Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Energy Conservation versus Quantum and Relativistic Effects

  1. Mar 18, 2012 #1
    A mathematical statement of energy conservation can be given using the continuity equation in terms of the total energy

    [tex]\frac{\partial}{\partial t}\iiint_V\epsilon dV+\iint_S \epsilon \mathbf{V\cdot dS}=0[/tex]

    where t is time, V is a velocity vector, V is the volume of the system, dS is a point along the surface area, S, of the volume, and [tex]\epsilon[/tex] describes the total energy of a point in the volume which might be given as

    [tex]\epsilon=\rho (e+0.5v^2 +\Phi)+h\nu[/tex]
    where [tex]\rho[/tex] is the internal energy, v is the velocity, [tex]\Phi[/tex] is the potential energy of the body forces acting on the mass, h is plank's constant, and [tex]\nu[/tex] is light frequency.


    So, the first equation tells us that the total energy of a volume, V, can only change if energy is transported into or out of it. I derived the first equation as an application to classical systems, but I am wondering if it will become problematic when talking about quantum mechanics or relativity.

    My questions are:

    1. Can this formalism be applied to quantum mechanics? I imagine that the uncertainty principle, and phenomena like quantum fluctuations and the production of virtual particles are problematic for this formulation that is based on the continuity equation.

    Perhaps we must say, for instance, that you cannot consider a volume less than some length because of uncertainty, or something like that?

    2. How do relativistic effects, such as time dilation, factor into the dependence on time in a formulation like this? It would seem problematic, for instance, to consider a large volume in which time proceeds at different rates due to their relative velocities.
     
  2. jcsd
  3. Mar 19, 2012 #2
  4. Mar 19, 2012 #3
    quantum mechanics doesnt say that energy itself isnt precisely conserved it just says that you cant measure it to arbitrary precision.

    sure, virtual particles do spontaneously pop in and out of existence but they arent real.
     
    Last edited: Mar 19, 2012
  5. Mar 19, 2012 #4
    According to the uncertainty principle, the shorter the amount of time, the more possible it is for conservation laws to be violated, due to the high uncertainty.
     
  6. Mar 19, 2012 #5

    Jano L.

    User Avatar
    Gold Member

    Hello Hypatio,

    your equations do not make much sense. How did you arrive at them?

    Equation for energy can be reliably introduced only from equations of motion (or, from a variational principle). Your equation of continuity holds only for dust with no pressure between its constituents. Otherwise the pressure will contribute to the surface integral.

    Another problem is your expression for the energy density [itex]\epsilon[/itex]. Quantity [itex]h\nu[/itex] has units of energy, so it does not make sense to say it contributes to volume density of energy.
     
  7. Mar 19, 2012 #6
    Well I'm trying to find an equation of energy conservation which shows that the energy is conserved regardless of transport mechanisms, and then to be able to understand how the equation can be or cannot be applied to quantum mechanical or relativistic problems due to the derivatives in space and time. I think I agree that the one I first posted might not be good.

    I am primarily working from Bird, Stewart and Lightfoot (1960), Anderson (1996), the wiki article on Reynolds transport theorem, a here: https://dcwind.sslcert35.com/books/EFM-Ch6.pdf (page 8 in the pdf gives an integral form of the energy equation).

    In the linked text, the integral form looks like mine except they state that the rate of change of energy inside the volume is dE/dt, whereas I said it was zero. I suppose theirs is correct so I would write:

    [tex]\frac{\partial}{\partial t}\iiint_V\epsilon dV+\iint_S \epsilon \mathbf{V\cdot dS}=\frac{d E}{dt}[/tex]

    where E is total energy of the volume. I am hoping this is not inconsistent with an "applied" form such as (paraphrased from Bird et al., 1960):

    rate of increase of total energy = rate in by convection - rate out by convection + rate in by heat conduction - net work done by system on surroundings

    Can't the information which tells us how the energy is transmitted (from conduction, radiation, internal heating, as well as work done by pressure, viscous forces, and body forces) be contained in the description of epsilon, as long as the dimensions of my volume are held constant?

    Also, I made another mistake: \rho is a mass, not density or energy density. I think h\nu should still be correct, it only needs to be realized that you are integrating over a field of particles.
     
    Last edited: Mar 19, 2012
  8. Mar 20, 2012 #7

    Jano L.

    User Avatar
    Gold Member

    If you derive the conservation theorem from the equations of motion, you will see that the surface integral has to contain the pressure force of the surrounding liquid, so answer to your question is no; the exact form of the equation is important, the [itex]\epsilon[/itex] does not contain everything.

    Your formula for epsilon still does not seem to make much sense. Try to formulate it more carefully. Does your matter consist of point-like particles or liquid? You can find some info on this in first chapters of Lifgarbagez & Landau, Fluid Mechanics.
     
  9. Mar 20, 2012 #8
    Thanks for helping me think about this.

    One thing I am uncertain about is the use of two surface integrals (one in terms of the velocity of the boundary, and another in terms of the velocity of the fluid relative to the boundary) in the reynolds theorem in terms of energy given here: http://en.wikipedia.org/wiki/Reynolds_transport_theorem

    However, in the pdf linked earlier (https://dcwind.sslcert35.com/books/EFM-Ch6.pdf [Broken]), around page 10 they give an integral form (Eq. 6.31) which looks similar but only uses one surface integral with the velocity relative to the boundary.

    Why are they able to get rid of the surface integral describing the velocity of the boundary? Frankely, I don't see how this surface integral is needed anyway.

    Anyways, it seems that since a full energy equation can be given as

    [itex]\frac{\partial}{\partial t}\epsilon = -(\nabla\cdot v\epsilon)-(\nabla\cdot q)+\rho(v\cdot g)-(\nabla \cdot p v)-(\nabla \cdot [\tau \cdot v])[/itex]

    I can write an integral form as

    [itex]\frac{\partial}{\partial t}\int_V\epsilon dV-\int_V\rho(g\cdot v)dV+\int_S q+(\epsilon+p)(v_r\cdot \mathbf{n})+[(\tau\cdot v_r)\cdot \mathbf{n}]dS[/itex]=0

    where g is gravitational acceleration, v is velocity vector, v_r is velocity of fluid relative to the boundary, q is heat flux, p is pressure, and \tau is the viscous force.

    In this way \epsilon can remain as before but removing the potential field term and the term for light. That is: [itex]\epsilon=\rho (e+v^2/2)[/itex]

    Presumably, I could then introduce more potential field forces, such as electromagnetism using another term as with the gravity.

    This should be correct for a fluid. I'm unsure how to incorporate a term for radiation since it will behave independent of the forces acting on the liquid. Perhaps this one should be an easy volume integral of radiation in the volume with a surface integral for input and output, as with the heat.
     
    Last edited by a moderator: May 5, 2017
  10. Mar 26, 2012 #9

    Jano L.

    User Avatar
    Gold Member

    I guess they want to have expression giving the rate of increase of something inside the surface. The rate of income naturally depends only on the mutual velocity of the surface and the field.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook