Relationship between Newtons 1st and 3rd law

[yaseen shah]

does Newtons first law discribes Newtons third law.

 

[Pythagorean]

Well, I can't see how 1 could work without 3. I wouldn't be surprised if someone could argue a case for 1 without 3, but until then I agree with you.

 

[K^2]

First law holds in an inertial coordinate system. Third law holds in any coordinate system.

 

[Dale]

You certainly could imagine a physics where momentum is not conserved. Then you could have a physics with the same first law and a different third law.

 

[Pythagorean]

I'm having difficulty imagining momentum not being conserved. What would the picture look like?

 

[yaseen shah]

My friends i think first law that describes about inertia that Inertia is property of a body by virtue of it resists motion but in the form of force that he exerts equal on that body which apply force.
In my opinion 3rd law describes quantitative behavior of inertia.

 

[Dale]

For example, the forces could act in the same direction. Or the accelerations could be equal instead of the forces. Or the Lagrangian could depend on position.

 

[K^2]

My friends i think first law that describes about inertia that Inertia is property of a body by virtue of it resists motion but in the form of force that he exerts equal on that body which apply force.
In my opinion 3rd law describes quantitative behavior of inertia.

Yes, 3rd law essentially describes the "resistance" attributed to inertia, but there is no connection to 1st law there. 1st law doesn't say that body resists motion. It says that whenever you try to change the way it moves, you have to apply force.

 

[Pythagorean]

For example, the forces could act in the same direction.

So I sit on my couch and the surface of the couch pushes me down further into the couch, so then wouldn't I continue on my path, and Newton's 1 would be violated?

edit: nevermind, I thought of the off-normal case. And adding speed in the same direction would qualify for N1 anyway.

Or the accelerations could be equal instead of the forces.

Ok, that ones kind of mind-melting to think about assuming mass conservation still holds. I watch a ball fly by me and hit a fence that is stationary with respect to me. Once the ball and the fence contact, their instantaneous accelerations would have to somehow match up, so one would have to jump to the other or something, and the forces would bend around the will of the acceleration law.

Or the Lagrangian could depend on position.

The Lagrangian is terribly unintuitive to me.

 

[K^2]

The Lagrangian is terribly unintuitive to me.

Just keep this in mind.
F_q = \frac{\partial L}{\partial q}

The Newton's 2nd in terms of this F_q is also relatively easy to write down.

\frac{d}{dt}p_q = F_q

It's exactly the same as classical Newton's F=ma, except written with generalized momentum. And of course, it's easy to find the actual generalized momentum.

p_q = \frac{\partial L}{\partial \dot{q} }

Naturally, substituting this momentum into Newton's 2nd gives you a very familiar result.

\frac{d}{dt} \frac{\partial L}{\partial \dot{q}} - \frac{\partial L}{\partial q} = 0

And that's all there is to it.

 

[Pythagorean]

Just keep this in mind.
F_q = \frac{\partial L}{\partial q}

The Newton's 2nd in terms of this F_q is also relatively easy to write down.

\frac{d}{dt}p_q = F_q

It's exactly the same as classical Newton's F=ma, except written with generalized momentum. And of course, it's easy to find the actual generalized momentum.

p_q = \frac{\partial L}{\partial \dot{q} }

Naturally, substituting this momentum into Newton's 2nd gives you a very familiar result.

\frac{d}{dt} \frac{\partial L}{\partial \dot{q}} - \frac{\partial L}{\partial q} = 0

And that's all there is to it.

That's a good refresher, but the quantity L is still somewhat vague to me. From what you've formulate here, I'm eager to compare it to Work because it's a difference in energies (T-V) and the calculus form of it with regard to force is similar.

 

[K^2]

Well, the actual definition is Σpq-H. That way, you can build L even when you aren't sure exactly which energies are which. As long as you can write total energy in terms of generalized coordinates, you can define it. And don't try to think of it as a physical quantity. Just think of it as a generator for generalized forces and generalized momenta.

 

[Dale]

Ok, that ones kind of mind-melting to think about assuming mass conservation still holds. I watch a ball fly by me and hit a fence that is stationary with respect to me. Once the ball and the fence contact, their instantaneous accelerations would have to somehow match up, so one would have to jump to the other or something, and the forces would bend around the will of the acceleration law.

Yes, such a universe would not look like ours at all. A head on collision between a fly and a train would be as disastrous for the train as the fly. I don't know if extended objects could even form.

But the point is that such a universe could be consistent with Newton's 1st law, despite not being consistent with the 3rd law. They are not redundant.

 

[Pythagorean]

Well, the actual definition is Σpq-H. That way, you can build L even when you aren't sure exactly which energies are which. As long as you can write total energy in terms of generalized coordinates, you can define it. And don't try to think of it as a physical quantity. Just think of it as a generator for generalized forces and generalized momenta.

Thanks for your input on the matter, K*K, I'll have to let my hippocampus do some work on it.

Yes, such a universe would not look like ours at all. A head on collision between a fly and a train would be as disastrous for the train as the fly. I don't know if extended objects could even form.

I'm not sure... the train would experience very low forces wouldn't it? As a drops for the train, F would have to drop too. I'm still trying to imagine the effect on tensile strength.

But the point is that such a universe could be consistent with Newton's 1st law, despite not being consistent with the 3rd law. They are not redundant.

Yeah, I accept that. But I think exploring the alternative helps us to really internalize the meaning of the laws and their significance.

 

[nonequilibrium]

First law holds in an inertial coordinate system. Third law holds in any coordinate system.

What about fictitious forces?

 

[K^2]

Well, the way I stated it there are no fictitious forces, because first law is violated anyways. But yes, if you fix the first law by introducing fictitious forces, you end up breaking the 3rd law. Or rather, putting it in need of correction. You would then state that forces of interaction are always equal-and-opposite, which would then exclude fictitious forces.

 

[nonequilibrium]

But then the term "forces of interaction" would be defined ad hoc -- it seems you can only define that term with the use of the 3rd law, but I could be missing something.

But anyway, I don't see how you could say (in your first choice) that the 1ste law was broken but the 3rd law worked, because as you say F = ma is the definition (taking p = mv for simplicity), which automatically brings on the existence of fictitious forces.

But okay maybe I'm hammering on something that's not quite so important.

 

[K^2]

Well, yes, I mean, you can potentially define all forces as arbitrary external by-the-power-of-god kind of forces. Point is that you don't have to. Only fictitious force, that is, forces proportional to the mass and acceleration of the frame, will be left without equal-and-opposite.

 

[Dale]

if you fix the first law by introducing fictitious forces, you end up breaking the 3rd law

This is a good point, and already a kind of hint for the need of GR and tensors etc.