What if we redefine force in physics as a function of velocity and momentum?

  • Context: Graduate 
  • Thread starter Thread starter MathematicalPhysicist
  • Start date Start date
  • Tags Tags
    Force
Click For Summary

Discussion Overview

The discussion explores the idea of redefining force in physics as a function of velocity and momentum, specifically considering the implications of such a redefinition on the equations of physics. The scope includes theoretical considerations and conceptual clarifications regarding the nature of force and its relationship with momentum and velocity.

Discussion Character

  • Exploratory
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Some participants propose redefining force as F=F(p,q,˙p,˙q) and question how this would alter the equations of physics.
  • One participant argues that such a redefinition would not change the underlying physics, suggesting that the equations would remain mathematically identical despite different variable labels.
  • Another participant notes that momentum and velocity are typically interdependent, raising concerns about the generalization of defining force without assuming one derives from the other.
  • One contribution discusses a formulation of classical mechanics that treats momentum and velocity equally, leading to equations of motion derived from a quantity Q(p,˙p,q,˙q) that could align with traditional equations under certain conditions.
  • Another participant expresses skepticism about the consistency of defining force in terms of its own derivative, suggesting it leads to circular reasoning.
  • A later reply compares the redefinition of force to redefining mathematical concepts, arguing that while definitions can change, their utility and clarity in communication are important considerations.
  • Some participants call for clarification from the original poster regarding their intentions with the proposed redefinition.

Areas of Agreement / Disagreement

Participants express a range of views on the implications of redefining force, with no consensus reached on whether such a redefinition is meaningful or useful. Disagreement exists regarding the consistency and practicality of the proposed changes.

Contextual Notes

Participants highlight potential limitations in the proposed redefinition, including the interdependence of momentum and velocity, and the implications of circular definitions. The discussion remains open-ended with various assumptions and interpretations present.

MathematicalPhysicist
Science Advisor
Gold Member
Messages
4,662
Reaction score
372
What if we wouldn't define a force as F=dp/dt but instead as a function of

F=F(p,q,\dot{p},\dot{q})

How will this change the equations of physics?
Maybe there are cases where the force behaves as k\cdot \frac{dp}{dq} where 'k' is some constant to fix the dimensions.I am just tinkering with this idea, really.
 
Physics news on Phys.org
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.
 
Last edited:
MathematicalPhysicist said:
What if we wouldn't define a force as F=dp/dt but instead as a function of

F=F(p,q,\dot{p},\dot{q})

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 Q(p,\dot{p},q,\dot{q}) associated with the motion. The equations of motion are derived by the requirement that
\int Q dt is minimized. Then that leads to the equations of motion:

\dfrac{d}{dt} \dfrac{\partial Q}{\partial \dot{q}} = \dfrac{\partial Q}{\partial q}


\dfrac{d}{dt} \dfrac{\partial Q}{\partial \dot{p}} = \dfrac{\partial Q}{\partial p}

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

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

then the equations of motion become:

\dfrac{d}{dt} (-p) = \dfrac{\partial Q}{\partial q}


0= \dfrac{p}{m} - \dot{q}

Which is equivalent to the usual equations of motion.
 
I think that if there was such a force, then the definition of "force", i.e. F = \dot{p} woulld be inconsistent, since the magnitude that defines the force (\dot{p}) would be in what we want to define (F) . To me, it looks non-sence to define "something" using terms that invlolve that "something".
 
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.
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 21 ·
Replies
21
Views
2K
  • · Replies 19 ·
Replies
19
Views
4K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 27 ·
Replies
27
Views
3K
  • · Replies 2 ·
Replies
2
Views
844
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K