' dynamics which demonstrates I am right.
    For the argument that Relativity explains a large number of experiments I can say that I'm working in a new theory that explains everything Relativity explains but with a different better new point of view.
     
    Last edited: Aug 6, 2019
  14. paddoboy Valued Senior Member

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    22,846
  15. James R Just this guy, you know? Staff Member

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    Why do you consider it wrong?

    If $p=mv$ then
    $\frac{dp}{dt}=m\frac{dv}{dt} + v\frac{dm}{dt}$
    This is just maths, and I assume you accept it. Please tell me if you don't.

    So, I assume that what you don't accept is that the force on a system is equal to the rate of change of momentum of the system. Or, rather, you only accept it in the case that the system has constant mass (so that $dm/dt=0$). Is that correct?

    Tell me how, in general, you relate the net force on a system to the change in momentum of that system, for the non-constant mass case. If you can provide it, I would also be interested to see any experimental evidence that you are right and all professional physicists are wrong about this.

    Show me how you derive that F=ma even for a varying mass system.
     
  16. Q-reeus Valued Senior Member

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    3,417
    First, an erratum re #7: In first line there it should have read in part "...Where dm/dt is the mass of exhaust gas expelled per second..."

    Now, to reiterate from #5 - correct use of dp/dt = d(mv)/dt = mdv/dt + vdm/dt requires proper interpretation and application.
    We agree the propulsive force is F = -udm/dt (using your notation u for exhaust relative velocity). The reaction to that prescribed, constant thrust, is an instantaneous rocket acceleration
    dv/dt = F/(M-mt) - where v is the rocket velocity
    (It's assumed this is happening out in deep space where gravity and air-resistance don't complicate matters.)
    That dv/dt is time changing is irrelevant to the fact the rocket rate of change of momentum is constant and simply F = -udm/dt - the prescribed thrust.
    Confusing action and reaction, cause and effect will create a false issue.

    As far as relativity is concerned, it's an extremely well verified fact that particle accelerators absolutely work on the basis that for highly relativistic particles,
    vdm/dt completely dominates over mdv/dt. Your claim otoh implies that particles could easily be accelerated to superluminal speeds, but it never happens.
     
  17. przyk squishy Valued Senior Member

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    3,190
    So \(\boldsymbol{F} = \frac{\mathrm{d}\boldsymbol{p}}{\mathrm{d}t}\) only applies to closed systems, i.e., systems composed of matter that is not changing over time. This is well known, for example, at least enough to be pointed out on Wikipedia (emphasis added):

    Your conclusion that modern physics is "wrong" is contradicted by your own evidence:
    So look what you've done here. Your evidence that \(\boldsymbol{F} = \frac{\mathrm{d}\boldsymbol{p}}{\mathrm{d}t}\) is invalid for an accelerating rocket is that physicists behave like they believe it is invalid in that circumstance. So what is anyone actually wrong about?


    Like with \(\boldsymbol{F} = \frac{\mathrm{d}\boldsymbol{p}}{\mathrm{d}t}\), \(\boldsymbol{F} = m \boldsymbol{a}\) also isn't universally valid and it's easy to think of situations where it doesn't work.

    For example, consider a system S composed of all the matter that happens to be in some region of space R. Suppose region R is open and matter is flowing into and out of it all the time. Then the centre of mass of system S in general won't obey \(\boldsymbol{F} = m \boldsymbol{a}\) since it can move around in a complicated way even if no force is applied to anything.


    \(\boldsymbol{F} = \frac{\mathrm{d}\boldsymbol{p}}{\mathrm{d}t}\) is valid in special relativity under the same conditions it is valid in nonrelativistic mechanics, except with \(\boldsymbol{p} = \gamma m \boldsymbol{v}\) where \(\gamma = (1 - v^{2}/c^{2})^{-1/2}\) and \(m\) is the rest (invariant) mass. This means the analogue of \(\boldsymbol{F} = m \boldsymbol{a}\) in relativity is

    \(\begin{eqnarray*} \boldsymbol{F} &=& m \frac{\mathrm{d}\gamma}{\mathrm{d}t} \boldsymbol{v} + m \gamma \boldsymbol{a} \\ &=& \gamma^{3} m (\boldsymbol{a} \cdot \boldsymbol{v}) \boldsymbol{v}/c^{2} + \gamma m \boldsymbol{a} \\ &=& \gamma^{3} m \Bigl( \boldsymbol{a} + \frac{1}{c^{2}} (\boldsymbol{a} \times \boldsymbol{v}) \times \boldsymbol{v} \Bigr) \,. \end{eqnarray*}\)​

    (The last line can be derived using \(\gamma^{-2} = 1 - v^{2}/c^{2}\) and the general identity \(\boldsymbol{a} \times (\boldsymbol{b} \times \boldsymbol{c}) = \boldsymbol{b} (\boldsymbol{a} \cdot \boldsymbol{c}) - \boldsymbol{c} (\boldsymbol{a} \cdot \boldsymbol{b})\) for the vector and scalar products of Euclidean vectors.)
     
    Last edited: Aug 6, 2019
  18. martillo Registered Senior Member

    Messages:
    895
    paddoboy wrote "Nup, you are still wrong."
    Definition:

    p = mv

    By simple derivatives:

    dp/dt = mdv/dt + vdm/dt

    Defintion (not a derivation):

    F = ma = mdv/dt (even when mass varies). Rockets dynamics shows it Works fine.

    Deduced general relation of momentum and force:

    dp/dt = F + vdm/dt


    End note:
    My new theory explains everything Relativity Theory and Quantum Theory explain better and more just with a correction in the definition of the Fields and Forces and with the right structure for the elementary particles of the Universe. Certainly with a new point of view of course. It takes time to realize about this of course. Obviously many experiments have a different new interpretation, it could not be other way.
     
    Last edited: Aug 6, 2019
  19. martillo Registered Senior Member

    Messages:
    895
    For Q-reeus and przyk:
    I will only answer to you with a popular quote: "There's no more bline that who don't want to see".

    Note:
    My new theory explains everything Relativity Theory and Quantum Theory explain better and more just with a correction in the definition of the Fields and Forces and with the right structure for the elementary particles of the Universe. Certainly with a new point of view of course. It takes time to realize about this of course. Obviously many experiments have a different new interpretation, it could not be other way.
     
    Last edited: Aug 6, 2019
  20. martillo Registered Senior Member

    Messages:
    895
    NOTE:
    In my new theory some things have no sense like, Universe's Expansion", "Dark Matter", "Dark Energy", "Black Holes", "Wave-Particle Duality", "Uncertainty Principle", "Quantum Entanglement", "Quarks Theory", "Parallel Universes", "WormHoles"...

    APPROPIATED QUOTE:
    "Classical Phsics is comIng back, reloaded."
     
    Last edited: Aug 6, 2019
  21. exchemist Valued Senior Member

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    Too true, Squire.

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  22. Beer w/Straw Transcendental Ignorance! Valued Senior Member

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    You've gotta' setup an experiment.
     
  23. martillo Registered Senior Member

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    895
    Which one?
    I propose new versions of the "Davisson-Germer" and the "Double Slit" experiments I cannot setup on myself.
    For now well known experiments are explained in a different way with a totally new viewpoint.
     
    Last edited: Aug 6, 2019

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