# Gravitational Redshift, Temperature and Weins Law

Discussion in 'Alternative Theories' started by SimonsCat, Jan 20, 2017.

1. ### SimonsCatRegistered Member

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I shouldn't have said curl, because that involved the cross product, which we don't have. A very small mistake thank you for pointing it out, kind of.

What we have is the divergence of the gradient $\nabla \cdot \nabla f = \Delta f$

3. ### SimonsCatRegistered Member

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But I am sure you can understand, this is a lot to write out the brain will make mistakes. You trying to capitalize on them is slightly amusing.

5. ### Q-reeusValued Senior Member

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I don't have the time or inclination to wade through all your stuff here, but something further amusing is your 'correction' in #21. I will trust Wiki's take on that one:
And of course any reliable source e.g. any textbook on EM will give the same standard answer. Do have another stab. But here's a further thing about that 'correction' - you now in effect claim a divergence on LHS equates to a current density on the RHS. Well actually curl curl A equates to a current density in EM. So sadly even after correcting your #21 'correction', it still leaves everything in a mess.

7. ### SimonsCatRegistered Member

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are you for real? There are all sorts of different currents. In hydrodynamics, the countinuity is

$\nabla \cdot j_m = - \frac{\partial \rho}{\partial t}$

in electromagnetism, it is

$\nabla \cdot j = - \frac{\partial \rho}{\partial t}$

Do you know the difference of these equations?

8. ### SimonsCatRegistered Member

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Look, you don't need to go through all the work, just focus on the part which is confusing you.

Just to add some physics to this, the density is related to the RHS via

$\nabla \phi = \frac{4 \pi G}{c^2}\ j_m$

Just need to make a quick correction to this, because we use dimensionless form of $\phi$. This is simply averted by stating the previous equation, in a new way

$\nabla \phi = \frac{4 \pi G}{c^4}\ j_m$

and actually it seems very appealing to express the equation like this, as we have retrieved the Einstein factor

$k = \frac{4 \pi G}{c^4}$

Note

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

so we gain an extra factor of $c^2$ on the RHS and we have tweaked this.

but what we really have is

$(\nabla \cdot \nabla) \phi =\nabla \cdot \frac{4 \pi G}{c^4}\ j_m$

which is just

$\Delta \phi = \nabla \cdot \frac{4 \pi G}{c^4}\ j_m$

this derivation is very straightforward.

9. ### SimonsCatRegistered Member

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213
and more to the point, as far as I am aware, the derivation above is totally legal.

10. ### Q-reeusValued Senior Member

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2,497
The two expressions in #24 are correct in their own right, but have zilch to do with what you either wrote in #9, or as subsequently modified by your 'correction' in #21.
[and for good measure I note you now repeat that mathematical nonsense in #25. Sad.]

11. ### SimonsCatRegistered Member

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213

show what you mean. Clearly.

12. ### SimonsCatRegistered Member

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you mean this statement?

''$\nabla \phi = \frac{4 \pi G}{c^4}\ j$

from standard theory, but in our energy binding equation, we have the presence of an extra $\nabla \phi$ which is the same as $\phi \Delta \phi$ so I drew on these relationships to derive what I think would be the right form of the equation.''

Look at my energy binding equation and you can see the terms clearly match, which actually shouldn't be a surprise.

13. ### SimonsCatRegistered Member

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No need for my full binding energy equation anyway, it can be shortened to be worked with. In case you missed it, the energy equation was:

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

14. ### Q-reeusValued Senior Member

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Yes to question. To the rest - sigh.

15. ### SimonsCatRegistered Member

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Well its right, check the math.

$\phi \Delta \phi$ in the energy equation is the same thing as the squared component of

$\nabla \phi = k\ j_m$

square it and you will see.

$(\nabla \cdot \nabla) \phi = \Delta \phi$

and

$\phi (\nabla \cdot \nabla) \phi = \phi \Delta \phi$

Last edited: Jan 23, 2017
16. ### SimonsCatRegistered Member

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213

Looking at this post again, I hadn't noticed, but I squared my components by first distibuting $\nabla$ and then by distributing the second $\phi$. But this wasn't the only way to do this, clearly one can also do the following:

$\nabla^2 \phi^2 = \frac{16 \pi^2 G^2}{c^8}\ j^2_m$

Which is just as valid but I expect it may give a different form of the binding equation. I will look into what it gives later, unless someone else takes it to pen first. But I will do it later.

17. ### Q-reeusValued Senior Member

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sigh.sigh = sigh²

18. ### exchemistValued Senior Member

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5,862

I see no reason to be charitable on that score. Most of us extend readers the simple courtesy of checking what we write before posting it, to save them the arse-ache of continually being puzzled by errors and waiting for the poster to correct them.

You, however, seen unable even to attribute statements to the correct poster reliably. You also seem to be so mentally slack that you have several times completely inverted the sense of what people said, thus grossly misrepresenting their views.

If you go on like this, you are in no position to complain if people get cross with you.

Q-reeus likes this.
19. ### SimonsCatRegistered Member

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did you just come in here to attack me? Are you trying to troll me? Why are you bringing into a thread where we have had no confrontation, all the baggage from the other day?

Are you so fixated on me and bound by your own ego? grow up.

20. ### SimonsCatRegistered Member

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213

Here is another one who likes to troll.

Why didn't you reply to me the other day, did you realize you were wrong after all?

21. ### Q-reeusValued Senior Member

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Try and take it as wry humour maybe. Decided a while back there is no value in arguing over technicalities with you. So from now on I will just observe. OK?

22. ### SimonsCatRegistered Member

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It's kind of too late since I have reported both posts as off-topic and trolling.

23. ### SimonsCatRegistered Member

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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.