Finding Equations of Motion from the Stress Energy Tensor

[samalkhaiat]

So, ...
In GR, is divT=0 sufficient to find the FULL trajectory of particles moving in curved space-time?

Thanks.

When Einstein first presented his GR, he postulated the geodesic as the equation of motion. In 1927, he noted that the equations of motion are in fact contained with in the field equations and therefore do not need to be postulated separately. In 1951, Papapetrou managed to show that the equations of motion of spinning particles (which are not geodesics) can also be derived from Einstein’s field equations.
There have been two different approaches to the problem of obtaining the equations of motion in GR. The first (due to Einstein and Infeld) starts with left hand side of the field equations and is a very complicated geometrical treatment. The other approach (the one we are interested in) starts with right hand side of the field equations and is due to Papapetrou and Fock.
OK, let us begin. The field equations $G^{ab} = T^{ab}$ imply
$$\nabla_{b}T^{ab} = 0.$$
This can be written as
$$\partial_{b}\left( \sqrt{-g} T^{ab}\right) + \sqrt{-g}\Gamma^{a}{}_{bc}T^{bc} = 0. \ \ (1)$$
We integrate this over a small 3-volume and write
$$\int d^{3}x \partial_{b}\left( \sqrt{-g} \ T^{ab}\right) + \int d^{3}x \sqrt{-g}\Gamma^{a}{}_{bc}T^{bc} = 0, \ \ (2)$$
or
$$\frac{d}{dx^{0}}\int d^{3}x \sqrt{-g} \ T^{a0} + \int d^{3}x \sqrt{-g}\Gamma^{a}{}_{bc}T^{bc} = - F^{a}, \ \ (3)$$
where the object $F^{a}$ is defined by
$$F^{a} = \int d^{3}x \partial_{j}(\sqrt{-g} \ T^{aj}).$$
We will identify $F^{a}$ with external non-gravitational forces. So, for the case of gravitational forces alone, one sets $F^{a} = 0$; one writes $F^{a}$ as surface integral and take a large enough volume for which $T^{ab}$ vanishes on the surface.
Let us now expand the connection about a point $X^{a}$ which will be taken as the coordinates of the “particle”; a function of time or other single parameter,
$$x^{a} = X^{a} + \delta x^{a}, \ \ (4)$$
$$\Gamma^{a}{}_{bc}(x) = \Gamma^{a}{}_{bc}(X) + \delta x^{d}\partial_{d}\Gamma^{a}{}_{bc} + … \ \ (5)$$
Let us spell out what we mean by particle. We define a single pole (non-spining) particle by the followings;
i) For some a and b
$$\int d^{3}x \sqrt{-g} \ T^{ab} \neq 0. \ \ \ (6)$$
In particular; when evaluated in a locally inertial rest frame, the integral
$$\int d^{3}\bar{x} \sqrt{-\bar{g}} \ \bar{T}^{00} = m \ > \ 0, \ \ \ (6a)$$
represents the rest mass of our particle.

ii) For all a, b and c
$$\int d^{3}x \sqrt{-g} \ \delta x^{c} \ T^{ab} = 0. \ \ \ (7)$$
iii) The particle is sufficiently small so that the forces acting on it are large compared to the first moment of these forces on its surface,
$$\int d^{3}x \partial_{j}\left(\delta x^{c}\sqrt{-g} \ T^{aj}\right) = \int dS_{j} \delta x^{c}\sqrt{-g} \ T^{aj} \approx 0. \ \ \ (8)$$

For later convenience, let us write eq(2) and eq(3) taking into account eq(5) and eq(7)
$$\int d^{3}x \partial_{b}\left( \sqrt{-g} \ T^{ab}\right) + \Gamma^{a}{}_{bd}(X)\int d^{3}x \sqrt{-g} \ T^{bd} = 0, \ \ (9)$$
$$\frac{d}{dx^{0}}\int d^{3}x \sqrt{-g} \ T^{a0} + \Gamma^{a}{}_{bd}(X) \int d^{3}x \sqrt{-g} \ T^{bd} = - F^{a}. \ \ (10)$$
Now multiply eq(1) by $x^{c}$, integrate over 3-volume and write the result as
$$\int d^{3}x \{ \partial_{0}(\sqrt{-g} \ x^{c}T^{a0}) + \partial_{j}(\sqrt{-g} \ x^{c}T^{aj}) + \sqrt{-g} \ x^{c}\Gamma^{a}{}_{bd}T^{bd}\} = \int d^{3}x \sqrt{-g} \ T^{ac}.$$
Substituting eq(4) and eq(5) into the above equation gives, with the help of eq(7) and eq(8),
$$\frac{dX^{c}}{dx^{0}}\int d^{3}x \sqrt{-g} \ T^{a0} + X^{c}\int d^{3}x \partial_{b}\left( \sqrt{-g} \ T^{ab}\right) + X^{c}\Gamma^{a}{}_{bd}\int d^{3}x \sqrt{-g} \ T^{bd}= \int d^{3}x \sqrt{-g} \ T^{ac}.$$
Making use of eq(9), the above equation reduces to the following nice looking equation
$$\frac{dX^{c}}{dx^{0}}\int d^{3}x \sqrt{-g} \ T^{a0} = \int d^{3}x \sqrt{-g} \ T^{ac}. \ \ (11)$$
For $a = 0$, and since $T^{ab}=T^{ba}$, we find
$$\int d^{3}x\sqrt{-g} \ T^{a0} = \frac{dX^{a}}{dx^{0}}\int d^{3}x \sqrt{-g} \ T^{00}. \ \ (12)$$
Inserting eq(12) back into eq(11), we find
$$\int d^{3}x \sqrt{-g} \ T^{ac}= \frac{dX^{a}}{dx^{0}}\frac{dX^{c}}{dx^{0}}\int d^{3}x \sqrt{-g} \ T^{00}. \ \ (13)$$
Substituting eq(12) and eq(13) back into eq(10), we obtain
$$\frac{d}{dx^{0}}\left( \frac{dX^{a}}{dx^{0}}\int d^{3}x \sqrt{-g} \ T^{00}\right) + \Gamma^{a}{}_{bc}\frac{dX^{b}}{dx^{0}}\frac{dX^{c}}{dx^{0}}\int d^{3}x \sqrt{-g} \ T^{00} = - F^{a}. \ \ (14)$$
Using
$$\frac{d}{dx^{0}} = \frac{ds}{dx^{0}}\frac{d}{ds},$$
we rewrite eq(14) as
$$\frac{d}{ds}\left( \frac{dX^{a}}{ds}\frac{ds}{dx^{0}}\int d^{3}x \sqrt{-g} \ T^{00}\right) + \Gamma^{a}{}_{bc}(X)\frac{dX^{a}}{ds}\frac{dX^{c}}{ds}\frac{ds}{dx^{0}}\int d^{3}x \sqrt{-g} \ T^{00} = - F^{a}\frac{dx^{0}}{ds}. \ \ (15)$$
In this equation we recognize the rest mass of our particle,
$$m = \frac{ds}{dx^{0}}\int d^{3}x \sqrt{-g} \ T^{00} = \int d^{3}\bar{x} \sqrt{-\bar{g}} \ \bar{T}^{00},$$
Thus, eq(15) becomes
$$\frac{d}{ds}\left( m \frac{dX^{a}}{ds} \right) + m \Gamma^{a}{}_{bc}(X) \frac{dX^{a}}{ds}\frac{dX^{c}}{ds} = - F^{a}\frac{dx^{0}}{ds}. \ \ (16)$$
Finally, assuming that the rest mass does not change with time, we find the equation of motion a pole (non-spining) particle subjected to gravitational and non-gravitational forces,
$$\frac{d^{2}X^{a}}{d s^{2}} + \Gamma^{a}{}_{bc}\frac{dX^{b}}{ds}\frac{dX^{c}}{ds} = \frac{f^{a}}{mc^{2}}$$
where
$$f^{a} = - c^{2}\frac{dx^{0}}{ds}F^{a}$$
This method can be extended by relaxing the assumption (ii). Doing this leads to equations of motion for spinning particles.

Regards
Sam

 

[Matterwave]

Why did you include the sqrt(-g) terms in the definition of the covariant derivative? T is a regular tensor right, so why are those terms there?

 

[pervect]

If you have Wald, try looking up "Tensor Densities" and see if it answers your question.

 

[George Jones]

Why did you include the sqrt(-g) terms in the definition of the covariant derivative? T is a regular tensor right, so why are those terms there?

See section 1.7 from Eric Poisson's notes,

http://www.physics.uoguelph.ca/poisson/research/agr.pdf,

which evolved into the excellent book, A Relativist's Toolkit: The Mathematics of black hole Mechanics.

 

[Matterwave]

See section 1.7 from Eric Poisson's notes,

http://www.physics.uoguelph.ca/poisson/research/agr.pdf,

which evolved into the excellent book, A Relativist's Toolkit: The Mathematics of black hole Mechanics.

Ok, but he specifically notes in equation 1.7.4 that the equation is for an anti-symmetric tensor, whereas T is symmetric. Is this important?

Also this equation has no Christoffel symbol terms now.

But I guess I can see where that equation would come from. Thanks.

 

[George Jones]

Also this equation has no Christoffel symbol terms now.

I am not sure what you mean.

$$\nabla_{b}T^{ab} = 0.$$
This can be written as
$$\partial_{b}\left( \sqrt{-g} T^{ab}\right) + \sqrt{-g}\Gamma^{a}{}_{bc}T^{bc} = 0. \ \ (1)$$

$$\begin{align} 0 &= \nabla_{b}T^{ab} \\ &= \partial_{b}T^{ab} + \Gamma^a_{db} T^{db} + \Gamma^b_{db} T^{ad} \\ &= \sqrt{-g} \left( \partial_{b}T^{ab} + \Gamma^a_{db} T^{db} + \Gamma^b_{db} T^{ad} \right) \\ &= \partial_{b} \left( \sqrt{-g} T^{ab} \right) - \partial_b \left( \sqrt{-g} \right) T^{ab} + \sqrt{-g} \left( \Gamma^a_{db} T^{db} + \Gamma^b_{db} T^{ad} \right) \\ &= \partial_{b} \left( \sqrt{-g} T^{ab} \right) - \sqrt{-g} \Gamma^c_{cb} T^{ab} + \sqrt{-g} \left( \Gamma^a_{db} T^{db} + \Gamma^b_{db} T^{ad} \right) \\ \end{align}$$
In the last line: equation 1.7.2 from Poisson has been used; the second and fourth terms cancel.

 

[juanrga]

No, not by definition.
The Penrose–Hawking singularity theorems are not definitions.

You pretend to answer something totally unrelated to what I have said. I never said that a theorem was a definition, evidently...

 

[juanrga]

I am using rather standard terminology: «Event Horizon» and «Apparent Horizon» (http://relativity.livingreviews.org/Articles/lrr-2007-3/ ). Call it X and Y if you prefer, that is unimportant. The point is when you mix X and Y.

BHs have an event horizon associated to a spacetime singularity (by definition) and the spacetime that you cited (which is not even Minkowski equivalent) has only a coordinate singularity and, thus, no event horizon.

Of course, I am not saying that «GR is ultimately wrong» (your own words), but something different. I am saying that my bet is that Nature does not like spacetime singularities and event horizons (both unrelated to gravitational lensing) and that GR is better viewed as the limiting case of an underlying theory without singularities.

Of course you can put your own money where you want. I know that mine is safe .

Ignoring terminology, that reference discussion agrees with my understanding; in particular, in a closed spacetime, all horizons are apparent horizons (there is no null infinity; there is no total causal disconnection between the singular region and any other region of spacetime). The problematic feature is the singularity, not the horizon.

As stated spacetime singularities and event horizons are different beasts than coordinate singularities and apparent horizons, although you want to mix them.

I will just finish by emphasizing, again, that a Rindler spacetime in SR has not a spacetime singularity nor an event horizon and, therefore, has nothing to see with the predictions of FTG about spacetime singularities and event horizons using gravitons.

 

[juanrga]

When Einstein first presented his GR, he postulated the geodesic as the equation of motion. In 1927, he noted that the equations of motion are in fact contained with in the field equations and therefore do not need to be postulated separately. In 1951, Papapetrou managed to show that the equations of motion of spinning particles (which are not geodesics) can also be derived from Einstein’s field equations.
There have been two different approaches to the problem of obtaining the equations of motion in GR. The first (due to Einstein and Infeld) starts with left hand side of the field equations and is a very complicated geometrical treatment. The other approach (the one we are interested in) starts with right hand side of the field equations and is due to Papapetrou and Fock.
OK, let us begin. The field equations $G^{ab} = T^{ab}$ imply
$$\nabla_{b}T^{ab} = 0.$$
This can be written as
$$\partial_{b}\left( \sqrt{-g} T^{ab}\right) + \sqrt{-g}\Gamma^{a}{}_{bc}T^{bc} = 0. \ \ (1)$$
We integrate this over a small 3-volume and write
$$\int d^{3}x \partial_{b}\left( \sqrt{-g} \ T^{ab}\right) + \int d^{3}x \sqrt{-g}\Gamma^{a}{}_{bc}T^{bc} = 0, \ \ (2)$$
or
$$\frac{d}{dx^{0}}\int d^{3}x \sqrt{-g} \ T^{a0} + \int d^{3}x \sqrt{-g}\Gamma^{a}{}_{bc}T^{bc} = - F^{a}, \ \ (3)$$
where the object $F^{a}$ is defined by
$$F^{a} = \int d^{3}x \partial_{j}(\sqrt{-g} \ T^{aj}).$$
We will identify $F^{a}$ with external non-gravitational forces. So, for the case of gravitational forces alone, one sets $F^{a} = 0$; one writes $F^{a}$ as surface integral and take a large enough volume for which $T^{ab}$ vanishes on the surface.
Let us now expand the connection about a point $X^{a}$ which will be taken as the coordinates of the “particle”; a function of time or other single parameter,
$$x^{a} = X^{a} + \delta x^{a}, \ \ (4)$$
$$\Gamma^{a}{}_{bc}(x) = \Gamma^{a}{}_{bc}(X) + \delta x^{d}\partial_{d}\Gamma^{a}{}_{bc} + … \ \ (5)$$
Let us spell out what we mean by particle. We define a single pole (non-spining) particle by the followings;
i) For some a and b
$$\int d^{3}x \sqrt{-g} \ T^{ab} \neq 0. \ \ \ (6)$$
In particular; when evaluated in a locally inertial rest frame, the integral
$$\int d^{3}\bar{x} \sqrt{-\bar{g}} \ \bar{T}^{00} = m \ > \ 0, \ \ \ (6a)$$
represents the rest mass of our particle.

ii) For all a, b and c
$$\int d^{3}x \sqrt{-g} \ \delta x^{c} \ T^{ab} = 0. \ \ \ (7)$$
iii) The particle is sufficiently small so that the forces acting on it are large compared to the first moment of these forces on its surface,
$$\int d^{3}x \partial_{j}\left(\delta x^{c}\sqrt{-g} \ T^{aj}\right) = \int dS_{j} \delta x^{c}\sqrt{-g} \ T^{aj} \approx 0. \ \ \ (8)$$

For later convenience, let us write eq(2) and eq(3) taking into account eq(5) and eq(7)
$$\int d^{3}x \partial_{b}\left( \sqrt{-g} \ T^{ab}\right) + \Gamma^{a}{}_{bd}(X)\int d^{3}x \sqrt{-g} \ T^{bd} = 0, \ \ (9)$$
$$\frac{d}{dx^{0}}\int d^{3}x \sqrt{-g} \ T^{a0} + \Gamma^{a}{}_{bd}(X) \int d^{3}x \sqrt{-g} \ T^{bd} = - F^{a}. \ \ (10)$$
Now multiply eq(1) by $x^{c}$, integrate over 3-volume and write the result as
$$\int d^{3}x \{ \partial_{0}(\sqrt{-g} \ x^{c}T^{a0}) + \partial_{j}(\sqrt{-g} \ x^{c}T^{aj}) + \sqrt{-g} \ x^{c}\Gamma^{a}{}_{bd}T^{bd}\} = \int d^{3}x \sqrt{-g} \ T^{ac}.$$
Substituting eq(4) and eq(5) into the above equation gives, with the help of eq(7) and eq(8),
$$\frac{dX^{c}}{dx^{0}}\int d^{3}x \sqrt{-g} \ T^{a0} + X^{c}\int d^{3}x \partial_{b}\left( \sqrt{-g} \ T^{ab}\right) + X^{c}\Gamma^{a}{}_{bd}\int d^{3}x \sqrt{-g} \ T^{bd}= \int d^{3}x \sqrt{-g} \ T^{ac}.$$
Making use of eq(9), the above equation reduces to the following nice looking equation
$$\frac{dX^{c}}{dx^{0}}\int d^{3}x \sqrt{-g} \ T^{a0} = \int d^{3}x \sqrt{-g} \ T^{ac}. \ \ (11)$$
For $a = 0$, and since $T^{ab}=T^{ba}$, we find
$$\int d^{3}x\sqrt{-g} \ T^{a0} = \frac{dX^{a}}{dx^{0}}\int d^{3}x \sqrt{-g} \ T^{00}. \ \ (12)$$
Inserting eq(12) back into eq(11), we find
$$\int d^{3}x \sqrt{-g} \ T^{ac}= \frac{dX^{a}}{dx^{0}}\frac{dX^{c}}{dx^{0}}\int d^{3}x \sqrt{-g} \ T^{00}. \ \ (13)$$
Substituting eq(12) and eq(13) back into eq(10), we obtain
$$\frac{d}{dx^{0}}\left( \frac{dX^{a}}{dx^{0}}\int d^{3}x \sqrt{-g} \ T^{00}\right) + \Gamma^{a}{}_{bc}\frac{dX^{b}}{dx^{0}}\frac{dX^{c}}{dx^{0}}\int d^{3}x \sqrt{-g} \ T^{00} = - F^{a}. \ \ (14)$$
Using
$$\frac{d}{dx^{0}} = \frac{ds}{dx^{0}}\frac{d}{ds},$$
we rewrite eq(14) as
$$\frac{d}{ds}\left( \frac{dX^{a}}{ds}\frac{ds}{dx^{0}}\int d^{3}x \sqrt{-g} \ T^{00}\right) + \Gamma^{a}{}_{bc}(X)\frac{dX^{a}}{ds}\frac{dX^{c}}{ds}\frac{ds}{dx^{0}}\int d^{3}x \sqrt{-g} \ T^{00} = - F^{a}\frac{dx^{0}}{ds}. \ \ (15)$$
In this equation we recognize the rest mass of our particle,
$$m = \frac{ds}{dx^{0}}\int d^{3}x \sqrt{-g} \ T^{00} = \int d^{3}\bar{x} \sqrt{-\bar{g}} \ \bar{T}^{00},$$
Thus, eq(15) becomes
$$\frac{d}{ds}\left( m \frac{dX^{a}}{ds} \right) + m \Gamma^{a}{}_{bc}(X) \frac{dX^{a}}{ds}\frac{dX^{c}}{ds} = - F^{a}\frac{dx^{0}}{ds}. \ \ (16)$$
Finally, assuming that the rest mass does not change with time, we find the equation of motion a pole (non-spining) particle subjected to gravitational and non-gravitational forces,
$$\frac{d^{2}X^{a}}{d s^{2}} + \Gamma^{a}{}_{bc}\frac{dX^{b}}{ds}\frac{dX^{c}}{ds} = \frac{f^{a}}{mc^{2}}$$
where
$$f^{a} = - c^{2}\frac{dx^{0}}{ds}F^{a}$$
This method can be extended by relaxing the assumption (ii). Doing this leads to equations of motion for spinning particles.

Regards
Sam

Which is not THE equation of motion but an approximated equation of motion for a particle. Moreover, the whole procedure fails when one consider the motion of a second particle, by essentially the same reasons that Classical Electrodynamics fails also to provide the equations of motion for two charges (See Jackson book).

 

[PAllen]

As stated spacetime singularities and event horizons are different beasts than coordinate singularities and apparent horizons, although you want to mix them.

I will just finish by emphasizing, again, that a Rindler spacetime in SR has not a spacetime singularity nor an event horizon and, therefore, has nothing to see with the predictions of FTG about spacetime singularities and event horizons using gravitons.

I have never conflated coordinate singularities and spacetime singularities. I do believe that differences between apparent horizons and true horizons are not very significant because a true horizon can be changed or eliminated (converted to apparent) by boundary conditions at infinity and I don't believe that real physics is that non-local. Further, it is not established that all black holes have horizons at all - it is a hypothesis with known exceptions. The only remaining unknown is whether there are physically plausible exceptions. Finally, since the horizon region of a supermassive black hole is in every way locally unexceptional, I find it implausible that this is where QG differs from GR.

But the great virtue of the papers you referenced is they provide a near term testable alternative to GR, which is fantastic however the tests come out. If you and I met in some other context, I would enjoy making a healthy bet about the likely outcome (of whether the visible surface of galactic BH is where GR predicts, with the properties predicted by GR).

 

[Sam Gralla]

Seems I'm getting some conflicting answers here, but it seems the general idea is that this cannot be done for generic matter fields, but only for special cases such as a perfect fluid? o.o

Yes, that's correct. I haven't read all the responses, but if people are disputing your basic statement above then they are wrong. What you wrote is 100% non-controversial. Just keep in mind that you have to specify an equation of state to relate p to rho before having deterministic equations for the perfect fluid.

(Boy was this thread hijacked, lol...)

 

[Passionflower]

You pretend to answer something totally unrelated to what I have said. I never said that a theorem was a definition, evidently...

Well you wrote:

"BHs have an event horizon associated to a spacetime singularity (by definition) "

That is simply wrong as there is nothing that defines that.

However there is a theorem that shows that when there is an event horizon then there must be a singularity in general relativity.

 

[juanrga]

You pretend to answer something totally unrelated to what I have said. I never said that a theorem was a definition, evidently...

Well you wrote:

"BHs have an event horizon associated to a spacetime singularity (by definition) "

Which confirms that I never said that a theorem was a definition.

 

[Matterwave]

I am not sure what you mean.


$$\begin{align} 0 &= \nabla_{b}T^{ab} \\ &= \partial_{b}T^{ab} + \Gamma^a_{db} T^{db} + \Gamma^b_{db} T^{ad} \\ &= \sqrt{-g} \left( \partial_{b}T^{ab} + \Gamma^a_{db} T^{db} + \Gamma^b_{db} T^{ad} \right) \\ &= \partial_{b} \left( \sqrt{-g} T^{ab} \right) - \partial_b \left( \sqrt{-g} \right) T^{ab} + \sqrt{-g} \left( \Gamma^a_{db} T^{db} + \Gamma^b_{db} T^{ad} \right) \\ &= \partial_{b} \left( \sqrt{-g} T^{ab} \right) - \sqrt{-g} \Gamma^c_{cb} T^{ab} + \sqrt{-g} \left( \Gamma^a_{db} T^{db} + \Gamma^b_{db} T^{ad} \right) \\ \end{align}$$
In the last line: equation 1.7.2 from Poisson has been used; the second and fourth terms cancel.

Got it, thanks. =]

 

[samalkhaiat]

(See Jackson book).

Thank you for the advice!