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

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SUMMARY

The discussion centers on the 'Power' equation for a single 'Planck Energy Photon', which is defined as P_p = \frac{c^5}{G}, resulting in a value of approximately 3.6E52 watts. This power is significantly larger than the total power consumption of human civilization, which is around 17 terawatts. The participants clarify that the Planck power unit is derived from fundamental constants, specifically Planck's constant, and is used in conjunction with other Planck units such as length and mass. The conversation also touches on the comparison of Planck power to current global energy consumption.

PREREQUISITES
  • Understanding of Planck units, including Planck length, mass, and time.
  • Familiarity with fundamental physics constants such as the speed of light (c) and gravitational constant (G).
  • Knowledge of energy and power equations, specifically P = \frac{E}{t}.
  • Basic grasp of quantum mechanics and the significance of Planck's constant.
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  • Research the derivation of Planck units and their applications in theoretical physics.
  • Explore the implications of Planck power in cosmology and high-energy physics.
  • Investigate the relationship between Planck energy and other energy scales in physics.
  • Learn about the current global energy consumption metrics and their significance in energy policy.
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Physicists, researchers in quantum mechanics, energy policy analysts, and anyone interested in the fundamental scales of energy in the universe.

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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:
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Planck power quantum = 6.626069 E -27 erg second or E -34 joule second etc etc --
Cheers, Jim
 
NEOclassic said:
Planck power quantum = 6.626069 E -27 erg second or E -34 joule second etc etc --
Cheers, Jim

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
 
Orion1 said:
...
How does this 'Planck Photon Power' compare with the current power consumption of the world (at any instantaneous moment)?

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

and it is

\text{Planck power} = \frac{c^5}{G}

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.
 

NEOclassic said:
Planck power quantum = 6.626069 E -27 erg second or E -34 joule second etc etc --
Cheers, Jim

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

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

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

dE = \frac{\hbar}{dt}
dP = \frac{dE}{dt} = \frac{\hbar}{dt^2}

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

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

marcus said:
just take Planck mass and multiply by c^2 to get Planck energy, then the power is that amount of energy per unit time
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}
\boxed{P_w = \sqrt{\frac{\hbar c^5}{G}} \frac{}{dt}}

marcus said:
the Power I am talking about is what would be delivered by a one Planck photon arriving each Planck time unit.
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}
\boxed{P_p = \frac{c^5}{G}}

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?

[/color]
Reference:
http://www.answers.com/topic/planck-units
 
Last edited:
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.
 
I apologize

marcus said:
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

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
 
Jim's Joule Jam...


NEOclassic said:
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)
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...
[/color]
 
Last edited:
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
 
  • #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 2 \pi 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:
r_p = \overline{\lambda_p}
\boxed{\overline{\lambda_p} = \sqrt{\frac{\hbar G}{c^3}}}

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

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?
[/color]
Reference:
https://www.physicsforums.com/showpost.php?p=481704&postcount=21
http://www.poemsinc.org/factsenergy.html
 
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