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Generalization of force.

by MathematicalPhysicist
Tags: force, generalization
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MathematicalPhysicist
#1
Feb2-13, 07:29 AM
P: 3,215
What if we wouldn't define a force as F=dp/dt but instead as a function of

[tex]F=F(p,q,\dot{p},\dot{q})[/tex]

How will this change the equations of physics?
Maybe there are cases where the force behaves as [tex]k\cdot \frac{dp}{dq}[/tex] where 'k' is some constant to fix the dimensions.


I am just tinkering with this idea, really.
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Simon Bridge
#2
Feb2-13, 08:06 AM
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Please define your terms - i.e. can you express the second equation in words?
It looks far too general for any sensible answer.

In general - it would not change the physics ... the equations would look different, they'd have different letters in them, but would be mathematcally identical.
If you wanted to use some function of p and dp/dt for force, and you wanted to use that to get an equation of motion, then you will find yourself only needing the dp/dt part.

You realize that you can define any word to mean anything you like? All you are doing is assigning the label to a different object.
stevendaryl
#3
Feb2-13, 08:38 AM
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Quote Quote by MathematicalPhysicist View Post
What if we wouldn't define a force as F=dp/dt but instead as a function of

[tex]F=F(p,q,\dot{p},\dot{q})[/tex]

How will this change the equations of physics?
That's sort of a strange generalization, because "momentum" really only has a meaning relative to the equations of motion. Typically, momentum is computed from the velocity (or vice-versa) so they aren't independent dynamical variables.

However, now that I think about it, there is a formulation of classical mechanics that puts momentum and velocity on equal footing, without assuming one is derivable from the other.

Assume that there is a quantity [itex]Q(p,\dot{p},q,\dot{q})[/itex] associated with the motion. The equations of motion are derived by the requirement that
[itex]\int Q dt[/itex] is minimized. Then that leads to the equations of motion:

[itex]\dfrac{d}{dt} \dfrac{\partial Q}{\partial \dot{q}} = \dfrac{\partial Q}{\partial q}[/itex]


[itex]\dfrac{d}{dt} \dfrac{\partial Q}{\partial \dot{p}} = \dfrac{\partial Q}{\partial p}[/itex]

If you choose [itex]Q[/itex] carefully, this is equivalent to the usual equations of motion. For example, if you let:

[itex]Q = \dfrac{p^2}{2m} + V(q) - p \dot{q}[/itex]

then the equations of motion become:

[itex]\dfrac{d}{dt} (-p) = \dfrac{\partial Q}{\partial q}[/itex]


[itex]0= \dfrac{p}{m} - \dot{q}[/itex]

Which is equivalent to the usual equations of motion.

cosmic dust
#4
Feb2-13, 02:09 PM
P: 123
Generalization of force.

I think that if there was such a force, then the definition of "force", i.e. F = [itex]\dot{p}[/itex] woulld be inconsistent, since the magnitude that defines the force ([itex]\dot{p}[/itex]) would be in what we want to define (F) . To me, it looks non-sence to define "something" using terms that invlolve that "something".
Simon Bridge
#5
Feb2-13, 11:07 PM
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I still think it's like asking what our calculations would be like if we defined quadratics as third order polynomials and lines as second order. It would be a funny thing to do - but there's nothing stopping anyone. Math would be just the same, only we'd say that ballistic motion in the absence of air resistance is "linear" (it would just mean something different to what we are used to.)

That's why it doesn't matter that the Newtonian definition of force is inconsistent with the above definition.... it's a different definition. Some definitions are just more useful than others - with one of the considerations being communication.

I think we need OP to clarify what was meant.


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