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Energy in co-moving co-ordinates?

  1. Oct 20, 2012 #1
    Consider the energy of a quantum system

    [itex] E_t = h f [/itex]

    [itex] E_t = \frac{h}{\Delta t} [/itex]

    where [itex]\Delta t[/itex] is the period of the quantum system in cosmological time [itex]t[/itex].

    What is the energy of the system in co-moving co-ordinates?

    In co-moving co-ordinates time is measured in conformal time [itex]\tau[/itex] given by

    [itex] \Delta \tau = \frac{\Delta t}{a(t)} [/itex]

    Thus the energy of the co-moving quantum system is given by

    [itex] E_\tau = \frac{h}{\Delta \tau} [/itex]

    [itex] E_\tau = a(t) \frac{h}{\Delta t} [/itex]

    [itex] E_\tau = a(t) E_t [/itex]

    Is this correct?
     
    Last edited: Oct 20, 2012
  2. jcsd
  3. Oct 20, 2012 #2

    BillSaltLake

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    No. If you impose a time-dependent time (and/or length) scale, then h is also changing, as is the relative meaning of E.
     
  4. Oct 21, 2012 #3
    But surely one can use units such that h is unity.

    One can understand the cooling CMB in the comoving frame in this manner. The CMB photon energy is fixed (no expansion of space) but the energy of comoving systems is increasing. The effect is that a photon emitted from some type of atom in the past is effectively redshifted relative to a comoving atom of the same type in the present epoch as the atom's energy has increased whereas the photon has the same energy as the atom in the past.
     
    Last edited: Oct 21, 2012
  5. Oct 21, 2012 #4

    BillSaltLake

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    h has units of (mass)(length2)(time-1). If the time measurement unit is changing, then h must become variable, whether you let l units change in ratio with t units or keep l units constant. Similarly, E has units of ml2t-2, so if you keep c constant by letting l and t units vary in ratio, then E is unchanged. However, if c is a constant, then E = E' = hf = h'f ' other than a trivial scaling constant.
     
    Last edited: Oct 21, 2012
  6. Oct 22, 2012 #5
    But in the standard proper co-ordinates in which the Universe is expanding it is understood that the momentum of photons is inversely proportional to the scale factor. As far as I can see there is no assumption that Planck's constant changes as the length scale changes. So why should Planck's constant change in co-moving co-ordinates?
     
  7. Oct 23, 2012 #6

    BillSaltLake

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    As with most things, one must be careful of definitions. I assume your coordinate system defines v=0 locally as 'at rest with respect to CMB'. However, there are two different ways that lengths (and time intervals) can then be defined. The "normal" way (such as CGS or MKS, for example) interprets atomic lengths, such as wooden yardsticks, as constant. Then the distance between distant objects increases [itex]\propto[/itex] a. Of course h is a constant in this case.
    If your lengths and time intervals instead scale with a, so that distant objects appear to maintain ~constant distance, then the value of h will decrease over time (h [itex]\propto[/itex] 1/a).
     
    Last edited: Oct 23, 2012
  8. Oct 24, 2012 #7
    Ok, in a co-moving frame in which units are such that atomic lengths are constant, I presume I can define a small interval of time [itex]d\tau[/itex] as the time it takes light to travel along a constant length element dr so that

    [itex]d\tau = dr / c [/itex]

    Now the Freidmann metric says that light obeys the relation

    [itex] c \ dt = a \ dr [/itex]

    Therefore

    [itex] d\tau = dt / a(t) [/itex]

    Therefore in a co-moving frame, in which atomic lengths are constant, one uses conformal time where an interval of conformal time [itex]d\tau \propto 1/a[/itex].

    Have I got this correct?

    If the above is right then one can say the energy of a quantum system with period [itex]d\tau[/itex] is given by

    [itex] E = h / d\tau [/itex]

    [itex] E \propto a(t) [/itex]

    as Planck's constant is actually constant.

    If the above was true then the energy density of matter would go like

    [itex] \rho \propto \frac{a}{a^3} [/itex]

    [itex] \rho \propto \frac{1}{a^2} [/itex]

    instead of [itex] \rho \propto 1/a^3[/itex]

    This would give a linearly expanding Universe.
     
    Last edited: Oct 24, 2012
  9. Oct 25, 2012 #8

    Chronos

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    Under GR, a global definition of energy defies explanation.
     
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