Canonical energy momentum tensor: Quick question

[binbagsss]

This is probably a stupid question but say I am in Minkowski s-t and I want to find $T_{ii}$,

(Assuming signature $(+,-,-,-)$), I get :

$(firstterm) + L = T^{ii}$

$T^{ii}$ is a vector but $L$ is a scalar (obviously I appreciate for a more complicated metric the vector form will be apparent) but in this case it does not look right/ how should I think of it to understand it?


Thanks

 

[ergospherical]

$T_{ii}$ presumably with no sum (i.e. fixed value of i)? That is a number.

($\sum_i T_{ii}$ is a different number)

 

[Orodruin]

$T^{11}$ is not a vector. It is a component of a rank 2 tensor. So is $\delta_{11} = 1$.

 

[binbagsss]

I thought $T_{00}$ denotes a scalar, $T_{i0}$ denotes a vector, as does $T_{ii}$? e.g the energy density, energy/momentum flux density,and $T_{ii}$ the momentum flux? (since $i$ runs over1,2,3)

* T_{ii} a rank 2 apologies, just not a scalar.

 

[binbagsss]

$T_{ii}$ presumably with no sum (i.e. fixed value of i)? That is a number.

($\sum_i T_{ii}$ is a different number)

$\eta^{ii}$ let denote the Minkowski metric. Takes the same value for all $i$. so it wouldn't make sense to write $-\eta^{ii}L$ (although then the rank-two form is apparent) , but to simplify it as $-L$...

 

[Orodruin]

Don’t confuse tensor components with the tensor itself.

Unless you are using abstract index notation, $T^{\mu\nu}$ are the components of $T$.

The expression $T_{ii}$ makes little sense at all. By the summation convention this should be summed over $i$, but it seems to not be what you intend. I think you need to take a step back and collect yourself in terms of tensor notation.

 

[binbagsss]

Don’t confuse tensor components with the tensor itself.

Unless you are using abstract index notation, $T^{\mu\nu}$ are the components of $T$.

The expression $T_{ii}$ makes little sense at all. By the summation convention this should be summed over $i$, but it seems to not be what you intend. I think you need to take a step back and collect yourself in terms of tensor notation.

ok, to give context, I have expressions for $T_{00}, T_{ii}, T_{i0}$ by identifying them with the 'energy density' , 'momentum flux ' etc. I want to use the canonical expression for the energy-momentum tensor (a kind of inverse Noether's theorem) to solve for $L$ in terms of the fields.

I therefore take the canonical expression for $T^{\mu \nu}$ and want to get expressions for $T_{ii}$, $T_{i0}$

$T_{ij}$ perhaps makes more sense then, just the components not involving the time components where $i \neq j$

 

[binbagsss]

$T_{i0}=T_{0i}$ the momentum flux /energy flux is a vector...?

 

[martinbn]

$T_{i0}=T_{0i}$ the momentum flux /energy flux is a vector...?

This is irrelevant. It was already said, the formula in your question gives the components of the tensor. They are numbers.

 

[dextercioby]

$T_{ii}|_{i\in\{1,2,3\}}$ are the three elements of the trace of the stress tensor. More here: https://en.wikipedia.org/wiki/Stress–energy_tensor, precisely point c. under Components.

 

[binbagsss]

This is irrelevant. It was already said, the formula in your question gives the components of the tensor. They are numbers.

The formula gives $T^{\mu nu}$. Isn't that just like saying $T^{\mu \nu}$ are components, rather than a rank - 2 tensor? Why can't $T^{ij}$ denote the 3d matrix running over the indices 1,2,3, rather than the case $\mu =1,2,3,4$? what else should I use to write this then?

So you're telling me its impossible to get an expression for the vector for energy flux by choosing indices to get $T^{i0}$ from the canonical energy-momentum tensor expression?

 

[martinbn]

The formula gives $T^{\mu nu}$. Isn't that just like saying $T^{\mu \nu}$ are components, rather than a rank - 2 tensor? Why can't $T^{ij}$ denote the 3d matrix running over the indices 1,2,3, rather than the case $\mu =1,2,3,4$? what else should I use to write this then?

So you're telling me its impossible to get an expression for the vector for energy flux by choosing indices to get $T^{i0}$ from the canonical energy-momentum tensor expression?

Of course you can. What exactly is the problem?

 

[binbagsss]

Of course you can. What exactly is the problem?

OP was the question. If consdiering $T_{ij}$ where $i=j$ then on the first term the two rank tensor will be apparent in the index noration, for the lsat term $\eta^{ij}$ the rank 2 tensor is not apparent as $eta_{ij}=\delta^{ij}$, would just get $-L$

 

[martinbn]

OP was the question. If consdiering $T_{ij}$ where $i=j$ then on the first term the two rank tensor will be apparent in the index noration, for the lsat term $\eta^{ij}$ the rank 2 tensor is not apparent as $eta_{ij}=\delta^{ij}$, would just get $-L$

No, eta is not a number, it has components.

 

[binbagsss]

No, eta is not a number, it has components.

yes !! but $\eta_{uv}=diag(1,-1,-,1,-1)$ . has the same value for the spatial compoentns so one would plug it in rather than leaving it in component form...

 

[Ibix]

I think you are starting with the expression $$T^{\mu\nu}=\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\partial^\nu\phi-\eta^{\mu\nu}\mathcal{L}$$Then you are saying that at present you are only interested in the three spacelike on-diagonal components in an Einstein coordinate system, which are$$T^{ii}=\frac{\partial\mathcal{L}}{\partial(\partial_i\phi)}\partial^i\phi-\eta^{ii}\mathcal{L}$$where $i=1,2,3$ and no summation convention is assumed. Note that this is not a tensor expression. It is simply three expressions stating a relationship between $\mathcal{L}$, $\phi$ and specific components of $T$. That is, this is explicitly these three expressions:$$\begin{eqnarray*}
T^{11}&=&\frac{\partial\mathcal{L}}{\partial(\partial_1\phi)}\partial^1\phi-\eta^{11}\mathcal{L}\\
T^{22}&=&\frac{\partial\mathcal{L}}{\partial(\partial_2\phi)}\partial^2\phi-\eta^{22}\mathcal{L}\\
T^{33}&=&\frac{\partial\mathcal{L}}{\partial(\partial_3\phi)}\partial^3\phi-\eta^{33}\mathcal{L}
\end{eqnarray*}$$Because these are boring old differential equations, you may now substitute values for the components of $\eta^{ij}$. You no longer expect the indices to match, any more than you do when you say something like $\eta^{22}=-1$.

Your problem is that you substituted in some components of the metric, then expected the result not to be an expression about components.

 

[binbagsss]

I think you are starting with the expression $$T^{\mu\nu}=\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\partial^\nu\phi-\eta^{\mu\nu}\mathcal{L}$$Then you are saying that at present you are only interested in the three spacelike on-diagonal components in an Einstein coordinate system, which are$$T^{ii}=\frac{\partial\mathcal{L}}{\partial(\partial_i\phi)}\partial^i\phi-\eta^{ii}\mathcal{L}$$where $i=1,2,3$ and no summation convention is assumed. Note that this is not a tensor expression. It is simply three expressions stating a relationship between $\mathcal{L}$, $\phi$ and specific components of $T$. That is, this is explicitly these three expressions:$$\begin{eqnarray*}
T^{11}&=&\frac{\partial\mathcal{L}}{\partial(\partial_1\phi)}\partial^1\phi-\eta^{11}\mathcal{L}\\
T^{22}&=&\frac{\partial\mathcal{L}}{\partial(\partial_2\phi)}\partial^2\phi-\eta^{22}\mathcal{L}\\
T^{33}&=&\frac{\partial\mathcal{L}}{\partial(\partial_3\phi)}\partial^3\phi-\eta^{33}\mathcal{L}
\end{eqnarray*}$$Because these are boring old differential equations, you may now substitute values for the components of $\eta^{ij}$. You no longer expect the indices to match, any more than you do when you say something like $\eta^{22}=-1$.

Your problem is that you substituted in some components of the metric, then expected the result not to be an expression about components.

but is it not true for #ii# general expression you get for the second term $-L$. Why wouldn't one sub that in?

"Note that this is not a tensor expression. It is simply three expressions". That is like saying obtainin an expression for $T_{i0}$ is not a vector. Obviously it is the energy flux and so it is a vector?

 

[Ibix]

but is it not true for #ii# general expression you get for the second term $-L$. Why wouldn't one sub that in?

It's only true in Einstein coordinates. You are picking those coordinates and then picking some of the components.

That is like saying obtainin an expression for $T_{i0}$ is not a vector.

You've only got three components. It may or may not be a three vector. It's certainly not a four vector.

 

[binbagsss]

It's only true in Einstein coordinates. You are picking those coordinates and then picking some of the components.

You've only got three components. It may or may not be a three vector. It's certainly not a four vector.

pretty sure i said a hundredf times i dont think its a four vector. i said in miskowski s-t. i.e. what most textbooks or lit would regard as minkowski. question still hasn;t really been answered

 

[Ibix]

pretty sure i said a hundredf times i dont think its a four vector.

Right. So why are you surprised the expression doesn't work like a four vector?

question still hasn;t really been answered

It has, several times. You have expressions for some components of a tensor. They do not function as tensor equations.

You understand that $\eta^{22}=-1$ is not a tensor expression, right? You just have a more algebraically complex version of this.

 

[Ibix]

no, i really struggled to grasp that concept when it was replied to me about 10 posts ago

Think about a pair of ordinary three vectors $\vec u$ and $\vec v$. You can think of them as arrows in space, and you can write their inner product $\vec u\cdot\vec v=A$. That is a vector equation. It's true in any coordinate system.

There exists a coordinate system in which the first component of $\vec u$ is zero - i.e., $u^1=0$. That is not a vector expression because it's not a general truth about the vector. It's only true in some coordinate systems. And you can see that the indices don't match up, but that's fine because it's not an expression about the tensor, just one of its components.

For a slightly more complex example, consider the electric field three vector of a plane EM wave. We can write $E^x=A\sin(kz-\omega t)$ and $E^y=E^z=0$. Again, note that none of the indices match up, but that's fine because these are components. But we can write plenty of vector equations that are true no matter how we represent the vectors, like $\nabla\cdot\vec E=\rho$.

The expression you started with in #1 can be interpreted as a single statement about the tensor $T$, or as sixteen statements about the components of that tensor. When you picked a particular form of $\eta$ and picked out three particular components to look at, you came down on the side of interpreting it as components.

 

[ergospherical]

The reason that the indices match in tensorial equations is because tensors are functions, so you have to be able to act both sides of the equation on the same set of arguments. Taking components = evaluating the function on the relevant basis vectors (T_ij := T(e_i, e_j)), and now you have an equation relating numbers with other numbers rather than tensors with other tensors. In which case there's no longer any reason for the indices to match (e.g. "η₂₂ = -1").

Which is why Sir Roger Penrose put forth abstract index notation: because mathematicians are fearful creatures who can't possibly run the risk of conflating tensorial equations with their coordinate representations.

 

[binbagsss]

The energy flux expression is given by the above in some text. Why isn't this wrote as $S^{\alpha}_0$? Because $0$ is a specified index right so not an index as such, but,

E.g when one refers to the components of the energy-momentum tensor as follows , one would write $T^{i0}$, not $T^i$.. this is kind of what I am getting at:

 

[Orodruin]

Because energy flux is a spatial 3-vector. It is forms a vector under the three-dimensional subgroup of rotations to the Lorentz group. In terms of the full Lorentz group you obviously have $T^{i0}$ but the particular choice of O(3) specifies a reat frame.

 

[binbagsss]

Because energy flux is a spatial 3-vector. It is forms a vector under the three-dimensional subgroup of rotations to the Lorentz group. In terms of the full Lorentz group you obviously have $T^{i0}$ but the particular choice of O(3) specifies a reat frame.

can you explain why it selects a rest frame/ point me to some links? many thanks

 

[Ibix]

can you explain why it selects a rest frame/ point me to some links? many thanks

Because what is a pure rotation in one frame is a mix of rotation and boost in another. Or to put it another way, to pull out the spatial components of a four vector/tensor you have to have picked a definition of "space". Another frame will not pick out the same values.