I Why is the Energy Momentum Tensor of a Perfect Fluid a Tenso

[klpskp]

The energy momentum tensor of a perfect relativistic fluid is given by

$$T^{\mu\nu} = (\rho + p)u^\mu u^\nu + p g^{\mu\nu}$$

I don't understand why this is a tensor, i.e. why it transforms properly under coordinate changes.

$u^\mu u^\nu$ and $g^{\mu\nu}$ are tensors, so for $T^{\mu\nu}$ to be a tensor, $\rho$ and $p$ must be scalar functions. However, $\rho$ is not a scalar function, because under the coordinate transform
$$t'=t $$ $$ x'=2x $$ $$ y'=2y $$ $$ z'=2z$$

we get $\rho' = \frac{\rho}{8}$

so $\rho$ does not transform like a scalar function.

So why is the energy momentum tensor of a perfect fluid a tensor anyway?
Thank you for your help.

 

[Orodruin]

$\rho$ is a scalar. It is the physical density in the rest frame of the fluid, not the coordinate density.

 

[klpskp]

Dear Ordruin,

what do you mean by physical density?

I understand that we mean the density in some frame in which the fluid is at rest, i.e $u^0=1$ and $u^i=0$. However, such a frame is not unique, the coordinate transformation from my original post takes one rest frame to another rest frame. But this transformation changes the density.

 

[PeterDonis]

what do you mean by physical density?

He means the actual physical observable "density in the frame in which the fluid is at rest". That's a measurable number at every event in spacetime: $\rho$ is simply the scalar function on spacetime that outputs that measurable number at every event. (Similar remarks apply to $p$, by the way.)

 

[PeterDonis]

such a frame is not unique, the coordinate transformation from my original post takes one rest frame to another rest frame. But this transformation changes the density

It "changes the density" in the sense that an observer at rest in the new frame will measure a different density. But the density the observer at rest in the new frame measures is not $\rho$; $\rho$ is defined as the measurable number "density measured by an observer at rest relative to the fluid". Changing frames does not change measurable numbers; an observer at rest in the new frame will still agree that the observer at rest relative to the fluid measures the density $\rho$.

 

[klpskp]

Unfortunately I still don't understand the answer, maybe I clarify my problem:

Suppose we have observer 1, who is at rest relative to the fluid. Now suppose we have observer 2, whose worldline is the same as that of observer 1. In particular, observer 2 is also at rest relative to the fluid. However, observer 2 decides to measure all his length in in different unit than observer 1, say he uses a meter stick half as long as that of observe 1. This corresponds to the coordinate change of my original post. Observer 1 and observer 2 will measure different densities of the fluid.

Now look at some observer 3, who is not at rest relative to the fluid. He tries to measure the physical density of the fluid. Does he choose the density which observer 1 measures, or the density which observer 2 measures?

 

[Dale]

It is the same density. 1000 kg/m^3 is the same density as 1 kg/L.

 

[PeterDonis]

Suppose we have observer 1, who is at rest relative to the fluid. Now suppose we have observer 2, whose worldline is the same as that of observer 1. In particular, observer 2 is also at rest relative to the fluid. However, observer 2 decides to measure all his length in in different unit than observer 1, say he uses a meter stick half as long as that of observe 1. This corresponds to the coordinate change of my original post.

No, it doesn't. See below.

Observer 1 and observer 2 will measure different densities of the fluid.

No, they won't. Observer 2's unit of volume will be smaller, but the amount of mass he measures to be in that volume will also be smaller by the same amount.

 

[PeterDonis]

Now look at some observer 3, who is not at rest relative to the fluid. He tries to measure the physical density of the fluid. Does he choose the density which observer 1 measures, or the density which observer 2 measures?

Neither. He picks his own units of mass and volume and measures the mass of the fluid per unit volume.

 

[klpskp]

I found the source of my confusion: Of course all observers who are at rest with respect to each other can agree on the volume of a given object, because the metric allows them to agree on the length of spacetime intervals. Then the density is the mass of the object, divided by this common volume, not divided by the product of the three coordinates that describe that object.

Thanks to everyone to help clarifying :)

 

[PeterDonis]

all observers who are at rest with respect to each other can agree on the volume of a given object, because the metric allows them to agree on the length of spacetime intervals.

You don't need the part in bold. All observers, whether they are at rest with respect to each other or not, will agree on the lengths of all spacetime intervals.

What they won't agree on are which spacetime intervals describe the "volume" of the object in their rest frame, if they are in relative motion. See further comments below.

Then the density is the mass of the object, divided by this common volume, not divided by the product of the three coordinates that describe that object.

"The product of three coordinates" is just another way of saying "the product of three particular spacetime intervals". The key is which three spacetime intervals describe the appropriate "volume", since, as noted above, different observers that are moving relative to each other will define different spacetime intervals as describing the sides of a cube at rest relative to them. The correct answer is the three spacetime intervals that describe the sides of a cube at rest relative to the fluid at some instant of time according to clocks at rest relative to the fluid. The scalar $\rho$ is just the rest mass of fluid inside that cube divided by its volume. All observers, whatever their state of motion, will agree on these numbers, because they are defined in terms of spacetime intervals, which are invariant for all observers.

So what makes $\rho$ a scalar is not that all observers will agree on the volume of a given fluid element--they won't, because observers in relative motion will use different spacetime intervals to describe the volume. What makes $\rho$ a scalar is that all observers will agree on the lengths of a particular set of three spacetime intervals, the ones that describe the volume of a fluid element for an observer at rest relative to the fluid. "At rest relative to the fluid" is the physical criterion that picks out that particular set of spacetime intervals.