# Force=mass times acceleration, or time derivative of momentum?

Which is more generally corrrect,

F=ma

or

F=dp/dt

?

Pengwuino
Gold Member
Aren't they mathematically equivalent?

Some particles, such as photons, don't have mass but they still have a momentum. F=ma won't work in these cases.

Depends on the situation, are you dealing with an impulse, where you need to consider momentum transfer, or a longer duration event, where acceleration acts over a longer time duration. F=ma or F=dp/dt is situational.

Aren't they mathematically equivalent?
Not in general, if you consider the product rule.

HallsofIvy
Homework Helper
F= ma is the same as F= dp/dt only if m is a constant. F= dp/dt is more general.
(In fact, one can consider "rate of change of momentum" to be the DEFINITION of force.)

Loren Booda said:
Which is more generally corrrect,

F=ma

or

F=dp/dt

?
Let = mv. Then

$$F = \frac{dp}{dt}$$

is a definition of F. F = ma is an equality between the quantities F, m and a when m is constant.

Incorrectly assuming that F = ma is a definition has gotten people really mixed up when going to relativity.

Pete

My question arises from the rocket problem, where the object expends mass.

arildno
Homework Helper
Gold Member
Dearly Missed
Hi, Loren:
The crucial difference between the rocket+remaining fuel system and "normal" systems is that it is an "open" system (or geometric control volume) that loses momentum-carrying particles over time, whereas material systems (or material control volumes) keeps all momentum-carrying particles through all times.

To take a silly example:
Have a frictionless table of finite length, and let your system be that part of a book which happens to remain on top of the table, the book sliding along with some constant velocity V.
As the book gets to the edge, the momentum of your chosen system decreases because the parts of the book that travels beyond the table edge is not within your chosen system.
That is, material particles leave your observed system and carry their own momentum with themselves. This is called momentum flux.
But the correspomding decrease of momentum in your system cannot be related to any net force acting upon the material particles remaining in your system!!

Thus, for open systems, we need to take care of the momentum flux term in order not to get wrong answers.

It is for this reason that the rocket equation looks somewhat differently than the version of Newton's 2.law for a MATERIAL system.

arildno
Homework Helper
Gold Member
Dearly Missed
pmb_phy said:
Let = mv. Then

$$F = \frac{dp}{dt}$$

is a definition of F. F = ma is an equality between the quantities F, m and a when m is constant.

Incorrectly assuming that F = ma is a definition has gotten people really mixed up when going to relativity.

Pete

Just a comment to this:
Pre-relativistically speaking, your fundamental laws for material systems were those of F=ma and mass conservation, and F=dp/dt was a derived law.
It is Einstein who deserves the credit for bringing about a more productive conceptual switch.

Last edited: