General Theory of Relativity or the theory of relativistic gravitation is the one which describes black holes, gravitational waves and expanding Universe. The goal of the course is to introduce you into this theory. The introduction is based on the consideration of many practical generic examples in various scopes of the General Relativity. After the completion of the course you will be able to solve basic standard problems of this theory. We assume that you are familiar with the Special Theory of Relativity and Classical Electrodynamics. However, as an aid we have recorded several complementary materials which are supposed to help you understand some of the aspects of the Special Theory of Relativity and Classical Electrodynamics and some of the calculational tools that are used in our course. Also as a complementary material we provide the written form of the lectures at the website: https://math.hse.ru/generalrelativity2015
Energy–momentum pseudo–tensor for gravity
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From the course by National Research University Higher School of Economics
Introduction into General Theory of Relativity
72 оценки
National Research University Higher School of Economics
72 оценки
From the lesson
Gravitational waves
With this module we start our study of gravitational waves. We explain the important difference between energy-momentum conservation laws in the absence and in the presence of the dynamical gravity. We define the gravitational energy-momentum pseudo-tensor. Then we continue with the linearized approximation to the Einstein equations which allows us to clarify the meaning of the pseudo-tensor. We end up this module with the derivation of the free monochromatic gravitational waves and of their energy-momentum pseudo-tensor. These waves are solutions of the Einstein equations in the linearized approximation.
Meet the Instructors
Emil AkhmedovAssociate Professor
Faculty of Mathematics
[MUSIC]
Next two lectures we're going to devote to the study of the gradational radiation.
To define what means gradational radiation,
we have to specify what means energy flux in the presence of gravity.
That is a problem we are going to address in this lecture.
Well to understand what is the problem.
Let us start with the standard situation of Minkowski spacetime.
In Minkowski spacetime,
the energy momentum conservation condition is expressed like this.
With a short derivative rather than than covariant long derivative.
And because of this, if we integrate both sides of this equation
over a surface sigma and take 0 component.
For nu equals to 0, we obtain the following equation.
That integral
over dp0/ dt +
integral over sigma over boundary of this sigma.
D 2 write as three dimensional.
Because it's three dimensional.
D 2 sigma i T i 0 equals to 0.
Where P mu is the falling full vector.
When it's a function of time, it's an interval over sigma T mu 0 t,
x d3 sigma.
So this is three dimensional Kushi surface.
0 is a long time.
So we have a Kushi surface sigma.
And we have a normal direction to this Kushi surface,
which is designated by t or 0 component.
So this is what we obtain.
And P0 as one can see, P0 is nothing but the energy.
Because it's integral over the energy density over the three dimensional
hypersurface.
And this is volume element, area element at the boundary of this surface sigma.
So this law establishes conservation, actual conservation of this quantity.
There is no energy flux through the boundary.
This is conserved in this region.
If there is energy flux going out, energy is reduced.
If in, energy is increased.
But in case of the gravity.
What we have been calling as the energy momentum conservation
doesn't have this form.
It does have the following form.
It has a form like this with the covariant long derivative.
And as a result we have the following situation.
So 0 is equal to, we just write explicitly this covariant derivative.
It's T mu nu we express it in this form, sorry.
It was one up and this lower index, and this is upper index.
Plus gamma, mu beta,
mu T, beta nu.
Minus gamma beta nu mu.
T mu beta.
And as a result, this is equivalent 1 over square root of the modulus of
the determinant of the metric, d mu acting on T mu nu.
Square root of the determinant
of the metric, minus 1/2
of d mu g mu alpha T mu alpha.
To obtain this expression, we have used explicit form.
Explicit form for gamma mu nu alpha through the metric.
It's symmetry properties.
And also the fact that energy momentum tensor is symmetric.
T mu nu, T nu mu and finally we have
used that gamma, this gamma.
The trace of gamma, gamma mu nu mu with two indices.
And some summation of the indices is 1 over square root
of g d nu square root of modulus of g.
This is, it follows from the definition of the gamma matrices.
So now what we have is that there is no such conservation.
Well, we have something similar to that expression here, but
we have additional term.
This conservation law, although we call it conservation law,
it doesn't express any explicit conservation of some quantity.
And it should be expected, on general grounds,
that if we are dealing with a theory containing gravity,
the energy momentum tensor of matter shouldn't be simply conserved.
Because there can be energy transmission from matter to gravity and
from gravity to matter.
So the question is what is the quantity which specifies what
means energy, momentum for the gravity.
How to find it and that's the issue we going to address now.
So now we're going to specify what we mean by the energy momentum for the gravity.
And what is the total energy momentum of gravity plus matter?
So let me first write this relation that we have ended with.
So this is a relation like this.
G mu nu square root of modulus of
the determinant of the metric minus
1/2 d mu g mu alpha T mu alpha.
And so let us fix the following reference frame.
Set your reference frame around
arbitrary point x0 such that in this
frame d alpha g mu nu of x0 is equal to 0.
It doesn't mean that g mu nu is equal
to eta mu nu, it's not equal.
It's constant but at this point it's derivative of 0.
But it doesn't mean that it is equal to this.
But it is important that at this point d mu of square root of modulus
of g at the point x0 is equal to 0.
It's important.
As a result from this equation because
this is true we obtain that at this
point d mu T mu nu of x0 is equal to 0.
So in this particular frame, in this particular point, we have this relation.
So now let us see how this relation changes in another reference system.
So let us use Einstein equation to express energy
momentum tensor through the Ricci tensor.
So from Einstein equations, we have this relation
g mu nu R.
So in this frame, we have that
because, derivative gi = 0.
We also have that Christoffel symbols, mu nu alpha(X0) = 0.
But, d beta gamma mu nu alpha
(X0) is not equal to 0.
As a result Ricci tensor at this point and
in this frame is as follows.
It's one-half g
mu alpha g nu gamma g
sigma delta [d alpha
d delta g sigma gamma
+ d sigma d gamma g
alpha delta- d
alpha d gamma g sigma
delta- d sigma d delta
g sigma gamma].
So this is a form Ricci tensor has in this frame.
As a result, putting this together,
we obtain from this equation and
this equation,
after straightforward calculation,
that T mu nu (X0) = d (alpha)
{1/16 pi kappa
1/modulus of g d beta {
modulus of g, (g mu,
g alpha beta- g mu alpha
g mu beta)]}.
And finally closing this.
And now, we going to define this as follows,
we going to designate this as follows.
It's 1/ modulus of g d alpha tensor eta mu nu alpha.
And this eta mu nu alpha
= 1/16 pi kappa d
beta acting on modulus
of g (determinant of
metric g mu nu g alpha beta-
g mu alpha g nu beta)]
To obtain this we have extracted this modulus of the determinant
from the derivative because recall that in this frame we have this fact.
And of course, in obtaining all these relations
we had in mind the following thing.
That around the point, so
g mu nu, around the point (X) 0 can be expressed like this.
It's g mu nu X(0).
First derivative is absent but
there are second derivatives d alpha,
d beta, g mu nu (X0),
(X- X0) alpha, (X- X0) beta.
So in expressing this like this,
we kept the terms in this expansion.
So we have obtained that at the point X0,
in the gauge d alpha g mu nu (X0) = 0.
That T mu nu, first of all,
d mu T mu nu = 0, in this frame or gauge.
But at the same time T mu nu
(X0) = 1/ modulus of
determinant of the metric,
d alpha eta mu nu alpha.
And eta mu nu alpha is the following quantity.
It's 1/16 pi kappa
d beta [modulus of g( g
mu nu g alpha beta-
g mu alpha g mu beta)].
And one can see that this tensor has the following symmetry properties that
under the exchange of the first two indices, it is symmetric.
But under the exchange of the alpha index with any of mu and
nu, it is anti-symmetric.
So if we keep the first one untouched, but exchange the third and
the second index, it changes the sign, as it's seen from this property.
And now because in this
frame d mu l g l = 0.
One can immediately see that from this property of this tensor if we apply to
this relation d mu, immediately we obtain this.
Now let us see what happens beyond this gauge.
If we change the frame.
Then of course, it's not zero.
Let us denote.
So d alpha eta mu nu alpha-
modulus of the determinant T of T mu nu is not any more zero.
As you can see because of this.
It's some quantity which we denote like this, t mu nu.
Let us find it, let us find this quantity.
This quantity one can find from the following relation.
That it is the difference between
what follows from here and
Einstein equations.
Let me see what happens.
It's 1/16 pi.
Kappa d alpha d beta
acting on modulus of
g times g mu nu g alpha
beta- g mu alpha g mu
beta- modulus of g
times 8 pi kappa (R
mu nu- one-half g mu nu R).
Well, this just follows from this,
the definition of this is this,
the definition of this is this while this follows from here.
And this follows from equation.
So in this gauge, they cancel each other and this is 0.
But beyond this gauge, we have often tedious but
straightforward calculation were find the following.
Well, it will be a long expression.
So let me start writing here.
So modulus of g times t mu nu is equal to the following quantity.
It's 1 over 16 pi kappa.
I first write the expression and then explain the indentation.
G mu nu comma alpha G alpha
beta comma beta- G mu
alpha comma alpha G nu
beta comma beta + one-half
g mu nu g alpha beta.
G alpha gamma comma sigma G
sigma beta comma gamma -,
( g mu gamma g alpha
beta G nu beta comma
sigma G alpha sigma
comma gamma + g nu
alpha g beta gamma,
G mu gamma comma sigma
G beta sigma comma alpha
) + g alpha beta.
g gamma sigma G mu
alpha comma gamma G
nu beta comma sigma
+ one-eighth.
( 2 g mu alpha g nu
beta- g mu nu g alpha
beta ) times ( 2 g
gamma sigma g delta
psi- g sigma delta
g gamma psi times G
gamma psi alpha G,
this is comma,
G sigma delta comma beta ).
So this is a tedious relation for
this quantity where G mu nu.
G mu nu, by definition,
is just square root of g times g mu nu.
And I remind you that this comma of any quantity is just,
well anyway, G mu nu comma alpha is just d alpha of G mu nu.
Well, one can immediately see that in the gauge, in the frame, where this is 0.
This just vanishes, because it contains only derivatives of the metric.
But beyond this gauge, this is not 0.
Well, it's a tedious calculation to obtain from here this quantity.
So what we have obtained is the following.
That d alpha eta mu nu alpha- modulus
of g T mu nu = modulus of g T mu nu.
And we have obtained a rather complicated expression for this quantity.
The reason for introducing this quantity, which is we did that after Landau and
Lifshitz and this quantity is referred to as pseudotensor energy momentum,
tensor for gravity.
Why it is called pseudotensor,
because it is not a tensor quantity under the coordinate transformations.
Because here from its definition, one can see that this expression contains short
derivatives rather than covariant derivatives.
And as a result, the transformation law for
this quantity is not appropriate for the tensor.
But the reason for
introducing this pseudo-energy momentum tensor was as follows.
As you can sequence of this relation and
the symmetric properties of this guy under the exchange
of index alpha with any of this, that is antisymmetric.
As you can sequence, we have the following conservation law.
T mu nu + T mu nu = 0.
So this is a real conservation law with short derivative and
this quantity is conserved.
So we can define for momentum as follows.
d3 sigma nu modulus of g T mu nu + t mu nu.
And this is for vector which is perpendicular to the surface sigma.
So we integrate, it is perpendicular to surface sigma.
And the modulus of it is equal to the elementary volume on this surface.
Elementary volume on this surface.
So in the absence of the gravitational field, this is 1, this is 0,
and we just go back to the same quantity as space.
In the presence, that was the gravitational field,
this is the quantity we have to consider.
So in the presence of the gravitational field to have a conservation of some
quantity, we have to fix a bare ground and consider.
Well, this is the issue we're going to discuss right now.
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[MUSIC]
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