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F=dp/dt proof?

  1. Nov 27, 2006 #1
    Ok, so I have a provblem with this.


    Where did i go wrong?
  2. jcsd
  3. Nov 27, 2006 #2
    Well, for one thing:

    [tex]\frac {dP}{dt} = M\frac{d(V)}{dt}+ V \frac{d(M)}{dt} [/tex]

    If the mass of the body does not change with time that term drops to zero.

    Then you are left with:

    [tex]\frac {dP}{dt} = M\frac{d(V)}{dt} [/tex]


    [tex]\frac{d(V)}{dt} = A[/tex]


    [tex]F=MA [/tex]

    Your flaw: DO NOT take derivatives of units.
    Last edited: Nov 27, 2006
  4. Nov 27, 2006 #3
    ok, so product rule is used here.
  5. Nov 27, 2006 #4
    Yes, it is always used.
  6. Nov 28, 2006 #5
    You have gone wrong here too.
  7. Nov 28, 2006 #6
    I didn't realize you could prove Newton's second law? Why would you go about proving something that's simply stated as fact?
  8. Nov 28, 2006 #7
    He's not proving anything.

    What he's doing is deriving the simpler


    from the more general [itex]F=dp/dt[/itex], assuming constant mass.
  9. Nov 28, 2006 #8
    If we allow the mass to change in some way depending on the velocity, does the "force" correspond to anything that looks like a relativistic "force"? What I mean is, can Newton's law in the form [tex]\frac {dP}{dt} [/tex] be shown to give the correct relativistic increase from the rest mass if [tex]\frac{d(M)}{dt} [/tex] is done correctly (ie, with some special version of M)?
  10. Nov 28, 2006 #9
    To be honest, I don't know a damn thing about relativity. I'm not studying physics, sorry. Doc Al or Van Esh can give you an answer to that question.
  11. Nov 28, 2006 #10
    @Einstein McFly: the answer to your question is no.

    Accounting for variable mass is correct still only non-relativistically.

    To look at the force law for relativistic mechanics, you have to look at four-force and it's relation to four-momentum.
  12. Dec 11, 2006 #11
    s=vt is used for a body moving with constant speed.by taking derivative your are implying that acceleration is already present i.e. speed is not constant.
  13. Dec 11, 2006 #12

    If you use relativistic mass then Newton's [itex]F=\frac{dp}{dt}[/itex] still holds, just with the modified form for momentum: [itex]p=m_{r}v[/itex], where [itex]m_{r}[/itex] is the relativistic mass [itex]\gamma m_{0}[/itex], where [itex]m_0[/itex] is the rest mass and [itex]\gamma[/itex] is the usual Lorentz factor.

    Doubt this will help. Relativistic mass equations just confuse things in my experience... it is much better to use four-vectors for everything as suggested above, it then becomes inherently more difficult to break things with bad Newtonian logic.
  14. Dec 11, 2006 #13
    Taking the derivative w/respect to time, not velocity.
  15. Dec 13, 2006 #14
    I am saying that s and t will change in such a manner that s/t is constant.so taking its derivative with respect to anything wont affect it.
    Therefore p=ms/t.dp=m(d(s/t))for a non changing mass.
    dp/dt=m(d(s/t))/dt.since d(s/t)=0,f=dp/dt=0.which is very true since the formula s=vt is used(valid only for constant speed motion).
  16. Dec 18, 2006 #15
    My question is similar. I hope I'm posting this in the right place.

    If F = ma, we can integrate to get P=mv, and integrate again
    to get Ek = 1/2 mv^2 . Since a=dv/dt, and v=ds/dt, where s
    is distance, can we replace P=mv with P=m(ds/dt) and integrate
    to get Ek = ms? This would make kinetic energy a function of distance,
    not time. Mathematically, I can't see any errors in those
    assumptions, but my instinct tells me that we can't just get rid
    of the time variable altogether. It would make sense if it were potential energy as a function of distance, though. Is there some connection?
    Does the law of conservation of energy come in somewhere
  17. Dec 19, 2006 #16
    No. In the first instance you have integrated with respect to velocity and in the second instance you have integrated with respect to time. There is no reason why these should be equal.Explicitly:

    mv &=& m\frac{ds}{dt} \\
    \int mv\,dv &\ne& \int m\frac{ds}{dt}\,dt
    Last edited: Dec 19, 2006
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