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## Main Question or Discussion Point

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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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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russ_watters

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F=ma

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The first law is the principle of inertia :|F=ma

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russ_watters

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Yes...The first law is the principle of inertia :|

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Can you give its formulation? Then we can look for a mathematical statement.The first law is the principle of inertia :|

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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##).Is there a mathematic model for the first law of dynamics?

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$$\mathrm{d \vec{p}}{\mathrm{d} t}=\vec{F}.$$

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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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.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.

- #9

S.G. Janssens

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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?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.

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$$\vec{x}=\vec{x}'-\frac{\vec{a}}{2} t^2.$$

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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Yes, sure.I think this utmost simple example is understandable even without this calculation!

That is why I found post #8 confusing, which led to my remark in #9.

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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].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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Without the third law this just means that there is a force acting in the accelerated frame.from which

$$\vec{x}'=\frac{\vec{a}}{2} t^2 + \vec{v}_0' t + \vec{x}_0', \quad \vec{v}'=\vec{a} t+\vec{v}_0' \neq \text{const}.$$

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In general relativity, ageodesicgeneralizes the notion of a "straight line" to curved spacetime. Importantly, the world line of a particle free from all external, non-gravitational force, is a particular type of geodesic. In other words, a freely moving or falling particle always moves along a geodesic.

https://plato.stanford.edu/entries/spacetime-iframes/#IneFra20tCenSpeGenRel

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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.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!

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.You feel the force pushing you forward!

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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.Only if you are not a physicist knowing that these are inertial forces

- #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

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

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

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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That's not quite correct. The logical equivalent of the first expression isThere 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

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

[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

What's wrong with that? If there is acceleration there must be a net force.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]

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There is nothing wrong with that. It is just not equivalent with your second expression.What's wrong with that?

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A Lazy Shisno

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.There is nothing wrong with that. It is just not equivalent with your second expression.

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