Physics Asked by user92155 on December 18, 2020
If there are $n$ independent coordinates for a system of particles, but all of them are functions of time, is it correct to say that the degree of freedom is 1? If we know the instant of time we’re looking at the system, we would be able to describe it completely, wouldn’t we? But this doesn’t seem right to me. Can someone explain this?
The configuration of a deterministic system is a function of time and the initial configuration and velocities of the components of the system. Degrees of freedom refers to the number of independent parameters required to define the initial configuration of the system, which is the dimension of its configuration space. The term is also sometimes used to refer to the dimension of a system’s phase space, which can cause confusion.
A point particle that can initially be anywhere in a volume of three dimensional space has a configuration space with three degrees of freedom. It it is confined to move on a two-dimensional surface then it has two degrees of freedom. If it is confined to move along a fixed rail or track then it has only one degree of freedom.
Answered by gandalf61 on December 18, 2020
Time $t$ (or proper time $tau$) is one and only true independent quantity of any physical system. By definition, we describe any process and how other quantities vary with time. How many other quantities we include in a process is called the degrees of freedom of the system.
Some people may argue the position and momentum for example are independent of each other and the degrees of freedom should be 2× the position quantities, or $mathrm{dof}=2n$. But you can argue that time is part of the configuration space (the state) since it needs to be included to understand all the properties of a particle and thus with the same logic $mathrm{dof}= 2n+1$. So just counting the rows in the state vector $boldsymbol{Y}$ in a simulation is a stretch for the term degrees of freedom.
Below is an example of three competing system states for a system with $n$ particles
$$begin{array}{r|c|c|c} text{system} & text{dof}=n & text{dof}=2n & text{dof}=2n+1 hline text{state} & boldsymbol{Y}=begin{pmatrix}x_{1} vdots x_{n} end{pmatrix} & boldsymbol{Y}=begin{pmatrix}x_{1} vdots x_{n} dot{x}_{1} vdots dot{x}_{n} end{pmatrix} & boldsymbol{Y}=begin{pmatrix}t x_{1} vdots x_{n} dot{x}_{1} vdots dot{x}_{n} end{pmatrix} hline text{dynamics} & begin{aligned}dot{t} & =1 boldsymbol{dot{Y}} & =d(t,boldsymbol{Y}) boldsymbol{ddot{Y}} & =f(t,boldsymbol{Y},boldsymbol{dot{Y}}) end{aligned} & begin{aligned}dot{t} & =1 boldsymbol{dot{Y}} & =f(t,boldsymbol{Y}) end{aligned} & boldsymbol{dot{Y}}=f(boldsymbol{Y}) hline text{conditions} & g(t,boldsymbol{Y},boldsymbol{dot{Y}})=0 & g(t,boldsymbol{Y})=0 & g(boldsymbol{Y})=0 end{array}$$
In the world of mechanical systems where I come from, I prefer the first option with $text{dof}=n$ since you can argue that $dot{x} = d(t,x)$ is part of the dynamics of the system just as $ddot{x} = f(t,x,dot{x})$. Or if $x(t)$ is known, then so is $dot{x}(t)$ and all the higher derivatives. This is definitely a deterministic viewpoint which is relevant for my physics, and not a reflection of reality.
Answered by John Alexiou on December 18, 2020
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