Gravitational Redshift, Temperature and Weins Law

[exchemist]

I did the calculation in post 7 again and I have uncovered a mistake. Since you enjoy spotting mistakes, I will leave this one for you. Its a bit tricky, in the meanwhile I need to find the consistent calculation.

......[click]......

 

[wellwisher]

If you look at the gravitational red shift, as energy leaves a source of gravity, where does the energy difference go, if we assume conservation of energy. A red shift means less potential energy per quanta, therefore, gravity needs to absorb the energy potential difference, or energy is not conserved. How does this show up?

Let me illustrate this with a working example. Say we have hydrogen atoms, on the surface of a large dense star, being ionize to the highest energy level. This photon energy leaves the large star and red shifts, until it looks like energy from the lowest energy level of hydrogen. Where does this energy difference go, if energy is conserved?

Alternately, if energy is entering the same gravitational source, it will blue shift. This means the photons gains energy potential, therefore conservation of energy would mean the mass/matter has to lower potential, since it generated the field that makes this possible. How does the loss show up in the mass?

 

[Q-reeus]

If you look at the gravitational red shift, as energy leaves a source of gravity, where does the energy difference go,...

GR stands for general relativity. That last word is your clue. You do know that clocks (and therefore all processes) run slower the deeper down into a gravitational potential they are, right? Not their proper i.e locally measured time, but in relation to clocks further out.

So, if light of some frequency f1 is emitted down at radius r1 up to an observer at radius r2, the latter will locally measure a lower frequency (thus energy) f2 in direct inverse ratio to the relevant clock rates.
As elaborated on here: https://en.wikipedia.org/wiki/Gravitational_redshift
And as given in the lower RHS formulae in table shown here: https://en.wikipedia.org/wiki/Redshift#Redshift_formulae

Many interpret that as the photons 'losing energy' in 'climbing out against gravity', but it's not the best interpretation. Better to simply see it as a manifestation of gravitational time dilation - which is relative.
You may still be asking if energy is overall conserved. Yes (neglecting maybe a few contentious subtleties not worth pursuing here). What's missing from your picture is the total balance involved in forming a body such as a planet in the first place. Aggregation of initially loose and dispersed matter results in the release of heat radiation that largely escapes to space as collapse proceeds. Even if the total number of atoms before and after is identical, the latter reside in a lower gravitational potential hence each has a lower net energy than when initially dispersed as dust, grains etc. That 'lost' energy was radiated away as heat in the initial planet forming phase (an ever decreasing amount was retained as internal planetary heat). Nothing has really disappeared overall. Does this now all make sense?

 

[James R]

SimonsCat:

I have some questions, mostly to get at what your motivation is for this thread.

We need to first derive a new definition of binding energy density, which is what the first part consists of.

Why? What are you trying to achieve from your new definition of binding energy density?

And this is the binding energy density of what, exactly?

The second part consists of interpreting the results in terms of Wein's displacement law, which would translate to how redshift (wavelengths of radiation) is effected by the temperature of the system characterized by Weins constant.

What system are you referring to?

Part One

We need to establish the new metric coefficient, which is just a reinterpretation of the Schwarzschild factor:

$$1 - \frac{2GM}{E} \frac{M}{R} = 1 - \frac{E_g}{E}$$

Why do we need to do this? What is achieved by it?

The gravitational field inside a radius $$r = r(0)$$ is given as

$$\frac{dM}{dR} = 4 \pi \rho R^2$$

and the total mass of a star is

$$M_{total} = \int 4 \pi\rho R^2 dR$$

Ah, ok. So you're talking about stars here?

Specifically, what kind of star are you modelling?

...and so can be understood in terms of energy (where $$g_{tt}$$ is the time-time component of the metric),

$$\mathbf{M} = 4 \pi \int \frac{\rho R^2}{g_{tt}} dR = 4 \pi \int \frac{ \rho R^2}{(1 - \frac{2Gm}{E}\frac{M}{R})} dR$$

The difference of those two mass formula is known as the gravitational binding energy:

$$\Delta M = 4 \pi \int \rho R^2(1 - \frac{1}{(1 - \frac{2Gm}{E}\frac{M}{R})}) dR$$

Please explain this in more detail. Which two mass formulae are you using, and why is the difference equal to the gravitational binding energy?

And so, an energy can be obtained by the distribution of the speed of light squared:

$$Mc^2 = 4 \pi \int \frac{\rho c^2}{(1 - \frac{2Gm}{E}\frac{M}{R})} dV = 4 \pi \int \frac{T_{00}}{(1 - \frac{2Gm}{E}\frac{M}{R})} dV$$

The integration is worked out from the following equation:

$$T_{00} dV = \frac{c^4}{8 \pi G} \int dV\ \nabla^2 \phi^2 = \frac{c^4}{8 \pi G} \int dV\ \phi \Delta \phi$$

Plugging this into our mass formula we have a new equation and the final one for this post:

$$E(density) = \frac{c^4}{2G} \int \frac{\phi \Delta \phi}{(1 - \frac{2Gm}{E}\frac{M}{R})}$$

Where direct substitution has simplified the equations presence of the pi-symbol and the remaining quantity $$\frac{c^4}{2G}$$ is exactly 1/2 the classical upper limit of both electromagneism and gravitation.

What is $$c^4/2G$$ the classical upper limit of, exactly?

And what is the relevance of your derived formula for $$E(density)$$? What is this telling us about stars?

Part Two

It wasn't explained previously, but note that in the integration:

$$T_{00} dV = \frac{c^4}{8 \pi G} \int dV\ \nabla^2 \phi^2 = \frac{c^4}{8 \pi G} \int dV\ \phi \Delta \phi$$

... we use a dimensionless gravitational potential

$$\phi = -\frac{Gm}{c^2R}$$

which means

$$\delta E_{binding} = \frac{c^4}{2G} \int dV\ \phi \Delta \phi - \frac{c^4}{2G} \int dV\ \frac{\phi \Delta \phi}{(1 - \frac{2Gm}{E}\frac{M}{R})}$$

where

$$1 - \frac{2GM}{E} \frac{M}{R} = 1 - \frac{E_g}{E}$$

I don't understand why you are using a dimensionless gravitational potential, or why your expression is useful.

Can you explain?

In previous work, the gravitational binding energy (density) of a distant object, like a star or even a black hole was given in previous work as:

$$\delta E_{binding} = \frac{c^4}{2G} \int \phi \Delta \phi - \frac{c^4}{2G} \int \frac{\phi \Delta \phi}{(1 - \frac{2Gm}{E}\frac{M}{R})}$$

Whose previous work? Where can we find that work?
What do you mean by "distant object"? Distant from what?

Keep in mind, this full extension involving the Schwarzschild factor (written in terms of energy) is the correct way to describe the physics, but to keep this work nice and simple for the eye, we will work with only the first term (so if you want the full equation, just write it out), the equation we use in this work is a simple energy equation:

$$E = \frac{c^4}{2G} \int \phi \Delta \phi$$

What physics are you talking about here? The correct way to describe the physics of what?
And please explain your "extension" of the Schwarzschild factor in more detail, so I can understand it. Why are you extending it? How are you extending it? What is gained by extending it?

Dividing the mass on both sides, and then using the relationship $$a_g = \frac{c^4}{2Gm}$$

$$\frac{1}{V}c^2 = a_g \int \phi \Delta \phi$$

Is this still the mass of a star you're dealing with, or something else?
What is the significance of $$a_g$$ and your motivation for introducing it here?

distribute $$\frac{\hbar}{2 \pi k_B c}$$

$$\frac{\hbar c}{2 \pi k_B} = \frac{\hbar a_g}{2 \pi k_B c} \int dV\ \phi \Delta \phi$$

Due to equivalence principle, the temperature is

$$T = \frac{\hbar a_g}{2 \pi k_B c}$$

The temperature of what? The star?

And how does this follow from the equivalence principle? Please explain.

$$\frac{\hbar c}{k_B} = 2 \pi T \int dV\ \phi \Delta \phi$$

Which is equivalent to the second radiation law.

Which second radiation law?

The Hawking Bekenstein relationship for entropy is:

$$\frac{S}{k_B} = \frac{1}{\hbar c} Gm^2$$

The entropy of what?

if on the RHS $$\int dV\ \phi \Delta \phi$$ corresponds to a gravitational wavelength then

$$\frac{Gm^2}{S} = 2 \pi T \lambda_g$$

Explain why the RHS corresponds to a gravitational wavelength.

and a work equation

$$W = Gm^2 = 2 \pi S T \lambda_g$$

What is working on what, here?

Which is a form of Weins displacement law (for gravitation)...

I don't understand. What did Wein say about gravity? Also, isn't it "Wien"?

...which states that a blackbody radiation curves at different temperatures for a system like a star or a black hole, has been written gravitationally for fittingly, the entropy of the system.

Please link me to Wien's displacement law for gravitation, so I can check this for myself.

Clearly entropy and temperature have already had profound relationships established for them.

$$\lambda^{max}_{g} = \frac{\hbar c}{x} \frac{1}{k_BT}$$

$$\lambda^{max}_{g}T = \frac{\hbar c}{k_B x}$$

The entropy and temperature of what? Stars?

Thanks in advance for answering my naive questions.

 

[RajeshTrivedi]

SimonsCat, you have posted many threads. I am yet to decipher the motive behind. Are you proposing something alternative, or you just want to express yourself to find flaw in your understanding.

 

[SimonsCat]

SimonsCat:

I have some questions, mostly to get at what your motivation is for this thread.


Why? What are you trying to achieve from your new definition of binding energy density?

And this is the binding energy density of what, exactly?


What system are you referring to?


Why do we need to do this? What is achieved by it?


Ah, ok. So you're talking about stars here?

Specifically, what kind of star are you modelling?


Please explain this in more detail. Which two mass formulae are you using, and why is the difference equal to the gravitational binding energy?


What is $$c^4/2G$$ the classical upper limit of, exactly?

And what is the relevance of your derived formula for $$E(density)$$? What is this telling us about stars?


I don't understand why you are using a dimensionless gravitational potential, or why your expression is useful.

Can you explain?


Whose previous work? Where can we find that work?
What do you mean by "distant object"? Distant from what?


What physics are you talking about here? The correct way to describe the physics of what?
And please explain your "extension" of the Schwarzschild factor in more detail, so I can understand it. Why are you extending it? How are you extending it? What is gained by extending it?


Is this still the mass of a star you're dealing with, or something else?
What is the significance of $$a_g$$ and your motivation for introducing it here?


The temperature of what? The star?

And how does this follow from the equivalence principle? Please explain.


Which second radiation law?


The entropy of what?


Explain why the RHS corresponds to a gravitational wavelength.


What is working on what, here?


I don't understand. What did Wein say about gravity? Also, isn't it "Wien"?


Please link me to Wien's displacement law for gravitation, so I can check this for myself.


The entropy and temperature of what? Stars?

Thanks in advance for answering my naive questions.

I'm going to take all those questions and do a new write-up for you James so that it will answer all your questions (and more). Since I am quite sure the preliminary OP is correct, I am confident in writing it with more clarity.

 

[wellwisher]

GR stands for general relativity. That last word is your clue. You do know that clocks (and therefore all processes) run slower the deeper down into a gravitational potential they are, right? Not their proper i.e locally measured time, but in relation to clocks further out.

So, if light of some frequency f1 is emitted down at radius r1 up to an observer at radius r2, the latter will locally measure a lower frequency (thus energy) f2 in direct inverse ratio to the relevant clock rates.
As elaborated on here: https://en.wikipedia.org/wiki/Gravitational_redshift
And as given in the lower RHS formulae in table shown here: https://en.wikipedia.org/wiki/Redshift#Redshift_formulae

Many interpret that as the photons 'losing energy' in 'climbing out against gravity', but it's not the best interpretation. Better to simply see it as a manifestation of gravitational time dilation - which is relative.
You may still be asking if energy is overall conserved. Yes (neglecting maybe a few contentious subtleties not worth pursuing here). What's missing from your picture is the total balance involved in forming a body such as a planet in the first place. Aggregation of initially loose and dispersed matter results in the release of heat radiation that largely escapes to space as collapse proceeds. Even if the total number of atoms before and after is identical, the latter reside in a lower gravitational potential hence each has a lower net energy than when initially dispersed as dust, grains etc. That 'lost' energy was radiated away as heat in the initial planet forming phase (an ever decreasing amount was retained as internal planetary heat). Nothing has really disappeared overall. Does this now all make sense?

Sorry about not responding, but I move around so much, I often forget where I wrote, last.

If we take any star, the energy output is red shifting in all directions, since the space-time well is expanding in all directions from the core. I am not looking at this from the POV of an observer, near the star. Instead, I looking outward, in all directions, from the center of the star, where I can see the photons red shift, in all directions, as they climb out of the space-time well.

If we explain the loss of potential, due to the red shift, as being due to the light gaining gravitational potential, shouldn't the increase in gravitational potential blue shift the light, since it is gaining potential energy? If I lifted a mass, 1000 feet into the air, we would add this potential to the mass, not subtract it.

 

[origin]

Sorry about not responding, but I move around so much, I often forget where I wrote, last.

If we take any star, the energy output is red shifting in all directions, since the space-time well is expanding in all directions from the core. I am not looking at this from the POV of an observer, near the star. Instead, I looking outward, in all directions, from the center of the star, where I can see the photons red shift, in all directions, as they climb out of the space-time well.

If we explain the loss of potential, due to the red shift, as being due to the light gaining gravitational potential, shouldn't the increase in gravitational potential blue shift the light, since it is gaining potential energy? If I lifted a mass, 1000 feet into the air, we would add this potential to the mass, not subtract it.

Look at the wiki pager for gravitational redshift. Your questions are answered in the first couple of paragraphs.

Edited to add: I see that Q-reeus already answered your questions and directed you to the wiki page. Did you not read the write up or do you just disagree with it?

 

[wellwisher]

Look at the wiki pager for gravitational redshift. Your questions are answered in the first couple of paragraphs.

Edited to add: I see that Q-reeus already answered your questions and directed you to the wiki page. Did you not read the write up or do you just disagree with it?

Instead of pointing to a link, explain to us how you interpret this? If we I raised a mass 1000 feet above the earth, there is a force potential between the earth and object. The mass has potential energy added to the rest mass, which can converted to kinetic energy if we release it. It take work to raise the mass up, which goes into the mass. The two add. With energy, why do the two subtract, resulting in weaker energy quanta? To get a red shift you need to subtract energy not add energy.

Let me approach this was another angle, based on the wikipedia time shift of the photons. The explanation says that loss of energy; red shift, by being induced by the time shift, implies this is adding potential to time? This logic has to do with energy conservation. Energy can convert between various states, but it can't be created or destroyed. If the loss of energy; red shift, is connected to the time shift, then we have conversion of energy, into time, so energy is conserved. If time gets the energy and time is speeding up, in parallel to the energy red shift, is the faster time being induced by energy?

The way do the energy balance is connected to entropy. If we had two factories, side by side, both make the exact same object. Each factory makes with 1 error per hour, with the error is a measure of entropy. If we placed one of the two identical factories near the core of the star; bottom of the space-time well, and the other beyond the surface; end of the space-time well, since time moves faster beyond the surface, that factory will make errors more frequently, relative to the core, since one hour appears sooner the farther from the core you are. Entropy needs energy to increase, so to support the entropy, implicit of expanded spacetime, we will need energy. This is comes from the red shift difference.

If we started with UV, this is the energy of chemical bonding, which is quantized as one photon per bond. In chemical reactions, not all are reversible so entropy becomes more limited to a forward direction. If we red shift this to IR, IR is more about vibrational energy levels, which is more interactive, since it is based on reversible collisions. In this example, the red shift moves the energy to where things get busier; more entropy.

 

[origin]

Instead of pointing to a link, explain to us how you interpret this? If we I raised a mass 1000 feet above the earth, there is a force potential between the earth and object. The mass has potential energy added to the rest mass, which can converted to kinetic energy if we release it. It take work to raise the mass up, which goes into the mass. The two add. With energy, why do the two subtract, resulting in weaker energy quanta? To get a red shift you need to subtract energy not add energy.

I think I see your problem. First you should read some of the discussions that James R had about energy.
You are looking at energy like it is a substance that is added to the mass - it is not. There is work done to move a mass to a higher position but that does not 'go into' the mass. The mass is said to have a higher PE energy but that is in reference to the gravitational well, it does not mean that the mass has more energy.
If I take a 1 kg mass and raise it to 1000 ft you would say the mass has more energy, if I raise it to 10 miles above the earth you would say it has even more energy. This is not correct. Transport that mass 50 light years away and you would think that the mass has the highest 'energy', however if you were to compare the 1 kg mass that you transported 50 ly to a 1 kg mass that was already at that location you would find that they have the same amount of 'energy'.
The red shift is a result of time dilation.

 

[Q-reeus]

Sorry about not responding, but I move around so much, I often forget where I wrote, last.

That's fine.

Instead, I looking outward, in all directions, from the center of the star, where I can see the photons red shift, in all directions, as they climb out of the space-time well.

You are seeing no such thing. At best, if there is some fog say that simply scatters part of the outgoing light, what comes back to you will be seen as having the same frequency as when initially emitted. Imagining redshift of outgoing light is only a valid pov in the limited relative sense of inferring what observers/detectors further out are seeing - where clocks are ticking faster.

If we explain the loss of potential, due to the red shift, as being due to the light gaining gravitational potential, shouldn't the increase in gravitational potential blue shift the light, since it is gaining potential energy? If I lifted a mass, 1000 feet into the air, we would add this potential to the mass, not subtract it.

Origin covered that mistaken viewpoint in #50. I will add though there is no single correct position covering all situations. For instance it is legitimate to say that a compressed spring or spun-up flywheel possesses potential energy and kinetic energy respectively, in an absolute not relative sense. Such energies having been added in the centre of mass rest frame in each case. Similarly for thermal energy. Whereas when moving a mass against gravity or a charge against an electric field, no such change exists within the mass or charge respectively. Energy being merely redistributed within the system as a whole, or in other situations added to or withdrawn from the system in an exchange with an external system or agency . Each case or category must be treated according to the particulars applying.