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A vector is a quantity that transforms just the way the coordinates transform under rotation (while a scalar remains invariant under rotation).

In FLP, he says suppose $F$ is a vector and probably later wanted to expound on why $F$ is a vector but I think he forgets to explain it.

Also if $F$ defined to be $ma$, still, if the axes rotate with an angular speed whose second derivative is not 0, we find $F = ma$ to not hold. (Therefore we get introduced to the concept of a pseudoforce but how did force become a vector in the first place).

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Ok, I think this is the answer you're probably looking for. Everyone else is giving mathematically correct answers, but I think they've forgotten that Feynmann has a funny definition of vectors; he defines them to be a three component object that transforms like $\vec{r}$ when you rotate the system.

So: We want to show that $\vec{F}$ transforms like $\vec{r}$ does under rotations.

We note, first, that $\vec{v}$ is given by $\frac{d\vec{r}}{dt}$, so it transforms like $\vec{r}$. More precisely, if $R$ is a rotation matrix, and $\vec{v}'$ is our transformed velocity, we have

$$ \begin{array}{rcl} \vec{v}' & = & \frac{d\vec{r}'}{dt}\\ &=& \frac{d R\vec{r}}{dt}\\ &=& R\frac{d \vec{r}}{dt}\\ &=&R\vec{v} \end{array} $$

Similarly, $\vec{a}$ transforms like $\vec{r}$, since it is the second derivative of $\vec{r}$. Then since $\vec{F}=m\vec{a}$, it too transforms like $\vec{r}$.

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  • $\begingroup$ These are my exact doubts: 1. First of all, this argument assumes F = ma is a definition . I may be wrong but, if F = ma is a law, one could say the LHS is something and the RHS is another thing and by this law, backed up by empirical evidence, LHS = RHS. 2. Lets say F = m*d2x/dt2...and rotate the axis by θ(t) = t^3...then applying the transformations x->x' we get extra terms!!! $\endgroup$ – novak Mar 9 '16 at 20:02
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    $\begingroup$ $F=ma$ IS a definition. It defines what we mean by force. The fact that $F=ma$ is well defined--aka, that a smaller mass and a larger acceleration produce the SAME force--is the content of Newton's law $\endgroup$ – Jahan Claes Mar 9 '16 at 21:45
  • $\begingroup$ I'm not sure why you're so concerned with rotating frames of reference. When Feynman says a vector transforms a certain way under rotations, he's referring to a single rotation, constant in time. In other words, these transformations are between two inertial reference frames, rotated with respect to each other. He DOESN'T say anything about how they transform into non-inertial reference frames. $\endgroup$ – Jahan Claes Mar 9 '16 at 21:46
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I should warn you that you're conflating here two concepts:

1)whether a certain object is a vector or not a vector or a pseudo-vector.

2)Whether a given force is a real force or a pseudo/fictitious force.

regarding point 1): -The formal definition of a vector is that it's an element- $n$-tuple - of a vector space which satisfies a group of axioms.

I won't list the whole axioms(you can check them in the wiki article on vector spaces), but some include stuff like, if you scale a vector, it should remain a vector(closure under scaling), If you add two vectors, the resultant is a vector too(closure under addition) and etc.

Also This vector space has to be equipped with an inner product(dot product) that enables us to measure its magnitude and direction.

Equivalently one can define just as you stated a vector as an object which transforms properly under coordinate rotation say, like the displacement vector. Those which does not satisfy the aforementioned axioms are not vectors.

But there's a third category called pseudo-vectors, they're almost like vectors in everything, except that they behave differently under coordinate inversion. For example if you have a vector $\mathbf{A}$ pointing the positive x direction, and now if you inverted your coordinates then it'll become $- \mathbf{A}$ in the new system which makes sense.

If you consider the object $$\mathbf{C} =\mathbf{M} \times \mathbf{N}$$

If you applied a coordinate inversion, how $\mathbf{C}$ will look like in the new system? like this $$-\mathbf{M} \times -\mathbf{N}=\mathbf{M} \times \mathbf{N}= \mathbf{C} $$

So $\mathbf{C}$ does not change signs under inversion! Such an object is called a pseudo-vector.

Force is nothing but the acceleration vector scaled by a factor of $m$(remember the closure under scaling axiom) so it inherits from acceleration everything that makes it a proper vector. So what is the deal with the pseudo-force talk? this brings me to point 2):

-Even in the realm of classical mechanics, Newton's laws of motion are not always valid. They're valid only for observers that are in constant velocity motion.Therefore for Observers that are accelerating(rotating or moving non-uniform translational motion and so on) the laws break down.

A very peculiar thing called Fictitious force arises. For example when a bus suddenly stops, you and everything else experience a push forward. We call this pseudo/fictitious for a good reason. Because for observers on the ground, the people in the bus are not "pushed" by some force, rather its their inertia combined with the fact that the bus is decelerating(picking speed in the direction opposite to its direction of motion) that gives them the illusion of being pushed.

So a pseudo-force is a force that arises for accelerating observers, and once we switch coordinates to observers in constant velocity motion, such force disappears.

The thing is that a force is a vector, whether a particular force is a real force or a pseudo/fictitious force is another story.

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This is an old question, and one that used to be taken seriously by textbooks.

From a dynamics point out view you could assume $F=m\vec a$ or $\mathrm d \vec p/\mathrm d t$ and then a force has to be a vector because acceleration and momentum are, and that's because ultimately displacement (at least ones over small time intervals) are vectors (or at least the limits of their changes are).

But you can have forces without dynamics. So if you consider statics as a separate subject worthy of its own foundations and explanatory power, then you need a completely separate reason for forces to be a vector. But now you can make a symmetry argument. The lack of a velocity allows you to argue that it forces must add like vectors. And statics does require an ability to add forces to get a total force.

There are many different arguments for the statics symmetry argument. Some are easier or cleaner but only work for contact forces. Others are more drawn out. But it does come down to how you think forces should be added in a way where the result respects the symmetries you expect from nature. It isn't derived from something else in statics because in statics you don't have $\vec F=m\vec a$ and in fact statics should be compatible with any alternative to the second law.

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Force is a vector because it obeys the law of superposition via the parallelogram rule, which also works for geometric vectors. This result was known prior to Newton, for the case of static forces, and he included a "proof" when he introduced the parallelogram of force into his Principia Mathematica.

See https://en.wikipedia.org/wiki/Parallelogram_of_force; the first link leads to Newton's proof of Corollary I: https://en.wikisource.org/wiki/The_Mathematical_Principles_of_Natural_Philosophy_(1729)/Axioms,_or_Laws_of_Motion#Cor1

Note: It's often more convenient to describe vectors in terms of vector space properties, as shown in linear algebra, than the coordinate transformation rules that were used in older works.

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  • $\begingroup$ So what exactly is a vector. Is it a transformation thing described by Feynman or an experimental fact. $\endgroup$ – novak Mar 8 '16 at 20:31
  • $\begingroup$ A vector is an element of a vector space; it will, of course, satisfy the coordinate transformations that Feynman describes. Experimentally one can show that force behaves as a vector; for example, on an air table, or by testing with ropes and pulleys for statics. $\endgroup$ – Peter Diehr Mar 8 '16 at 20:43
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    $\begingroup$ I'm not qualified to say what is the definition of "vector" (if there even is a preferred definition), but it seems as if it was invented by proto-scientists to describe the kinematics of physical objects. In other words, force is a vector because vectors were invented to describe force, position, momentum, etc. $\endgroup$ – Solomon Slow Mar 8 '16 at 20:49
  • $\begingroup$ For the history of vector analysis see: math.mcgill.ca/labute/courses/133f03/VectorHistory.html $\endgroup$ – Peter Diehr Mar 8 '16 at 22:41
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A scalar in any number of dimensions needs only one number to define it completely. Commonly known scalars are mass, temperature and pressure.

A vector in N dimensions needs N numbers to completely specify it , but these can be chosen in many ways : x,y,z or r,theta,phi or r,theta,z.

The co-ordinate system does not even have to be orthogonal , however orthogonal axes are often chosen to avoid annoying cross terms.

The level above vectors are tensors of rank 2,3 and 4 . Stress(rank 2 )and strain ( rank 2) are related by elastic stiffness a rank 4 tensor with 81 components in 3-D space, however many of the components are 0.

Working in 2 dimensions Force needs 2 numbers to specify it and 3 in 3 dimensions.

As stated above force follows the additive and parallelogram laws for vectors.

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Force needs to be a vector, that is, to have components in the $d$ directions of space, because it is in duality relation with displacements in these $d$ directions of space.

Similarly, stress is a tensor because it is in a duality relation with deformations.

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Force is only a concept in Newtonian mechanics. It is said the concept of force is not even needed in Hamiltonian mechanics; yet it is equivalent to Newtonian mechanics .

The formula f = ma is a definition of force; therefore the property of force as a vector is acquired only through a definition of it. So the vectorial nature of force is nothing but the vectorial nature of acceleration.

The answer to "Why is acceleration a vector?" would be the same answer to the question "Why is force a vector?"

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