Density of energy from the stress-energy tensor

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fluidistic
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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?
 

Answers and Replies

  • #2
WannabeNewton
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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
TSny
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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.
 
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  • #4
fluidistic
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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.
 
  • #5
WannabeNewton
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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
TSny
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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##?
 
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  • #7
fluidistic
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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)##
[tex]=\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[/tex]
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?
 
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  • #8
fluidistic
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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.
 
  • #9
WannabeNewton
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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)##
[tex]=\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[/tex]
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}##.
 
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  • #10
TSny
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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.
 
  • #11
fluidistic
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##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?
 
  • #12
WannabeNewton
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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.
 
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  • #13
fluidistic
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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.
 
  • #14
WannabeNewton
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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
fluidistic
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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 :approve:, but here it goes:
[tex]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 ][/tex]
[tex]= (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) [/tex]
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
WannabeNewton
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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##.
 

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