Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

In his The Principles of Natural Knowledge, Alfred North Whitehead writes that famous $f=ma$ is an identity:

It has been popular to define force as the product of mass and acceleration. The difficulty to be faced with this definition is that the familiar equation of elementary dynamics, namely,

$mf = P$

now becomes

$mf = mf$

It is not easy to understand how an important science can issue from such premises.

How does he obtain the identity?

share|improve this question
2  
Unless Whitehead is using $P$ to mean force and $f$ to mean acceleration, I have no idea what he's talking about. –  David Z Jan 14 '12 at 23:45
    
@DavidZaslavsky: It looks like it because he says force (his P) is the product of mass and acceleration (his m and f). So how does he get the identity? –  Zeynel Jan 15 '12 at 2:59

3 Answers 3

Certainly, you can say that F=ma is a definition rather than a law, i.e. the force on an object is defined to be the acceleration it would produce (in the absence of any other forces) times the mass of the object.

A definition, in itself, cannot say anything substantive about the universe. However, there's something else: Force, defined this way, is a useful concept. For example, if two forces are applied independently to the same object then you can get the net force by vector addition; the force of a compressed spring pushing on an object is independent of the color or size or shape or mass of the object it is pushing on; etc. etc. If you defined force in some stupid way, like

force = (mass)×(RGB color of the object)×(Greenwich mean time)

or whatever, then force would be a dead-end not leading towards anything useful, and it would not have any of these properties that we expect forces to have.

In other words, all the "obvious" properties of force, too obvious to say explicitly, like the fact that the force of a spring can be fruitfully discussed without knowing what the spring is pushing on, are really part of the broad interpretation of Newton's law F=ma. And they are truly substantive facts about the physical universe.

share|improve this answer
    
I agree with you but Whitehead is saying that $F=ma$ is an identity not a definition because $ma$ is defined as itself, or $ma=ma$. How does he get this identity? –  Zeynel Jan 16 '12 at 13:02
    
@Zeynel: Your friend Alice says to you: "I'll tell you what the word 'force' means: Force is defined as mass times acceleration". The next day, your friend Bob says to you: "Listen to this great new law of physics I just learned in class: Force equals mass times acceleration!" Then you would say to Bob, "That's not a great new law of physics, that's a trivial identity, you're just saying that something equals itself." –  Steve B Jan 17 '12 at 14:12
    
Maybe there is an intended joke here that I don't get but I prefer to see how Whitehead obtains the identity by showing his missing steps. So far no one came up with one. –  Zeynel Jan 18 '12 at 3:02
    
@Zeynel, that wasn't a joke, that was me answering your question. I don't see how I can make it any clearer. I'll just say it again. First you say, "The definition of F is ma." Then you say, there is a law of physics "F=ma". If you take the law and substitute the definition of F, you get "ma=ma", an identity. You are not missing any steps, there is nothing to miss. The very simple point (described in a somewhat obscure way) is that F=ma cannot be simultaneously a definition of F and a substantive law of physics. –  Steve B Jan 19 '12 at 14:11

Repeated measurements and observations have shown the "identity" $f=ma$. This may be the reason Whitehead considers $f=ma$ as an "identity".

Nowadays $f=ma$ is not considered an identity, AFAIK. It's actually a special case of Newton's second law (Newton's second law with constant masses).

It simply comes from the general definition of force: rate of change of momentum with respect to time ($\mathbf{F}=\frac{d\mathbf{p}}{dt}$), assuming constant mass and negligible relativistic effects.

share|improve this answer
    
Whitehead is saying that $ma=ma$. This identity does not lend itself to observation or measurement. I am trying to understand how he obtains $ma=ma$ from $F=ma$ and why he thinks $F=ma$ is $ma=ma$. –  Zeynel Jan 16 '12 at 13:06
    
Take time reading his words very carefully. What he's trying to write is probably "Please don't define force as product between force and acceleration...". He meant that if we define $f$ as $ma$, the problem is, the "fundamental" equation $f=ma$ becomes $ma=ma$, which is trivial and meaningless. (Whitehead was not a physicist, he was a mathematician turned into a philosopher. Read an article about him, wiki) –  Sawi Jan 17 '12 at 14:23

Newton's three laws of motion are really based on circular reasoning: In an inertial reference frame, an object in motion stays in motion and an object at rest stays at rest (when not acted upon by a force). But what is an inertial reference frame? An inertial reference frame is a frame in which the preceeding statement is true. Remember, it is pointless to speak of and/ or to quantify- describe any motion without specifying the reference frame to which the motion is related.

To break this circular reasoning, one needs to remove the privileged status of "inertial reference frames" and describe equations of motion in tensor form.

share|improve this answer
    
Newton's first law is not circular reasoning if you add "...and there is at least one inertial reference frame in the universe." (Tensor form equations are not necessary.) –  Steve B Jan 15 '12 at 18:07
    
Whitehead is saying $F=ma$ is an identity: $ma=ma$. This is not circular reasoning, it is an identity. Why is he saying this? –  Zeynel Jan 16 '12 at 13:09

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.