## Main Question or Discussion Point

Hi there, i have a question about the energy-momentum tensor T of a perfect fluid, en the relations between pressure en density. I've read something about the Dominant Energy condition, which states that the contraction between T and 2 null- or timelike vectors is always positive. This I understand, but it also implies some relations between the pressure and density, and I don't see this. T contains 2 timelike vectors also, can those 2 be identified with the 2 vectors which are used in the contraction, when the T of a perfect fluid is filled in this condition? ( so the normalisation of timelike vectors can be used?)

The book of d'Inverno speaks about turning the question into an eigenvalue problem, in which he turns the eigenvectors in a tetrad. I don't understand this also.....and least but not last, I look for a derivation of the dominant energy condition of Hawking and Ellis ( maybe they use the way which I described earlier? )

If someone could help me, I would be very pleased! I'm busy right now with a little research in cosmology and the cosmological constant, and I'm a little stuck right now :)

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hellfire
hellfire said:
Ok, thank you for your time, but then still my questions hold if, in the energy condition, you may contract the velocities in the tensor with the velocities used in the condition and use the normalisation for such timelike vectors.

If I calculate the contraction between the mixed En-mom tensor and a timelike vector, I get w*p*U, with U the contravariant timelike vector and rho=w*p. How can I conclude from this that w<-1 ?

And i didn't quite understate your statement about the velocity of a sound wave ( dp/d(rho) )

I hope I'm not to much a pain in the arse :)

hellfire
If you calculate $$T^{\mu}_{\nu}A^{\nu}$$ you get

$$( - \rho A^0, p A^1, p A^2, p A^3)$$

For the DEC this must be timelike, therefore the following must hold considering $$p = w \rho$$:

$$(A^0)^2 > w^2 ((A^1)^2 + (A^2)^2 + (A^3)^2))$$

If w2 > 1, this is not true for every timelike A.

Ok, very clear :) But I've another question now.

From the standard action you can derive an Enery momentum tensor, and identify that T(00)=rho, en pg(ij)=T(ij). From this, I read that rho=1/2*(timederivative of scalarfield)^2 + V, and p=/2*(timederivative of scalarfield)^2 -V.

The first is clear, but the expression for p is not.... i need this for an examination of the relation p=rho*w, where w is than variable. Anyone? :)

hellfire
haushofer said:
From the standard action you can derive an Enery momentum tensor, and identify that T(00)=rho, en pg(ij)=T(ij). From this, I read that rho=1/2*(timederivative of scalarfield)^2 + V, and p=/2*(timederivative of scalarfield)^2 -V.
What action? Are you talking about a scalar field? In such a case you have different equations of state w = -1, 1 depending on whether you neglect the kinetic term or the potential term. Otherwise, for the general case, I don't think there must be an equation of state of type $$\inline{p = \omega \rho}$$... (but I am not sure).

Sorry for the late reply. Well, w is then just defined as the ratio between rho and p. And rho an p are something you can extract from the general energy momentum tensor of the scalar. But those are different bij a term 2*V, and they both contain the time derivative of the scalarfield... And I'm heavily traumatised, cause I can't see this step :(

hellfire
If you take the Lagrangian for the scalar field $$\inline{L \approx \frac{1}{2} \dot \phi^2 + V = K + V}$$ and apply Noether's theorem to compute the energy-momentum tensor, you will get $$\inline{\rho = K + V}$$ and $$\inline{p = K - V}$$. Now, depending on whether you neglect K or V you will have w = -1 or w = 1. If you do not neglect any of them, then I guess you cannot have a simple relation $$\inline{p = w \rho}$$.

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hellfire said:
If you calculate $$T^{\mu}_{\nu}A^{\nu}$$ you get

$$( - \rho A^0, p A^1, p A^2, p A^3)$$

For the DEC this must be timelike, therefore the following must hold considering $$p = w \rho$$:

$$(A^0)^2 > w^2 ((A^1)^2 + (A^2)^2 + (A^3)^2))$$

If w2 > 1, this is not true for every timelike A.
But this doesn't work for a general, symmetric energy-momentum tensor; only for a tensor in the form T=diag(rho,-p,-p,-p,). How does one generalize this for a general symmetric T(ab) ?

I also saw that you can construct in this way a contravariant vector an a covariant vector, and take the innerproduct and set this always bigger than 0. Does anyone know this derivation, because I can't get the right results...

But the rest is clear now.

hellfire
I am sorry but I cannot help you with this. I know about the relation $$\inline{p = \omega \rho}$$ only in the context of perfect fluids for which the energy-momentum tensor is diagonal. I guess (but I don’t know) that there will not exist such a simple relation for fluids with viscosity and heat conduction and you will need to consider other state functions or other variables to find an expression for the pressure of the fluid.

George Jones
Staff Emeritus
Gold Member
For now, I just want to comment that I hope to get involved in this thread sometime soon, but I want to do some other things first, so this won't happen until maybe later today, maybe tomorrow, maybe ... .

The only references that I have available are the books by Hartle and Carroll, but I think this will be enough to allow us to work through some things together. In the meantime, you might find Matt Visser's paper, Twilight for the energy conditions?, interesting (and not entirely relevant).

http://www.arxiv.org/abs/gr-qc/0205066

Regards,
George

hellfire said:
I am sorry but I cannot help you with this. I know about the relation $$\inline{p = \omega \rho}$$ only in the context of perfect fluids for which the energy-momentum tensor is diagonal. I guess (but I don’t know) that there will not exist such a simple relation for fluids with viscosity and heat conduction and you will need to consider other state functions or other variables to find an expression for the pressure of the fluid.
Well, you know, the problem is more this:

I've got a perfect fluid, and with this you get T(00)=rho and T(ij)=-pg(ij). With the dominant energy condition one can reveal that -1<w<1, that is to say, I read it all over the place. But no-one gives a normal derivation of this result! I've read Carroll, Inverno, Wald, Cheng, and a lot of articles, but everyone thinks it is trivial or something...

What I've done is this: I contracted the perfect fluid-tensor with a time-like vector ( 2 times ofcourse, to create a contra and a covariant vector) . Then i took the innerproduct, and stated that it should be bigger than 0. And that's were I got stuck.

Anyway, I really appreciate the effort from you guys :) It's just a little frustrating that something that seems to be a little algebra causes so much headaches...

hellfire
haushofer said:
What I've done is this: I contracted the perfect fluid-tensor with a time-like vector ( 2 times ofcourse, to create a contra and a covariant vector) . Then i took the innerproduct, and stated that it should be bigger than 0. And that's were I got stuck.
I think I explained in post #4 how to proceed then. Actually, after reading your last post I am not sure to understand what your problem is.

George Jones
Staff Emeritus
Gold Member
According to Carroll, the energy-momentum tensor for a perfect fluid is

$$T_{\mu\nu} = (\rho + p)U_\mu U\nu + p g_{\mu\nu}.$$

The dominant energy con requires that for all timelike 4-vectors $\mathbf{t}$:

1) $T_{\mu\nu} t^\mu t^nu \geq 0$

George Jones
Staff Emeritus
Gold Member
Sorry, I clicked Submit Reply when I meant to click Preview Post. Here's what seems to me to be a complete argument.

According to Carroll, the energy-momentum tensor for a perfect fluid is

$$T_{\mu\nu} = (\rho + p)U_\mu U_\nu + p g_{\mu\nu}.$$

The dominant energy con requires that for all timelike 4-vectors $\mathbf{t}$:

1) $T_{\mu\nu} t^\mu t^nu \geq 0$;
2) $v^\nu = T^{\mu\nu} t_\mu$ is timelike or lightlike.

$\mathbf{U}$ is timelke, and using this in 1) gives

$$\begin{equation*} \begin{split} 0 &\leq T_{\mu\nu} U^\mu U^\nu\\ &= [(\rho + p)U_\mu U_\nu + p g_{\mu\nu}] U^\mu U^\nu\\ &= (\rho + p)(-1)(-1) + p(-1)\\ &= \rho, \end{split} \end{equation*}$$

so $\rho \geq 0$.

The information that is needed from 2) is obtained by keeping $\mathbf{t}$ arbitrary. $v^\nu = T^{\mu\nu} t_\mu$ timelike or lightlike means

$$\begin{equation*} \begin{split} 0 &\geq v^\nu v_\nu\\ &= T^{\mu\nu} t_\mu T^\lambda{}_\nu t_\lambda\\ &= [(\rho + p)U^\mu U^\nu + p g^{\mu\nu}][(\rho + p)U^\lambda U_\nu + p \delta^\lambda_\nu]t_\mu t_\lambda\\ &= [-(\rho + p)^2 U^\mu U^\lambda + p(\rho + p)U^\lambda U^\mu + (\rho + p)pU^\mu U^\lambda + p^2 g^{\mu\lambda}] t_\mu t_\lambda\\ &= [(p^2 - \rho^2)U^\mu U^\lambda + p^2 g^{\mu\lambda}] t_\mu t_\lambda\\ &= (p^2 - \rho^2)(U^\mu t_\mu)^2 + p^2 t_\mu t^\mu. \end{split} \end{equation*}$$

Because $\mathbf{t}$ is an arbitrary timelike vector, the magnitude of the second term in the last line above can be made arbitrarily small, i.e. for very (no matter how small) $\epsilon > 0$, there is a $\mathbf{t}$ with $0 > p^2 t_\mu t^\mu \geq -\epsilon$. Thus, neglect this second term. Now, divide by the positive scalar $(U^\mu t_\mu)^2$ to give $\rho^2 \geq p^2$. Because $\rho$ is positive,

$$-\rho \leq p \leq \rho.$$

I have tried to be fairly explicit, but if anything is unclear, let me know.

Regards,
George

George Jones said:
Sorry, I clicked Submit Reply when I meant to click Preview Post. Here's what seems to me to be a complete argument.

According to Carroll, the energy-momentum tensor for a perfect fluid is

$$T_{\mu\nu} = (\rho + p)U_\mu U_\nu + p g_{\mu\nu}.$$

The dominant energy con requires that for all timelike 4-vectors $\mathbf{t}$:

1) $T_{\mu\nu} t^\mu t^nu \geq 0$;
2) $v^\nu = T^{\mu\nu} t_\mu$ is timelike or lightlike.

$\mathbf{U}$ is timelke, and using this in 1) gives

$$\begin{equation*} \begin{split} 0 &\leq T_{\mu\nu} U^\mu U^\nu\\ &= [(\rho + p)U_\mu U_\nu + p g_{\mu\nu}] U^\mu U^\nu\\ &= (\rho + p)(-1)(-1) + p(-1)\\ &= \rho, \end{split} \end{equation*}$$

so $\rho \geq 0$.

The information that is needed from 2) is obtained by keeping $\mathbf{t}$ arbitrary. $v^\nu = T^{\mu\nu} t_\mu$ timelike or lightlike means

$$\begin{equation*} \begin{split} 0 &\geq v^\nu v_\nu\\ &= T^{\mu\nu} t_\mu T^\lambda{}_\nu t_\lambda\\ &= [(\rho + p)U^\mu U^\nu + p g^{\mu\nu}][(\rho + p)U^\lambda U_\nu + p \delta^\lambda_\nu]t_\mu t_\lambda\\ &= [-(\rho + p)^2 U^\mu U^\lambda + p(\rho + p)U^\lambda U^\mu + (\rho + p)pU^\mu U^\lambda + p^2 g^{\mu\lambda}] t_\mu t_\lambda\\ &= [(p^2 - \rho^2)U^\mu U^\lambda + p^2 g^{\mu\lambda}] t_\mu t_\lambda\\ &= (p^2 - \rho^2)(U^\mu t_\mu)^2 + p^2 t_\mu t^\mu. \end{split} \end{equation*}$$

Because $\mathbf{t}$ is an arbitrary timelike vector, the magnitude of the second term in the last line above can be made arbitrarily small, i.e. for very (no matter how small) $\epsilon > 0$, there is a $\mathbf{t}$ with $0 > p^2 t_\mu t^\mu \geq -\epsilon$. Thus, neglect this second term. Now, divide by the positive scalar $(U^\mu t_\mu)^2$ to give $\rho^2 \geq p^2$. Because $\rho$ is positive,

$$-\rho \leq p \leq \rho.$$

I have tried to be fairly explicit, but if anything is unclear, let me know.

Regards,
George
You are my hero, but that last sentence, that the arbitrary timelike t can be made arbitrary small isn't quite clear yet. I don't see yet how you can pick such a freedom to talk t that small; you want the DEC to be true for all vectors t, also for vectors with a norm bigger than epsilon. But this is indeed the derivation I was looking for !

And I get a feeling that I should learn how to use LaTeX :') So excuse me if I wasn't that clear in my formulation of the problem.

George Jones
Staff Emeritus
Gold Member
Actually, it's important to note that $b^2 := -t_\mu t^\mu$ can be made arbitrarily small with respect to $a^2 := (U^\mu t_\mu)^2$. Dividing through by $a^2$ without neglecting $b^2$ gives

$$\rho^2 \geq p^2 (1 - (b/a)^2 ).$$

$(b/a)^2$ is a scalar-valued function of $\mathbf{t}$ that can take on any value in the range $0<(b/a)^2<\infty$. The idea is that since this is true for any $\mathbf{t}$, why not choose a $\mathbf{t}$ that gives the tightest constraint?

For exampe, take

$$t^\mu = U^\mu + \sqrt{1 - \delta}s^\mu,$$

where $\mathbf{s}$ is a spacelike unit vector orthogonal to $\mathbf{U}$. Then

$$\begin{equation*} \begin{split} t^\mu t_\mu &= (U^\mu + \sqrt{1 - \delta}s^\mu)(U_\mu + \sqrt{1 - \delta}s_\mu)\\ &= -1 + (1 - \delta)\\ &= -\delta, \end{split} \end{equation*}$$

so, $\mathbf{t}$ is timelike as long as $0 < \delta \leq 1$. Clearly, $U^\mu t_\mu = -1$. Thus, $(b/a)^2 = \delta$, and

$$\rho^2 \geq p^2 (1 - \delta)$$.

Choosing $\delta$ as close to zero as possible makes $(1 - \delta)$ as large as possible, which gives the tightest constraint.

You should give LateX a try! I had never used LateX before I joined Physics Forums a few days ago. Before this, I had always used Scientific Workplace, a front end for Latex that also includes a basic Maple package. I have found the thread

to be invaluable.

Regards,
George

Ok, there comes a light bubbling into my head, but I oversee something. I would guess that, in order to represent a flow of energy, you would contract the energy momentum tensor with a velocity vector. So t would be a velocity vector, and the innerproduct of velocity vectors with themselfs are normalised as -1. ( with metric convention -,+,+,+) So I would say that b2 is always -1...

I guess I'm a little confused by the definition of the vectors...

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George Jones
Staff Emeritus
Gold Member
The condition

2) $v^\nu = T^{\mu\nu} t_\mu$ is timelike or lightlike for all timelike $\mathbf{t}$

is completely equivalent (in a mathematical sense) to the condition

2') $v^\nu = T^{\mu\nu} t_\mu$ is timelike or lightlike for all $\mathbf{t}$ such that $t^\mu t_\mu = -1.$

Exercise: show this.

Therefore, using 2) to establish a result, e.g., our desired inequality, is equivalent to using 2') to establish the same result. However, it seems that you would like to see the result established starting directly from 2'). OK.

The result

$$\rho^2 \geq p^2 (1 - (b/a)^2 ),$$

with $a^2 := (U^\mu t_\mu)^2$ and $b^2 := -t_\mu t^\mu$ is obtained exactly as before. Now, $b^2 := 1$, so the idea is to find a $\mathbf{t}$ that makes $a$ as large as possible.

To this end, choose $\mathbf{t}$ such that

a) $t^0 = 1/(\sqrt{\delta}U^0 )$;
b) $U^1 t^1 + U^2 t^2 + U^3 t^3 = 0$.

Then, $a^2 = 1/\delta$ and $\rho^2 \geq p^2 (1 - \delta)$ as before. Let $\delta$ go to zero. This will mean that $(t^0)^2$ goes to infinity, but this is OK as long as $((t^1)^2 + (t^2)^2 + (t^2)^2)$ goes to infinity such that $-(t^0)^2 + ((t^1)^2 + (t^2)^2 + (t^2)^2) = -1$ is maintained.

Regards,
George

George Jones said:
The condition

2) $v^\nu = T^{\mu\nu} t_\mu$ is timelike or lightlike for all timelike $\mathbf{t}$

is completely equivalent (in a mathematical sense) to the condition

2') $v^\nu = T^{\mu\nu} t_\mu$ is timelike or lightlike for all $\mathbf{t}$ such that $t^\mu t_\mu = -1.$

Exercise: show this.

Therefore, using 2) to establish a result, e.g., our desired inequality, is equivalent to using 2') to establish the same result. However, it seems that you would like to see the result established starting directly from 2'). OK.

The result

$$\rho^2 \geq p^2 (1 - (b/a)^2 ),$$

with $a^2 := (U^\mu t_\mu)^2$ and $b^2 := -t_\mu t^\mu$ is obtained exactly as before. Now, $b^2 := 1$, so the idea is to find a $\mathbf{t}$ that makes $a$ as large as possible.

To this end, choose $\mathbf{t}$ such that

a) $t^0 = 1/(\sqrt{\delta}U^0 )$;
b) $U^1 t^1 + U^2 t^2 + U^3 t^3 = 0$.

Then, $a^2 = 1/\delta$ and $\rho^2 \geq p^2 (1 - \delta)$ as before. Let $\delta$ go to zero. This will mean that $(t^0)^2$ goes to infinity, but this is OK as long as $((t^1)^2 + (t^2)^2 + (t^2)^2)$ goes to infinity such that $-(t^0)^2 + ((t^1)^2 + (t^2)^2 + (t^2)^2) = -1$ is maintained.

Regards,
George
0k, clear! But a last wondering. Isn't it so that U and t both are independant of each other, and that you simply could state that both terms in the derivation should be bigger then 0 ? Because the normalisation of t, and p^2 is always bigger then 0 this would also imply that -1>w>1.

Or am i talking crap now?

George Jones
Staff Emeritus
Gold Member
haushofer said:
Isn't it so that U and t both are independant of each other, and that you simply could state that both terms in the derivation should be bigger then 0 ? Because the normalisation of t, and p^2 is always bigger then 0 this would also imply that -1>w>1
Sorry - you've lost me here. Which 2 terms do you mean?

Regards,
George

George Jones said:
Sorry - you've lost me here. Which 2 terms do you mean?

Regards,
George

I'm sorry. I meant the terms with a and b. The term a is a contraction between U and t, and b can be normalised to give 1 ( or -1, that isn't that important ). Because U and t are independant, both terms in front of a and be should be bigger then 0.

Is this correct?

So I meant the terms

$$(\rho^2 - p^2)(U^\mu t_\mu)^2 + p^2 t_\mu t^\mu > 0.$$

Because $$(U^\mu t_\mu)^2$$ and $$p^2 t_\mu t^\mu$$ are independant, I thought that both terms could be set to be bigger then 0. If you then divide the first by $$(U^\mu t_\mu)^2$$ , you get

$$(\rho^2 - p^2) >0$$

and if you divide the second term by $$t_\mu t^\mu$$, you get

$$p^2 >0$$ , which is trivial. Because you know that $$\rho>0$$,

this would also give the constraint that -1<w<1.

If I made any sign-errors, excuse me, because I'm using the (+,-,-,-) signature of the metric :)

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