# Density of energy from the stress-energy tensor

1. Sep 29, 2013

### fluidistic

1. The problem statement, all variables and given/known data
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.

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

3. 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?

2. Sep 29, 2013

### 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.

3. Sep 29, 2013

### 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.

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.

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

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.

4. Sep 29, 2013

### fluidistic

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

5. Sep 29, 2013

### 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.

6. Sep 29, 2013

### 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 $t^0, t^1, t^2, t^3$?

Last edited: Sep 29, 2013
7. Sep 29, 2013

### 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?

Last edited: Sep 29, 2013
8. Sep 29, 2013

### fluidistic

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.

9. Sep 29, 2013

### WannabeNewton

$I$ 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 $T_{\mu\nu}t^{\mu} t^{\nu}$.

Last edited: Sep 29, 2013
10. Sep 29, 2013

### TSny

Yes, that looks good.

11. Sep 29, 2013

### fluidistic

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.

No, really I don't see why.
I see.

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

12. Sep 29, 2013

### WannabeNewton

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

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?

$\epsilon_{\mu\nu\alpha\beta}$ 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.

Indeed no difference.

13. Oct 1, 2013

### fluidistic

Ok I see.

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

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"?

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

14. Oct 1, 2013

### 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?

15. Oct 1, 2013

### fluidistic

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 $J=\underbrace{(F_{\mu\nu} t^\nu t_\nu -F_{\nu \mu } t^\mu t_\mu )}_{=0} \varepsilon ^{\mu\nu\alpha\beta } t_\alpha B_\beta+\varepsilon _{\mu\nu\alpha\beta } t^\alpha B^\beta \underbrace{(F^{\mu\nu} t_\nu t^\nu - F^{\nu \mu}t_\mu t^\mu)}_{=0}$.

16. Oct 1, 2013

### 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 $\alpha$.