# Thread: Einstein's constant

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## Einstein's constant

I have a \$100 question related to Einstein's constant:
https://en.wikipedia.org/wiki/Einstein's_constant

I can see linear algebra and the Poisson's equation but it's not clear to know what are the real input parameters to this calculation?

Is the Poisson's equation used to calculate the mass density distribution of the solar system?

Thanks a lot,
philippeb8

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The "real input parameters" are in theCalculation section
In the following, the value of Einstein's constant will be calculated. To do so, at the beginning a field equation where the cosmological constant Λ is equal to zero is taken, with a steady state hypothesis. Then we use the Newtonian approximation with hypothesis of a weak field and low velocities with respect to the speed of light.
I would think that Poisson's equation is not used for solar systems. The solar system is generally treated as the Sun + planets, i.e. point sources.

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Originally Posted by Reality Check
The "real input parameters" are in theCalculation section
I meant: what are the calculations based on?

I would think that Poisson's equation is not used for solar systems. The solar system is generally treated as the Sun + planets, i.e. point sources.
In terms of "mass density", I assume the Newtonian gravitational potential could be used to calculate the "mass density"? This constant is not very well explained anywhere.

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Originally Posted by philippeb8
In terms of "mass density", I assume the Newtonian gravitational potential could be used to calculate the "mass density"? This constant is not very well explained anywhere.
But we agree:
G = m^3*kg^-1*s^-2
c = m/s

Thus:
G/c^2 = m/kg

Is a gravitational potential measurement. I just don't understand what it's based on.

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Originally Posted by philippeb8
I have a \$100 question related to Einstein's constant:
https://en.wikipedia.org/wiki/Einstein's_constant

I can see linear algebra and the Poisson's equation but it's not clear to know what are the real input parameters to this calculation?
What do you mean "the real input parameters"? It is derived from theory, it is independent of any measurements. As Reality Check says the only 'input' is the assumption that the cosmological constant is approximately zero.

Is a gravitational potential measurement. I just don't understand what it's based on.
It is a constant of proportionality. Gravitational potential has units of J/kg, not m/kg
Last edited by Shaula; 2017-Mar-13 at 06:26 AM.

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Originally Posted by philippeb8
But we agree:
G = m^3*kg^-1*s^-2
c = m/s

Thus:
G/c^2 = m/kg

Is a gravitational potential measurement. I just don't understand what it's based on.
That's not quite right. The wiki article points out that G/c^4 is also an invariant(therefore valid) choice for kappa. Which gives meters/Joule

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Originally Posted by Shaula
What do you mean "the real input parameters"? It is derived from theory, it is independent of any measurements. As Reality Check says the only 'input' is the assumption that the cosmological constant is approximately zero.
Ok thanks. So it's a constant found with some calculation process that isn't based on anything and if we were to change the cosmological constant then kappa will have to change as well; which I haven't seen changing when Einstein decided to add the cosmological constant back into his equations.

It is a constant of proportionality. Gravitational potential has units of J/kg, not m/kg
Sorry I meant: the gravitational potential / - the gravitational constant

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Originally Posted by ShinAce
That's not quite right. The wiki article points out that G/c^4 is also an invariant(therefore valid) choice for kappa. Which gives meters/Joule
Yeah that's confusing when it depends on which metric that is being used. But let's keep it simple for now and use G/c^2.

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Originally Posted by philippeb8
Ok thanks. So it's a constant found with some calculation process that isn't based on anything and if we were to change the cosmological constant then kappa will have to change as well; which I haven't seen changing when Einstein decided to add the cosmological constant back into his equations.
No on all counts. First off it is based on something. It is based on theory. Second the value of the cosmological constant changing won't change kappa. That assumption was made to simplify the calculation such that kappa could easily be derived. Kappa is then true for general cases where lambda is not zero.

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Originally Posted by Shaula
No on all counts. First off it is based on something. It is based on theory. Second the value of the cosmological constant changing won't change kappa. That assumption was made to simplify the calculation such that kappa could easily be derived. Kappa is then true for general cases where lambda is not zero.
If I were to write simple code in C, I would have something like this:

Code:
```float const pi = 3.1416;

float theory(float G, float c, float lambda)
{
return G/c^2 * 8 * pi;
}

float kappa(float G, float c, float lambda)
{
return theory(G, c, lambda);
}```
So what I mean here is that the function theory() is not based on any external variable other than G, c and lambda.

I believe my original question was answered but is it possible to have the derived value of kappa without the simplifications? I never liked "simplifications" because it's a source of error.

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Originally Posted by philippeb8
I believe my original question was answered but is it possible to have the derived value of kappa without the simplifications? I never liked "simplifications" because it's a source of error.
When I say simplification I mean selecting a situation that makes the calculation tractable. Say you wanted to work out the strength of gravity at the Earth's surface. You could work it out in an experiment on the Earth's equator (making the calculation as simple as possible). Or you could do it on a rollercoaster, in the dark, at high latitudes while giant lumps of leads are spun around you. Which would you pick?!

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Originally Posted by Shaula
When I say simplification I mean selecting a situation that makes the calculation tractable. Say you wanted to work out the strength of gravity at the Earth's surface. You could work it out in an experiment on the Earth's equator (making the calculation as simple as possible). Or you could do it on a rollercoaster, in the dark, at high latitudes while giant lumps of leads are spun around you. Which would you pick?!
Yes but let's not forget GR was written at a time they only knew the dimension of the Universe to be the dimension of the Milky Way (1915) so they could have simplified a lot of stuff. When applied to nowadays Universe we need to adjust the cosmological constant because it doesn't work. So perhaps there is a mistake in kappa at these scales, that's all.

I don't think I can ask for the full mathematics of the non-simplified kappa here but if I could have any pointers that would be great, thanks.

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Originally Posted by philippeb8
I don't think I can ask for the full mathematics of the non-simplified kappa here but if I could have any pointers that would be great, thanks.
I'll try to ask the same question to: http://physics.stackexchange.com because the question goes into excessive details for this forum.

But thanks again, I really appreciate.

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Originally Posted by philippeb8
I meant: what are the calculations based on?
What calculations are you referring to?
You cited a Wikipedia article Einstein's constant that does explain the constant. The article simplifies the derivation by setting the cosmological constant to zero.

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Originally Posted by Reality Check
What calculations are you referring to?
You cited a Wikipedia article Einstein's constant that does explain the constant. The article simplifies the derivation by setting the cosmological constant to zero.
I'm looking for the version of the derivation where the cosmological constant is not null.

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Originally Posted by philippeb8
I'll try to ask the same question to: http://physics.stackexchange.com because the question goes into excessive details for this forum.
For the record, the follow up for this question will be here:

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Originally Posted by philippeb8
For the record, the follow up for this question will be here:
... Kappa does not depend on the Cosmological Constant. You have misrepresented the way it is derived.

18. Originally Posted by philippeb8
Yes but let's not forget GR was written at a time they only knew the dimension of the Universe to be the dimension of the Milky Way (1915) so they could have simplified a lot of stuff. When applied to nowadays Universe we need to adjust the cosmological constant because it doesn't work. So perhaps there is a mistake in kappa at these scales, that's all.
Wouldn't that mean a more general error/change in GR? (Such as those proposed for MOND-like theories).

I don't think I can ask for the full mathematics of the non-simplified kappa here but if I could have any pointers that would be great, thanks.
I am curious: can you follow the derivation on the wikipedia page? (I certainly can't - not even in general terms)

Presumably, the version including non-zero lambda is significantly more complex.

(I also thought that someone who could follow that derivation would then be able to work out how to do it with non-zero lambda. But that may be unreasonable.)

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Originally Posted by Shaula
... Kappa does not depend on the Cosmological Constant. You have misrepresented the way it is derived.
I apologize if I misinterpreted it but this is the first thing that is written here:
https://en.wikipedia.org/wiki/Einste...nt#Calculation

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Originally Posted by Strange
Wouldn't that mean a more general error/change in GR? (Such as those proposed for MOND-like theories).
Yes but that would be called: MOED (Modified Einsteinian Dynamics) but applied to dark energy instead of dark matter. I'm surprised I'm the first one questioning this constant after 100 years.

I am curious: can you follow the derivation on the wikipedia page? (I certainly can't - not even in general terms)
Vaguely but now I understand it is not based on any measurements but only on theory, which is also very vague.

Presumably, the version including non-zero lambda is significantly more complex.

(I also thought that someone who could follow that derivation would then be able to work out how to do it with non-zero lambda. But that may be unreasonable.)

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Originally Posted by philippeb8
I apologize if I misinterpreted it but this is the first thing that is written here:
https://en.wikipedia.org/wiki/Einste...nt#Calculation
We can also see that his assumptions originally were based on the steady state hypothesis which is today completely obsolete.

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Originally Posted by philippeb8
I apologize if I misinterpreted it but this is the first thing that is written here:
https://en.wikipedia.org/wiki/Einste...nt#Calculation
No, that is not what is written there. What is written there is:
In the following, the value of Einstein's constant will be calculated. To do so, at the beginning a field equation where the cosmological constant Λ is equal to zero is taken, with a steady state hypothesis. Then we use the Newtonian approximation with hypothesis of a weak field and low velocities with respect to the speed of light.
That does not mean kappa is dependent on the cosmological constant, merely that the calculations used to derive it in this specific case use a field equation where lambda is zero (which allows them to collapse to the Newtonian equations). Solving it for other cases should lead to the same value for kappa.

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Originally Posted by philippeb8
We can also see that his assumptions originally were based on the steady state hypothesis which is today completely obsolete.
Again. Only for that calculation. Kappa is a general feature and its value is not dependent on the steady state hypothesis.

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Originally Posted by Shaula
That does not mean kappa is dependent on the cosmological constant, merely that the calculations used to derive it in this specific case use a field equation where lambda is zero (which allows them to collapse to the Newtonian equations). Solving it for other cases should lead to the same value for kappa.
Ok it looked like Einstein was solving kappa based on the "assumption" the cosmological constant was null but if you say that kappa will always be the same regardless of the cosmological constant then I guess it is what it is.

I'll notify this thread if Quora is able to answer my question mathematically. It's been a day now and they still haven't answered; I hope my question was clear enough.

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Originally Posted by philippeb8
I'll notify this thread if Quora is able to answer my question mathematically. It's been a day now and they still haven't answered; I hope my question was clear enough.
So the follow ups are:

Let's see who answers first mathematically... Thanks again CQ, I appreciate!

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Originally Posted by philippeb8
So the follow ups are:
Kappa (Einstein's constant) itself depends on G and c, not the cosmological constant.
The derivation in that Wikipedia article (maybe not of Einstein) of kappa sets the cosmological constant to zero to allow comparison to Newton's law in the appropriate limits. That article cites textbooks from 1975 and 1992 . We now have good evidence for dark energy and so a non-zero value of the cosmological constant. However that does not change the definition of kappa. We can still write the Einstein field equations by replacing the multiplier of the stress–energy tensor with kappa.

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Originally Posted by Reality Check
Kappa (Einstein's constant) itself depends on G and c, not the cosmological constant.
The derivation in that Wikipedia article (maybe not of Einstein) of kappa sets the cosmological constant to zero to allow comparison to Newton's law in the appropriate limits.
Ok thanks.

That article cites textbooks from 1975 and 1992 . We now have good evidence for dark energy and so a non-zero value of the cosmological constant. However that does not change the definition of kappa. We can still write the Einstein field equations by replacing the multiplier of the stress–energy tensor with kappa.
But if we were to refactor the theory behind kappa, which apparently I'm the first one in a 100 years to suggest, then we wouldn't need dark energy.

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Dark energy is an observation. Kappa is a definition, maybe defined 100 years ago.
Alternately forget about the article calculation - look at the Einstein field equations and replace the multiplier of the stress–energy tensor with the symbol kappa. Does that make the cosmological constant vanish?
Last edited by Reality Check; 2017-Mar-15 at 02:44 AM.

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Originally Posted by Reality Check
Dark energy is an observation. Kappa is a definition, maybe defined 100 years ago.
I understand that. What I am saying is if we were to replace the constant "kappa" with a function "foo(scale)" then the mathematics will match the observations once and for all. I'm just saying kappa is a definition based on some 3rd party theory so it might as well be refactored. I'll refer to this as: MOED.

Alternately forget about the article calculation - look at the Einstein field equations and replace the multiplier of the stress–energy tensor with the symbol kappa. Does that make the cosmological constant vanish?
The question seems to turn into a debate and this is just a QA forum so I'll keep it short but I am able to draw my own conclusions now, thanks.

30. Originally Posted by philippeb8
I understand that. What I am saying is if we were to replace the constant "kappa" with a function "foo(scale)" then the mathematics will match the observations once and for all.
Can you show that, mathematically? Or is it just a guess?

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