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Planck Power

  1. Mar 27, 2005 #1
    Planck Photon Power...


    What is the 'Power' equation for a single 'Planck Energy Photon'?

    What is the numerical value for the 'Planck Photon Power' solution?

    How does this 'Planck Photon Power' compare with the current power consumption of the world (at any instantaneous moment)?
     
    Last edited: Mar 27, 2005
  2. jcsd
  3. Mar 27, 2005 #2
    Planck power quantum = 6.626069 E -27 erg second or E -34 joule second etc etc --
    Cheers, Jim
     
  4. Mar 27, 2005 #3

    marcus

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    this is not a quantity of power, it is a quantity of angular momentum or action
    Jim it looks like you just copied in the value of the Planck h constant


    there is a unit of power in the system of Planck units
    it is very large wattage
    much larger than the average total power consumed by human civilization

    the planck unit power is about 3.6E52 watts
     
  5. Mar 27, 2005 #4

    marcus

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    Orion I think you are asking for the definition of the planck unit power

    and it is

    [tex]\text{Planck power} = \frac{c^5}{G}[/tex]

    and if you plug in the usual values for c and G you get 3.6E52 watts

    and that is roughly E26 times the total power output of the sun.

    so it is many orders bigger than human civilization requirement

    this Planck power unit is the unit that goes along with conventional Planck length, Planck mass, Planck time.
    just take Planck mass and multiply by c^2 to get Planck energy, then the power is that amount of energy per unit time

    or if you imagine a "planck photon" with the Planck unit energy, then the Power I am talking about is what would be delivered by a one Planck photon arriving each Planck time unit.
     
  6. Mar 28, 2005 #5

    NEOclassic, the (SI) unit for Power is:
    [tex]\text{1 Watt} = \frac{\text{1 Joule}}{\text{1 second}}[/tex]

    Planck's Constant has (SI) units:
    [tex]\hbar = 1.054 \cdot 10^{-34} \text{Joules second}[/tex]

    In order to derive 'Power' from Planck's Constant, it must first be converted into an 'energy constant' and then a 'power constant'.

    [tex]dE = \frac{\hbar}{dt}[/tex]
    [tex]dP = \frac{dE}{dt} = \frac{\hbar}{dt^2}[/tex]

    'Planck Constant Power' for a single (SI) second:
    [tex]dP_c = \frac{\hbar}{dt^2} = \frac{\hbar}{\text{1 second}^2} = 1.054 \cdot 10^{-34} \; \text{Watts}[/tex]
    [tex]\boxed{dP_c = 1.054 \cdot 10^{-34} \; \text{Watts}}[/tex]

    In Planck Units:
    [tex]P_p = \frac{\hbar}{t_p^2} = \hbar \left( \frac{c^5}{\hbar G} \right) = \frac{c^5}{G}[/tex]
    [tex]\boxed{P_p = \frac{c^5}{G}}[/tex]

    [tex]P_w = \frac{dE}{dt} = \frac{M_p c^2}{dt} = \sqrt{\frac{\hbar c}{G}} \cdot \frac{c^2}{dt} = \sqrt{\frac{\hbar c^5}{G}} \frac{}{dt}[/tex]
    [tex]\boxed{P_w = \sqrt{\frac{\hbar c^5}{G}} \frac{}{dt}}[/tex]

    [tex]P_p = \frac{dE}{dt} = \frac{E_p}{t_p} = \sqrt{\frac{\hbar c^5}{G}} \cdot \sqrt{\frac{c^5}{\hbar G}} = \frac{c^5}{G}[/tex]
    [tex]\boxed{P_p = \frac{c^5}{G}}[/tex]

    Marcus, that is correct, however, how much 'power' is delivered by a single 'Planck Energy Photon' in a single (SI) second?

    What is the numerical value for the 'Planck Photon Power' solution for a single (SI) second?

    How does this 'Planck Photon Power' compare with the current power consumption of the world in a single (SI) second?


    Reference:
    http://www.answers.com/topic/planck-units
     
    Last edited: Mar 28, 2005
  7. Mar 28, 2005 #6

    dextercioby

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    Instead of dividing by 10^{-43} seconds,divide by 1 second...I'm sure the # won't be that huge...

    As for the comparation,it would be less than mankind uses every SI second...

    Daniel.
     
  8. Mar 28, 2005 #7
    I apologize

    Hi Marcus,
    My problem was that I knew from having referred to Planck units of Joule-seconds from the NIST Codata table of constants, so that when I picked up my 37th Ed of C and P handbook (because its easier to handle than my heavy 82nd Ed) and turned to the Power section of "Units and Conversion factors" had included in that section four sub sections one of which was titled "action". When I noticed that the units given were Joule seconds I erred in a bad assumption. Looking back I found that power was mv^2/sec while the action sub section units were mvl which is momentum. there were seven listed momentum item one of which was specifically Planck's quantum.
    Please forgive, I really thought I was being helpful. Jim
     
  9. Mar 29, 2005 #8
    Jim's Joule Jam...


    NEOclassic, what is the equation for the 'Planck Energy Photon' wavelength?

    NEOclassic, what is the numerical value for the 'Planck Energy Photon' wavelength solution?

    This is your chance to redeem yourself for the Physics Forums Science Advisors...
     
    Last edited: Mar 29, 2005
  10. Apr 6, 2005 #9
    Hi Orion,
    According to the 1998 Codata table, the eqn of Planck length is (h-barG/c^3)^0.5 G=1.5 E -3 (Nast t-of-constants,1998) Its numerical value is 1.616 E -35 meters
    The energy equivalence of such length is 7.67 E 22 MeV.
    Cheers, Jim
     
  11. Apr 6, 2005 #10
    Solar Nexus...


    NEOclassic, your 'Planck Energy Photon' wavelength numerical value is correct, however, your energy equivalence of length is not correct, for some reason your numerical value contains an unexplained [tex]2 \pi[/tex] factor. When stating the Planck Energy numerical value, although the prefix is relative (Ev, Mev, Gev, etc), I believe it is more meaningful and standard now to state it in either 'ev' or 'Gev' prefix.

    Please consult with reference 1 below for the correct Planck Energy numerical value solution. Note that nobody except for me was able to derive the correct numerical value for Planck Energy on that thread.

    Planck Energy Photon wavelength solution:
    [tex]r_p = \overline{\lambda_p}[/tex]
    [tex]\boxed{\overline{\lambda_p} = \sqrt{\frac{\hbar G}{c^3}}}[/tex]

    Planck Energy Photon power solution:
    [tex]\boxed{P_w = \sqrt{\frac{\hbar c^5}{G}} \frac{}{dt}}[/tex]
    [tex]P_{\gamma} = 1.956 \cdot 10^9 \; \text{Watts} \cdot \text{s}^{-1} = 1.956 \; \text{Gigawatts} \cdot \text{s}^{-1}[/tex]
    [tex]\boxed{P_{\gamma} = 1.956 \; \text{GW} \cdot \text{s}^{-1}}[/tex]

    The current power consumption of the USA and EU is 5.612 Gigawatts per second.

    How many 'Planck Energy Photons' is this equivalent to every second?

    According to reference 2, the current power consumption of the world is 17 Terawatts.

    How many 'Planck Energy Photons' is this equivalent to every second?

    How does this compare to the total amount of solar power reaching Earth from the sun?

    How many 'Planck Energy Photons' is this equivalent to every second?

    Reference:
    https://www.physicsforums.com/showpost.php?p=481704&postcount=21
    http://www.poemsinc.org/factsenergy.html
     
    Last edited: Apr 6, 2005
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