Is action and reaction instantanious?

Andrew Mason

It may be more useful to use energy rather than force in some situations. In nuclear physics that is exactly what is done: we speak of binding energies and collision energies rather than forces.

I don't follow you here. Kinetic energies are always greater than zero. So a system experiencing no external force may still have significant energy. Think of a star that undegoes a supernova.

AM

granpa

I dont even know how to begin to answer that. Its clear to me that you are making it all much much more complicated than it really is. stop trying so hard and maybe you will get what I'm saying

are you familiar at all with Kirchoffs law?

Andrew Mason

This is not equivalent. A person on the surface of the earth can measure atmospheric pressure by doing a local experiment. An orbiting astronaut cannot detect gravity by doing a local experiment.

AM

granpa

once again, I dont even know how to begin to answer. I see no relevance and no contradiction to anything I've said. I think you think I'm saying something complicated when all I'm saying is really very simple.

Andrew Mason

I am not sure how you are using it. Kirchoff's law does not have anything to do with inertia or its electrical equivalent.

AM

granpa

Kirchoffs law has nothing to do with inductance??

DaleSpam

OK, first the action=reaction part is simply wrong. Action and reaction refer to Newton's 3rd law and action-reaction pairs act on different bodies. F=ma refers to Newton's 2nd law and acts on the same body. So even if you want to call ma a force it is not a reaction force to f since it is acting on the same body that f is acting on.

Now, the remainder of what you did is mathematically correct and leads to the trivially true assertion 0=0. This algebraic manipulation can be done for any formula.

W = f.d
W = W
0 = 0

x(t) = 1/2 a t² + v0 t + x0
x(t) = x(t)
0 = 0

Yes you can always do it, but in doing so you completely lose the meaning of the original expression.

It looks like you are using the D'Alembert approach which can be useful in certain circumstances, but you need to understand what it is doing. Here is a thread on the subject. It should generally be avoided because of the conceptual confusion it causes, and it should only be applied when the specific problem demands it.

In the end, if you are talking about inertial reference frames then ma is not a force and the body accelerates. If you are talking about the non-inertial rest frame of an accelerating body (as D'Alembert does) then there is a fictitious inertial force (like the Coriolis force) of magnitude ma and the body does not accelerate. This force that exists only in the non-inertial frame does not follow Newton's 3rd law because its source is the non-inertial reference frame and not an interaction with another object.

DaleSpam

There is a net force on each object (Newtonian gravity). So the earth pulls the moon "down" and the moon pulls the earth "up". The force on the moon is equal and opposite to the force on the earth, per Newton's 3rd law.

DaleSpam

This is correct. Inertia is only a force in non-inertial reference frames where it appears as a so-called ficticious force. In non-inertial reference frames the momentum of an isolated system is not conserved (hence the designation "non-inertial").

DaleSpam

Because they are not zero everywhere. You can choose a non-inertial frame where any one object is at rest, but in that reference frame an object which is not experiencing any real forces will still experience the inertial force and will therefore accelerate. The net effect is a violation of the conservation of momentum for an isolated system, or equivalently, a violation of Newton's 3rd law. This is simply what happens in non-inertial frames.

By the way, none of this has anything to do with relativity, this is all just Newtonian mechanics.

atyy

I have no idea what inertia is (and don't wish to find out), so I am not addressing that. But I think some subscripts are missing on the equations.

Consider a two particle system:
m₁, v₁, a₁: mass, velocity and acceleration of particle 1
m₂, v₂, a₂: mass, velocity and acceleration of particle 2
F₁₂: force on particle 1 due to the effect of particle 2
F₂₁: force on particle 2 due to the effect of particle 1

F₁₂=m₁a₁ [E1: Newton's 2nd law for particle 1]
F₂₁=m₂a₂ [E2: Newton's 2nd law for particle 2]
F₁₂=-F₂₁ [E3: Newton's 3rd law, action and reaction are equal and opposite]

We can rearraange E3 as you did: F₁₂+F₂₁=0 [E3b]

From E3 or E3b, neither F₁₂ nor F₂₁ is necessarily zero, only equal and opposite. From E1 and E2, assuming m₁ and m₂ positive and constant, F₁₂ and F₂₁ must be zero only if a₁ or a₂ are zero.

But we can combine E3b with E1 and E2:
F₁₂+F₂₁=m₁a₁+m₂a₂=d(m₁v₁+m₂v₂)/dt=0,
which says that there is no net force on both particles considered together (not separately) and that total momentum does not change over time, ie. momentum is conserved.

Andrew Mason

You may be thinking of Lenz's law.

AM

granpa

nobody has addressed my very simple question. what about a hypothetical massless charged particle? it has self inductance so it would accelerate under the influence of an external field in exactly the same way that a massive particle would. the force due to self inductance exactly balancing the force due to the external field. net force is zero yet it still accelerates.

thats all I'm saying. I'm comparing the behavior of mass to the behavior of self inductance.

as for defining force/mass, I can imagine a video game like universe in which time distance and velocity are all well defined but possessing nothing that we would recognize as force or mass. so I would guess that both arise simultaneously if the system possess some kind of symmetry. possibly related to conservation laws and Noethers theorem.

I'm just guessing at this point.

atyy

I think this is not quite what you had in mind, but is it close?
Xiaochao Zheng's notes on "Radiation reaction and electron's self energy -- an unsolved problem" http://www.jlab.org/~xiaochao/teachi...x/chap11-6.pdf

granpa

this hypothetical massless charged particle isnt a point change. it has finite diameter. and no I dont know whits holding it together. it doesnt matter. maybe superglue.

Andrew Mason

There is, of course, no such thing as a massless charged particle. A massless particle travels at the speed of light relative to all inertial frames, so it cannot accelerate.

A particle with charge q in an electric field [itex]\vec E[/itex] experiences a force [itex]\vec F[/itex] = q[itex]\vec E[/itex]. The only thing that affects its acceleration is its mass. [itex]\vec F[/itex] = q[itex]\vec E[/itex] = ma. There is no other force.

AM

granpa

whoa. a moving charge has energy in its magnetic field. this energy must be supplied by the external force. it would not move at the speed of light.