Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Can one derive Newton's

second and third laws from the first law or

first and third laws from the second law or

first and second laws from the third law

I think Newton's laws of motions are independent to each other. They can not be derived from one another. Please share the idea.

share|improve this question

5 Answers 5

up vote 3 down vote accepted

The modern interpretation of Newton's First Law is about the existence of inertial reference frames, mainly to solidify the idea that such coordinate systems exist and are important.

However, I sincerely doubt this is what Newton himself had in mind when he postulated his laws. Historically, Newton probably introduced his first law to put emphasis on the fact that moving bodies do not slow down by their own accord (as was common wisdom at the time). So yes, the first law in this context can certainly be derived from the second law by setting F=0. In fact I doubt Newton even had an idea of what an inertial frame was (though it probably could have been explained to him with relative ease).

share|improve this answer
    
The second law does not imply the first. The second law only says that F=0 implies a=0, but that does not mean that the velocity is constant, merely that the acceleration is zero, but if you have a nonzero jerk, then the acceleration can change. Jumping from a pointwise zero acceleration to a constant velocity is just like a student analyzing projectile motion, noting that the velocity is zero at the top and then assuming the projectile stays there forever. The student ignored the possibility of a nonzero acceleration, you ignored the possibility of a nonzero jerk –  Timaeus Dec 29 '14 at 1:50
1  
@Timaeus Here in the real world we need only consider accelerations. Realistic equations of motion are always first or second-order. –  jld Dec 29 '14 at 18:13
    
Even second order differential equations (i.e. F=ma) are not enough to tell you that a=0 implies that v is constant. Just like v=0 doesn't imply that x is constant, v=0 only means that the changes in x are smaller than first order in time. You can bring up whatever additional principles you want to get Newton's first law (you can even use Newton's first law itself to get Newton's first law), but don't say F=ma implies the first law. Dhar gave counterexamples in "Nonuniqueness in the solutions of Newton’s equation of motion" Am. J. Phys. 61, 58 (1993); dx.doi.org/10.1119/1.17411 –  Timaeus Dec 30 '14 at 7:01
2  
@Timaeus Why the strawman? I clearly never said "a=0 at a single instant is enough to tell you v=const." When I say "set F=0" I'm implicitly doing this for all time, and that is indeed enough to tell you that v=const. You're arguing against a strange strawman that I clearly never stated nor implied in any way. –  jld Dec 30 '14 at 19:13
2  
@Timaeus x(t)=t^3 does not have a(t)=0. It has a(t)=x''(t)=6t. The only moment where a=0 is at t=0. It is implicit in my answer (and I have now explicitly said it) that I'm setting a(t)=0 for all t, not just at a single instant. a(t)=0 does in fact imply v(t)=const. –  jld Dec 30 '14 at 19:59

Newton's laws of motion cannot be derived from each other. They are the building blocks of Newtonian mechanics and if fewer were needed, Newton would simply formulate fewer.

The first law postulates the existence of an inertial reference frame in which an object moves at constant velocity if the net force acting on it is zero. Although it might seem you can derive it from the second law (if the net force is zero, there is no acceleration and the velocity is constant) but in fact, both second and third law assume that the first law is valid. If an observer is in a non-inertial reference frame, she will observe that the second and third laws are not valid (when you sit in an accelerating car, the Earth accelerates in the opposite direction without any force acting on it).

You also cannot derive the second law from the first one because all you know from the first law is that when an object accelerates, there is a force acting but the first law says nothing about the relation between the force and the acceleration. That's what second law is for, to say that there is a linear relationship.

The third law adds something more to the first and second laws. It deals with interactions and states that two bodies exert same but opposite forces o each other. That is something you cannot see from the first or second law and similarly, there is no way to use this to derive the second law (you cannot derive the first law because that is assumed to be valid in order to postulate the third law).

share|improve this answer

No, they're not independent, because the first can be deduced from the second. Newton's second law says that $F=ma$. The first law says that $a=0$ when $F=0$, which clearly follows from $F=ma$. The purpose of the first law is not to be an independent postulate from the second law, but just to emphasise this particular special case, which presumably would have been counterintuitive to many of the contemporary readers of Newton's work.

Other answers have tried to claim that the first law is really about the existence of an inertial reference frame, and of course you're free to interpret it that way if you want, but what it actually says is not independent of the second law.

From a modern point of view, all of newton's laws follow from the conservation of momentum. For example, for two bodies in one dimension, the total momentum is $m_1v_1 + m_2v_2$. If it doesn't change over time then its derivative must be zero, i.e. $$ \frac{d}{dt}(m_1v_1 + m_2v_2) = m_1a_1 + m_2a_2 = 0. $$ If we define $F_1 = m_1 a_1$ and $F_2 = m_2 a_2$ then this becomes $F_1 = -F_2$, which is Newton's third law. The second law is just the definition of $F$, and the first law comes from noting that if you just have one body then $mv$ can't change, so $v$ has to be constant.

share|improve this answer
    
The second law does not imply the first. The second law only says that F=0 implies a=0, but that does not mean that the velocity is constant, merely that the acceleration is zero, but if you have a nonzero jerk, then the acceleration can change. Jumping from a pointwise zero acceleration to a constant velocity is just like a student analyzing projectile motion, noting that the velocity is zero at the top and then assuming the projectile stays there forever. The student ignored the possibility of a nonzero acceleration, you ignored the possibility of a nonzero jerk. –  Timaeus Dec 29 '14 at 1:41
    
@Timaeus if there's a nonzero jerk the acceleration changes, and the velocity changes because the acceleration is not zero any more. As long as the acceleration is zero, the velocity stays constant. –  Nathaniel Dec 29 '14 at 3:25
    
@Nathaniel If jerk is bounded away from zero in the past then Newton's 2nd law is totally OK with jerk being continuous and the body moving on. But Newton's 1st law would insist that the jerk drop down discontinuously to 0 and the object remain at rest, merely because the position, velocity, and force all instantaneously were 0 (at rest, no force, so stay at rest). The fact that the second law allows multiple solutions is the whole reason the 2nd can't imply the 1st. Just because the acceleration is 0 for an instant doesn't mean the velocity won't change. Consider the projectile example again –  Timaeus Dec 29 '14 at 5:23
    
@Timaeus Newton's second law says the body stays at rest if there is no force. But if there's a non-zero jerk then at time $t+\delta t$ there is a force. There has to be one in order to cause the now non-zero acceleration. So there's no contradiction, and no problem with multiple values. This is all basic calculus, which Newton understood well, but his laws are written that way to explain it to people who don't. –  Nathaniel Dec 29 '14 at 8:16
    
Newton's 2nd law says that F=ma, i.e. if you have a function x=x(t) then firstly the function x should have a second derivative a=a(t) and secondly given an F=F(x,v,t), then F(x(t),v(t),t)=m a(t) should hold for every t. This does not mean v(0)=0 and F(x(0),v(0),0) together require that x(t)=x(0). It doesn't require it even if F=F(x,v) or even if F=F(x) or even if F(x)=- grad V(x). Dhar gave counterexamples in "Nonuniqueness in the solutions of Newton’s equation of motion" Am. J. Phys. 61, 58 (1993); dx.doi.org/10.1119/1.17411 F=ma allows multiple and acausal solutions. –  Timaeus Dec 30 '14 at 7:13

You cannot derive any of the laws from each other.

Imagine a universe with two bodies of equal finite mass m. One exerts a constant force on the other that pulls it towards the origin with a force proportional to how far away it is from the origin and the other exerts no force on the one. The one has motion $x(t)=100$ and the other has motion $x(t)=\sin(\omega t)$. The first two laws are satisfied, the third is not.

Now imagine a universe with three bodies of equal finite mass m. The first two exert a constant nonzero external force on the other, each of the forces are equal and opposite. The motions are $x(t)=100$ and $x(t)=50$ and $x(t)=0$. The first law is satisfied, as is the third. The second law is not.

Now imagine a universe with three bodies $1$, $2$, and $3$ of equal finite mass m. Let $d$ and $C$ be positive nonzero constants with the appropriate units. Suppose the universe has a potential energy function $V(x_1,x_2)=-C(\frac{x_1-x_3-d}{2})^{4/3}$, so $1$ and $3$ exert equal and opposite forces on each other $F_{13}=\frac{dV}{dx_1}=-\frac{dV}{dx_3}=F_{31}$. Suppose $x_1(0)=d$ and $x_2(0)=d/2$ and $x_3(0)=0$. Further suppose that $v_1(0)=v_2(0)=v_3(0)=0$. Obviously we can satisfy all three of Newton's laws by taking as solutions $x_i(t)=x_i(0)$, however, instead suppose the particles move as $x_1(t)=d+(Kt)^3$, $x_2(t)=d/2$ and $x_3(t)=-(Kt)^3$, for $K= \sqrt{\frac{2C}{9m}}$. Then third law holds because $F_{ij}=-F_{ji}$ and the second holds because $ma_2(0)=0=F_2$ and

$ma_1=mK^36t=mK^26Kt=m\frac{2C}{9m}6(K^3t^3)^{1/3}=\frac{4C}{3}(\frac{2K^3t^3}{2})^{1/3}=\frac{4C}{3}(\frac{x_1(t)-x_2(t)-d}{2})^{1/3}=F_1$

and

$ma_3=-mK^36t=-mK^26Kt=-m\frac{2C}{9m}6(K^3t^3)^{1/3}=-\frac{4C}{3}(\frac{2K^3t^3}{2})^{1/3}=-\frac{4C}{3}(\frac{x_1(t)-x_2(t)-d}{2})^{1/3}=F_3.$

So the second and third laws are upheld, but the first law says that if no net external force acts, then the velocity is constant. This is not a property of the solution given, the velocities are all zero at $t=0$, as is the force, but yet the velocity is never constant, it is always changing, it's just changing there so slowing that $a=0$. A zero acceleration is different than an unchanging velocity. The solution $x_1(t)=d+(Kt)^3$ has a zero acceleration, but the velocity is changing. Note that the solutions $x_1(t)=d$ and $x_2(t)=d/2$ and $x_3(t)=0$ are also solutions to $F=ma$, so Newton's 2nd law allows multiple solutions with the same initial position and velocities, but the first law can pick a unique solution.

So there is an example where the 2nd and 3rd laws hold, but the 1st does not.

So none of the three can be derived from each other.

Edit I'd like to credit Abhishek Dhar's paper "Nonuniqueness in the solutions of Newton’s equation of motion" Am. J. Phys. 61, 58 (1993); http://dx.doi.org/10.1119/1.17411 for inspiring the example force law with nonunique solutions that I gave.

Ten years later Norton introduced his dome and noticed that you can have the stay-at-rest solution persist either forever, or for any finite amount of time and then spontaneous start to move. I added the symmetric force so that you can clearly see the third law unaffected. Norton disagrees with me about the meaning of the first law. Since Newton also intended to include uniform rotation as inertial motion (that's why he talks about bodies having their own force), to me Newton clearly meant zero net external force as the case for the first law and was attempting to make distinctions between an external force applied to a body and a body exerting its own preference for inertial motion. And that body's own inertia is the causal agent in what selects the solution of constant velocity in my example as opposed to one of the many solutions where the velocity changes, but merely changes in a way slowly enough where $a=0$ as it starts changing. The merely having $a=0$ approach, using the second law without the first, would say that $F=ma$ is all that matters and the bodies own inertia has no say about whether to have a uniform motion or whether to move. That allows multiple solutions if you really want to throw away the first law, plus you get Norton's motion that happens after any random amount of time. Throw out the first law and there are consequences.

share|improve this answer

You can only deduce the third law from the second one, its long and can be found here.

The first cannot be deduced from the second because it talks about initial reference frames. And with that you can see that you can't deduce any others from each other (without creating a dependence circle).

However you must keep in mind that this was not that trivial in the 17th century. You can't state that clearly that one is a derivation of the other, because you are assuming something about how the world works, and that it follows some deep mathematical consistency. It could have done other stuff too, however today those kind of actions would not be considered 'physical'. But note, Newton also invented calculus which is in a way the tool we used to derive the first and second law.

share|improve this answer

protected by Qmechanic Mar 2 '14 at 18:34

Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.