Newton’s Notion of Space:
According to sir Newton, “Absolute space in its own nature without relation to anything external remains always similar and immovable.” This indicates that space is independent of the existence of anybody, i.e. space describes something, which distinct from anybody.
Properties of the space:
In order to describe the motion of a body or a particle, the knowledge of the relationship of three independent quantities, such as, mass, length and time are required. The study of the motion of a body also involves the relationship of space and time. The apparent from of these relationship depends on the co-ordinate system
According to sir Newton the main properties of space are:
Space is three-dimensional:
Space and time is the fundamental concept to study of the mechanics. When an object is placed, it occupies some space because the object has a length, a breadth and a height. So we can say that the space is three dimensional. As a result the position of a point can be described completely by the three co-ordinates. In general three co-ordinate systems are used : (i) Rectangular or Cartesian co-ordinate, (ii) Spherical polar co-ordinate and (iii) Cylindrical co-ordinate.
The positon of point in rectangular or Cartesian co-ordinate system can be described by three co-ordinates \( (x,y,z) \)
The positon of point in spherical polar co-ordinate system can be described by three co-ordinates \( (r,\ \theta,\ \phi) \).
The positon of point in cylindrical co-ordinate system can be described by three co-ordinates \( (\rho,\ \phi,\ z) \).
Space is flat:
The shortest distance of any two points in space is a straight line. This means that the space is flat which indicates that it possesses Eucledian Flatness. In order to form a triangle, when we take three points in space, the sum of the angles is equal to \( \pi \). The three sides of the triangle are related by the Pythagoras Theorem, when the triangle is a right angle triangle. According to this theorem, \( {Hypotenuse}^2={Base}^2+{Altitude}^2 \).
But according to the latest theory, space is not exactly flat, it is somewhat curved. Since the departure of the flatness is very small, so in the study of classical mechanics it can be ignored.
Space is homogeneous:
Space is homogeneous, i.e. space is everywhere alike. Translational Invariance of the properties of the free space is due to the homogeneity of free space when there is no fields and forces. In simple word, it implies that the result of an experiment remains unchanged due to the linear displacement of the co-ordinate system.
Let us consider two inertial frame of reference S and S’ with their X-axis along the same line and the Y and Z axes parallel to each other. Here S frame is represented by XYZ co-ordinate system and the S’ frame is represented by the X’Y’Z’ co-ordinate system. The origin O of the S frame is displaced by the distance d from the origin O’ of the S frame along the X-axis.
Let us consider two particles A and B lying on the X-axis. The co-ordinates of the point A and B with respect to the origin O of the S frame are \( x_1 \) and \( x_2 \) respectively. And the co-ordinates of the point A and B with respect to the origin O’ of the frame S’ are \( x_{1}^{’} \) and \( x_{2}^{’} \) respectively.
Let’s consider the particles at the points A and B exert gravitational force on each other. So the force between them is only due to the mutual interaction, i.e. the force is conservative.
Now the conservative force \( F_x \) acting on the particle can be expressed as,
\( \displaystyle{F_x=-\frac{\delta{U}}{\delta{x}}} \)
where, U is the potential energy of the particle and x is the co-ordinate of the particle on the x-axis in equation.
Now here the potential energy U should obey the following conditions:
- The form of the function of potential energy U should be taken in such that the force derivative of this function will be independent of the frame of reference. Here also the frame of references S and S’ are taken in such a way that the Y and Z axes are parallel to each other and the origins O and O’ of the frames S and S’ respectively are displaced by a distance d along the x-axis. When the force is independent of the displacement of the frame of reference then it indicates the linear uniformity of the space. Hence \( \displaystyle{F_x=-\frac{\delta{U}}{\delta{x}}=-\frac{\delta{U’}}{\delta{x’}}} \), where U and U’ are the potential energies in the two frame of references S and S’.
- The potential energy due to the interaction between the two particles has to depend on their co-ordinates. The potential energy U is a function of \( x_1 \) and \( x_2 \) in the frame of reference S. So here \( U=U(x_1,\ x_2) \) should be such that the equation \( \displaystyle{F_x=-\frac{\delta{U}}{\delta{x}}} \) holds good. Similarly, the potential energy U’ is a function of \( x_{1}^{‘} \) and \( x_{2}^{‘} \) in the frame of reference S’. So here \( U’=U’(x’_1,\ x’_2) \) should be such that the equation \( \displaystyle{F_x=-\frac{\delta{U’}}{\delta{x’}}} \) holds good.
Now, Since the potential energy U in the frame of reference S is a scalar quantity, therefore the form of \( U(x_1,\ x_2)=(x_1^2-x_2^2) \), because even if \( (x_1-x_2) \) is a vector but \( {(x_1-x_2)}^2 \) is a scalar quantity.
Similarly in case of the frame of reference S’, The potential energy U’ will be, \( U’(x’_1,\ x’_2)={(x’_1-x’_2)}^2 \).
\( U’(x’_1,\ x’_2)={(x’_1-x’_2)}^2 \)
\( or,\ U’(x’_1,\ x’_2)={\left((x_1+d)-(x_2+d)\right)}^2 \)
\( or,\ U’(x’_1,\ x’_2)= {(x_1-x_2)}^2 \)
\( or,\ U’(x’_1,\ x’_2)=U(x’_1,\ x’_2) \)
This shows that the potential energy function U is independent of the frame of reference or the co-ordinate system.
Space is isotropic:
Let us consider a point P, now if we move this point in any direction, then the properties of that point remains unchanged. In other word the properties are the same and there is nothing, which can distinguish one direction from the other. Therefore in space, there is no preferred direction, the isotropy of space indicates the rotational invariance of the free space.