Vector Definition: Magnitude & Direction Effects

  • Thread starter Thread starter hokhani
  • Start date Start date
  • Tags Tags
    Definition Vector
AI Thread Summary
Vectors cannot be solely defined as quantities with magnitude and direction because their mathematical definition is independent of coordinate systems. They are elements of a vector space, emphasizing their nature as geometrical objects rather than mere numerical arrays. The behavior of vector components under coordinate transformations, such as rotations, highlights their intrinsic properties. This perspective reinforces that vectors exist beyond the choice of any specific coordinate system. Understanding vectors in this way allows for a more robust and universal application in mathematics and physics.
hokhani
Messages
561
Reaction score
18
Why we can not define a vector as a quantity which has magnitude and direction? Why we define the vectors according to behavior of its components in rotated coordinate-frames?
 
Physics news on Phys.org
Vectors are, by definition, the elements of a vector space. In my opinion, other definitions are corollaries in disguise.
 
Vectors and tensors in general are mathematically defined independent of coordinate systems. This emphasizes the fact that they are geometrical objects which don't care about your preference for a certain coordinate system. If you regard a vector via its components as an array {a,b,...} which you can assign a length to, it looks like vectors cannot exist without the choice of coordinates.

From the coordinate-free definition it follows that the components (!) have a certain behaviour under coordinate transformations, like the rotations you mentioned.
 
  • Like
Likes hokhani
Thread 'Gauss' law seems to imply instantaneous electric field propagation'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Maxwell’s equations imply the following wave equation for the electric field $$\nabla^2\mathbf{E}-\frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2} = \frac{1}{\varepsilon_0}\nabla\rho+\mu_0\frac{\partial\mathbf J}{\partial t}.\tag{1}$$ I wonder if eqn.##(1)## can be split into the following transverse part $$\nabla^2\mathbf{E}_T-\frac{1}{c^2}\frac{\partial^2\mathbf{E}_T}{\partial t^2} = \mu_0\frac{\partial\mathbf{J}_T}{\partial t}\tag{2}$$ and longitudinal part...
Thread 'Recovering Hamilton's Equations from Poisson brackets'
The issue : Let me start by copying and pasting the relevant passage from the text, thanks to modern day methods of computing. The trouble is, in equation (4.79), it completely ignores the partial derivative of ##q_i## with respect to time, i.e. it puts ##\partial q_i/\partial t=0##. But ##q_i## is a dynamical variable of ##t##, or ##q_i(t)##. In the derivation of Hamilton's equations from the Hamiltonian, viz. ##H = p_i \dot q_i-L##, nowhere did we assume that ##\partial q_i/\partial...
Back
Top