I The Tensor & Metric: Spacetime Points & Momentum Flux

[dsaun777]

The components of the energy tensor are defined sometimes as the flux of the ith component of the momentum vector across some component jth of constant surface. But isn't the tensor a function of points of spacetime just as the metric? How can you evaluate a surface of j when the tensor is a function of a single 4d point?

 

[PeterDonis]

isn't the tensor a function of points of spacetime just as the metric?

A tensor has a particular value at each point of spacetime, yes.

How can you evaluate a surface of j when the tensor is a function of a single 4d point?

The surface in question is, strictly speaking, a surface in the tangent space at the point of spacetime where you are evaluating the tensor.

 

[dsaun777]

A tensor has a particular value at each point of spacetime, yes.
The surface in question is, strictly speaking, a surface in the tangent space at the point of spacetime where you are evaluating the tensor.

What defines the boundaries of the tangent space of surface?

 

[PeterDonis]

What defines the boundaries of the tangent space of surface?

The tangent space has no boundary. It isn't part of the spacetime. It's an abstract space in which vectors and tensors at a particular spacetime point are defined.

 

[dsaun777]

The tangent space has no boundary. It isn't part of the spacetime. It's an abstract space in which vectors and tensors at a particular spacetime point are defined.

So how does one measure flux across an abstract space with no boundaries?

 

[PeterDonis]

how does one measure flux across an abstract space with no boundaries?

You don't. You measure a flux across an actual surface in an actual small patch of spacetime. You mathematically model it using vectors and tensors in the tangent space at a chosen point on which the actual small patch of spacetime is centered. The mathematical model works because it gives you numbers that correspond to the numbers you get from the actual measurement you make in actual spacetime.

 

[PeterDonis]

You mathematically model it using vectors and tensors in the tangent space at a chosen point on which the actual small patch of spacetime is centered.

Another way of thinking about this is that the tangent space is an approximation to the actual small patch of spacetime which ignores any spacetime curvature in the small patch. So a particular surface in the tangent space is an approximation to the actual surface in spacetime across which you measure the flux. How good the approximation is depends on how small the patch of spacetime is and how large the spacetime curvature is.

 

[dsaun777]

Another way of thinking about this is that the tangent space is an approximation to the actual small patch of spacetime which ignores any spacetime curvature in the small patch. So a particular surface in the tangent space is an approximation to the actual surface in spacetime across which you measure the flux. How good the approximation is depends on how small the patch of spacetime is and how large the spacetime curvature is.

Would the x surface be like evaluating the partial of the 4d position vector with respect to your x coordinate?

 

[PeterDonis]

Would the x surface be like evaluating the partial of the 4d position vector with respect to your x coordinate?

The best way to think about components of the stress-energy tensor is that $T_{ab}$ gives the flux of $a$ momentum in the $b$ direction. The "surface" you are thinking about is just the surface perpendicular to the $b$ direction: if you think of a vector pointing in the $b$ direction, the surface is perpendicular to that vector. You can also think of the surface as a surface of constant coordinate $x^b$.

 

[dsaun777]

The flux represented by the $T_{0a}$ components is not an invariant quantity due to gamma factors correct? In relativity are the fluxes scalars or vectors or a mix. in some non relativistic texts I've seen flux as vectors and\or scalars i.e

The best way to think about components of the stress-energy tensor is that $T_{ab}$ gives the flux of $a$ momentum in the $b$ direction. The "surface" you are thinking about is just the surface perpendicular to the $b$ direction: if you think of a vector pointing in the $b$ direction, the surface is perpendicular to that vector. You can also think of the surface as a surface of constant coordinate $x^b$.

 

[Nugatory]

The flux represented by the $T_{0a}$ components is not an invariant quantity due to gamma factors correct?

It is not invariant because it depends on the coordinate system you’re using.

You can choose two coordinate systems that are related by the Lorentz transforms, and then the $\gamma$ factor will appear but that’s a special case resulting from choosing to transform the SE tensor from one coordinate system to another one that just happens to work that way.

In general, the components of a tensor depend on the choice of coordinate system (this will be clear if you look at the tensor transformation rule) so will not be invariant.

 

[dsaun777]

It is not invariant because it depends on the coordinate system you’re using.

You can choose two coordinate systems that are related by the Lorentz transforms, and then the $\gamma$ factor will appear but that’s a special case resulting from choosing to transform the SE tensor from one coordinate system to another one that just happens to work that way.

In general, the components of a tensor depend on the choice of coordinate system (this will be clear if you look at the tensor transformation rule) so will not be invariant.

But there are factors that remain invariant such as total energy that passes through the modeled tangent space of a small patch of spacetime correct?

 

[Nugatory]

But there are factors that remain invariant such as total energy that passes through the modeled tangent space of a small patch of spacetime correct?

Nothing passes through the tangent space - it’s a mathematical abstraction that allows us to define and manipulate vectors at a point. But that’s a digression from the main thrust of your question, for which the answer is...

The various components of a tensor transform in such a way that any invariant quantity calculated from the tensor (such as the energy flux in a particular direction at a given point) comes out the same. That’s why we use tensors in the first place - because they allow us to work with any coordinate system.

There is a subtlety around that phrase “in a particular direction” though. $T_{0a}$ is the momentum flux in the $a$ direction (for a suitably chosen coordinate system - @PeterDonis's "$T_{ab}$ is the flux of $a$ momentum in the $b$ direction" is more complete and accurate) but a given direction is only "the $a$ direction" if we're using coordinates such that the $x_a$ axis points in that direction.

 

[dsaun777]

Nothing passes through the tangent space - it’s a mathematical abstraction that allows us to define and manipulate vectors at a point. But that’s a digression from the main thrust of your question, for which the answer is...

The various components of a tensor transform in such a way that any invariant quantity calculated from the tensor (such as the energy flux in a particular direction at a given point) comes out the same. That’s why we use tensors in the first place - because they allow us to work with any coordinate system.

There is a subtlety around that phrase “in a particular direction” though. $T_{0a}$ is the momentum flux in the $a$ direction (for a suitably chosen coordinate system - @PeterDonis's "$T_{ab}$ is the flux of $a$ momentum in the $b$ direction" is more complete and accurate) but a given direction is only "the $a$ direction" if we're using coordinates such that the $x_a$ axis points in that direction.

The $T_{0a}$ for a=0123 is the energy flux, is this flux invariant?

 

[pervect]

The $T_{0a}$ for a=0123 is the energy flux, is this flux invariant?

No. $T^{00}$ can be alternatively interpreted as the density of energy per unit volume.

First a comment on the "flux" interpretation. Consider a 4d cubical region of space-time, containing a lump of stationary matter. So it's a 3-d cube of matter, for a duration of one time unit. Then $T^{00}$ is the flux of energy entering the 4-d volume form the past, and exiting at the future, the flux of energy in the "time" direction. This is just the amount of energy in the cube.

Let's do an example. Suppose say we have an object in a rest frame, that's a piece of matter with some density $\rho$. And we'll use geometric units, where c=1. Because of this, we can use the terms "matter" and "energy" as being equivalents, because the rest energy matter is E=mc^2, and c=1. Then we can write the compnents of the stress energy tensor T in the usual (t,x,y,z) Minkowskii basis as:

$$T = \begin{bmatrix} \rho & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$

This means that $T^{00} = \rho$, the density of the object, and the other components are all zero.

No, we boost this to a moving frame in the -x direction, with an index of "1". The tensor equation for how the components transform is:

$$T^{cd} = T^{ab} \Lambda^c{}_a \Lambda^d{}_b$$

where $\Lambda$ is the Lorentz transform matrix.

$$\Lambda = \begin{bmatrix} \gamma & \beta \gamma & 0 & 0 \\ \beta \gamma & \gamma & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$

Since I boosted in the -x direction, $\beta$ is positive, but feel free to make it negative if you wish. The important thing is the result of this boost, which is the stress-energy tensor of the same piece of matter moving in the +x direction. Let's call the boosted stress-energy tensor T'.

Because the only nonzero term is T is $T^{00} = \rho$, we can write
$$T'^{00} = \Lambda^0{}_0 \Lambda^0{}_0 T^{00} \quad T'^{01} = \Lambda^0{}_0 \Lambda^1{}_0 T^{00} \quad T'^{10} = \Lambda^1{}_0 \Lambda^0{}_0 T^{00} \quad T^{11} = \Lambda^1{}_1 \Lambda^1{}_1 T^{00}$$

$$T = \begin{bmatrix} \gamma^2 \rho & \beta \gamma^2 \rho & 0 & 0 \\ \beta \gamma^2 \rho & \beta^2 \gamma^2 \rho & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$

So - what has happened? $T^{00}$ the density of energy, has increased by a factor of $\gamma^2$. There is one factor of gamma for the increase in energy due to the motion of the mass, and another factor of gamma as this is the density of energy / unit volume, and Lorentz contraction means that the volume of the moving object is smaller by a factor of $\gamma$.

We can clearly see that $T^{00}$ is not invariant, as it depends on the frame.

Similarly, $T^{01} = T^{10}$ is the density of x-momentum. The mildly tricky thing to interpret is $T^{11}$. It's absolutely necessary to have the term though, for the tensor transformation properties to remain valid. I view the matter as a fluid, and $T^{11}$ as the dynamic pressure, but this may be somewhat of a personal interpretation.

I don't use your flux picture much, though I've seen it in writing (Caroll & Baez). Let me try and put the results in those terms.

$T^{00}$ is the flux of energy in a 4-volume from the past to the future, the flux in the "time" direction. $T^{01}$ is the flux in a 4-volume of energy in the x direction, i.e the momentum. $T^{11}$ is the flux of x-momentum in the x direction, which is a sort of dynamic pressure.

It might be helpful to consider the 1-dimensoinal case. So we imagine we have only one space dimension, and one time dimension. And we have a piece of string in this simple example. The piece of string has some rest frame, with basis vectors $u$ and $w$. u represents the time basis vector for the rest frame of our piece of string, and $w$ represent the spatial basis vector.

Our piece of matter in 1 dimension has only two physical quaities of interest. Those are it's density, and it's pressure / tension. If the piece of string isn't under stress, the tension/pressure term is zero. But if it is under tension (say the string is supporting a weight), it's under tension, and $T^{11}$ is negative. If it was in compression, $T^{11}$ would be positive. In the rest frame of the string, there is no momentum, so we can set that to zero.

We can write the stress-energy tensor of this string in it's rest frame in tensor notation in a couple of different ways.

with index free notation

$$T = \rho u \otimes u + P w \otimes w$$

In component language, i.e. abstract index notation, we'd write

$$T^{ab} = \rho u^a u^b + P w^a w^b$$

Because this is a tensor equation, if it's valid in the rest frame of the string, it's valid in any frame. So, we now have the stress-energy tensor of our string in any frame of reference.

The 3d case is more confusing, we just have 2 more spatial dimensions. The simple idea of tension/pressure in the string gets replaced by the classical 3d stress tensor.

 

[dsaun777]

I appreciate this well thought out response thank you. The reason I like the flux approach is that what I'm imagining is that I am at rest with the 4d cube of spacetime and I'm measuring how much energy passes through my 4d cube in a given time. If there is another observer that is boosted along some axis then he will perceive a different value because of volume contraction and relativistic momentum. However, there is still an agreed upon value of how much is flowing through the 4d box, it's just that there are different flow rates due to gamma.
What is the contracted energy momentum tensor? Is that a quantity that all observers agree upon?

 

[Nugatory]

If there is another observer that is boosted along some axis then he will perceive a different value because of volume contraction and relativistic momentum.

He will not.

The flux through the surface of a solid region of spacetime is the same for all observers, so as long as you're taking about the same region of spacetime you'll both get the same result. Your disagreement will be about the shape of that volume: You will say that it is a 4d hypercube with edges $dx$, $dy$, $dz$ and $dt$; because they are using different coordinates they will describe the same region of spacetime as some more complicated shape with no correspondingly simple description in their coordinates. The components of the tensor will be different in his coordinates, but so will be the components of the vector normal to the surface, as well as the coordinate description of the surface across which we're integrating. All of these changes together combine in such a way that the final answer comes out the same.

Now, if we compare the flux across the surface of you call a 4d hypercube at a given point and what they call a 4d hypercube at that point, we will get different results. But that's to be expected - now we're talking about different volumes with different surfaces.

 

[dsaun777]

He will not.

The flux through the surface of a solid region of spacetime is the same for all observers, so as long as you're taking about the same region of spacetime you'll both get the same result. Your disagreement will be about the shape of that volume: You will say that it is a 4d hypercube with edges $dx$, $dy$, $dz$ and $dt$; because they are using different coordinates they will describe the same region of spacetime as some more complicated shape with no correspondingly simple description in their coordinates. The components of the tensor will be different in his coordinates, but so will be the components of the vector normal to the surface, as well as the coordinate description of the surface across which we're integrating. All of these changes together combine in such a way that the final answer comes out the same.

Now, if we compare the flux across the surface of you call a 4d hypercube at a given point and what they call a 4d hypercube at that point, we will get different results. But that's to be expected - now we're talking about different volumes with different surfaces.

What your saying is my hypercube gets contracted into something different when compared to an outside boosted observer's cube? But we both the measure the same amount of energy after 4d integration?
just on a side not, is the contracted energy momentum tensor invariant? I know it equals the minus ricci scalar but what is it? I think for dust it is simply rho-3p