4
$\begingroup$

Classical physics models events occuring in the spacetime $\mathcal E\times \mathcal T$ where $\mathcal E$ is a dimension 3 euclidean point space and $\mathcal T$ is an interval of $(\mathbb R, <)$ (an ordered set).

An observer is a fictitious human being or sensor that can infinitely precisely describs events.

A reference frame is made of a frame $R=(O, \vec e_1, \vec e_2, \vec e_3)$, an initial time $t_0$ and a coordinate system uniquely mapping a triplet $(x,y,z)$ to a generic point $M$ in $\mathcal E$. Point $O$ is in $\mathcal E$ and is chosen as the origin and $(\vec e_1, \vec e_2, \vec e_3)$ is a basis of the euclidean vector space $\vec E$ attached to $\mathcal E$.

It seems to me that the first purpose of such a reference frame is to quantify (I mean with values not with symbols) the location and the instant of any event. Without it a generic point $M$ still qualifies a unique location of $\mathcal E$ and a real $t$ an instant in $\mathcal T$ and the inner product a tool for geometry...

Since we can define arbitrarily a reference frame given a point and a basis let's suppose that $O$ is the location of the observer. The observer moves with point $O$.

It is often said that Newton's first law postulates the existence of privilegied reference frames in which a body experiencing null resultant of forces $\vec F$ have a constant velocity, that is null acceleration $\vec a$, such frames are called inertial reference frames or galilean reference frames. Then Newton's second law as the simple form $\vec F=m \vec a$ only in inertial reference frames.

In mathematics, it is taught that a vector is intrinsic in the sense that it's existence preceeds the one of basis and it is not dependent of the basis in which it is quantified.

So my question is why the notion of reference frame is usefull in the statement of Newton's second law?

It seems to me that only point $O$ is needed, we actually don't care about the basis. Point $O$ should be an "inertial point" the basis can rotate, vectors change in norm, angles between two vectors can change as long as they stay linearly independent because we still can quantity $\vec F$ in every such basis.

Do you have an opinion on that? Any book that consider vectors as intrinsic objects?

$\endgroup$
  • $\begingroup$ Newton's First Law of Motiondefines the inertial reference frame. Since constant motion, in any direction, satisfies the conditions, there are an infinite number of valid choices for Newton's Second Law. The choice of an observer fixes this choice. The choice of a basis is thus secondary, but matematically convenient. From this you can see that there is no privileged frame. $\endgroup$ – Peter Diehr Mar 22 '16 at 0:05
  • $\begingroup$ The usefulness of the concept of reference frames in Newtonian physics arises when we deal with systems where calculations become really difficult if we evaluate the motion of a body from the perspective of an inertial frame (which we ALWAYS do without realising it). It becomes easier to work in the frame of the accelerating body (non inertial reference), and we introduce a 'pseudo force' on each body we observe. (mass of the body times the acceleration of frame). This is where most people realise the point of reference frame, which we always take for granted. $\endgroup$ – GRrocks Mar 22 '16 at 4:55
  • $\begingroup$ @PeterDiehr, I totally agree with you, it is mathematically convenient, but when using tensors in an intrinsic way (without using indices), which is the case in continuum mechanics, there is no need, in my opinion to define the basis. $\endgroup$ – KevMoriarty Mar 22 '16 at 17:36
  • $\begingroup$ @GRrocks, I agree as well with you, but what I wonder is if 'reference of frame' can be replaced by 'observer location' in Newton's second law since it is enough, in my opinion, to describe the motion in an affine point space. Then we can choose an observer attached to the accelerating body that we are studying if we want to simplify calculus. Isn't the 'pseudo force' always determined with respect to a reference frame? I never tried to compute one. $\endgroup$ – KevMoriarty Mar 22 '16 at 17:47
  • $\begingroup$ @GRrocks, In my question, in previous comment you have to read 'inertial reference frame' not 'reference frame'. $\endgroup$ – KevMoriarty Mar 22 '16 at 18:02
2
$\begingroup$

You have a wrong idea of classical spacetime $V^4$. It is not the Cartesian product $\mathbb E^3 \times \mathbb R$.

It is instead a fiber bundle $$T: V^4 \to \mathbb R$$ such that each fiber $\Sigma_t = T^{-1}(t)$, the absolute space at fixed time $t$, is isomorphic to a three dimensional Euclidean space (*) $\mathbb E^3$. The basis of the bundle, the axis $\mathbb R$, is the range of the absolute time $T$ which is defined up to an additive constant.

The difference between this notion of spacetime as a bundle $T: V^4 \to \mathbb R$ and a trivial scalar product $\mathbb E^3 \times \mathbb R$ is fundamental: Here there is no canonical representation of $V^4$ as a Cartesian product $\mathbb E^3 \times \mathbb R$.

More precisely, every choice of a reference frame defines such a representation.

A reference frame is nothing but a (smooth) surjective map $$\pi : V^4 \to \mathbb E^3$$ such that $\pi|_{\Sigma_t} : \Sigma_t \to \mathbb E^3$ is an isomorphism of Euclidean spaces (i.e., a surjective affine isometry), for every $t\in \mathbb R$.

In this picture, $\mathbb E^3$ is viewed as the rest space of the reference frame.

This way the spacetime $V^4$ is identified with the Cartesian product $\mathbb R \times \mathbb E^3$ by means of

$$V^4 \ni p \mapsto (T(p), \pi(p)) \in \mathbb R \times \mathbb E^3$$

There are however infinitely many such identifications depending on the choice of the reference frame.

Consider two reference frames $\pi$ and $\pi'$ and fix Cartesian orthonormal coordinates in the respective rest spaces $\mathbb E^3$ and $\mathbb E'^3$, and use the absolute time defined up to an additive constant as time coordinate.

Using the fact that $\pi'|_{\Sigma_t}\circ (\pi|_{\Sigma_t})^{-1} : \mathbb E^3 \to \mathbb E'^3$ is a surjective affine isometry, you easily see that the transformation of coordinates must be of the form $$t'=t+c \:,\quad x'_i = \sum_{j=1}^3 R_{ij}(t) x_j + b_i(t) \tag{1}$$ where $R(t) \in O(3)$ and $b(t) \in \mathbb R$ for every $t$.

These are the most general transformation of coordinates between Cartesian coordinates at rest with different reference frames.

To define the velocity of a section $$\mathbb R \ni t \mapsto \gamma(t) \in \Sigma_t $$ you need a full reference frame as indicated not only a reference point. Indeed, the velocity of $\gamma$ with respect to $\pi$ is computed as $$\vec{V}_\pi(t) := \frac{d}{dt}\pi(\gamma(t))$$ and, with this definition, it is a vector in the rest space $\mathbb E^3$ of $\pi$. However it can be seen as a vector in $\Sigma_t$ using the inverse of the isomprphism $\pi|_{\Sigma_t} : \Sigma_t \to \mathbb E^3$. Exploiting this identification, velocities of the same section but referred to different reference frames can be compared in the absolute space $\Sigma_t$.

REMARK. It is not enough to fix a reference point, that is a section $\mathbb R \ni t \to O(t) \in \Sigma_t$ to define the velocity of another section $\mathbb R \ni t \to \gamma(t) \in \Sigma_t$. Your idea is to take the limit $$\lim_{h \to 0} \frac{1}{h}\left[\left(\gamma(t+h) -O(t+h)\right) - \left(\gamma(t) -O(t)\right)\right]\:.$$ The point is that the difference $$\left(\gamma(t+h) -O(t+h)\right) - \left(\gamma(t) -O(t)\right)$$ does not make sense as the two vectors$\gamma(t+h) -O(t+h)$ and $\gamma(t) -O(t)$ belong to different vector spaces. In order to make sensible that difference is necessary to isometrically identify the spaces. This is exactly what the notion of reference frame does.

Inertial reference frames are defined as reference frames where every isolated body moves with constant velocity. It is easily proved that this constraint imposes a strong restriction to the form of the coordinate transformation (1) between inertial frames which therefore specialises to

$$t'=t+c \:,\quad x'_i = \sum_{j=1}^3 R_{ij} x_j + tv_i + b_i \tag{2}$$

that is a generic transformation of Galileo's group. It is nice to observe that, up to isomorphisms, there is only one affine structure in the classical spacetime such that Cartesian coordinates at rest with inertial reference frames together with absolute time as forth coordinate define affine coordinate systems of that structure. The sections of the spacetime which are right lines (geodesics) of that affine structure are all possible inertial evolutions of isolated matter points. To this respect classical physics and GR are not so different.

Statements as Newton's second law are formulated with this notion of reference frame (if one wants to be completely rigorous).


(*) An Euclidean space $\mathbb E^n$ is an affine space whose $n$-dimensional space of vectors $T^n$ describing translations in $\mathbb E^n$ is equipped with a positive scalar product.

$\endgroup$
  • $\begingroup$ Your answer is really interesting and remembers me that i am still far away of understanding manifolds. Do these defs are based on experiments? I mean, I understand why we requires the axioms of an affine point space, from origin and translation vectors the observer can "points" the location of any points. I understand the inner product as a modelling tool to measure distances and angles. What do we have in addition with fiber? The ability to deform the space, like if we see the inner product as covariant order 2 tensor field, then distance and angles measurement are location dependent? $\endgroup$ – KevMoriarty Mar 24 '16 at 19:17
  • $\begingroup$ The overall idea of the picture I illustrated is to combine the fact that metric properties of physical bodies are reference frame independent but velocities, accelerations depend on the reference frame. Fibers are the spaces (affine spaces with a scalar product nothing further is necessary)where everything happens and carry the absolute metric properties. Relative kinematics is encapsulated in the way differnt fibers are related. This relation is the intrinsic nature of the notion of reference frame. $\endgroup$ – Valter Moretti Mar 24 '16 at 19:51
2
$\begingroup$

It seems to me that only point O is needed, we actually don't care about the basis.

We actually do care about the basis, very much so. In Newtonian mechanics, the displacement vector between one point and another in $\mathbb R^3$ is indeed frame independent. This displacement vector might well have different representations in different bases, but all such displacement vectors are substantively the same vector.

The same does not apply to the time derivatives of these vectors. The time derivative of a displacement vector is a frame-dependent quantity, depending on the linear and angular velocities of the observers. This also applies to the second time derivative of a displacement vector. Since Newton's second law is a statement about second time derivatives, that displacement vectors are substantively the same to all observers is irrelevant.

$\endgroup$
  • $\begingroup$ I don't quite understand why since a time derivative is a limit of the difference between two displacement vectors taken at different instants divided by the duration. By axioms of the vector space, the difference belongs to the same vector space and also the time derivative. $\endgroup$ – KevMoriarty Mar 22 '16 at 17:50
  • $\begingroup$ @KevMoriarty - Consider the tip of a plastic horse's nose on a merry-go-round, and two observers of the horse's nose, both situated at the center of the merry-go-round, one fixed with respect to the rotating merry-go-round, the other fixed with respect to the earth. While the two observers see the same displacement vector to the tip of the horse's nose, they see very different time derivatives. The rotating observer sees the horse's nose as stationary. The non-rotating observer sees the horse's nose as undergoing uniform circular motion. $\endgroup$ – David Hammen Mar 22 '16 at 20:14
  • $\begingroup$ Your description make me realise that the location of the observer is not sufficient in order to give the frame of reference. Some researchers say that a frame of reference is a rigid body, it is doted of a motion, like the rotation of the merry-go-round. I like such definition, it keeps only physics, mathematics come later, and your are free to choose a frame to your affine point space, this frame rotates with the rigid body... $\endgroup$ – KevMoriarty Mar 24 '16 at 19:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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