# Scalar and Vector Field Functionality#

## Introduction#

### Vectors and Scalars#

In physics, we deal with two kinds of quantities – scalars and vectors.

A scalar is an entity which only has a magnitude – no direction. Examples of scalar quantities include mass, electric charge, temperature, distance, etc.

A vector, on the other hand, is an entity that is characterized by a magnitude and a direction. Examples of vector quantities are displacement, velocity, magnetic field, etc.

A scalar can be depicted just by a number, for e.g. a temperature of 300 K. On the other hand, vectorial quantities like acceleration are usually denoted by a vector. Given a vector $$\mathbf{V}$$, the magnitude of the corresponding quantity can be calculated as the magnitude of the vector itself $$\Vert \mathbf{V} \Vert$$, while the direction would be specified by a unit vector in the direction of the original vector, $$\mathbf{\hat{V}} = \frac{\mathbf{V}}{\Vert \mathbf{V} \Vert}$$.

For example, consider a displacement of $$(3\mathbf{\hat{i}} + 4\mathbf{\hat{j}} + 5\mathbf{\hat{k}})$$ m, where , as per standard convention, $$\mathbf{\hat{i}}$$, $$\mathbf{\hat{j}}$$ and $$\mathbf{\hat{k}}$$ represent unit vectors in the $$\mathbf{X}$$, $$\mathbf{Y}$$ and $$\mathbf{Z}$$ directions respectively. Therefore, it can be concluded that the distance traveled is $$\Vert 3\mathbf{\hat{i}} + 4\mathbf{\hat{j}} + 5\mathbf{\hat{k}} \Vert$$ m = $$5\sqrt{2}$$ m. The direction of travel is given by the unit vector $$\frac{3}{5\sqrt{2}}\mathbf{\hat{i}} + \frac{4}{5\sqrt{2}}\mathbf{\hat{j}} + \frac{5}{5\sqrt{2}}\mathbf{\hat{k}}$$.

### Fields#

In general, a $$field$$ is a vector or scalar quantity that can be specified everywhere in space as a function of position (Note that in general a field may also be dependent on time and other custom variables). In this module, we deal with 3-dimensional spaces only. Hence, a field is defined as a function of the $$x$$, $$y$$ and $$z$$ coordinates corresponding to a location in 3D space.

For example, temperate in 3 dimensional space (a temperature field) can be written as $$T(x, y, z)$$ – a scalar function of the position. An example of a scalar field in electromagnetism is the electric potential.

In a similar manner, a vector field can be defined as a vectorial function of the location $$(x, y, z)$$ of any point in space.

For instance, every point on the earth may be considered to be in the gravitational force field of the earth. We may specify the field by the magnitude and the direction of acceleration due to gravity (i.e. force per unit mass ) $$g(x, y, z)$$ at every point in space.

To give an example from electromagnetism, consider an electric potential of form $$2{x}^{2}y$$, a scalar field in 3D space. The corresponding conservative electric field can be computed as the gradient of the electric potential function, and expressed as $$4xy\mathbf{\hat{i}} + 2{x}^{2}\mathbf{\hat{j}}$$. The magnitude of this electric field can in turn be expressed as a scalar field of the form $$\sqrt{4{x}^{4} + 16{x}^{2}{y}^{2}}$$.

## Implementation of fields in sympy.physics.vector#

In sympy.physics.vector, every ReferenceFrame instance is assigned basis vectors corresponding to the $$X$$, $$Y$$ and $$Z$$ directions. These can be accessed using the attributes named x, y and z respectively. Hence, to define a vector $$\mathbf{v}$$ of the form $$3\mathbf{\hat{i}} + 4\mathbf{\hat{j}} + 5\mathbf{\hat{k}}$$ with respect to a given frame $$\mathbf{R}$$, you would do

>>> from sympy.physics.vector import ReferenceFrame
>>> R = ReferenceFrame('R')
>>> v = 3*R.x + 4*R.y + 5*R.z


Vector math and basic calculus operations with respect to vectors have already been elaborated upon in other sections of this module’s documentation.

On the other hand, base scalars (or coordinate variables) are implemented as special SymPy Symbols assigned to every frame, one for each direction from $$X$$, $$Y$$ and $$Z$$. For a frame R, the $$X$$, $$Y$$ and $$Z$$ base scalar Symbols can be accessed using the R[0], R[1] and R[2] expressions respectively.

Therefore, to generate the expression for the aforementioned electric potential field $$2{x}^{2}y$$, you would have to do

>>> from sympy.physics.vector import ReferenceFrame
>>> R = ReferenceFrame('R')
>>> electric_potential = 2*R[0]**2*R[1]
>>> electric_potential
2*R_x**2*R_y


In string representation, R_x denotes the $$X$$ base scalar assigned to ReferenceFrame R. Essentially, R_x is the string representation of R[0].

Scalar fields can be treated just as any other SymPy expression, for any math/calculus functionality. Hence, to differentiate the above electric potential with respect to $$x$$ (i.e. R[0]), you would have to use the diff function.

>>> from sympy.physics.vector import ReferenceFrame
>>> R = ReferenceFrame('R')
>>> electric_potential = 2*R[0]**2*R[1]
>>> from sympy import diff
>>> diff(electric_potential, R[0])
4*R_x*R_y


Like vectors (and vector fields), scalar fields can also be re-expressed in other frames of reference, apart from the one they were defined in – assuming that an orientation relationship exists between the concerned frames. This can be done using the sympy.physics.vector.vector.Vector.express method, in a way similar to vectors - but with the variables parameter set to True.

>>> from sympy.physics.vector import ReferenceFrame
>>> R = ReferenceFrame('R')
>>> electric_potential = 2*R[0]**2*R[1]
>>> from sympy.physics.vector import dynamicsymbols, express
>>> q = dynamicsymbols('q')
>>> R1 = R.orientnew('R1', rot_type = 'Axis', amounts = [q, R.z])
>>> express(electric_potential, R1, variables=True)
2*(R1_x*sin(q(t)) + R1_y*cos(q(t)))*(R1_x*cos(q(t)) - R1_y*sin(q(t)))**2


Moreover, considering scalars can also be functions of time just as vectors, differentiation with respect to time is also possible. Depending on the Symbols present in the expression and the frame with respect to which the time differentiation is being done, the output will change/remain the same.

>>> from sympy.physics.vector import ReferenceFrame
>>> R = ReferenceFrame('R')
>>> electric_potential = 2*R[0]**2*R[1]
>>> q = dynamicsymbols('q')
>>> R1 = R.orientnew('R1', rot_type = 'Axis', amounts = [q, R.z])
>>> from sympy.physics.vector import time_derivative
>>> time_derivative(electric_potential, R)
0
>>> time_derivative(electric_potential, R1).simplify()
2*(R1_x*cos(q(t)) - R1_y*sin(q(t)))*(3*R1_x**2*cos(2*q(t))/2 -
R1_x**2/2 - 3*R1_x*R1_y*sin(2*q(t)) - 3*R1_y**2*cos(2*q(t))/2 -
R1_y**2/2)*Derivative(q(t), t)