Mathematical model of Newton's first law

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Is there a mathematic model for the first law of dynamics? If no, do you think that this law can be modellized with maths?
 

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  • #2
russ_watters
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F=ma
 
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  • #5
dextercioby
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The first law is the principle of inertia :|
Can you give its formulation? Then we can look for a mathematical statement.
 
  • #6
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Is there a mathematic model for the first law of dynamics?
No, the first law is the qualitative part of the definition of force (force is the only cause for changes in motion). Therefore there is no corresponding mathematical model. But it should be possible to find a logical expression. The quantitative part of the definition is given in the second law (##F = \dot p##) and the third law(##F_1 = - F_2##).
 
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  • #7
vanhees71
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The 1st and 2nd law are summarized by the statetment that there exists an inertial reference frame such that
$$\mathrm{d \vec{p}}{\mathrm{d} t}=\vec{F}.$$
If there's no force you have
$$\vec{p}=\text{const}.$$
This is the 1st Law: For a point-particle like body with constant mass you have ##\vec{p}=m \vec{v}## and thus ##\vec{v}=\text{const}##, i.e., a point particle moves with constant velocity against an inertial reference frame, if no force is acting on the body.
 
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  • #8
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This is the 1st Law: For a point-particle like body with constant mass you have ##\vec{p}=m \vec{v}## and thus ##\vec{v}=\text{const}##, i.e., a point particle moves with constant velocity against an inertial reference frame, if no force is acting on the body.
No, it just means that a body with constant mass moves with constant velocity if no force is acting on it. Without the third law this is not limited to inertial frames.
 
  • #9
S.G. Janssens
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No, it just means that a body with constant mass moves with constant velocity if no force is acting on it. Without the third law this is not limited to inertial frames.
I don't understand this. Are you saying that the first law by itself holds in non-inertial frames as well? Do you have fictitious forces in mind?
 
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  • #10
vanhees71
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In, e.g., a reference frame ##(t,\vec{x}')## which is accelerated with uniform acceleration against an inertial frame ##(t,\vec{x})##, you have
$$\vec{x}=\vec{x}'-\frac{\vec{a}}{2} t^2.$$
If no force is acting, you have
$$\ddot{\vec{x}}=\ddot{\vec{x}}'-\vec{a}=0,$$
from which
$$\vec{x}'=\frac{\vec{a}}{2} t^2 + \vec{v}_0' t + \vec{x}_0', \quad \vec{v}'=\vec{a} t+\vec{v}_0' \neq \text{const}.$$
I think this utmost simple example is understandable even without this calculation!
 
  • #11
S.G. Janssens
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I think this utmost simple example is understandable even without this calculation!
Yes, sure.
That is why I found post #8 confusing, which led to my remark in #9.
 
  • #12
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Are you saying that the first law by itself holds in non-inertial frames as well? Do you have fictitious forces in mind?
Yes, without the third law there is no difference between interactive forces and fictitious forces. In order to make all three laws valid in non-inertial frames you just need to remove the first sentence of the third law. Newton did that in his personal copy of the Principia [http://cudl.lib.cam.ac.uk/view/PR-ADV-B-00039-00001/49].
 
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  • #13
vanhees71
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The 3rd law is more about homogeneity of space, i.e., momentum conservation rather than inertial frames. I'm still puzzled about your statement that Newton's laws read all the same in non-inertial frames. The very basis of Newtonian mechanics from a modern point of view is Galilei invariance, which leads to the inertial frames as a kind of preferred frames.
 
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from which
$$\vec{x}'=\frac{\vec{a}}{2} t^2 + \vec{v}_0' t + \vec{x}_0', \quad \vec{v}'=\vec{a} t+\vec{v}_0' \neq \text{const}.$$
Without the third law this just means that there is a force acting in the accelerated frame.
 
  • #15
robphy
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  • #16
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Well, my question come from a ordinary life problem. When I am on the train, and it starts going foward, I feel pushed backward because I tend to maintain my state of quiet (let's call state ##A##). But after some while I don't feel a force pushing me because I am now in a state of steady speed (state ##B##). So my inertia switched from quiet to steady speed. And my question is: is there a mathematic model that correlates all the variables (time, my mass, the acceleration of the train, etc...)? Because if I say the simple equation F=ma or v =at, I know my speed at any time, but I don't know if I feel a force pushing me backwards!
 
  • #17
vanhees71
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You feel the force pushing you forward! It's most simply understood if you look at your acceleration inside the train from an inertial frame (e.g., (to a good approximation) the frame of an observer at rest on the platform): Then you see that there must be a force acting on you (or your center of mass) such that you get accelerated, and Newton's 2nd Law tells you the this force is given by ##\vec{F}=m \vec{a}##.
 
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  • #18
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Because if I say the simple equation F=ma or v =at, I know my speed at any time, but I don't know if I feel a force pushing me backwards!
If you know your mass and speed at any time you would also know the force acting on you if there would be no third law. It becomes tricky when you distinguish between interactive forces and fictious forces. Then you need to find a frame of references where total momentum is conserved. That can be the rest frame of the platform as vanhees71 already suggested.

You feel the force pushing you forward!
That depends on the frame of reference your brain is currently using for the processing of the cognitive input. Inside a train this may be the rest frame of the train (depending on where you are looking at) and then you feel a force pushing you backwards.
 
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  • #19
vanhees71
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Only if you are not a physicist knowing that these are inertial forces (sometimes called fictitious forces, but that's a misnomer in my opinion, because as you rightfully say you "feel" them and they have of course an impact on the motion of matter, e.g., the Coriolis force on earth for the motion of the air).
 
  • #20
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Only if you are not a physicist knowing that these are inertial forces
Beeing a physicist is neither required nor sufficient to know that these are inertial forces. It can be very hard or even impossible to distinguish inertial frames of reference from non-inertial frames.
 
  • #21
A Lazy Shisno
There is no mathematical model so to speak, it's just a principle. Mathematically, the first law can be described as the following.

$$\text{If}\quad\sum \vec F=0\quad\text{then}\quad\frac{\text{d}\vec v}{\text{d}t}=0$$

and conversely

$$\text{If}\quad\sum \vec F\neq 0\quad\text{then}\quad\frac{\text{d}\vec v}{\text{d}t}\neq 0$$
 
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  • #22
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There is no mathematical model so to speak, it's just a principle. Mathematically, the first law can be described as the following.

$$\text{If}\quad\sum \vec F=0\quad\text{then}\quad\frac{\text{d}\vec v}{\text{d}t}=0$$

and conversely

$$\text{If}\quad\sum \vec F\neq 0\quad\text{then}\quad\frac{\text{d}\vec v}{\text{d}t}\neq 0$$
That's not quite correct. The logical equivalent of the first expression is

[itex]{\rm If}\quad \frac{{d\vec v}}{{dt}} \ne 0\quad {\rm then}\quad \sum {\vec F \ne 0}[/itex]
 
  • #23
A Lazy Shisno
That's not quite correct. The logical equivalent of the first expression is

[itex]{\rm If}\quad \frac{{d\vec v}}{{dt}} \ne 0\quad {\rm then}\quad \sum {\vec F \ne 0}[/itex]
What's wrong with that? If there is acceleration there must be a net force.
 
  • #24
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What's wrong with that?
There is nothing wrong with that. It is just not equivalent with your second expression.
 
  • #25
A Lazy Shisno
There is nothing wrong with that. It is just not equivalent with your second expression.
What I was saying is that if the sum of all forces is zero, then acceleration is zero, and conversely, if the sum of all forces is not zero, then acceleration is not zero.
 

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