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I am trying to connect two different representations of the energy-momentum-tensor, which I found in different books.

In Weinberg's Gravitation one finds for the energy-momentum-tensor of a point particle

\begin{equation}T^{\mu\nu}(\vec{x},t)=\frac{1}{\sqrt{-g}}m\frac{\text{d}x_m^\mu(t)}{\text{d}t}\frac{\text{d}x_m^\nu(t)}{\text{d}t}\frac{\text{d}t}{\text{d}\tau_m}\delta^3(\vec{x}-\vec{x}_m(t))\end{equation}

Where $g$ ist the determinant of the metric tensor, $x^\mu_m$ ist the coordinate of the Mass $m$ and $\tau_m$ is the particle's proper time.

In Creighton and Anderson's Gravitational Waves Physics and Astronomy one finds for a perfect fluid

\begin{equation} T^{\mu\nu}=(\rho+p/c^2)u^\mu u^\nu+pg^{\mu\nu} \end{equation}

For a point particle it is

\begin{equation} \rho(x)=m\frac{\delta^4(x-x_m(\tau))}{\sqrt{-g}}~~~~~\text{and}~~~~~p=0 \end{equation}

which leads to

\begin{alignat}{2} T^{\mu\nu}&=m\frac{\delta^4(x-x_m(\tau))}{\sqrt{-g}}\frac{\text{d}x^\mu}{\text{d}\tau}\frac{\text{d}x^\nu}{\text{d}\tau}\\ &= m\frac{\delta^4(x-x_m(\tau))}{\sqrt{-g}}\frac{\text{d}x^\mu}{\text{d}t}\frac{\text{d}x^\nu}{\text{d}t}\Bigl(\frac{\text{d}t}{\text{d}\tau}\Bigr)^2 \end{alignat}

Now I have a few problems, of which some are conceptual while others might be just computational:

1. Why isn't just

$$\rho=\frac{m}{\sqrt{-g}}\delta^3(\vec{x}-\vec{x}_m(t))$$

correct to describe the density of a point particle? This would have been my guess. Also the 0th component of the delta function which has a timelike argument, would give a $T^{-1}$ to the Dimension of the density, while I think we need a Dimension of $L^{-3}M$?

2. Part of the reason for me trying to do this derivation is, that I don't really understand the meaning of parts of those formulas. For example I cannot really see the meaning of an object like $u^0$ for a perfect fluid. I can understand $u_m^0=\text{d}(ct_m)/\text{d}\tau$ where one parametrizes the coordinate time of a particle in terms of its poper time. But I have problems to think in similar terms with formulas like the one for the perfect fluid. How can I understand $\text{d}x^\mu/\text{d}\tau$ where $x^\mu$ is not the spacetime position of a moving particle, but just an abstract coordinate? To whoms proper time am I differentiating this coordinate $x^\mu$?

3. The last problem is, that I don't really know, how to get rid of the time-part of the delta-function, while simultaneously producing a factor $\text{d}\tau/\text{d}t$ which seems to be what needs to happen, to get connect the two formulas.

Additional Question in respons to void's answers:

Maybe it is better to forget about the notion of the fluid for the moment and start later. In Creighton's and Andersons's book one can read that the energy-momentum-tensor for a point particle has the form

$$T^{\mu\nu}(x')=\rho(x')u^\mu u^\nu$$

where

$$\rho(x')=m\frac{\delta^4(x'-x(\tau))}{\sqrt{-g}}$$

I think this cannot be correct is Weinbergs formulas are, because in this one there is no integral. Therefore I don't see how to get rid of the time-part of the delta-function and even the dimensions don't agree. My guess would be that at least one of those formulas just isn't a 100 procent correct, as there are quite some mistakes in Creighton and Andersons book.

My Idea would be that it has to be either

$$T^{\mu\nu}(x')=\int\rho(x')u^\mu u^\nu\text{d}\tau$$

where $\rho(x')$ is defined as above, being a density in time and space. This would be equivalent to the energy-momentum-tensor given by Void.

However I would not know then how to build a bridge to the coninous description of matter as a perfect fluid or as dust because I wouldn't know what proper time to use to integrate, as every little dust particle should experience a different proper time.

My second guess would be:

$$T^{\mu\nu}=\rho(x')u^\mu u^\nu$$

is correct, but the density is a density only in space (not in time) and given by

$$\rho(x')=\int m \frac{\delta^4(x'-x(\tau))}{\sqrt{-g}}\text{d}\tau$$

which could give (in Voids notation)

$$T^{\mu\nu}(x^\mu)=\int \frac{m}{\sqrt{-g(x^\mu)}}\delta^4(x^\mu-\gamma^\mu(\tau))\text{d}\tau\frac{\text{d}x^\mu}{\text{d}\tau}\frac{\text{d}x^\nu}{\text{d}\tau}$$

This is nearly what I would want, but the problem is, that the $\text{d}x^\mu/\text{d}\tau$ are outside of the integral which means I don't know how to justify $\text{d}x^\mu/\text{d}\tau\rightarrow\text{d}\gamma^\mu/\text{d}\tau$

I think in both my Ideas the problem is, that I cannot really connect the continous case with the point-particle case. But in principle that should be possible, shoudn't it? Also if it weren't possible, it would make things very difficult because everey formulae which was derived for a continuum would have to be derived again for point masses. This would be quite a hustle and normaly one should be able to express one as a special case of the other... right?...

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2 Answers 2

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The missing piece of information is that the point-particle stress-energy tensor is actually $$T^{\mu\nu} (x^\mu) = \int \frac{m}{\sqrt{-g(x^\mu)}} \frac{d \gamma^\mu}{d \tau}(\tau) \frac{d \gamma^\nu}{d \tau}(\tau)\delta^{4}[x^\mu - \gamma^\mu(\tau)] d\tau $$ where $\gamma^\mu(\tau)$ is the world-line of the particle and the integration bounds goes through the entire history of the particle. (Note that I made sure to be explicit on what is a function of what here!) When you think about it, it makes a lot of sense, since you need to place your matter on the entire worldline in the space-time and that is what the integral achieves.

Now you can get rid of the integral by making a 3+1 split $x^\mu \to t, x^i$ and assuming that you are able to express $\tau$ as a function of coordinate time. For good time coordinates this can usually be done. We can then choose the coordinate time as an alternate integration variable to obtain $$T^{\mu\nu}(t,x^i) = \int \frac{m}{\sqrt{-g(t,x^i)}} \frac{d \gamma^\mu}{d \tau}(t') \frac{d \gamma^\nu}{d \tau}(t') \delta(t - t') \delta^3[x^i - \gamma^i(t')] \frac{d \tau}{d t}(t') d t' $$ Notice that there are two "t" coordinates, one is the time you are evaluating your stress-energy tensor at ($t$), and the other parametrizes the worldline ($t'$). Also, the term that transforms the differential could also be written as $d\tau/dt = 1/(d \gamma^0/d\tau)$. Now it is easy to use the delta function to integrate out the $t'$ parametrization to obtain $$T^{\mu\nu}(t,x^i) = \frac{m}{\sqrt{-g(x^i,t)}} \frac{d \gamma^\mu}{d \tau}(t) \frac{d \gamma^\nu}{d \tau}(t) \delta^3[x^i - \gamma^i(t)] \frac{d \tau}{d t}(t) $$ To obtain Weinberg's formula, you can use the property $$\frac{d}{d\tau} = \frac{d t}{d \tau} \frac{d}{dt}$$


Our common abuse of notation of writing $\gamma^\mu(\tau) \to x^\mu(\tau)$ is what causes most confusion here really but if you are careful about that and the integration, things start making a lot of sense. In the case of the fluid, inserting a delta function in there kind of defies the point of having the continuum description at all. However you can understand the four-velocity of the fluid as the four-velocity of a Lagrangian fluid element. Furthermore, the proper time measures e.g. the rate at which chemical reactions, heating, cooling etc. are happening within this element.


Why should the $u^\nu(x^\mu)$ of a "single-particle fluid" be equal to the particle four-velocity $d\gamma^\mu/d\tau$?

The "single-particle fluid" as you introduce it in the OP is somewhat artificial, but we can work with it as well. So we have the mass density

$$\rho(x^\mu) = \int \frac{m}{\sqrt{-g(x^\mu)}}\delta^{4}[x^\mu - \gamma^\mu(\tau)] d\tau$$

If you think about it, the stress-energy tensor $T^{\nu\lambda}(x^\mu) = \rho(x^\mu) u^\nu u^\lambda$ is zero everywhere apart from the worldline $\gamma^\mu(\tau)$ no matter what are the actual values of $u^\nu(x^\mu)$ away from it. So the only important question is, is it necessary to have $u^\nu(\gamma^\mu(\tau)) = d \gamma^\nu/d\tau$?

To show that, we will use the conservation of rest-mass density. For $P=0$ the total mass-energy density $\rho$ is equal to rest-mass density. The rest-mass conservation then reads $(\rho u^\mu)_{;\mu} = 0$. The divergence of a vector field fulfills $V^\mu{}_;\mu = (V^\mu \sqrt{-g})_{,\mu} /\sqrt{-g}$. The mass conservation is thus equivalent to $(\rho u^\nu \sqrt{-g})_{,\nu} = 0$ which reads $$\left( \int m u^\nu(x^\mu)\delta^{4}[x^\mu - \gamma^\mu(\tau)] d\tau \right)_{\!,\nu} = \int m u^\nu(x^\mu)_{,\nu}\delta^{4}[x^\mu - \gamma^\mu(\tau)] d\tau + \int m u^\nu(x^\mu)\delta^{4}[x^\mu - \gamma^\mu(\tau)]_{,\nu} d\tau$$ This is a distributional equation, so I am going to use a test function to figure out what the distribution is on the worldline: $$\int f(x^\mu) (\rho u^\nu\sqrt{-g})_{,\nu} d^4 x = 0$$ Using properties of distributions this implies $$\int f(x^\mu) (\rho u^\nu\sqrt{-g})_{,\nu} d^4 x =-\int m f_{,\nu} u^\nu|_{x^\mu = \gamma^\mu(\tau)} d\tau = 0$$

Now the expression above should disappear for any test function disappearing sufficiently fast at early/late times. It can be shown that this is only possible if $u^\mu = d \gamma^\mu/d \tau$ on the worldline because then the integral is $\propto \int (d f/ d\tau)\, d \tau$ is zero up to a boundary term (vanishing because of the asymptotic behaviour of the test function). If this is not fulfilled, then there will always be test functions for which the integral does not disappear. (I leave the exploration of all the details to the dear reader.)

In summary, as shown above, mass conservation implies the alignment of the four-velocity with the worldline for a "single-particle fluid".

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    $\begingroup$ Thanks! This already helps me. I still have some questions thow. I still struggle with the problem, that I can't connect the two representations. What I would like to do, is derive the formula for a point particle as a special case of the one for a perfect fluid by neglecting any forces between the particles ($p=0$) and inserting the correct density for a point particle. This would give me the opportunity to use formulas for the perfect fluid and at some point just insert the correct density to get my point particles. I ammended my original question, to explain my further problems. $\endgroup$ Jun 9, 2021 at 18:59
  • $\begingroup$ @BenitoMcLanbeck I edited the question to answer your new additions. It is a bit more involved but it can also be done. $\endgroup$
    – Void
    Jun 10, 2021 at 13:28
  • $\begingroup$ Thanks a lot! This helps me :) $\endgroup$ Jun 10, 2021 at 14:04
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The question: 1. Why isn't just $$ \rho=\frac{m}{\sqrt{-g}}\delta^3(\vec{x}-\vec{x}_m(t)) $$ correct to describe the density of a point particle?

The scalar fields, like the densities, have an invariant meaning in General Relativity, as was stated by Schutz (A first course in general relativity, second edition, pag 94): “All scalar quantities associated with a fluid element in relativity (such as number density, energy density and temperature, are defined to be their values in the MCRF”. The energy density is a (0,0) tensor defined in the MCRF, the Momentarily Comoving Refrerence Frame, which is a local Lorentz frame instantaneously comoving with the particle. Consider the expression: $$\rho(x^\alpha) = \int \frac{m}{\sqrt{-g}}\delta^{4}[x^\alpha - x^\alpha_m(\tau)] d\tau$$ where $x^\alpha_m(\tau)$ is the particle worldline. Integration in $x^0_m$ (change of variable from $\tau$ to $x^0_m$) gives: $$\rho(x^\alpha)=\frac{m}{U^0_m(x^0_m)\sqrt{-g}} \delta^3(x^i-x^i_m(x^0_m))$$ where $U^0_m=dx^0_m/d\tau$. The volume contraction $V= V_{MCRF}/\gamma$ with $\gamma=U^0_m$ is implicit in the last $\rho$ expression. In the MCRF, where $U=(1,0,0,0)$, we have $\rho=\frac{m}{\sqrt{-g}}\delta^3(x^i-x^i_m(t))$. This is the expression proposed in the question. It is important to remark that it is not a tensor equivalence. It only works in the MCRF.

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  • $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$ Mar 28, 2022 at 15:17

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