Relationship between momentum and kinetic energy in the real world?

  • Thread starter nhmllr
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  • #1
nhmllr
185
1
I was thinking about the 1/2 in KE = 1/2*m*v2
I thought about how strange it was, and then I realized that as a function on velocity (v), kinetic energy is the antiderivative function of momentum.

However, I am very confused as to what that implies. My question is very vague, but what exactly is the relationship between momentum and kinetic energy in the real world? Is 1/6*m*v3 meaningful?
 

Answers and Replies

  • #2
LostConjugate
848
3
The energy is the dot product of the velocity and an infinitesimal change in momentum which happens to be

[tex]d(\frac{1}{2}mv^2)[/tex]

The integral just drops the d.
 
  • #3
nhmllr
185
1
The energy is the dot product of the velocity and an infinitesimal change in momentum which happens to be

[tex]d(\frac{1}{2}mv^2)[/tex]

The integral just drops the d.

The dot product is just the magnitudes of the two vectors multiplied, right?
I still don't understand what the relationship between momentum and kinetic energy really means.

I understand the whole derivative thing with the relationship between distance, velocity, and acceleration. How is this similar?
 
  • #4
RedX
970
3
Just to add another step:

[tex]d(\frac{1}{2}mv^2)=vdp=v(mdv)=d(\frac{1}{2}mv^2) [/tex]

Basically this equation:

[tex]\frac{dE}{dp}=\frac{dT}{dp}=v [/tex]

where T is the kinetic energy.

So the rate at which the energy varies with momentum is the velocity of the particle.

(As an example consider special relativity:

E^2=p^2+m^2
2EdE=2pdp
dE/dp=p/E=v )

So I guess you would take d of both sides

[tex]dv=d\frac{dE}{dp}=d\frac{dT}{dp} [/tex]

Multiply both sides by p and integrate (the RHS by parts):

[tex]\int pdv= p\frac{dT}{dp}-T [/tex]

So I think basically if the kinetic energy is a quadratic function in momentum, then the RHS will be equal to T. To prove this just solve the differential equation:


[tex]p\frac{dT}{dp}-T=T [/tex]

[tex]dT/T=2dp/p [/tex]

[tex]logT=2*Log[p] [/tex]

[tex]T=p^2 [/tex]

I don't know where the units went (I usually screw up units alot), but if the kinetic energy is quadratic in momentum, then the integral of momentum with respect to velocity will equal the kinetic energy!

Hopefully I didn't screw up somewhere.

Actually, now that I think about it, I think I did screw up somewhere. As an exercise try to find out where!

If you define momentum as p=dT/dv, then it automatically follows that [tex]\int pdv=\int dT=T [/tex]

Anyways, if F*distance=work, and F=dp/dt, and work equals change in energy, then:

F*distance=dp/dt*(vdt)=vdp=dT

That's how you get the expression dT=vdp. This should be completely general.
 
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  • #5
Dyson
21
0
My views are as follows:
I first regard the kinetic energy is a kind of energy which can show the magnitude of work resulted from the total external force. This definition, to a certain extend, also fits to our intuitions. Then, we may write the formula which may appear frequently in most textbooks:
[tex]\int F^{\rightarrow}\cdot d r^{\rightarrow}=\Delta E_k[/tex]
The left side equals:
[tex]\int F^{\rightarrow}\cdot v^{\rightarrow}dt=\int \frac{d p^{\rightarrow}}{dt}\cdot v^{\rightarrow}dt[/tex]
Then:
[tex]\int v^{\rightarrow}\cdot d p^{\rightarrow}=\Delta E_k[/tex]
So,for a simple 1-D condition:
[tex]\frac{dE_k}{dp}=v[/tex]
This may be the best relation between kinetic energy and momentum.
If the mass can be thought of a constant during motion, this corresponds a low velocity.
If the mass changes with the magnitude of the velocity, in fact, a real condition in our world, this can correspond the formula in special relativity.
 
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  • #6
Dyson
21
0
OH, sorry, my reply actually repeated the RedX's....
To my surprise!
 
  • #7
RedX
970
3
OH, sorry, my reply actually repeated the RedX's....
To my surprise!

That's my fault because I edited my post. I like your explanation better anyhow.

(oh and if you want to put vectors on top of symbols, just use \vec{symbol} )

Anyways, for those of you who know the Lagrangian and Hamiltonian formulation of classical mechanics, I can't figure out where I went wrong. It seems in the Lagrangian approach, the kinetic energy can have any form, and the integral of momentum with respect to velocity should give you kinetic energy.

For the Hamiltonian approach however, the integral of momentum with respect to velocity seems to only give you kinetic energy if the kinetic energy is quadratic in momentum or velocity.
 
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