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The definition is simply p = mv p = m v. In classical mechanics, this quantity is conserved in absence of an external net force, since the second principle of dynamics reads. dp dt =Fext d p d t = F e x t, and thus p(t) = const. p (t) = const. if Fext =0 F e x t = 0. In order to understand the meaning of the mass in the momentum, you need to ...
1. The momentum operator, like other operators in quantum mechanics, acts on a given wave function (state). Multiplication of operators in linear algebra is the same as them "acting on" a mathematical object. Originally quantum mechanics was called 'matrix mechanics'; so when you study linear algebra you're really studying quantum mechanics too ...
The canonical momentum p is just a conjugate variable of position in classical mechanics, in which we have the relation p = ∂L ∂˙r. When making the transition to quantum mechanics, we substitute p with an operator − iℏ∇ in the Hamiltonian; similarly, we substitute r by iℏ∇p in momentum representation. The kinetic momentum is ...
To supplement the previous answers, consider the Lagrangian for a particle in a 1D potential V(q) with a speed v = ˙q and mass m: L = 1 2m˙q2 − V(q). Then the generalized momentum is: p = ∂L / ∂˙q = m˙q. This matches the expression for momentum of a classical particle mv. Lagrangian mechanics, though, is much more general as pointed ...
The change in an objects momentum is equal to the force applied and the time interval over which the force is applied. Or, Δp = FΔt Δ p = F Δ t. Starting from rest, this will give you the total momentum of the object, p p. Considering inertia, objects with a higher mass will exhibit more resistance to velocity change.
$\begingroup$ This is an oversimplified definition. Momentum is the integral of mass * acceleration with respect to time, and of course F=ma, so momentum is the integral of force WRT time. $\endgroup$ –
Furthermore momentum is conserved $\Sigma m_{i}\vec{v_{i}}=const.$ Then you have the definition that force is the change of momentum with respect to time $\vec{F}=\dfrac{d(m\vec{v})}{dt}$. I've read the chapters concerning mechanics of Physics for scientists and engineers by Giancoli and the Feynman Lectures.
In introductory classical mechanics, we come across the quantity momentum defined as p = mv p = m v. There seems to be no other "definition" of momentum, (apart from the vague "measure of quantity of motion") and the only role it seems to play is in defining the force F F in Newton's second law F = dp dt F = d p d t.
If you're referring to the quantum mechanical momentum, then you are probably referring to the operator. To give you more context, what you wrote as $-i\nabla$ is the definition of the canonical or conjugate momentum of the particle, which is a mathematical object constructed by quantization of the classical phase space. In the case of a free ...
the etymology of the word "momentum" says it's been in use since the 1690's. That's when Newton was coming up with his laws. That means Newton had a quantity that was mass times velocity and wanted a name for it. He chose momentum. So you shouldn't ask why we say momentum is mv m v, you should ask why mv m v is momentum.