Math Is Fun Forum / Conclusions and Benchmarks

LQ · 2010-12-13 10:42:06

Today I have a new experience to relate. I have come to understand what I am more deep then I thought possible.
The information in my head is not stored as what I thought was ordinary matter composition. It is stored in a frozen quantum entanglement. All entanglements that enters this substance are frozen in it forever. The substance having once been entangled with a black hole, can never be of entangled. I recon that with the endless signals that enters my head from the impresions will cause the substance to grow each day feeding upon the quantumtraces it obtains. I have a feeling it grows faster then I first thought, even thought the entropy of a photon is really small, I am bombarded with what seems to be an endless precision of them. This worries me a little, even though I seem to grow physically and not remain within a small radius. What will become of me?

bobbym · 2010-12-14 05:37:58

Hi LQ;

Sorry for the late response. I predict an expansion of your head. Or maybe not...

LQ · 2010-12-14 08:28:31

I've been counting on it. if my entropy gain is 0.3 calories per recieved mole, then I divide that with c^2 and times 4.184 for joule.
I get 1.4 e-5 nanogram per recieved mole (mass)

Last edited by LQ (2010-12-14 08:44:44)

bobbym · 2010-12-14 08:43:30

A mole, back in the days when I went to school ( 1635 - 1651 ) was defined in terms of gram molecular weight. Now they define it in terms of Avogadro's number 6.022 * 10^23 particles of some substance. So first you must tell me what substance is bombarding you. Then I will begin work on an electromagnetic shield or umbrella that will keep them off you.

LQ · 2010-12-14 08:49:28

I think I absorb all quantum entanglements that I am in connection with.

bobbym · 2010-12-14 09:03:30

Do not worry about them.Sticks and stones make break our bones but quantum kaboobly doo will never hurt us. And you can take that to the bank.

LQ · 2010-12-14 09:06:40

Thanks. So, how many mole quantum entanglement entropy passes through a regular particle per day?

bobbym · 2010-12-14 09:14:58

Back from the bank already? That was fast.

Let's say you are being pelted by carbon atoms. Why carbon, because Carl Sagan likes carbon, and that is good enough for me. A mole of them would weigh 12 grams, less than 1 / 2 an ounce. So every time a half an ounce of carbon rams into you, you gain .004 nanograms.

LQ · 2010-12-14 09:18:06

wait, we missed something. We are talking about moles of photons (unless I got something wrong) they also carries quantum entanglement.
I get much higher figures, we can share them and see.

bobbym · 2010-12-14 09:26:00

I see now, LQ stands fpr La Quantique. A mole of photons, the quanta of light! In the old definition a mole of photons would be meaningless. But in the new definition it makes perfect sense. Why did I not see this earlier?

Show thy calculations, so that I may see them.

LQ · 2010-12-14 09:27:37

So if my heat is allways the same, 300k, I get that a small percentage of that temperature is the entanglement. Per mole photons there is .3 calories of energy that is quantum entanglement, if the energy bounced only one time on me, same temperature, the same energygain would apply, I get that kelvin is in some way proportional to joule per second.

Thereby we get our question. If the quantumentanglement is 0.3 calories per mole photons, and a mole photons carries so and so much energy, and all photons are entangled with something, then for the photons in 300 kelvin, what is the energy bound in form of quantum entanglement per second?

bobbym · 2010-12-14 09:33:11

300k, you say. That is a little on the chilly side. Do you feel okay?

LQ · 2010-12-14 09:34:29

Winter in sweden you know. Allthough I'm not that cold. Point taken.

Last edited by LQ (2010-12-14 09:35:27)

bobbym · 2010-12-14 09:38:20

LQ wrote:
Thereby we get our question. If the quantumentanglement is 0.3 calories per mole photons, and a mole photons carries so and so much energy, and all photons are entangled with something, then for the photons in 300 kelvin, what is the energy bound in form of quantum entanglement per second?

Beats me man. I have no idea how to calculate how many photons are passing a given point. Let's change it to electrons passing a given point, something to do with coulombs.

LQ · 2010-12-14 09:41:03

okay, I think I'm closer to the answer, let's say have the index, say 5, which means that it bounces atleast 5 times, every photon. so every mole photons delivers 1.5 calories. with 300K how many calories is that?

LQ · 2010-12-14 09:45:29

Well, we have E = Joule = hf and for visible light... (does anyone remember this?)

for green light the frequency is 612,5–522,5 THz

If all the light was green... wait... looking up tera again.

n*612*10^12*h = the energy of 300 kelvin... What is n??

Last edited by LQ (2010-12-14 09:52:30)

bobbym · 2010-12-14 09:50:50

Hi;

A calorie if I remember is the amount of heat needed to raise 1 gram of water, 1 degree celsius. Anyway, it is outdated, the calorie I mean. What do you weigh?

LQ · 2010-12-14 09:53:01

around 120 kg m8.

LQ · 2010-12-14 09:57:28

okay, 30 celcius is 10 newton, what gives?

bobbym · 2010-12-14 10:01:04

120 000 grams. Assuming the density of water and human tissue are around the same.

If you got into a tub of water that had 31 gallons in it at 40F. You would raise the temperature of the water by 20 degrees F!

LQ · 2010-12-14 10:03:49

oh, sorry, that's the temperature newton. buggars.

LQ · 2010-12-14 10:08:44

Probably, allthough I wonder how many photon bounces I get if n*612*10^12*h = 30*120000 and n is the number of photons...

h is plancks constant, mind you!

h = 6,6261·10−34 Js

n = 8877549299250646063650253 photons

Last edited by LQ (2010-12-14 10:14:54)

bobbym · 2010-12-14 10:10:29

Okay, I get 1 320 000 calories.

LQ · 2010-12-14 10:19:15

well I get 14 mole and how many times does photons bounce before they leave me you think?

I guess that depends on how well you can see right through me?

Last edited by LQ (2010-12-14 10:29:49)

bobbym · 2010-12-14 10:25:01

The concept of photonic bounce is not one that I have thought about recently. Let's see, now I think you are asking for the average number of bounces off of you by a high energy photon. Is that it?

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#1 2010-12-13 10:42:06

Conclusions and Benchmarks

#2 2010-12-14 05:37:58

Re: Conclusions and Benchmarks

#3 2010-12-14 08:28:31

Re: Conclusions and Benchmarks

#4 2010-12-14 08:43:30

Re: Conclusions and Benchmarks

#5 2010-12-14 08:49:28

Re: Conclusions and Benchmarks

#6 2010-12-14 09:03:30

Re: Conclusions and Benchmarks

#7 2010-12-14 09:06:40

Re: Conclusions and Benchmarks

#8 2010-12-14 09:14:58

Re: Conclusions and Benchmarks

#9 2010-12-14 09:18:06

Re: Conclusions and Benchmarks

#10 2010-12-14 09:26:00

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Re: Conclusions and Benchmarks

#13 2010-12-14 09:34:29

Re: Conclusions and Benchmarks

#14 2010-12-14 09:38:20

Re: Conclusions and Benchmarks

LQ wrote:

#15 2010-12-14 09:41:03

Re: Conclusions and Benchmarks

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Re: Conclusions and Benchmarks

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Re: Conclusions and Benchmarks

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