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Questions about forces Newton's Laws (first law)

  1. Feb 6, 2010 #1
    Hi, it is known that all bodies moving with a velocity will remain moving with a constant velocity unless a force is applied to a body. (according to newton's laws). But I thought about this:

    F = Ma. But if we had a very very very very small object (hypothetically), we could virtually take M = 0 and F = 0, but this small object may still have an acceleration. This is probably okay, since M = 0 can only an approximation - but it does seem like it goes against the law stated above.. (that the acceleration may be somewhat independent of the force) . I just wanted to point this out and I hope someone will set me right. thanks
    Last edited: Feb 6, 2010
  2. jcsd
  3. Feb 6, 2010 #2
    Acceleration due to what?
    Why should this object be accelerating?
  4. Feb 6, 2010 #3
    but if force is equated as F = Ma, zero force doesn't imply zero acceleration , zero acceleration implies zero force, but not the other way around
  5. Feb 6, 2010 #4
    All you are saying is:
    - There is a very very small mass M close to 0.0
    - Oh by the way it is accelerating....at a

    Therefore if F=Ma, then F is close to 0.0...!

    So what about it?? For a very very small mass M, it does not take much force to create an acceleration a :) ...so can't you you can imagine that an F close to 0 can accelerate a M close to 0, at a....?
  6. Feb 6, 2010 #5
    right, that's true. I was just taken aback because I was looking at a problem solution that assumed that the acceleration was zero, because the force was zero (the object was stated to be really really small in the question) thanks
  7. Feb 7, 2010 #6
    First of all, you need to realize that Isaac Newton wrote

    [tex]F=\frac{d p}{d t}[/tex]

    Before Relativity showed that not all time is the same for all observers, you could use

    [tex]p = m v[/tex]

    to get

    [tex]F = \frac{d p}{d t} = \frac{d \left( m v \right)}{d t} = m \frac{d v}{d t} = m a[/tex]

    Now, if [tex]m \neq 0[/tex] no matter how small, then for non-relativistic type motions with F=0, then clearly

    [tex]F = 0 = m a[/tex]

    means that a=0 also (since [tex]m \neq 0[/tex]). Hence, v=constant; and so Newton's first law remains valid.

    For m=0 and relativistic motion, the situation becomes more complicated, but I don't think you were asking about that.

    73s and clear skies.
    Last edited: Feb 7, 2010
  8. Feb 7, 2010 #7
    If we allow the possibility that m = 0, then, as far as I can see, Holezch is right to say that "if force is equated as F = Ma, zero force doesn't imply zero acceleration, zero acceleration implies zero force, but not the other way around". In that case,

    [tex]F = ma \text{ and } F = 0[/tex]

    don't, on their own, imply [itex]a = 0[/itex], although Newton's first law does say no acceleration without force in an inertial frame, so maybe that's what the problem solution was relying on. I suppose that implies no acceleration for a massless particle, since no mass implies no force, and no force implies no acceletation.

    But are Newton's laws valid in general for m = 0?
  9. Feb 7, 2010 #8
    Yes, an example is the force-free photon. From Special Relativity we have

    [tex]E^2=p^2 c^2 + m^2 c^4[/tex]

    and for the photon from experiment


    so that [tex]m=0[/tex] for the photon.
    Newton's second law is

    [tex]F\equiv\frac{d p}{d t}=\frac{1}{c}\cdot\frac{d E}{d t}=\frac{P}{c}[/tex]
    where P=power dissipated by the photon.

    When no power is dissipated, then

    [tex]F\equiv\frac{d p}{d t}=0[/tex]

    which means that p is constant. Thus the p-vector never changes and so the photon continues in a straight line with speed c just as Newton's first law says.
    Last edited: Feb 7, 2010
  10. Feb 7, 2010 #9
    Here's my attempt, but I didn't really get anwhere. Let F be 3-force, and p 3-momentum, then

    [tex]\textbf{F} = \frac{\mathrm{d} \textbf{p}}{\mathrm{d} t} = m_0 \frac{\mathrm{d} }{\mathrm{d} t}\left ( \gamma \textbf{v} \right ) = m_0 \frac{\mathrm{d} \gamma}{\mathrm{d} t} \textbf{v}\right + m_0 \gamma \frac{\mathrm{d} \textbf{v}}{\mathrm{d} x},[/tex]

    which, unless I've made a mistake (quite possible!), gives

    [tex]\textbf{F} = m_0 \gamma \left ( \textbf{a} - \frac{\gamma v \textbf{v}}{\sqrt{c^2 - v^2}} \right ).[/tex]

    But as [itex]v \rightarrow c, \enspace \gamma \rightarrow \infty[/itex], so we'd have

    [tex]\textbf{F} = 0 \infty \left ( \textbf{a} - \frac{\infty v \textbf{v}}{\infty} \right ).[/tex]

    Argh... unless there's some meaningful way to find the limit?

    [tex]\lim_{m_0 \rightarrow 0, v \rightarrow c} m_0 \gamma \left ( \textbf{a} - \frac{\gamma v \textbf{v}}{\sqrt{c^2 - v^2}} \right ).[/tex]
  11. Feb 7, 2010 #10
    Can a photon be affected by a force and change in energy, and if so, is the effect of force on a photon only to change its energy, rather than its trajectory?
  12. Feb 7, 2010 #11
    Either way. Note that if [tex]\stackrel{\rightarrow}{p}[/tex] changes then it can change either in magnitude or direction, or even both.

    73s and clear skies.
  13. Feb 7, 2010 #12
    I tried to reply earlier and evidently the reply got bounced off somewhere. So if there is another answer, I apologize.

    Except for a minor typo, it's good.
    I got

    [tex]\textbf{F} = m_0 {\left ( \gamma \textbf{a} + {\gamma}^3 \textbf{v}\frac{\textbf{v}\cdot\textbf{a}}{c^2} \right )}[/tex]

    This can be decomposed into

    [tex]\textbf{F} = \textbf{F}_\perp + \textbf{F}_\parallel[/tex]

    where [tex]\textbf{F}_\perp}=\gamma m_0 \textbf{a}[/tex] corresponds to [tex]\textbf{v}\cdot\textbf{a}=0[/tex] and [tex]\textbf{F}_\parallel={\gamma}^3 m_0 \textbf{a}[/tex] corresponds to [tex]\textbf{v}\cdot\textbf{a}={v a}[/tex]

    In either case, Newton's first law remains valid.
    Last edited: Feb 7, 2010
  14. Feb 7, 2010 #13
    Is that 4-momentum or 3-momentum? The magnitude of a photon's 4-momentum, being its rest mass, is always 0, right? So, I guess you mean 3-momentum here.
  15. Feb 7, 2010 #14
    Take two:

    [tex]\frac{\mathrm{d} \mathbf{p}}{\mathrm{d} t} = m_0 \frac{\mathrm{d} }{\mathrm{d} t}\left ( \gamma \mathbf{v} \right ) = m_0 \left [ \frac{\mathrm{d} \gamma}{\mathrm{d} t} \mathbf{v} + \gamma \frac{\mathrm{d}\mathbf{v}}{\mathrm{d} t} \right ].[/tex]


    [tex]\frac{\mathrm{d} \gamma}{\mathrm{d} t} = \frac{\mathrm{d}}{\mathrm{d}t}\left ( \frac{1}{\sqrt{1-\left ( \frac{v}{c} \right )^2}} \right ).[/tex]


    [tex]z=\frac{v}{c}, \enspace y=1-z^2, \enspace x=y^{-1/2},[/tex]

    so that

    [tex]\frac{\mathrm{d} \gamma}{\mathrm{d} t}=\frac{\mathrm{d} x}{\mathrm{d} y} \frac{\mathrm{d} y}{\mathrm{d} z} \frac{\mathrm{d} z}{\mathrm{d} v} \frac{\mathrm{d} v}{\mathrm{d} t} = \left ( -y^{-3/2} \right ) \cdot \left ( -2z \right ) \cdot \frac{1}{c} \cdot \frac{\mathrm{d} v}{\mathrm{d} t}[/tex]

    [tex]= \frac{2v \gamma^3}{c^2} \frac{\mathrm{d} v}{\mathrm{d} t} = \frac{2av \gamma^3}{c^2}.[/tex]

    That gives

    [tex]\textbf{F} = \frac{\mathrm{d} \textbf{p}}{\mathrm{d} t}= \gamma m_0 \left [ \frac{2av \gamma^2}{c^2} \textbf{v} + \textbf{a} \right ].[/tex]

    And the part of force parallel to acceleration is

    [tex]\textbf{F}_\parallel = \gamma m_0 \textbf{a},[/tex]

    so the part perpendicular to acceleration must be

    [tex]\textbf{F}_\perp = \frac{2avm_0 \gamma^3}{c^2} \textbf{v},[/tex]

    which, I think, implies [itex]\textbf{v} \cdot \textbf{a} = 0[/itex] or [itex]m_0 = 0[/itex], in which latter case [tex]\textbf{F} = 0[/itex]. So acceleration is always perpendicular to velocity in an inertial frame? But we know that it's not, so I must have got something wrong.

    I don't understand what happened to the 2 in your version, and I'm not sure how you equate [itex]av[/itex] with [itex]\textbf{a} \cdot \textbf{v}[/itex]. This will be true when velocity and acceleration are parallel, which will be true when both are the zero vector. And velocity and acceleration could be parallel when [itex]\textbf{F} = \textbf{0}[/itex]. But I don't understand how it can be true in general, given that the equation says mutually orthogonal parts of [itex]\textbf{F}[/itex] point in the directions of [itex]\textbf{v}[/itex] and [itex]\textbf{a}[/itex].

    Is this equation for a frame comoving with this body being accelerated? If so, I suppose it can't be applied to body moving at c, and hence not to a massless particle if we start with the assumption that a massless particle must move at c.
  16. Feb 9, 2010 #15
    If it is known that, all bodies moving with a velocity remain moving with a constant velocity unless a force is applied to a body. (According to Newton’s laws). WHAT force {strength} is moving these bodies ??? In addition, has this theory ever been done in a real live air tight vacuum, without and gravity effecting it ??? I would think not, as it would have to be as big as our World. Thus how do we know it is true, and as I said, after the acceleration, what force is then being applied to keep the body in a constant speed ??? As nothing moves without a force.

    If the Moon is able to move the tides, why are we weightless and hardily move in the space, I mean the space between us and the Moon ??? As if the gravity of the Moon pulls Millions of tons of water, why does it not pull us ???

  17. Feb 9, 2010 #16


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    This question makes no sense because it contradicts what you said in your first sentence. Do you see it?

  18. Feb 9, 2010 #17
    But I said "if" its known.

  19. Feb 9, 2010 #18


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    So Newton's laws are still in doubt with no verifiable experimental evidence? Run out of your house immediately!

  20. Feb 9, 2010 #19
    Zero force! It goes against our earthbound intuitions, but it's proved a very useful point of view. You can just as well choose a reference frame (coordinate system) that's moving at the same constant velocity as the object; in other words, a coordinate system in which the object isn't moving, and Newton's laws will work just the same.

    Newtonian mechanics is very well tested, as you'll find if you do a bit of reading. It works well within its area of application. According to the definitions it uses, force is not necessary for an object to move, only for the object's velocity to change.

    The tides are caused by the difference in the moon's gravitational influence on the far side of the earth (where it's weaker) and the near side (where it's stronger). The moon's gravity doesn't differ much in strength from one side of a little space capsule to another.
  21. Feb 9, 2010 #20
    If that pesky moon hasn't already dragged you out of it...
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