Newton's third law of motion - why?

[Ledgeknow]

Greetings everyone, I'm new here!

I tried searching the forums for my problem, but most deal more with the common misconceptions, not the question why - if there is such a question. Lately, I've been trying to gain an in-depth look into Newton's work, moreover the third law which has been bugging me for quite a while.

It states, that whenever an object exerts a force on another object, the second one will also exert a force on the first, equal in magnitude, collinear / opposite in direction.

Why?

Why does the other object respond with an equal, opposite force? Is there a causality link I'm missing? I'd really like to understand this more. For example, I can understand that when a bird flaps its wings, the wings exert a force on the air, pushing it out of the way, but at the same time the air exerts a force on the wings, giving them an upward force, lift.

Perhaps it sounds like a stupid question, but why and how does it happen... When you push on the air, how does it push back on your wings? Is it an intrinsic, observational fact or is there more to it?

The more ambiguous case to me is the example of the gravitational interaction between our planet and the Sun. The Sun exerts a gravitational force on Earth and Earth returns the "favor", but considering the mass ratio, the center of rotation is pretty much in the Sun.

Why? Why would Earth exert a force on the Sun, equal in magnitude, opposite in direction? Why wouldn't the Sun just push Earth around?

Can anybody see what's bothering me? Can you give me a way of looking at this, to find a logical explanation to this... Any insight is helpful! I hope I am making some sense.

Thanks everyone! You don't have to give me an answer, just point me in the right direction, books, web resources, anything. I am willing to learn, but sometimes stuff doesn't want to be learned by me.

 

[xts]

Look at the topic historically:
1. Tycho Brache made thousands of observations: at such date/time planets are seen in such positions, he never cared about any theory behind those observations;
2. Kepler got confused by Ptolemeian/Copernican concept of epicycles, as he would have to complicate it even more in order to reflect accurate Tycho's observations. He looked for something simpler. He made a lot of calculations, concluding them: if planets obey Kepler laws, their positions are like Tycho spotted. And Kepler laws occurred to be much simpler than multiple epicycle model, although it was hard to accept, as the circular motion was the heavenly perfect one, while ellipses were not so ellegant for 17th cent taste. Kepler just chose: better one non-ellegant ellipse with non-uniform motion, than ugly combination of 40+ perfectly circular epicycles.
Kepler gave no justification for his laws. Or rather he gave: angels are pushing planets that way;
3. Newton made next step: Kepler laws, but also other earthy phenomena, like falling apples, may be explained by his principles and his law of gravity. We don't need Kepler's angels any longer.

But Newton gave no further justification. Neither for 3rd law, neither for 1st one. You may equally well ask why if the body is not a subject to external forces its velocity remain constant? Newton did not answer it. He just reduced experimental knowledge about our world to smaller number of simpler laws, treating them as fundamental.

``Can anybody see what's bothering me?''
Not quite. You are either bothered why Newton stopped at that point on the way to find more fundamental causes, or you are asking us (21st century physicists) about thouse causes. I believe I answered you in the first case. If it is the second - I could direct you to Einstein's theories, going further on the way. But (unless you have some religious explanation) the path never ends - we are making just simpler and more ellegant theories, explaining how the world acts, but those theories always must start from some axioms or assumptions, which are taken arbitrarily, just to make our theory fitting to observations by Tycho (and MtPalomar).Sample of modern explanation of your question: if the forces wouldn't be balanced, the total momentum of Sun-Earth system wouldn't be conserved. That is not so painful by itself, but there is a correspondence between conservation laws and symmetries. Momentum conservation (thus Newton's 3rd law) is bound to translational symmetry. If 3rd law would be violated, the same experiments would give different results if performed at different places. It is something contrary to our common-sense experience and would destroy the very foundations of the whole physics.

 

[Ledgeknow]

Yeah, I thought that was the case, I was just looking for a confirmation. I'd like to know more about the conservation of momentum (as I understand, it's a repercussion of the third law of motion), but I won't take anymore of your time, I'll try to find a good book and study it more in-depth. Thank you for the assist, it was helpful!

Just one more question, why is conservation of momentum important?

 

[xts]

Just one more question, why is conservation of momentum important?

Because it is equivalent to translational symmetry of our world...
Could you imagine the world having different laws of physics acting in Berlin and different in London? That would be an implication of breaking translational symmetry (or breaking 3rd law of dynamics, as it is equivalent to translational symmetry)

Regarding conservation laws <-> symmetry correspondence - try to read something about Noether's theorem, you may start from Wiki: http://en.wikipedia.org/wiki/Noether's_Theorem

 

[Ledgeknow]

Because it is equivalent to translational symmetry of our world...
Could you imagine the world having different laws of physics acting in Berlin and different in London? That would be an implication of breaking translational symmetry (or breaking 3rd law of dynamics, as it is equivalent to translational symmetry)

Regarding conservation laws <-> symmetry correspondence - try to read something about Noether's theorem, you may start from Wiki: http://en.wikipedia.org/wiki/Noether's_Theorem

Thank you very much for the link, I'll try to delve deeper into the subject!

 

[darkxponent]

@Ledgeknow : just keep in mind one thing in physics. When call something law in physics it simply means it is law . A law is not derived it is just abserved and accepted. We accept it just becoz it happens.A LAW CANT BE PROVED. A law might seem illogical to someone but it is just the truth. So we have to accept it. that's it.

People do find logic and stories that WHY A LAW IS THERE. but its not correct. It will only take you to another LAW.

for example i can prove NEWONS III LAW with the help of CONSERVATION OF LINEAR MOMENTUM but it is not a proof as it makes CONSERVATION OF LINEAR MOMENTUM a LAW.

You see i tried to proove it but it took me to another law.

Now someone told THE TRANSLATION SYMMETRY and u accepted it. WHY didnt u asked WHY TRANSLATION SYMMETRY. And not asking a 'WHY?' to a theory makes the theory LAW. Thats it. You tried to find a logic behind a LAW and it took you to a new LAW

 

[Philip Wood]

Try this informal argument...

Suppose that two particles have a potential energy E, which is a function of their separation,s. This relies on translational symmetry: if both particles are translated through space by the same displacement, their mutual PE doesn't change.

Choose a co-ordinate system such that both particles (1 and 2) lie on the x-axix, at x₁ and x₂ [x₁ < x₂].

Then s = x₂ - x₁.

Now, the force on particle 1 is F₁ = -$\partial$E/$\partial$x₁ = + dE/ds,

whereas the force on particle 2 is F₂ = -$\partial$E/$\partial$x₂ = - dE/ds.

So F1 = - F₂

 

[DaveC426913]

Why? Why would Earth exert a force on the Sun, equal in magnitude, opposite in direction? Why wouldn't the Sun just push Earth around?

I see this question has received much attention already but I'll add a tidbit of intuitive logic.

Think about what would happen if objects did not exert an equal and opposite force when a force is applied.

It would mean that, if I tried to use my own strength to push on the sun, it would provide no resistance. I could toss the sun about like a basketball!

 

[Philip Wood]

Can't resist having another crack at this, trying to boil the argument in my last post down to its essentials...

Suppose two particles, 1 and 2, attract each other with forces depending only on their separation. [This assumes translational symmetry: the forces stay the same even if we displace the particles, provided we keep their separation the same.]

This means that the system has a certain potential energy. The energy belongs to the whole system and can't be separated into energy belonging to particle 1 and energy belonging to particle 2. For me, this notion of mutual energy is intuitively acceptable, and more 'obvious' than Newton's Third Law. But I'm not claiming that it's more basic than Newton's third law, just that I like it better as a starting point. That said, the argument goes like this...

If we let particle 2 move by small distance $\delta$s
in a direction towards particle 1, particle 2 has work F₂ $\delta$s done on it by the attractive force from particle 1 and the system loses F₂ $\delta$s of potential energy.

But If we let particle 1 move by the same small distance $\delta$s
in a direction towards particle 2, particle 1 has work F₁ $\delta$s done on it by the attractive force from particle 2 and the system loses F₁ $\delta$s of potential energy.

But by our original postulate that the potential energy depends only on the separation of the particles, these energy changes are equal, so F₁ = F₂. [These are,of course, magnitudes of forces; their directions are opposite.]

 

[simpatico]

The more ambiguous case to me is the example of the gravitational interaction between our planet and the Sun.
Why? Why would Earth exert a force on the Sun, equal in magnitude, opposite in direction? Why wouldn't the Sun just push Earth around?
Can anybody see what's bothering me?.[/B]

No need to be worried ledge , you see...

...You are simply forgetting that law #3 follows law #2 : F = ma

The equality of the forces is only formal.When you apply the formulas
F(s) = GMm / r^2 F(e) = GMm/ r^2

You get the real values:
F(s) = GM/ r^ 2
F(e) = Gm/ r^ 2

so F(s) / F (e) = 333,000

cheer up! no more nightmares :zzz:

 

[Doc Al]

The equality of the forces is only formal.

What does that mean?

When you apply the formulas
F(s) = GMm / r^2 F(e) = GMm/ r^2

OK.

You get the real values:
F(s) = GM/ r^ 2
F(e) = Gm/ r^ 2

Nope. You had it right the first time.

so F(s) / F (e) = 333,000

Nope.

 

[Dale]

When you apply the formulas
F(s) = GMm / r^2 F(e) = GMm/ r^2

You get the real values:
F(s) = GM/ r^ 2
F(e) = Gm/ r^ 2

so F(s) / F (e) = 333,000

The variables highlighted in red are all accelerations, not forces. The forces are equal and opposite, the accelerations are not.

 

[simpatico]

Nope. You had it right the first time.
Nope.

The forces are equal and opposite, the accelerations are not.

I just applied the same explanation given in all scientific textbooks to justify g ! . . ()
Is the force of an apple equal to the force of the earth?
Doesn't Earth give the same acceleration to a feather and a cannonball?

So isn't F = a , disregarding mass (of the other body) ?

(My source was Encyclopaedia Britannica)

 

[Dale]

I just applied the same explanation given in all scientific textbooks to justify g ! . . ()
Is the force of an apple equal to the force of the earth?
Doesn't Earth give the same acceleration to a feather and a cannonball?

Yes, but you have severely misapplied the physical principle. In the absence of air, the Earth gives the same acceleration a1 to a feather and a cannonball, and the sun gives the same acceleration a2 to a feather and a cannonball, but that in no way implies that the acceleration from the sun, a2, is equal to the acceleration from the earth, a1.

You can derive this from the laws as follows:
f = ma
f = GMm/r²
ma = GMm/r²
a = GM/r²

So the acceleration of one object depends on the mass of the other object. You can divide out the mass of one object, but you must do so from both sides leaving you an acceleration on the left, not a force. You cannot just willy nilly drop out terms for no reason as you did above. It is mathematically incorrect (illogical) and is inconsistent with experiment.

So isn't F = a , disregarding mass (of the other body) ?

No, they cannot possibly be equal. The units are not the same. Always check your units.

(My source was Encyclopaedia Britannica)

Perhaps you should try a physics textbook instead.

 

[chingel]

Here is how I understand it.

When you push on an object, it moves because electrons have charge, the like electromagnetic charges repel each other and when you put them close the electrons push each other away. This force applies both ways. If you try to push a spring together with your fingers, it doesn't want to move and exerts a force, it just pushes in both directions. You can put pressure on with the top finger or the bottom finger, the spring just pushes the same way, meaning there is an equal force in both directions.

So basically there is a force between any two objects pushing them apart when you try to push them together and this force just pushes, it doesn't matter which body is causing it.

 

[simpatico]

1) that in no way implies that the acceleration from the sun, a2, is equal to the acceleration from the earth, a1.

2) So the acceleration of one object depends on the mass of the other object.
.

1) I never said that.Actually I did not even mention the Sun in that post

2)That is what I said, but not exactly

I said that the acc. of an object depends exclusively on the mass of the other object, and NOT on its own mass
so a feather (in vacuum or not) receives exactly the same acc. of a cannonball, a plane ...
is that correct?

(P.S. I need to make an example, could you, kindly, give me some reliable data:
1) a ball 1 kg is rolling at 1 m/s. what is the F(b) (push) it exerts on an obstacle?
2) how do you calc F(e) (gravity pull) of earth?
3)do they obey to exactly the same laws, can I compare freely Fb and Fe to explain my point?

much obliged)

 

[Dale]

I said that the acc. of an object depends exclusively on the mass of the other object, and NOT on its own mass

No, you said that the gravitational force on an object depends exclusively on the mass of the other object. That is not at all the same as saying that the accelerations do. If you had said "acceleration" instead of "force" then neither of us would have corrected you.

(P.S. I need to make an example, could you, kindly, give me some reliable data:
1) a ball 1 kg is rolling at 1 m/s. what is the F(b) (push) it exerts on an obstacle?

That depends on the Young's modulus and mass of the obstacle.

2) how do you calc F(e) (gravity pull) of earth?

f=GMm/r^2

3)do they obey to exactly the same laws, can I compare freely Fb and Fe to explain my point?

No. Fe obeys Newtons law of gravitation. Fb obeys Hookes law.

 

[simpatico]

1) f=GMm/r^2

2) Fb obeys Hookes law.

Hi Dalespam, I owe you an explanation,
and an apology for being too hasty
( when I said formal (DocAl) I meant the same as when I apparently dropped the unit.
that is m = 1 (the same as you get when you reduce 6/8 to 3/4*2/2)
so, when I say F = a, it is not through stupidity or ignorance but because I'm just meaning
F / (m=1) = a
I thought the feather and cannonbal had clarified this point

at the end of the discussion, if appropriate, I'd be glad to apologize

1) I know the formula: I want to see how you get your result
2) you mean that F = ma does not apply here?
could you please calculate F explicitly, and then calculate the acc. of a feather of 10 grams and a cannonball of 10 kilos?

Much obliged

 

[Dale]

that is m = 1 (the same as you get when you reduce 6/8 to 3/4*2/2)
so, when I say F = a, it is not through stupidity or ignorance but because I'm just meaning
F / (m=1) = a [/I]

That's fine, then you say something like "using units where m=1". However, what you wrote above is still incorrect even so, because there are no units where both the mass of the Earth and the mass of the sun are 1.

You could use two different systems of units for each formula, but again, you would need to say something like "now switching to a system of units where M=1".

Furthermore, you would not be able to simply divide out the two expressions to obtain the ratio of the forces, since they are in different units. You would have to put in the conversion factor between the different units, which, had you done so, would have demonstrated that the ratio of forces was 1.

1) I know the formula: I want to see how you get your result

I am traveling and posting on a mobile device, so I won't be doing any calculations. There is no mystery in plugging numbers into a formula.

2) you mean that F = ma does not apply here?

It certainly does apply, but what is a? Both f and a are unknown, so you have one equation in two unknowns. You need another equation, known as a force law (e.g. Newton's law of gravitation or Hooke's law), to determine the force.

Also, note that the f in f=ma is a net force. So you can only set it equal to the force law expression in the situation where that is the only force acting on the object. That is fine for the Earth and the sun, but not for a feather dropping in air, nor for a ball colliding with an obstacle.

 

[simpatico]

What does that mean?

.

Hallo Doc,
are you beginning,now, to see my point? It is a very strong point.

When you say that a feather gets the same a of a cannonball, you are implicitly admitting that
(a=k => ( a*m1 = F, a*m2 = F...) => (m1,m2...= 1) => (F = a)

we can manipulate the formula balancing the two sides (or in any other way),
but the gist will not change
Force is acting on all masses in the same way. mass is irrelevant!. This is a hard fact

You use 6/ 8, and you are right, because hou have the right to do so. And I know also why you have to do so
I use 3/ 4 and I'm right, too. Maybe a little more. And not only because it is more elegant and simple, but because 3/ 4 means => a= F= G * M (/...) and this
gives you a tremendous
insight on what G is (beyond the mere konstant), and most of all
of what the NATURE of F(G) is
...
I DO NOT want to convince you. I'm asking you just to reflect a while on it.
You, and my dear friend Dalespam, don't have to convince that it is as you say.
They already convinced me way back in 1959, when they first taught me the formulas.
I have reflected on it for half a century.I have just a little start

Good bye Sir, I have been greatly honoured by the attention you have given me.

(P.S.
I'll go on arguing ad libitum with Dalespam,
because is an old friend and has more patience with me)

 

[simpatico]

1) I am traveling and posting on a mobile device,...

2)It certainly does apply, but ...

2) Forget it, now I think is useless to put too many irons in the fire.
(if you feel like it, we will pick it up another time)

1) Don't worry ,Dalespam, I' ve got time on my hand. Take your time!, any time!

If you want to conclude the discussion,
I begged you to show me in a detailed way how you calculate ,not acc., mind you!
but F (e) at ground level and
THEN,
from this how you get a
This is fundamental.

(P.S. I hope you don't really think I am so ignorant I don't know these basic things
BTW, I did my homework in the first post!)

 

[Doc Al]

Hallo Doc,
are you beginning,now, to see my point? It is a very strong point.

No. You continue to conflate force with acceleration.

When you say that a feather gets the same a of a cannonball, you are implicitly admitting that
(a=k => ( a*m1 = F, a*m2 = F...) => (m1,m2...= 1) => (F = a)

Nonsense, as DaleSpam has explained.

Instead, find the acceleration using a = F/m and you'll find that 'm' drops out.

we can manipulate the formula balancing the two sides (or in any other way),
but the gist will not change
Force is acting on all masses in the same way. mass is irrelevant!. This is a hard fact

I think you are trying to say that acceleration (due to gravity) is independent of mass. That's true and well known.

But don't confuse the force of gravity, which depends on the mass of the object, with the resulting acceleration due to gravity, which does not.

 

[Dale]

2) Forget it, now I think is useless to put too many irons in the fire.
(if you feel like it, we will pick it up another time)

1) Don't worry ,Dalespam, I' ve got time on my hand. Take your time!, any time!

If you want to conclude the discussion,
I begged you to show me in a detailed way how you calculate ,not acc., mind you!
but F (e) at ground level and
THEN,
from this how you get a
This is fundamental.

(P.S. I hope you don't really think I am so ignorant I don't know these basic things
BTW, I did my homework in the first post!)

Then why don't we skip the details of the pointless arithmetic. In the expression f=GMm/r² I look up G, M of earth, r of Earth and assume a 5 kg bowling ball and a 5 g feather respectively for m. The gravitational force on the bowling ball will be 1000 times greater than that on the feather, but using f=ma for each (and assuming no other forces, i.e. free fall in vacuum) the accelerations will both be ~9.8 m/s². If you have done your homework you know how I would go from the formulas to those conclusions.

Given our discussion so far, is there anything that you would revise about your initial post to make it more clear or correct?

 

[Boy@n]

Here is how I understand it.

When you push on an object, it moves because electrons have charge, the like electromagnetic charges repel each other and when you put them close the electrons push each other away. This force applies both ways. If you try to push a spring together with your fingers, it doesn't want to move and exerts a force, it just pushes in both directions. You can put pressure on with the top finger or the bottom finger, the spring just pushes the same way, meaning there is an equal force in both directions.

So basically there is a force between any two objects pushing them apart when you try to push them together and this force just pushes, it doesn't matter which body is causing it.

I like this explanation, but not sure if it's correct... can anyone confirm it?

I can imagine an exception though, but just for fun, because it's not really against your explanation... if you push one magnet (+) towards the other (-), they won't move appart but "glue" together. Yet, these two magnets would never really touch due to electromagnetic charges repeling each other...

It's funny when we think about this in such a way, e.g. it means that when we sit on the chair in truth we are floating in the air...

 

[Dale]

I can imagine an exception though

Yes, there are lots of exceptions since he was specifically talking about contact forces between solid objects, but in that domain the explanation is valid.

The moon doesn't have an overall dipole magnetization like the earth.

 

[mathfeel]

Allow me to put an additional spin on this. Newton's third law is not satisfied for moving charge particle. Suppose particle A with charge q is at rest at the origin; particle B is at some distance r moving away from the origin with speed v. The force A exerts on B is the usual Coulomb law:
$$F_{B} = \frac{q^2}{r^2}$$
The force B exerts on A is reduced by the square of the Lorentz's factor:
$$F_{A} = \frac{q^2}{r^2}\left(1 - \frac{v^2}{c^2}\right)$$
This is a relativistic effect. It is due to finite speed of propagation of force. Momentum conservation is saved only if you also includes the momentum being carried by the field.

Here's an analogy I came up with. Suppose you are an astronaut and pulls on a string which is tucks on an object of mass M, resulting in its acceleration a.
The force you exerts (or transmit) through the string on the mass is M a, but the force that's pulling on you is (M + m)a, where m is the mass of the string. Of course, N3 is still satisfied "locally". If for some reason the string is invisible, and an observer is only counting You and the mass as "objects". He would proclaim that N3 is violated. The EM field which carries the electromagnetic force is analogous to the invisible string here.

 

[Boy@n]

The moon doesn't have an overall dipole magnetization like the earth.

Was thinking about that yes, thanks for clearing that up... and as I found out, moon is always facing Earth with the same side http://www.wisegeek.com/why-does-the-same-side-of-the-moon-always-face-the-earth.htm" "effects" on the long run...

 

[Dale]

Furthermore, you would not be able to simply divide out the two expressions to obtain the ratio of the forces, since they are in different units.

It looks like simpatico got himself banned, but I thought I would post a clarification anyway. This statement of mine is only strictly true if you are using a consistent set of units to define G. It is possible to use inconsistent systems of units such that M is in solar masses but f is still in Newtons. Of course, that makes G have a different numerical value and very strange units, but the net result is always that the forces are equal.

 

[D H]

In teaching
it for years, and for the purose of solving the classical problems, I summarized it to this:

http://www.thermospokenhere.com/wp/01_tsh/A219_About_f_ma/about_f_ma.html

I disagree on many fronts.

First off, you are not smarter than Isaac Newton. Neither am I, and (I suspect) neither is anyone on this forum. Newton wrote three laws, not two, not one. He did so for good reason.

The first law is distinct from the second.
The modern point of view is that the first law establishes the framework in which the other two laws apply. In particular, it defines the concept of an inertial frame of reference. An object that is subject to a null external force does not always travel along a straight line path at a constant velocity. When I'm on a merry-go-round I see the remote stars appear to be in orbit around me. The motion of those remote stars obeys Newton's first law if and only if the motion is described in / observed from the perspective of an inertial reference frame.

Is the second law really a law or is it just definitional?
One view is that Newton's second law just defines force: F=ma. Another point of view is that force is an undefined term in Newtonian mechanics, just as point, line, and plane are undefined terms in Euclidean geometry. In this point of view, Newton's second law truly is a law of nature.

Which is the correct form: F=ma or F=dp/dt?
Arguing whether the correct form of Newton's second law is F=ma versus F=dp/dt is rather pointless in the case of constant mass; the two forms are identical in this case. Newton wrote his laws with regard to particles (another undefined term in Newtonian mechanics) of constant mass. Two key concepts: Force is subject to the superposition principle (the net force on a particle is the sum of the individual forces on that particle) and force is frame invariant (the net force acting on an object is the same to all inertial observers).

There is a huge difference between F=ma and F=dp/dt when Newton's laws are generalized to systems of non-constant mass. Using F=ma gives some very nice results: Force is subject to the superposition principle, and force is frame invariant. Throw both of these out the window if you insist on F=dp/dt as being the correct form. People who deal with variable mass systems invariably use F=ma.

What about Newton's third law?
In a sense, Newton's third law is the only one of the three that qualifies as a law of nature. The first two can be viewed as definitional. The third law says something physical about forces (and about torques, depending on whether you are looking at the weak or strong form of Newton's third law). You can of course get to Newton's third law by conservation of linear and angular momentum, but you can also derive the conservation laws from Newton's third law. Which is more fundamental is in a sense another one of those "how many angels can dance on the point of a pin" arguments -- until you get to Noether's theorem.

 

[xts]

First off, you are not smarter than Isaac Newton. Neither am I, and (I suspect) neither is anyone on this forum. Newton wrote three laws...

Well... Have you tried to read Principia? I don't expect you reading Newton's Latin, but even modern translation?
To be honest - that was most difficult reading in my history of science expertise... Actually, Newton's theory is one of those few which had to be transformed and explained by others to become understandable.
What we today call 'Newtonian mechanics' is based on Newton's work, but its formulation has pretty little in common with his original work.

 

[D H]

Well... Have you tried to read Principia? I don't expect you reading Newton's Latin, but even modern translation?

Yes, I have, and yes, it is painfully difficult. Newton did not have the modern mathematics that makes the modern interpretation of Newton's laws easily comprehensible. Vectors? Vectors are a modern development, about a hundred years old. Reading any physics paper that dates from before the very end of the 19th century is extremely tedious. Vectors clean things up so very nicely. Algebra? Newton tended toward geometric reasoning rather than algebraic reasoning. Modern algebra was in its infancy in Newton's time. Calculus? While Newton and Leibniz are viewed as having independently developed the calculus, it is Leibniz form that we typically use nowadays, later modified extensively by Weierstrass. Newton's calculus was a bit (more than a bit) idiosyncratic.