Derivative of kinectic energy , dK/dv

  • Thread starter Thread starter c77793
  • Start date Start date
  • Tags Tags
    Derivative Energy
AI Thread Summary
The discussion centers on the relationship between kinetic energy and linear momentum, specifically how the derivative of kinetic energy with respect to velocity results in momentum. Kinetic energy is defined as Ek = 1/2mv², and its derivative with respect to velocity yields mv, which is the expression for linear momentum. Participants highlight the connection between these concepts through Newtonian physics, emphasizing that both are linked to force, as force is the derivative of momentum. The conversation also touches on the implications of these relationships in different inertial frames. Understanding these principles is crucial for grasping fundamental physics concepts.
c77793
Messages
10
Reaction score
0
i'm in my freshman year and I'm starting to learn Derivatives in Calculus, and I was wondering, once Ek (kinectic energy) = 1/2mv², then, the derivative of Ek in term of velocity would be mv, which is equal to the linear momentum... I'm finding hard to understand the idea that the variation of Kinect over a variable velocity is the linear momentum...can someone explain me this in a didactic way? I was just playing around deriving physics formulas haha
 
Physics news on Phys.org
hi c77793! :smile:
c77793 said:
I'm finding hard to understand the idea that the variation of Kinect over a variable velocity is the linear momentum...can someone explain me this in a didactic way?

it's because of Newtonian Relativity

the (Newtonian) laws of physics are the same in any inertial frame​

suppose you have bodies with ∑miui2 = ∑mivi2 in one frame

now choose another frame with relative velocity w in the k-direction …

∑mi(ui - wk).(ui - wk) = ∑mi(vi - wk).(vi - wk)
cancelling (and dividing by 2) gives us …

w∑miui.k = w∑mivi.k

since w is arbitrary, we can divide by w, and get conservation of momentum in in the k direction

however, instead of cancelling, we could have differentiated wrt w (in other words, exploiting the translational symmetry of Newtonian space), giving …

w∑mi(ui - wk).k = w∑mi(vi - wk).k

(and then cancelled, giving the same result)

(btw, this also works with Einsteinian energy and momentum, in Minkowski space)
 
c77793 said:
i'm in my freshman year and I'm starting to learn Derivatives in Calculus, and I was wondering, once Ek (kinectic energy) = 1/2mv², then, the derivative of Ek in term of velocity would be mv, which is equal to the linear momentum... I'm finding hard to understand the idea that the variation of Kinect over a variable velocity is the linear momentum...can someone explain me this in a didactic way? I was just playing around deriving physics formulas haha

Well I was also wondering about that and I can't still find the physical relationship between the two, however both are related to force (since force is the derivative of momentum and work is (force)(distance)).
The complete derivation you can find in this thread https://www.physicsforums.com/showthread.php?t=68682
 
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...
Thread 'Griffith, Electrodynamics, 4th Edition, Example 4.8. (First part)'
I am reading the Griffith, Electrodynamics book, 4th edition, Example 4.8 and stuck at some statements. It's little bit confused. > Example 4.8. Suppose the entire region below the plane ##z=0## in Fig. 4.28 is filled with uniform linear dielectric material of susceptibility ##\chi_e##. Calculate the force on a point charge ##q## situated a distance ##d## above the origin. Solution : The surface bound charge on the ##xy## plane is of opposite sign to ##q##, so the force will be...
Dear all, in an encounter of an infamous claim by Gerlich and Tscheuschner that the Greenhouse effect is inconsistent with the 2nd law of thermodynamics I came to a simple thought experiment which I wanted to share with you to check my understanding and brush up my knowledge. The thought experiment I tried to calculate through is as follows. I have a sphere (1) with radius ##r##, acting like a black body at a temperature of exactly ##T_1 = 500 K##. With Stefan-Boltzmann you can calculate...
Back
Top