How can l prove that Newton's laws are time invariant?

Click For Summary
The discussion centers on proving the time invariance of Newton's laws using mathematical expressions. A solution x(t) is presented, leading to the consideration of y(-t) and its relationship to x(-t). The key question raised is how the second derivative of x(-t) relates to the function f(x(-t)). Participants suggest using LaTeX for clarity in mathematical notation. The conversation emphasizes the need for precise mathematical formulation to demonstrate the invariance effectively.
stefano77
Messages
20
Reaction score
8
Misplaced Homework Thread -- Moved to the Schoolwork forums by the Mentors
how can l prove Newton's law is time invariant?

if x (t) is a solution of dd/ddx x(t) = f(x(t)) then if l put y(-t) dd/ddt y(t)=dd/ddt x(-t). Now how dd/ddt x(-t) is equal to f(x(-t))?dd/ddt is second derivative with respect to time
 
Physics news on Phys.org
You've previously used LaTeX on this site. Can I suggest you repost the above using LaTeX? If you have forgotten the syntax there is a guide linked below the reply box. If the LaTeX does not render when you try to preview it, refresh the page while in preview and it should work (you may wish to copy your text to clipboard first as a safety measure).
 
  • Like
Likes topsquark and vanhees71
stefano77 said:
how can l prove Newton's law is time invariant?

if x (t) is a solution of dd/ddx x(t) = f(x(t)) then if l put y(-t) such that dd/ddt y(t)=dd/ddt x(-t). Now how dd/ddt x(-t) is equal to f(x(-t))?dd/ddt is second derivative with respect to time
##\dfrac{d^2}{dt^2} y(-t) = \dfrac{d^2}{dt^2} x(t)##

or equivalently
##\dfrac{d^2}{d(-t)^2} y(t) = \dfrac{d^2}{d(-t)^2} x(-t)##

Can you finish?

-Dan

Addendum: Please note my addition of "such that" to your original post. You needed something to separate those two expressions.
 
Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...