Understanding Kinetic Energy in Special Relativity

[finchie_88]

In normal mechanics, the kinetic energy is the integral of the momentum with respect to the velocity ( [tex] \int m_0v.dv [/tex]. So, why is the kinetic energy not given by [tex] \int \frac{m_0v}{\sqrt{1 - v^2/c^2}}dv[/tex]?

Sorry it took a few edits to get the maths looking right.

 

[Hurkyl]

Because the actual definition of work is:

[tex]
W = \int \mathbf{F} \cdot d\mathbf{x}
[/tex]

and your expression for work in the Newtonian case is derived from the particular form that F takes in that theory.

 

[jtbell]

Working in one dimension for simplicity, non-relativistically we have

[tex]\Delta K = W = \int {Fdx} = \int {\frac{dp}{dt}dx} = m \int {\frac{dv}{dt}dx}[/tex]

whereas relativistically we have

[tex]\Delta K = W = \int {Fdx} = \int {\frac{dp}{dt}dx} = m_0 \int {\frac{d(\gamma v)}{dt}dx}[/tex]

 

[pervect]

Momentum is always the derivative of the Lagrangian, L, with respect to velocity. Thus the intergal of momentum with respect to velocity is always the Lagrangian.

Thus your intergal gives not the kinetic energy, but the relativistic lagrangian. In Newtonian theory the Lagrangian is equivalent to the kinetic energy when the potential energy is zero, as it is in this example (L=T-V and V=0), but not in relativisitc mechanics.

 

[pmb_phy]

In normal mechanics, the kinetic energy is the integral of the momentum with respect to the velocity ( [tex] \int m_0v.dv [/tex]. So, why is the kinetic energy not given by [tex] \int \frac{m_0v}{\sqrt{1 - v^2/c^2}}dv[/tex]?

Sorry it took a few edits to get the maths looking right.

The correct definition of kinetic energy in classical mechanics is as follows

Kinetic Energy: Defined by the requirement that the change in kinetic energy equals the work done on the body by external forces acting on the body (i.e. work-energy theorem).

(note: Work done equals change in kinetic energy)

In SR the same definition is used. I.e. dK = Fdx which yields [itex]K =(\gamma - 1)m_0c^2[/itex].

Pete

 

[pmb_phy]

Momentum is always the derivative of the Lagrangian, L, with respect to velocity.

That is the canonical momentum, not the (linear mechanical) momentum. In any case the canonical momentum is the positive derivative of the lagrangian.

Thus the intergal of momentum with respect to velocity ..

... is something not mentioned here.

In Newtonian theory the Lagrangian is equivalent to the kinetic energy (L=T), but not in relativisitc mechanics.

Huh? Since when? In classical mechanics the Lagrangian, L, is given by L = T - V, not as you gave.

Pete

 

[pervect]

That is the canonical momentum, not the (linear mechanical) momentum. In any case the canonical momentum is the positive derivative of the lagrangian

In this particular problem, the canonical momentum is equal to the mechanical momentum. If the system had fields, one might have to make the distinction. Since the canonical momentum is the only sort that's conserved, one would generally be interested in the canonical momentum of a system even in the case where the system has fields (for instance a moving charge).

... is something not mentioned here.

You may have missed the point. If the derivative of the Lagrangian is the momentum, then the intergal of the momentum is the Lagrangian.

Huh? Since when? In classical mechanics the Lagrangian, L, is given by L = T - V, not as you gave.

I'll revise and expand my original post to make my meaning more clear.

 

[Hans de Vries]

Momentum is always the derivative of the Lagrangian, L, with respect to velocity. Thus the intergal of momentum with respect to velocity is always the Lagrangian.

To give the Lagrangian explicitly:

[tex]-L\ = \ \sqrt{1-v^2/c^2}\ m_o c^2 \quad \approx \quad m_o c^2\ -\ \frac{1}{2}m_o v^2 \qquad . [/tex] (for v << c)

Which corresponds to the time dillation of an object moving at v.
In QM words: the number of phase changes (ticks) over the trajectory
of the particle (the t' axis) is less by a factor gamma.

While for the Hamiltonian we have:

[tex]H\ =\ \frac{1}{\sqrt{1-v^2/c^2}}\ m_o c^2 \quad \approx \quad m_o c^2\ +\ \frac{1}{2}m_o v^2 \qquad .[/tex] (for v << c)

In QM words: the number of phase changes (ticks) over the t axis is
higher by a factor gamma.

That's why -L = V-T while H = V+T in the non-relativistic limit.Regards, Hans

 

[pmb_phy]

In this particular problem, the canonical momentum is equal to the mechanical momentum. If the system had fields, one might have to make the distinction...

I neglected to comment on he questoners statement. I must have misread it since I didn't see it the first time. I don't understand why he thinks that kinetic energy is defined as the integral of the momentum with respect to the velocity. I guess some people confuse definitions with equalities.

Pete