# On Newton's first and second laws

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## Main Question or Discussion Point

I'm reading Scheck's book about Mechanics and it says that Newton's first law is not redundant as it defines what an inertial system is. My problem is that we could say the same about Newton's second law. Indeed, Newton's second law is only valid, in general, for inertial systems, so it also defines them.
Therefore, I think Newton's first law doesn't just define what an inertial system is, but, more importantly, states they exist (which is not obvious).
What do you think?

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I'm reading Scheck's book about Mechanics and it says that Newton's first law is not redundant as it defines what an inertial system is. My problem is that we could say the same about Newton's second law. Indeed, Newton's second law is only valid, in general, for inertial systems, so it also defines them.
Without the third law both the first and the second law would also be valid in non-inertial frames.

Without the third law both the first and the second law would also be valid in non-inertial frames.
According to Scheck's definition, inertial frames are frames with respect to which Newton's first law has analytic form $\ddot{\pmb{r}}(t)=0$.

According to Scheck's definition, inertial frames are frames with respect to which Newton's first law has analytic form $\ddot{\pmb{r}}(t)=0$.
You can also have $\ddot{\pmb{r}}(t)=0$ in non-inertial frames. That is obviously not sufficient.

You can also have $\ddot{\pmb{r}}(t)=0$ in non-inertial frames. That is obviously not sufficient.
Any example?

robphy
Homework Helper
Gold Member
For Newton's First Law, you need that it travels with constant velocity when there are no unbalanced forces.
Suppose you are in a rotating frame of reference (e.g. https://archive.org/details/frames_of_reference @17m05s )
and that a particle [somehow] moves along a straight line in that frame.
For the particle to move with constant velocity [e.g., zero] in the rotating frame, there must be unbalanced forces. (see @20m00s above )
(If there are no unbalanced forces, the particle would move with constant velocity in an inertial frame...
but not constant velocity [in a straight line] in the rotating frame... as in the video above.)

Marion and Thornton briefly discuss different ways to interpret Newton's Laws. Look here on (book page, not pdf page) number 49-51.

bhobba
Mentor
I'm reading Scheck's book about Mechanics and it says that Newton's first law is not redundant as it defines what an inertial system is.
True - but there are equivalent definitions that do not use it. For example IMHO a better definition is found in Landau - Mechanics. An inertial frame is one where the laws of physics are the same at all points in space, directions, and instants of time. It can be shown that any two inertial frames are moving at constant velocity relative to each other (hint on proof - divide time into the sum of infinitesimal times - in each infinitesimal instant the Taylor expansion has higher terms you can neglect hence the transformation is linear). Sum them up and you get a linear transformation from which constant velocity follows by looking at the transformation of a fixed point - say the origin. What it does not say is any frame travelling at constant velocity to an inertial frame is inertial - this is the actual assumption - the rest is just definitions and some math as explained before..

Now getting back to the original statement, place a particle at the origin and have nothing else in the frame. If it shoots off in any direction then the laws of physics are not the same in all directions. If you have a particle going at constant velocity then you can go to an inertial frame where it is at rest and from before must remain at rest. Hence, assuming, not acting on by a force means nothing is affecting the particle you have the first law - but this time from symmetry. You can make this argument more rigorous using the Principle Of Least Action as explained in Landau's book.

Why I think it is better, is its based on symmetry, the importance of which is essential to relativity:
http://www2.physics.umd.edu/~yakovenk/teaching/Lorentz.pdf

My problem is that we could say the same about Newton's second law. Indeed, Newton's second law is only valid, in general, for inertial systems, so it also defines them.Therefore, I think Newton's first law doesn't just define what an inertial system is, but, more importantly, states they exist (which is not obvious).
What do you think?
The symmetry argument I gave before make the existence of inertial frames very intuitive - but of course physics is an experimental science and it is an experimental fact internal frames with the rather obvious symmetry proprieties exist to a high degree of accuracy - especially in deep space. The second law strictly speaking is not a law - but the definition of what a force is. The assumption is the frame is not inertial if things start accelerating of their own accord - we naturally assume something must be making it do that. That something could be something in or outside the frame; and the frame inertial if that something was not present. So overall the frame can still be inertial but containing or influenced by something else. The definition of a force is a measure of that something. Why that definition - why not simply say acceleration or mass squared times acceleration - I am sure you can think of many others. The answer is its physical content - it says in analysing classical mechanics problems get thee to the forces as defined by the second law.

The third law is a statement whose validity is determined by experiment - it may be true or false. I assume you know it is equivilant to momentum conservation. For an advanced view of it look into Noethers Theorem:
http://applet-magic.com/noetherth.htm

Now what is the physical basis of Noether's Theorem? - the answer is Quantum Mechanics - but I will let you think about that - don't worry if you do not see it - you can do a post and me or someone else can explain it - but thinking about it will help develop your understanding.

As a further thing to think about, what is the actual basis of classical mechanics? The answer again is QM. Once you understand this it makes you laugh a bit when you see people say; are there any manifestations of QM here in the classical world. Well everything is actually quantum so there is no classical world separate from the quantum world - just a region where QM is very well approximated by classical laws.

Thanks
Bill

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Thank you all for your answers. As for QM I don't think I'll ever learn about it as I'm learning Mechanics to better understand underactuated robotics and locomotion in particular.

stevendaryl
Staff Emeritus
This is blatant historical revisionism, but it is possible to reformulate Newton's laws into an equivalent 4D spacetime theory where his laws are true in every coordinate system, not just inertial coordinate systems.

The key is to make the seemingly pointless distinction between time as a coordinate and elapsed time as a scalar parameter. We let $x^0$ be the time coordinate, and let $s$ be the path parameter for a moving test particle, and we impose the restriction: $\frac{d^2 x^0}{ds^2} = 0$, so that the time coordinate for a moving test particle increases smoothly.

$m \frac{d^2 x^j}{dt^2} = F^j$

This is three equations, as $j$ ranges over the three spatial dimensions. Now, we convert these to 4 equivalent equations:

$m \frac{d^2 x^j}{ds^2} = F^j$
$m \frac{d^2 t}{ds^2} = 0$

(Well, the 3D and 4D equations are equivalent if we assume the "initial" condition: $\frac{dt}{ds} = 1$)

This makes it into a 4-D equation:
$m \frac{d^2 x^\alpha}{ds^2} = F^\alpha$

where we let $x^0 = t$ and $F^0 = 0$

Now, we transform to a noninertial, non-Cartesian coordinate system $x^\mu$. Letting $L^\mu_\alpha = \frac{\partial x^\mu}{\partial x^\alpha}$ and $\overline{L}^\alpha_\mu = \frac{\partial x^\alpha}{\partial x^\mu}$, we can rewrite our equations in terms of the new coordinates $x^\mu$:

$\frac{dx^\alpha}{ds} = \overline{L}^\alpha_\mu \frac{dx^\mu}{ds}$
$\frac{d^2 x^\alpha}{ds^2} = \frac{\partial\overline{L}^\alpha_\mu}{\partial x^\nu} \frac{dx^\mu}{ds} \frac{dx^\nu}{ds} +\overline{L}^\alpha_\mu \frac{d^2 x^\mu}{ds^2}$

If we just define the components of the force in the new coordinate system to be: $F^\mu = L^\mu_\alpha F^\alpha$, then we have:

$F^\mu = L^\mu_\alpha F^\alpha = m L^\mu_\alpha \frac{d^2 x^\alpha}{ds^2} = m [ L^\mu_\alpha \frac{\partial\overline{L}^\alpha_{\mu'}}{\partial x^\nu} \frac{dx^{\mu'}}{ds} \frac{dx^\nu}{ds} + L^\mu_\alpha \overline{L}^\alpha_{\mu'} \frac{d^2 x^{\mu'}}{ds^2}]$

Since $L$ and $\overline{L}$ are inverses (viewed as matrices), we have:

$F^\mu = m [ \frac{d^2 x^\mu}{ds^2} + \Gamma^\mu_{\nu \mu'} \frac{dx^\nu}{ds} \frac{dx^{\mu'}}{ds} ]$

where $\Gamma^\mu_{\nu \mu'} \equiv L^\mu_\alpha \frac{\partial\overline{L}^\alpha_{\mu'}}{\partial x^\nu}$

This can then be considered a vector equation:

$F = m \frac{D}{Ds} V$

where $V$ is the vector with components $V^\mu$ and $\frac{D}{Ds}$ is the path derivative defined via the rules:

$\frac{D}{Ds} V =$ that vector $A$ such that $A^\mu = \frac{dV^\mu}{ds} + \Gamma^\mu_{\nu \mu'} V^\nu V^{\mu'}$

This can be understood as the ordinary derivative of $V$ if we introduce basis vectors $e_\mu$ with the covariant derivatives:

$\frac{\partial e_{\mu'}}{\partial x^\nu} \equiv \Gamma^\mu_{\nu \mu'} e_\mu$

Then we have, simply:

$F = m \frac{dV}{ds}$

valid in any coordinate system.

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haushofer
Any example?
You sitting on your computer in the co-rotating rest frame of Earth.

You sitting on your computer in the co-rotating rest frame of Earth.
That's because of gravity and friction. If you remove all forces, my acceleration won't be 0 in that frame.

That's because of gravity and friction. If you remove all forces, my acceleration won't be 0 in that frame.
First and second law say that there is no acceleration without a net force. That means if your acceleration isn't 0 than you didn't remove all forces.

First and second law say that there is no acceleration without a net force. That means if your acceleration isn't 0 than you didn't remove all forces.
... or your reference frame is not inertial, as in this case.

stevendaryl
Staff Emeritus
Some people take $F^j = \frac{d^2 x^j}{dt^2}$ as the definition of force in Newtonian physics. If you say that, then by definition, a rotating frame has forces that are not present in an inertial frame: centrifugal and Coriolis forces. On the other hand, if centrifugal force is a force, then Newton's third law is violated, because there is no equal and opposite force for the centrifugal force.

As I suggested in an earlier post, you can reformulate Newton's laws so that

$\mathbf{F} = m \dot{\mathbf{V}}$

is true as a vector equation in any frame.

... or your reference frame is not inertial, as in this case.
How do you know if the frame is inertial or not?

On the other hand, if centrifugal force is a force, then Newton's third law is violated, because there is no equal and opposite force for the centrifugal force.
Would there enything else be violated except the third law?

As I suggested in an earlier post, you can reformulate Newton's laws so that

$\mathbf{F} = m \dot{\mathbf{V}}$

is true as a vector equation in any frame.
All you need to do is deleting the first sentence in the third law. That's what Newton did in his personal copy of the principia:

stevendaryl
Staff Emeritus
How do you know if the frame is inertial or not?
By definition, the frame is inertial if force-free motion is constant velocity. The difficulty is knowing when the motion is force-free. You can eliminate all known forces, but you can never know for certain that there isn't some unknown force at work.

stevendaryl
Staff Emeritus
Would there enything else be violated except the third law?
No, but in my opinion, it's Newton's third law that gives $F=ma$ empirical content, and not just a tautology.