Density of energy from the stress-energy tensor

[fluidistic]

Homework Statement

Hi guys,
I would like to show that if $t^\mu$ is a temporal vector then $t^\mu t^\nu T_{\mu\nu}$ is the density of energy of the EM field measured by an observer with velocity $t^\mu$. And that it is greater or equal to 0.
Density of energy is proportional to $E^2+B^2$ if my memory doesn't fail me.

Homework Equations

Mostly tensor "operations"?
Def. of the S-E tensor: $T^{\mu \nu }=\frac{1}{4\pi} \left ( F^{\mu \sigma } F^{\nu \rho} \eta _{\sigma \rho}-\frac{1}{4} \eta ^{\mu \nu } F^{\sigma \rho} F_{\sigma \rho} \right )$

The Attempt at a Solution

I'm trying to grasp the basics of tensors through self study. I would like some feedback on this: $T_{\mu \nu}=\frac{1}{4\pi} \left ( F_{\mu\sigma}F_{\nu\rho}\eta^{\sigma\rho}-\frac{1}{4} \eta_{\mu\nu}F_{\sigma\rho}F^{\sigma\rho} \right )$. Is this expression correct? I just used the definition given and lowered upper indices and rised lower indices.
My idea is to perform/analyse first $t^\nu T_{\mu\nu}$ and then contract it with the velocity vector $t^\mu$.
So I wrote out the terms of $t^\nu T_{\mu\nu}$, and since $t^\nu$ is temporal I chose it as (1,0,0,0) in my mind. So the only non zero "terms" are $t^0T_{00}+t^0T_{10}+t^0T_{20}+t^0T_{30}=\sum _{\mu=0}^3 T_{\mu0}$.
I am not really sure what this sum is. It is a sum of what exactly? Of scalars? Can't be right... because then $t^\mu$ multiplied by a scalar is another vector and the density of energy is supposed to be a scalar...


More thoughts: Then $t^\mu t^\nu T_{\mu\nu}=t^\mu T_{\mu 0}=t^0 T_{00}+t^1 T_{10}+t^2 T_{20}+t^3 T_{30}=c\underbrace{t}_{\text {time}}T_{00}+t^1 T_{10}+t^2 T_{20}+t^3 T_{30}$.
Where I considered $t^\mu=(ct,t^1,t^2,t^3)$, the 4-velocity.
Now I guess I must evaluate $T_{\mu 0}$ and replace back into the expression I got for $t^\mu t^\nu T_{\mu\nu}$.
Is this correct so far?

 

[WannabeNewton]

O.k. knowing the nature of your professor's notes personally, you're going about it in a much more cumbersome way than needed given the tools in your notes. So let's start a bit differently i.e. in a frame independent manner (sometimes it is convenient to do calculations in a convenient choice of frame or coordinate system but in many other cases it is simply much more elegant and easier to do things in a frame/coordinate free way).

You know for starters that $F_{\mu\nu} = 2E_{[\mu} t_{\nu]} + \epsilon_{\mu\nu\alpha\beta}t^{\alpha}B^{\beta}$ right? Also, keep in mind the following identities (that are also probably in your notes) $\epsilon_{\mu\nu\alpha\beta}\epsilon^{\mu\gamma\sigma\rho} = -3!\delta^{[\gamma}_{\nu}\delta^{\sigma}_{\alpha}\delta^{\rho]}_{\beta}$ and $\epsilon_{\mu\nu\alpha\beta}\epsilon^{\mu\nu\sigma\rho} = -4\delta^{[\sigma}_{\alpha}\delta^{\rho]}_{\beta}$. Now calculate $F_{\sigma\rho}F^{\sigma \rho}$ and $F^{\mu\sigma}F_{\sigma \nu}t_{\mu}t^{\nu}$ using the above expression for $F_{\mu\nu}$ and the given identities.

 

[TSny]

fluidistic, I don't want to steer you away from WannabeNewton, but I just wanted to comment on your attempt so you can see where you have a mistake.

I'm trying to grasp the basics of tensors through self study. I would like some feedback on this: $T_{\mu \nu}=\frac{1}{4\pi} \left ( F_{\mu\sigma}F_{\nu\rho}\eta^{\sigma\rho}-\frac{1}{4} \eta_{\mu\nu}F_{\sigma\rho}F^{\sigma\rho} \right )$. Is this expression correct? I just used the definition given and lowered upper indices and rised lower indices.

Looks OK. Note that you didn't need to raise and lower the $\sigma$ and $\rho$ indices in the last term. You could have left it as

$T_{\mu \nu}=\frac{1}{4\pi} \left ( F_{\mu\sigma}F_{\nu\rho}\eta^{\sigma\rho}-\frac{1}{4} \eta_{\mu\nu}F^{\sigma\rho}F_{\sigma\rho} \right )$

But what you wrote is fine.

My idea is to perform/analyse first $t^\nu T_{\mu\nu}$ and then contract it with the velocity vector $t^\mu$.
So I wrote out the terms of $t^\nu T_{\mu\nu}$, and since $t^\nu$ is temporal I chose it as (1,0,0,0) in my mind.

Yes, that's right. For the observer, $t^\nu = (1,0,0,0)$

So the only non zero "terms" are $t^0T_{00}+t^0T_{10}+t^0T_{20}+t^0T_{30}=\sum _{\mu=0}^3 T_{\mu0}$.

There's a mistake here. In the expression $t^\nu T_{\mu\nu}$ you are summing over $\nu$ and not summing over $\mu$. So, $\nu$ takes the values 0, 1, 2, 3 while $\mu$ stays fixed.

 

[fluidistic]

Thanks guys.
WBN:I'll try "your" way and take for granted the formulae.

There's a mistake here. In the expression $t^\nu T_{\mu\nu}$ you are summing over $\nu$ and not summing over $\mu$. So, $\nu$ takes the values 0, 1, 2, 3 while $\mu$ stays fixed.

You're right, I misreported what I did on my draft which is somehow messy. But in the end I do reach that $t^\mu t^\nu T_{\mu\nu}=t^\mu T_{\mu 0}=t^0 T_{00}+t^1 T_{10}+t^2 T_{20}+t^3 T_{30}=c\underbrace{t}_{\text {time}}T_{00}+t^1 T_{10}+t^2 T_{20}+t^3 T_{30}$. I don't know why the notation uses t for both time and velocity, it can get pretty confusing to me.

 

[WannabeNewton]

You don't have to take the formulae for granted. I know your professor's lecture notes have the formulas themselves if I recall correctly but if they don't have derivations of them then you can always look at the notes I showed you recently as they have the derivations.

 

[TSny]

I'm not familiar with the phrase "temporal vector" as used in relativity. I suspect it just means that the vector is "time-like". If the vector also has norm 1: $t^\mu t_\mu = 1$, then the vector could represent the 4-velocity vector of some observer: $t^\mu = (\frac{dt}{d\tau}, \frac{dx^1}{d\tau}, \frac{dx^2}{d\tau}, \frac{dx^3}{d\tau}).$ (Assuming units where c = 1). So, the $t$ in $t^\mu$ does not stand for time, it's representing components of 4-velocity. In particular, $t^0$ does not represent time. It is the zeroth component of the 4-velocity of the observer.

Anyway, the problem statement says to interpret $t^\mu$ as the velocity of an observer. In the frame of the observer, this reduces to $t^\mu = (1, 0, 0, 0)$, as you noted.

You found $t^\mu t^\nu T_{\mu\nu}=t^\mu T_{\mu 0}=t^0 T_{00}+t^1 T_{10}+t^2 T_{20}+t^3 T_{30}$

What does this reduce to after substituting for $t^0, t^1, t^2, t^3$?

 

[fluidistic]

I'm going back to you in a bit TSny, I appreciate the help.
WNB: I do not really understand why I should focus on calculating $F_{\sigma\rho}F^{\sigma \rho}$ and $F^{\mu\sigma}F_{\sigma \nu}t_{\mu}t^{\nu}$.
I tried the first expression: $F_{\mu\nu}F^{\mu\nu}=(2E_{[\mu} t _ {\nu ]} +\varepsilon_{\mu\nu\alpha\beta } t^\alpha B^\beta )(2E^{[\mu} t ^ {\nu ]} +\varepsilon^{\mu\nu\alpha\beta } t_\alpha B_\beta)$
$$=\underbrace{ 4 E_{[\mu} t _ {\nu ]} E^{[\mu} t ^ {\nu ]} }_I+\underbrace{ 2E_{[\mu} t _ {\nu ]} \varepsilon ^{\mu\nu\alpha\beta } t_\alpha B_\beta +2\varepsilon _{\mu\nu\alpha\beta } t^\alpha B^\beta E^{[\mu} t ^ {\nu ]} }_J+\underbrace{\varepsilon_{\mu\nu\alpha\beta } t^\alpha B^\beta \varepsilon ^{\mu\nu\alpha\beta } t_\alpha B_\beta }_K$$
I calculated $I$ to be worth 0, it's 2 lines in my draft, I will post the details if I'm wrong.
Is this okay so far?

 

[fluidistic]

I'm not familiar with the phrase "temporal vector" as used in relativity. I suspect it just means that the vector is "time-like". If the vector also has norm 1: $t^\mu t_\mu = 1$, then the vector could represent the 4-velocity vector of some observer: $t^\mu = (\frac{dt}{d\tau}, \frac{dx^1}{d\tau}, \frac{dx^2}{d\tau}, \frac{dx^3}{d\tau}).$ (Assuming units where c = 1). So, the $t$ in $t^\mu$ does not stand for time, it's representing components of 4-velocity. In particular, $t^0$ does not represent time. It is the zeroth component of the 4-velocity of the observer.

Anyway, the problem statement says to interpret $t^\mu$ as the velocity of an observer. In the frame of the observer, this reduces to $t^\mu = (1, 0, 0, 0)$, as you noted.

You found $t^\mu t^\nu T_{\mu\nu}=t^\mu T_{\mu 0}=t^0 T_{00}+t^1 T_{10}+t^2 T_{20}+t^3 T_{30}$

What does this reduce to after substituting for $t^0, t^1, t^2, t^3$?

In such case I get $t^\mu t^\nu T_{\mu\nu}=t^0T_{00}=T_{00}$.
Which is worth $\frac{1}{4\pi}(F^{0\sigma }F^{0\rho } \eta_{\sigma \rho }-\frac{1}{4}\eta ^{00}F^{\sigma \rho }F_{\sigma \rho})$ where $\eta^{00}=1$ with the metric convention I'm using.

 

[WannabeNewton]

I'm going back to you in a bit TSny, I appreciate the help.
WNB: I do not really understand why I should focus on calculating $F_{\sigma\rho}F^{\sigma \rho}$ and $F^{\mu\sigma}F_{\sigma \nu}t_{\mu}t^{\nu}$.
I tried the first expression: $F_{\mu\nu}F^{\mu\nu}=(2E_{[\mu} t _ {\nu ]} +\varepsilon_{\mu\nu\alpha\beta } t^\alpha B^\beta )(2E^{[\mu} t ^ {\nu ]} +\varepsilon^{\mu\nu\alpha\beta } t_\alpha B_\beta)$
$$=\underbrace{ 4 E_{[\mu} t _ {\nu ]} E^{[\mu} t ^ {\nu ]} }_I+\underbrace{ 2E_{[\mu} t _ {\nu ]} \varepsilon ^{\mu\nu\alpha\beta } t_\alpha B_\beta +2\varepsilon _{\mu\nu\alpha\beta } t^\alpha B^\beta E^{[\mu} t ^ {\nu ]} }_J+\underbrace{\varepsilon_{\mu\nu\alpha\beta } t^\alpha B^\beta \varepsilon ^{\mu\nu\alpha\beta } t_\alpha B_\beta }_K$$
I calculated $I$ to be worth 0, it's 2 lines in my draft, I will post the details if I'm wrong.
Is this okay so far?

$I$ shouldn't be zero. In fact $4E_{[\mu}t_{\nu]}E^{[\mu}t^{\nu]} = (E_{\mu}t_{\nu} - E_{\nu}t_{\mu})(E^{\mu}t^{\nu} - E^{\nu}t^{\mu}) = -2E_{\mu}E^{\mu} = -2\left \| E \right \|^{2}$ because $E_{\mu}t^{\mu} = 0$ and $t_{\mu}t^{\mu} = -1$. Do you see that?

$J$ however is zero. Can you see why? Also in your last expression, you have dummy indices appearing more than twice again, which you can't have. It should be $\epsilon_{\mu\nu\alpha\beta}t^{\alpha}B^{\beta}\epsilon^{\mu\nu\sigma \rho}t_{\sigma}B_{\rho}$. Don't forget the identities for $\epsilon_{\mu\nu\alpha\beta}$.

We are calculating $F_{\sigma\rho}F^{\sigma \rho}$ and $F^{\mu\sigma}F_{\sigma \nu}t_{\mu}t^{\nu}$ for use in the expression for $T_{\mu\nu}t^{\mu} t^{\nu}$.

 

[TSny]

In such case I get $t^\mu t^\nu T_{\mu\nu}=t^0T_{00}=T_{00}$.
Which is worth $\frac{1}{4\pi}(F^{0\sigma }F^{0\rho } \eta_{\sigma \rho }-\frac{1}{4}\eta ^{00}F^{\sigma \rho }F_{\sigma \rho})$ where $\eta^{00}=1$ with the metric convention I'm using.

Yes, that looks good.

 

[fluidistic]

$I$ shouldn't be zero. In fact $4E_{[\mu}t_{\nu]}E^{[\mu}t^{\nu]} = (E_{\mu}t_{\nu} - E_{\nu}t_{\mu})(E^{\mu}t^{\nu} - E^{\nu}t^{\mu}) = -2E_{\mu}E^{\mu} = -2\left \| E \right \|^{2}$ because $E_{\mu}t^{\mu} = 0$ and $t_{\mu}t^{\mu} = -1$. Do you see that?

Hmm no I don't really see this. I understand that $t_{\mu}t^{\mu}$ is the norm of the time-like vector and in my case is worth 1-0-0-0=1 if I use the convention (+,-,-,-). Am I doing a sign error?
About $E_{\mu}t^{\mu} = 0$, I don't see why this hold. Wouldn't that mean that $E^\mu=(0,E^1,E^2,E^3)$? Why would the first component be zero?
Here is what I did: $I=4 \cdot \frac{1}{2} (E_\mu t_\nu -E_\nu t_\mu) \cdot \frac{1}{2} (E^\mu t^\nu -E^\nu t^\mu)=(E_{\mu}t_{\nu} - E_{\nu}t_{\mu})(E^{\mu}t^{\nu} - E^{\nu}t^{\mu})=E_\mu t_\nu E^\mu t^\nu -E_\mu t_\nu E^\nu t^\mu -E_\nu t_\mu E^\mu t^\nu +E_\nu t_\mu E^\nu t^\mu$ $=2(E_\mu t_\nu E^\mu t^\nu - E_\mu t_\nu E^\nu t^\mu )=2E_\mu t_\nu (E^\mu t^\nu -E^\nu t^\mu )=0$. Hmm I guess I can't make the last parenthesis worth 0, can't change mu by nu and nu by mu in the second term because of the term I factorized outside the parenthesis... but I'm not sure.

$J$ however is zero. Can you see why?

No, really I don't see why.

Also in your last expression, you have dummy indices appearing more than twice again, which you can't have. It should be $\epsilon_{\mu\nu\alpha\beta}t^{\alpha}B^{\beta}\epsilon^{\mu\nu\sigma \rho}t_{\sigma}B_{\rho}$. Don't forget the identities for $\epsilon_{\mu\nu\alpha\beta}$.

I see.

We are calculating $F_{\sigma\rho}F^{\sigma \rho}$ and $F^{\mu\sigma}F_{\sigma \nu}t_{\mu}t^{\nu}$ for use in the expression for $T_{\mu\nu}t^{\mu} t^{\nu}$.

Ok. I'm asked about $t^\mu t^\nu T_{\mu\nu}$, not really different from $T_{\mu \nu } t^{\mu } t^{\nu }$ right?

 

[WannabeNewton]

Hmm no I don't really see this. I understand that $t_{\mu}t^{\mu}$ is the norm of the time-like vector and in my case is worth 1-0-0-0=1 if I use the convention (+,-,-,-). Am I doing a sign error?

Based on your expression for the energy-momentum tensor, you should be using the (-,+,+,+) convention. In this convention, $t^{\mu}t_{\mu} = -1$.

About $E_{\mu}t^{\mu} = 0$, I don't see why this hold.

Recall that $E_{\mu} = F_{\mu\nu}t^{\nu}$ so $E_{\mu}t^{\mu} = F_{\mu\nu}t^{\mu}t^{\nu}$. Now $F_{\mu\nu}$ is antisymmetric and $t^{\mu}t^{\nu}$ is symmetric. What happens you contract a symmetric tensor with an antisymmetric tensor?

No, really I don't see why.

$\epsilon_{\mu\nu\alpha\beta}$ is antisymmetric in all of its indices. In $J$ you are contracting it with products of the form $t^{\mu}t^{\nu}$ (with appropriate indices in the context of the expression) right? So apply the same reasoning as above.

Ok. I'm asked about $t^\mu t^\nu T_{\mu\nu}$, not really different from $T_{\mu \nu } t^{\mu } t^{\nu }$ right?

Indeed no difference.

 

[fluidistic]

Based on your expression for the energy-momentum tensor, you should be using the (-,+,+,+) convention. In this convention, $t^{\mu}t_{\mu} = -1$.

Ok I see.

Recall that $E_{\mu} = F_{\mu\nu}t^{\nu}$ so $E_{\mu}t^{\mu} = F_{\mu\nu}t^{\mu}t^{\nu}$. Now $F_{\mu\nu}$ is antisymmetric and $t^{\mu}t^{\nu}$ is symmetric. What happens you contract a symmetric tensor with an antisymmetric tensor?

I didn't know this, but it is worth 0. So now I can fully follow you on your post 9.

$\epsilon_{\mu\nu\alpha\beta}$ is antisymmetric in all of its indices. In $J$ you are contracting it with products of the form $t^{\mu}t^{\nu}$ (with appropriate indices in the context of the expression) right? So apply the same reasoning as above.

Let me see here if what I've done is correct as I think I get that J=0 for a different reason.
Skipping some latex, I found out that $J=\underbrace{(F_{\mu\nu} t^\nu t_\nu -F_{\nu \mu } t^\mu t_\mu )}_{=0} \varepsilon ^{\mu\nu\alpha\beta } t_\alpha B_\beta+\varepsilon _{\mu\nu\alpha\beta } t^\alpha B^\beta \underbrace{(F^{\mu\nu} t_\nu t^\nu - F^{\nu \mu}t_\mu t^\mu)}_{=0}$. Except for the fact that I have mu and nu appearing 3 times in a single expression, is this "valid"?

Indeed no difference.

Ok I see why now I think. A simple multiplication where the order doesn't matter I guess.

 

[WannabeNewton]

I can't really see how you got that expression for $J$, mainly because the appearance of the dummy indices more than twice is really making it hard to parse the expression. Could you outline what you did there?

 

[fluidistic]

I can't really see how you got that expression for $J$, mainly because the appearance of the dummy indices more than twice is really making it hard to parse the expression. Could you outline what you did there?

Sure. I probably did something worse than dividing by zero , but here it goes:
$$J=2 \left [ \frac{1}{2} (E_\mu t_\nu -E_\nu t_\mu ) \varepsilon ^{\mu\nu\alpha\beta } t_\alpha B_\beta + \varepsilon _{\mu\nu\alpha\beta } t^\alpha B^\beta \cdot \frac{1}{2} (E^\mu t^\nu - E^\nu t^\mu) \right ]$$
$$= (E_\mu t_\nu -E_\nu t_\mu ) \varepsilon ^{\mu\nu\alpha\beta } t_\alpha B_\beta + \varepsilon _{\mu\nu\alpha\beta } t^\alpha B^\beta (E^\mu t^\nu - E^\nu t^\mu)$$
Now I used the relation $E_\mu=F_{\mu\nu}t^\nu$. And this is where I should have introduced 2 new Greeck letters instead of mu and nu.
This is how I got the line $J=\underbrace{(F_{\mu\nu} t^\nu t_\nu -F_{\nu \mu } t^\mu t_\mu )}_{=0} \varepsilon ^{\mu\nu\alpha\beta } t_\alpha B_\beta+\varepsilon _{\mu\nu\alpha\beta } t^\alpha B^\beta \underbrace{(F^{\mu\nu} t_\nu t^\nu - F^{\nu \mu}t_\mu t^\mu)}_{=0}$.

 

[WannabeNewton]

That doesn't work. This is why you should stick to the index convention. What you have is $E_{\mu}t_{\nu} - E_{\nu}t_{\mu} = F_{\mu\sigma}t^{\sigma}t_{\nu} - F_{\nu\sigma}t^{\sigma}t_{\mu}$ which does not vanish in general. You should use what I told you about the antisymmetry of $\epsilon^{\mu\nu\alpha\beta}$ in all of its indices. For example, $E_{\mu}t_{\nu}t_{\alpha}\epsilon^{\mu\nu\alpha\beta}B_{\beta} = 0$ because $t_{\nu}t_{\alpha}$ is symmetric in $\nu$ and $\alpha$ whereas $\epsilon^{\mu\nu\alpha\beta}$ is antisymmetric in $\nu$ and $\alpha$.