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

[Orion1]

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)?

 

[NEOclassic]

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

 

[marcus]

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

 

[marcus]

...
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.

 

[Orion1]

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

just take Planck mass and multiply by c^2 to get Planck energy, then the power is that amount of energy per unit time

$$P_w = \frac{dE}{dt} = \frac{M_p c^2}{dt} = \sqrt{\frac{\hbar c}{G}} \cdot \frac{c^2}{dt} = \sqrt{\frac{\hbar c^5}{G}} \frac{}{dt}$$
$$\boxed{P_w = \sqrt{\frac{\hbar c^5}{G}} \frac{}{dt}}$$

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?

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Reference:
http://www.answers.com/topic/planck-units

 

[dextercioby]

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.

 

[NEOclassic]

I apologize

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

 

[Orion1]

Jim's Joule Jam...

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...
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[NEOclassic]

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

 

[Orion1]

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