The first law of thermodynamics: What is it?

[Admin]

It started with the steam engine but applies to the universe too!

The first law of thermodynamics: What is it? : Read more

 

[rod]

"The first law of thermodynamics can be captured by the following equation: ΔU = Q — W, where ΔU is the change in the internal energy, Q is the heat added to the system, and W is the work done by the system."

Nice metric in this article. So does the 2nd Law or increasing entropy show W comes to an end where no work will be accomplished as time's arrow marches on?

 

[billslugg]

Yes, eventually everything will be at the same temperature thus no more flow of heat, no more Q can be added, no more work done, no more ΔU.

 

[Jzz]

It started with the steam engine but applies to the universe too!

The first law of thermodynamics: What is it? : Read more

A more conventional take on the first law of thermodynamics would be that energy can neither be created nor destroyed, only altered in form.

I have been thinking quite a bit about this problem of late. One of the thoughts I have had on the subject is: “What if all the energy that ever was, was created at the time of the Big Bang?” this would explain why the first law is so scrupulously maintained. Energy used at one place is immediately deducted at some other place and so on, so that equilibrium is always maintained.

Oh! Oh! the article that is referenced says almost exactly the same thing. It is quite gratifying to know that I was on the right track and vexing to think I could have got there faster.

 

[Jzz]

This in continuation of my previous post at #4

This maintaining of the equilibrium of the total energy within the Universe, would also fit in very well with the propagation of light as outlined in my paper “The Electromagnetic Universe” https://www.academia.edu/37258409/The_Electromagnetic_Universe_docx

Light is created, it travels through light and in the end returns into light. Therefore, according to this theory light is both transient and perpetual, in that it manifests as visible light but continues to exist as the ‘virtual photons’ that make up dark matter, which in turn permeates all of the Universe including the vacuum and matter itself.

This raises the question of how a photon can travel forever (according to quantum mechanics) until it meets some matter that can absorb it. Fine, there is the initial energy imparted to the photon when it is created (emitted) that can, when the photon is in its particle avatar, impart enough momentum to keep it moving at the speed of light. But, logically, how can the conservation of energy and momentum enter into this scenario. It is a fact that light (read photons) when it is propagating follows the inverse square law, meaning that it spreads out over an area equal to the square of the linear distance it has travelled. This cannot be denied. The alternative, is to completely ignore the facts (that light spreads out according to the inverse square law) and carry on as if it is inconsequential and can be explained away by allusion to the manner in which discrete particles would travel from a point source, the fact that the gap between individual photons would widen to mathematically unacceptable levels, being glossed over by another allusion to the number of photons created. However, practically speaking, it is obvious that every point in the solar system receives a signal from the voyager space craft. Given that we know the power output of the transmissions and relating that to the number of photons created, it is possible to calculate that the quantum mechanics version of events is patently false.

But the process of self-hypnosis under which present day quantum mechanics physicists exist is so strong, that little peccadilloes like this in the theory don't even deserve notice.

 

[billslugg]

"Given that we know the power output of the transmissions and relating that to the number of photons created, it is possible to calculate that the quantum mechanics version of events is patently false." - Jzz

Can you share those calculations with us?

 

[Jzz]

Can you share those calculations with us?

I will give it a try. These calculations are done using the GAT (Gestalt Aether Theory). The Voyager space craft, both have transmitters of about 25W power. According to GAT electricity is not conducted by electrons but by special photons called ‘conduction photons’. This explains the problem of Maxwell’s displacement current and also accurately explains the flow of current. Each ‘conduction photon’ has an energy of 1.6 x 10^-19 V. Therefore, in one watt there are 1.6 x 10 ^ 19 photons present and in 25 W there will be 4 x 10^20 photons. The frequency of the Voyager downlink is 8 GHz and the wave length would be 3.7 cm. At the distance of Voyager 1 which is about 20 billion km from earth, it can be considered to be an isotropic radiator and so follows the inverse square law, this would give a spread of the transmission over an area of 2000000000000000^2 sq cm. The strength of the recd. Signal would be 4 x 10 ^ 30 / 4 x 10^ 20 = 1 x 10 ^ -10 W which is more than sufficiently strong to be picked up on earth.

My bad, it was not a very good example but it does show that in a few more years Voyager 1 will be out of range. However, as one of mankind’s greatest achievements and as a verifiable signal source maybe it was not such a bad example after all.

 

[billslugg]

Thank you for your reply. Here is another way of looking at it:
My source lists their output at 22 Watts.
Both Voyagers use a 3.7 meter parabolic antenna with a .5° beam width.
Data is received by a 70 meter diameter dish at Deep Space Network.
Science data downlink is at 8.4 GHz, wavelength is .036 m
Distance to Voyager 1 is 155 AU.
Planck's constant (h) is 6.6x10^-34
Planck/Einstein equation is E = hc/λ

At 8.4 GHz, each photon has an energy of (6.6x10^-34 * 3x10^-8) / (.036) or 5.5x10^-24 J
A 22 Watt output produces 22 Joules per second or 4x10^24 photons per second.
At a distance of 155 AU (2.3x10^10 m) a .5° degree beam covers a width of 2x10^8 m and an area of 3.2x10^16 m^2.
A 70 meter dish has an area of 3.8x10^3 m^2 and can intercept 1.2x10-13 parts of the incident beam.
This equates to the reception at DSN antenna of 4.7x10^11 photons per second.
This is equivalent to 2.6X10-12 Watt, about 3% of your calculation. Close enough.

 

[Jzz]

Thank you for your reply. Here is another way of looking at it:

Thank you so much, both for your reply and for your consideration. This kind of logical reasoning is what has won quantum mechanics universal admiration and acclaim. The reasoning is based on pure logic, on past experiential data and on experimental evidence. Therefore, although the mathematics involved might be insignificant compared to the vastly complicated scenario of imaginary numbers and their statistical use; on an everyday level this result could be considered to be more satisfying. Calculate the total energy available for transmission, calculate the energy of the transmission wave. Divide the former by the latter to get the total number of photons available. Then use the inverse square law that is tried and time tested to calculate the area over which the transmission is received, divide the former by the latter. Put in a few extra factors such as the antenna diameter (which I didn’t do) and there is your answer: the most probable signal strength.

What is to me of great interest, is that although we used widely differing approaches, (I had used the Gestalt Aether Theory concept of conduction photons) we had arrived at a result that is 3% off from my calculation. If adequate provision were made for receiving and transmitting antenna gain, the result would be probably be closer.

I can use my concept of the conduction photon to derive the energy of a radio wave (photon). Divide the transmission wave-length by the conduction photon wave-length 1.24 x 10^-6 m . Therefore 0.03/ 1.24 x 10^-6 = 3.0 x 10^4. Divide the conduction photon energy by number of photons in the composite radio-wave and the photon energy of the transmission wave photon is found 5.33 x 10^ -24 J.

 

[Helio]

A 70 meter dish has an area of 3.8x10^3 m^2 and can intercept 1.2x10-13 parts of the incident beam.

You might check to see if you forgot to convert that angle to radians (not degrees) to calculate that area.

 

[billslugg]

I did the work in degrees, not radians. The beam is .5°. Tangent .5° is .0087. Take that times the distance, which is 2.3x10^10 meters and you get a circle of diameter 200 million meters. Area of that is 3.2x10^16 m^2. A 70 meter dish has an area of 3850 m^2 which is 1.2x10^-13 parts of the incident beam.

 

[Helio]

I did the work in degrees, not radians.

Odd that when I used Excel and did not convert to radians I thought I got your result, but I'm elsewhere now...

The beam is .5°. Tangent .5° is .0087. Take that times the distance, which is 2.3x10^10 meters and you get a circle of diameter 200 million meters. Area of that is 3.2x10^16 m^2. A 70 meter dish has an area of 3850 m^2 which is 1.2x10^-13 parts of the incident beam.

Check your 155AU conversion into meters.

 

[billslugg]

Yes, one AU is 150 million kilometers, not meters. Thank you for finding this error. The portion of the incident beam intercepted by the DSN antenna is thus 1.2x10^-16 of the 4x10^24 photons emitted by the transmitter or 4.8x10^8 photons per second. Each photon has 5.5x10-24 J of energy giving a power of 2.5x10-15 Watt. Jzz claims 1x10-10 Watt so we are now off from each other by a factor of 40,000.

 

[Jzz]

Yes, one AU is 150 million kilometers, not meters. Thank you for finding this error. The portion of the incident beam intercepted by the DSN antenna is thus 1.2x10^-16 of the 4x10^24 photons emitted by the transmitter or 4.8x10^8 photons per second. Each photon has 5.5x10-24 J of energy giving a power of 2.5x10-15 Watt. Jzz claims 1x10-10 Watt so we are now off from each other by a factor of 40,000.

I made the mistake of calculating the number of photons as (J/conduction photon energy i.e., 25/ 1.6 x 10^-19 ) in actual fact ALL radio waves possess an energy that is just some fraction of the energy of a single conduction photon and that gives a completely flawed answer. That is beside the point. A straight forward calculation gives an answer of 22/(2,3 x 10^12)^2) = 4.15 x 10^-24 if this is translated into photons : 5.33 x 10^24 x 22 = 1.176 x 10^26. So: 1.176 x 10^26/ 5.29 x 10^24 = 22 photos per sq m = 1.18 x 10^-22 W

Voyager 1 had a confirmed signal strength in 2017 of 1 atto watt or 10^-18 Watts by the time it reached earth. At the time the Voyager 1 space craft was 15 billion kilometres distant from earth. Today the Voyager 1 space craft is about 23 billion kilometres distance or roughly half again as distant as it was in 2017. So a signal strength close to 10^-22 W is not far off. A signal strength of just 10^-22 W or even less, is mind boggling to think off. Visions of elements cooled to almost zero kelvin begin to emerge.

However, this question has far deeper, path-breaking implications, such as the GAT postulate that all radio-waves are ‘composite’ waves i.e., made of ‘parallel linked virtual photons that share the energy of a single conduction photon. This discovery that the oscillation of a current in a circuit gives rise to micro-waves but exactly the same oscillation by an electron in an atom does not, raises the question of why not? When electrons in an atom oscillate in the optical frequency they emit photons of a certain frequency, wavelength and energy. Yet when those same electrons in that same atom oscillate (or are made to oscillate) at microwave frequencies, it does not result in microwaves being emitted by the atom, no it results in the atom going into a meta-stable state! One explanation is that these waves do not have adequate energy to leave the atom, which is a possible answer. When they can travel trillions of kilometers across space they can't leave the atom? Maybe radio-waves are fundamentally different. GAT is very clear about how this happens. Are you?