The non-relativistic behavior of particles

AI Thread Summary
The discussion focuses on the non-relativistic behavior of particles as described by the Schrödinger equation, which allows for the definitions of momentum as mv and energy as 1/2 mv². Participants clarify that these definitions are valid when particle velocities are much smaller than the speed of light (c). In contrast, relativistic behavior uses more complex formulas for momentum and energy that account for higher velocities. The conversation also touches on the relationship between kinetic energy and mass in the non-relativistic limit. Overall, the thread emphasizes the distinctions between non-relativistic and relativistic physics in particle behavior.
FRANCLI
Messages
12
Reaction score
0
hello
When I was studing about the schrodinger equation,I read that it describes the non-relativistic behavior of particles .
But I don't know what is the non-relativistic behavior of particles . S o I need somebody to explain this behavior .
 
Physics news on Phys.org
Hi FRANCLI! :smile:

It means that we can define momentum = mv and energy = 1/2 mv2.

(relativistic behavior of particles is more accurate, and defines momentum = mv/√(1 - v2/c2) and energy = m/√(1 - v2/c2), which are approximately the same as the non-relativistic definitions when v is very much smaller than c :wink:)
 
Last edited:
thank you ,
but if the energy is = 1/√(1 - v2/c2), and v is very much smaller than c , the energy will be 1 ?
 
oops!

oops! :redface: … i missed out the "m" :rolleyes:

it should have been energy = m/√(1 - v2/c2)

(or KE = m/√(1 - v2/c2) - m, which is approximately 1/2 mv2 when v is very small)

(i've edited my last post, so it's correct now :wink:)
 
OK, thanks .
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top