Einstein's Field Equations and Poisson's Equation

In summary, Einstein used Poisson's equation as a model to derive his Field Equations. However, he did not derive EFE from it.
  • #1
Mueiz
188
0
Einstein wrote in his book The Meaning of Relativity of 1921 p48 when deriving Field Equations :
" We must next attempt to find the laws of gravitational field .For this purpose ,Poisson's equation of the Newtonian theory must serve as a model.''
I have three question:
1\ How to derive Poisson's Equation from Newtonian theory?
2\How to use Poisson's Equation to derive Einstein's Field Equations?
3\ If Newtonian theory is not correct (as implied by GR) and poisson's Equation is derived from (or equavilant to) Newtonian theory ,then is it correct to use it to derive anything ?
 
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  • #2
Poisson's equations from Newton's law is dones here http://en.wikipedia.org/wiki/Electrostatics (Newton's law has the same form as Coulomb's law, so you can use the same logic.)

EFE is not derived from Poisson. It reduced to Poisson in the Newtonian limit. Einstein was sketching out how he guessed an equation that does reduce to Poisson in the Newtonian limit.
 
  • #3
Mueiz said:
1\ How to derive Poisson's Equation from Newtonian theory?

The procedure is identical to that used to derive gauss' law from coulomb's law: http://en.wikipedia.org/wiki/Gauss's_law#Deriving_Gauss.27s_law_from_Coulomb.27s_law

2\How to use Poisson's Equation to derive Einstein's Field Equations?
3\ If Newtonian theory is not correct (as implied by GR) and poisson's Equation is derived from (or equavilant to) Newtonian theory ,then is it correct to use it to derive anything ?

I haven't read the source you mention, but are you sure he was not discussing the weak field limit?
 
  • #4
atyy said:
Poisson's equations from Newton's law is dones here http://en.wikipedia.org/wiki/Electrostatics (Newton's law has the same form as Coulomb's law, so you can use the same logic.).
Poisson's equation is not the same as Gauss's law , in Poisson's equation there is the dinsity of matter (mass/volume) and in Gauss's law there is no the analogy (density of charge) there is the density in term of area rather than volume ..why did the wikipedia not make an independant page for poisson equation!
atyy said:
EFE is not derived from Poisson. It reduced to Poisson in the Newtonian limit. Einstein was sketching out how he guessed an equation that does reduce to Poisson in the Newtonian limit.

I did not say EFE is Derived from ... I did say Used to Derive
My question is about this ''sketching out "
 
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  • #5
Mueiz said:
Einstein wrote in his book The Meaning of Relativity of 1921 p48 when deriving Field Equations :
" We must next attempt to find the laws of gravitational field .For this purpose ,Poisson's equation of the Newtonian theory must serve as a model.''
I have three question:
1\ How to derive Poisson's Equation from Newtonian theory?
This page does a pretty good job:
http://en.wikipedia.org/wiki/Gauss'_law_for_gravity#General_case:_Mathematical_proof

Mueiz said:
2\How to use Poisson's Equation to derive Einstein's Field Equations?
3\ If Newtonian theory is not correct (as implied by GR) and poisson's Equation is derived from (or equavilant to) Newtonian theory ,then is it correct to use it to derive anything ?
He didn't use it to derive the EFE, he just used it as a model. I.e. as inspiration for his theory. Basically the line of similarity goes something like this:
1) the time-time component of the stress energy tensor is related to mass density
2) the time-time component of the metric is related to gravitational potential
3) in the non-relativistic limits only the time-time components will be significant
4) so we expect the relativistic law of gravitation to involve second derivatives of the metric and the stress-energy tensor
5) the curvature tensors involve second derivatives of the metric
6) therefore the EFE is a reasonable candidate for a relativistic theory of gravity
 
  • #6
Nabeshin said:
The procedure is identical to that used to derive gauss' law from coulomb's law: http://en.wikipedia.org/wiki/Gauss's_law#Deriving_Gauss.27s_law_from_Coulomb.27s_law

Poisson's equation is not analogy to Gauss's law , in Poisson's equation there is the dinsity of matter (mass/volume) and in Gauss's law there is no the analogy (density of charge)there is the density of flux in term of area rather than volume

Nabeshin said:
I haven't read the source you mention, but are you sure he was not discussing the weak field limit?

yes. I am sure
 
  • #7
Mueiz said:
Poisson's equation is not the same as Gauss's law , in Poisson's equation there is the dinsity of matter (mass/volume) and in Gauss's law there is no the analogy (density of charge) there is the density in term of area rather than volume ..why did the wikipedia not make an independant page for poisson equation!

There's a section on Poisson's equation right below the section on Gauss's law on http://en.wikipedia.org/wiki/Electrostatics .

Mueiz said:
I did not say EFE is Derived from ... I did say Used to Derive
My question is about this ''sketching out "

Poisson's equation contains the second derivative of the potential.

If by the principle of equivalence, we guess that the spacetime metric is the analogue of the gravitational potential, then we seek a second derivative of the spacetime metric - the Einstein tensor is such a second (covariant) derivative that preserves "energy conservation" in some form.
 
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  • #8
DaleSpam said:
He didn't use it to derive the EFE, he just used it as a model. I.e. as inspiration for his theory. Basically the line of similarity goes something like this:
1) the time-time component of the stress energy tensor is related to mass density
2) the time-time component of the metric is related to gravitational potential
3) in the non-relativistic limits only the time-time components will be significant
4) so we expect the relativistic law of gravitation to involve second derivatives of the metric and the stress-energy tensor
5) the curvature tensors involve second derivatives of the metric
6) therefore the EFE is a reasonable candidate for a relativistic theory of gravity
This is a good answer to my second and third question
but there is a difference between Poission Equation and EFE that seem to make the using of Poission Equation comfusing even as a model which is that in PE the density of matter means the ponderable matter (the mass of the source of the field /the volume of the sphere of radius from the center of the source to the point in cosideration)whilein EFE the density of matter means simply the density of matter in the point in which EFE is applied
 
  • #9
Mueiz said:
but there is a difference between Poission Equation and EFE that seem to make the using of Poission Equation comfusing even as a model
Yes. There are certainly differences and ample sources for confusion.

Mueiz said:
which is that in PE the density of matter means the ponderable matter ... whilein EFE the density of matter means simply the density of matter in the point in which EFE is applied
I don't know what you are saying here. "Ponderable matter" vs "matter"?
 
  • #10
Mueiz said:
This is a good answer to my second and third question
but there is a difference between Poission Equation and EFE that seem to make the using of Poission Equation comfusing even as a model which is that in PE the density of matter means the ponderable matter (the mass of the source of the field /the volume of the sphere of radius from the center of the source to the point in cosideration)whilein EFE the density of matter means simply the density of matter in the point in which EFE is applied

No. In Poisson's equation, the density is also a function of space.
 
  • #11
DaleSpam said:
I don't know what you are saying here. "Ponderable matter" vs "matter"?

I mean the way density is measured
In EFE density means local density .. the density of matter at a point one metre over the surface of the Earth is zero when we apply EFE
But in PE the density at a point one metre over the surface of the Earth means the mass of the Earth /the volume of the sphere covering the Earth one metre over.
 
  • #12
atyy said:
No. In Poisson's equation, the density is also a function of space.

If this were true then gravitational field outside any source would equal zero however close the position is
This is not true
the meaning of density in EFE if totally different from that of PE
 
  • #13
Mueiz said:
I mean the way density is measured
In EFE density means local density .. the density of matter at a point one metre over the surface of the Earth is zero when we apply EFE
But in PE the density at a point one metre over the surface of the Earth means the mass of the Earth /the volume of the sphere covering the Earth one metre over.
No. In PE the density is also local. The density at a point one meter over the Earth is zero in both the EFE and PE.
 
  • #14
DaleSpam said:
No. In PE the density is also local. The density at a point one meter over the Earth is zero in both the EFE and PE.
No.. now according to what you said poission equation must be useless because density is zero everywhere outside the source of gravitational field .. how can we use it as a law of gravitational field then.
(this problem does not exist in EFE because if stress-energy tensor equal zero there still a possibility for different point outside the source to have different geometrical properties)
 
  • #15
Mueiz said:
No.. now according to what you said poission equation must be useless because density is zero everywhere outside the source of gravitational field
Yes, the density is zero outside the Earth therefore the Laplacian (a second derivative) is zero. Just because a function's second derivative is 0 doesn't mean the function is 0.

Are you familiar with differential equations in general and the Laplacian in particular?
 
  • #16
DaleSpam said:
Yes, the density is zero outside the Earth therefore the Laplacian (a second derivative) is zero. Just because a function's second derivative is 0 doesn't mean the function is 0.

Are you familiar with differential equations in general and the Laplacian in particular?

Ok i will prove this before using math
According to your understanding of Poisson's equation one can change the second derevative of gravitational field in a point just by increasing the density of matter in the point (even if you put your hand in a point in empty space you will change its density from zero to more than 1000kg/m^3 ! can a Newtonian theory of gravitation talk about such great effect of a hand toch in gravitational field.
See the link which you posted to me to know that the density of matter in Poission Equation means the density of all the region containing the source and the point.
 
  • #17
Please read this very carefully. Note that a point charge corresponds to a delta function, which appears in Eq (223). Eq (223) is essentially Poisson's equation for the special case of a point charge, which hopefully makes it clear why, as he says at the end, "the Green's function has the same form as the potential generated by a point charge."

http://farside.ph.utexas.edu/teaching/em/lectures/node31.html
 
  • #18
Mueiz said:
See the link which you posted to me to know that the density of matter in Poission Equation means the density of all the region containing the source and the point.
No, this is a complete misunderstanding on your part.

I get the distinct impression that you are not very familiar with differential equations and that you have probably never actually worked any problem involving a Laplacian. Not knowing something is fine, this site is primarily educational in purpose. But when you are aware that you don't know something then why would you insist on being so pointlessly argumentative? You are not going to learn much that way. A much more productive path would be to ask questions about the parts that you don't understand.

Mueiz said:
According to your understanding of Poisson's equation one can change the second derevative of gravitational field in a point just by increasing the density of matter in the point
Yes.

Mueiz said:
(even if you put your hand in a point in empty space you will change its density from zero to more than 1000kg/m^3 ! can a Newtonian theory of gravitation talk about such great effect of a hand toch in gravitational field.
What makes you think this is a "such great effect"? Have you actually done the math to find out exactly how big the effect is?
 
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  • #19
DaleSpam said:
What makes you think this is a "such great effect"? Have you actually done the math to find out exactly how big the effect is?

derivative of GF=4 * pi *G *1000
but this derivative is not constant outside the source according to Newtonian theory
while it is constant according to your (incorrect) understanding of Poisson's Equation
(([edit] Deriving Gauss' law from Newton's lawGauss' law for gravity can be derived from Newton's law of universal gravitation, which states that the gravitational field due to a point mass is:


where

er is the radial unit vector,
r is the radius, |r|.
M is the mass of the particle, which is assumed to be a point mass located at the origin.
In this section, two alternative proofs of this fact are presented. The first proof is more visual and intuitive, while the second proof is more mathematical.))

This is a copy-paste from the pages of wikipeadia which you posted to me
you can see clearly that in derivation of the equation the mass is assumed to be a point mass located at the origin , then what does it mean to calculate density in any point according to your understanding to the meaning of density here
 
  • #20
DaleSpam said:
No, this is a complete misunderstanding on your part.

I get the distinct impression that you are not very familiar with differential equations and that you have probably never actually worked any problem involving a Laplacian. Not knowing something is fine, this site is primarily educational in purpose. But when you are aware that you don't know something then why would you insist on being so pointlessly argumentative? You are not going to learn much that way. A much more productive path would be to ask questions about the parts that you don't understand

"There must be no barriers for freedom of inquiry. There is no place for dogma in science. The scientist is free, and must be free to ask any question, to doubt any asssertion, to seek for any evidence, to correct any errors.” Robert Oppenheimer
This is the best way to learn physics ..To ask and discuss the answer even if you are wrong this will help your teacher to correct your erorr and misunderstanding.How can you make sure that your student understand your answer if prevent him from discussion.
In this thread I asked a question and start to discuss the answers according to what i know
and I did not give any answer to my own question .
 
  • #21
Mueiz said:
"There must be no barriers for freedom of inquiry...To ask and discuss the answer even if you are wrong this will help your teacher to correct your erorr and misunderstanding.
So then ask questions instead of making wrong assertions.
 
  • #22
Mueiz said:
derivative of GF=4 * pi *G *1000
but this derivative is not constant outside the source according to Newtonian theory
while it is constant according to your (incorrect) understanding of Poisson's Equation
It is, in fact, constant and equal to zero outside the source.

Before you proceed further why don't you actually work a bit of math? You know the gravitational potential outside of a spherically symmetric mass from Newton's law. Why don't you calculate the Laplacian of that function in the region outside the mass and see what you get? Check if it is equal to zero as I assert or equal to M as you assert.

Mueiz said:
M is the mass of the particle, which is assumed to be a point mass located at the origin.
And does M show up in the final derived result? If not, then what is the function that represents the density of a point particle? Does that show up in the derivation? Is that function zero outside of the point particle?
 
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  • #23
DaleSpam said:
And does M show up in the final derived result? If not, then what is the function that represents the density of a point particle? Does that show up in the derivation? Is that function zero outside of the point particle?

This is the argument which i used in my post #14 aganist your understanding of the meaning of density in Poisson Equation..if M is a point mass located at the orign (as said in wikipeadia page which you posted to me ) then according to you the density everywhere is zero and poisson Equation is unable to calculate any feature of gravitational field anywhere
But according to what I understand from the density there is no such a problem , because when we apply Poisson Equation in a point outside the source ,M is represented in the density because density means M/V where V is the volume of sphere whose radius is the distance from the source to the point.
 
  • #24
DaleSpam said:
It is, in fact, constant and equal to zero outside the source.

Before you proceed further why don't you actually work a bit of math? You know the gravitational potential outside of a spherically symmetric mass from Newton's law. Why don't you calculate the Laplacian of that function in the region outside the mass and see what you get? Check if it is equal to zero as I assert or equal to M as you assert.

I did not say it equal M what i said is that it is not zero , it equals =4*pi*G M/V
I said in my post #11 the meaning of density is different in Field Equations and Poission Equation
In EFE it means local density and it equal zero outside the source
In PE it means the density of pounderable matter an does not equal zero outside the source but equal 4*pi*G M/V
 
  • #25
Mueiz said:
if M is a point mass located at the orign (as said in wikipeadia page which you posted to me ) then according to you the density everywhere is zero
No, according to me the density is a delta function. It is zero everywhere except the origin. Are you familiar with the delta function? It is used explicitly in the derivation. If you don't understand something you should ask instead of making wrong assertions.

Please do the exercise I suggested above. Start with the known gravitational potential outside of a point mass and then calculate the Laplacian and see what you get.
 
  • #26
Mueiz said:
I did not say it equal M what i said is that it is not zero , it equals =4*pi*G M/V
Fine, do the exercise I suggested and see if in the region outside the mass you get 0 as I assert or 4 pi G M/V as you assert. Simply start with the known gravitational potential and calculate the Laplacian.
 
  • #27
DaleSpam said:
Fine, do the exercise I suggested and see if in the region outside the mass you get 0 as I assert or 4 pi G M/V as you assert. Simply start with the known gravitational potential and calculate the Laplacian.

you do the exersice ! to find that what you said is wrong
Why start with gravitational potential if there is the simple Equation of Poisson
If the right hand side is zero in an equation the left hand must be zero , see any book in math for biginners:biggrin:
 
  • #28
Mueiz said:
you do the exersice !
Sure, I did it twice this morning with Mathematica in two different coordinate systems:

Code:
In[1]:= << VectorAnalysis`

In[2]:= SetCoordinates[Spherical[r, \[Theta], \[Phi]]];

In[3]:= Laplacian[-((G M)/r)]

Out[3]= 0

In[4]:= SetCoordinates[Cartesian[x, y, z]];

In[5]:= Laplacian[- ((G M)/Sqrt[x^2 + y^2 + z^2])]

Out[5]= -((3 G M x^2)/(x^2 + y^2 + z^2)^(5/2)) - (
 3 G M y^2)/(x^2 + y^2 + z^2)^(5/2) - (3 G M z^2)/(x^2 + y^2 + z^2)^(
 5/2) + (3 G M)/(x^2 + y^2 + z^2)^(3/2)

In[6]:= Simplify[%]

Out[6]= 0

See Out[3] and Out[6] in spherical and Cartesian coordinates respectively. You really need to learn some basic differential equations.
 
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  • #29
Mueiz said:
you do the exersice ! to find that what you said is wrong
Why start with gravitational potential if there is the simple Equation of Poisson
If the right hand side is zero in an equation the left hand must be zero , see any book in math for biginners:biggrin:

While DaleSpam shows it via simple Mathematica entries, the Laplacian on 1/r is a trivial problem. Most texts will work out the homogeneous solution to the Laplacian in spherical coordinates and it is easy to see that the first mode is A_0 + B_0*r^{-1}.

I would also draw attention back to atyy's post above. If you must balk at the derivation from electrodynamics using the electrostatic Poisson equation, then you can look at atyy's link for a well developed derivation of the solution to the Poisson equation using a Green's function. This method, despite once again applied to electrostatics, is independent of the physics behind the values of \rho and \epsilon. They can be what you like them to be.
 
  • #30
DaleSpam said:
Sure, I did it twice this morning with Mathematica in two different coordinate systems:

Code:
In[1]:= << VectorAnalysis`

In[2]:= SetCoordinates[Spherical[r, \[Theta], \[Phi]]];

In[3]:= Laplacian[-((G M)/r)]

Out[3]= 0

In[4]:= SetCoordinates[Cartesian[x, y, z]];

In[5]:= Laplacian[- ((G M)/Sqrt[x^2 + y^2 + z^2])]

Out[5]= -((3 G M x^2)/(x^2 + y^2 + z^2)^(5/2)) - (
 3 G M y^2)/(x^2 + y^2 + z^2)^(5/2) - (3 G M z^2)/(x^2 + y^2 + z^2)^(
 5/2) + (3 G M)/(x^2 + y^2 + z^2)^(3/2)

In[6]:= Simplify[%]

Out[6]= 0

See Out[3] and Out[6] in spherical and Cartesian coordinates respectively. You really need to learn some basic differential equations.

Firstly in(3) is incorrect it should be GM/Sqrt r rather than GM/r
Secondly what is the result of out(5)? you did not write it !
But no problem that is easy suppose we want to apply it in the pont (1,0,0) which is outside the source;
Then we have out(5)=-3GM-3GM-3GM+3GM=-6GM not equal 0
-6GM equal 4 *pi *G *Density as I said
4* pi *G *density=-4 pi G M/(4/3 pi *1)=-6GM
I will not say you need to learn some basic DE but I think you just did a mistake.
 
  • #31
Mueiz said:
Firstly in(3) is incorrect it should be GM/Sqrt r rather than GM/r
GM/Sqrt r wouldn't even be the right units!
 
  • #32
Now the difference is clear between density in EFE which is local(so equal zero outside the source) and density in PE which is related to the region which include the point and the source(equal M/v outside the source)
I want to remind you my question; Is it not true that using poisson equation as a model to derive field equations is confusing ?
 
  • #33
Mueiz said:
But no problem that is easy suppose we want to apply it in the pont (1,0,0) which is outside the source;
Then we have out(5)=-3GM-3GM-3GM+3GM=-6GM not equal 0
No, you would get -3GM-0-0+3GM=0.
 
  • #34
lugita15 said:
GM/Sqrt r wouldn't even be the right units!

We want to calculate the lablacian of what?
of the gravitational field
How gravitational field is measured ?
GM/Sqrt r
 
  • #35
Mueiz said:
We want to calculate the lablacian of what?
of the gravitational field
How gravitational field is measured ?
GM/Sqrt r
Gravitational field is inversely proportional to r^2. Gravitational potential is inversely proportional to r.
 

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