Energy momentum tensor - off diagonal terms

  • #1
LCSphysicist
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Homework Statement:
A rod has cross sectional area A and mass per unit length \mu. Write down the stress energy tensor inside the rod, when it is under a tension F.
Relevant Equations:
.
Let's arrange the rod's axis parallel to the z axis.

##T_{00} = A/\mu## (since it represents the energy density)

##T_{03}=T_{30} = \frac{F\sqrt{\mu / F}}{A}## (It represents the flow of energy across the z direction)

##T_{33} = F/A## (pressure)



It seems that ##T_{33}## i have got has the wrong sign, and that ##T_{03} = T_{30}## should actually be zero.



i am a little confused on both cases: Where does the sign at ##T_{33} = (-F/A)## comes from? And why are my reasoning involving ##T_{30}## wrogn?
 
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  • #2
Orodruin
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T00=A/μ (since it represents the energy density)
I hope you mean the other way around …

I did not see any reasoning from you regarding ##T_{03}##, just a statement that it is the energy current. Why should the energy current be what you quoted?

Regarding ##T_{33}##, consider how T is defined.
 
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  • #3
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I hope you mean the other way around …

I did not see any reasoning from you regarding ##T_{03}##, just a statement that it is the energy current. Why should the energy current be what you quoted?

Regarding ##T_{33}##, consider how T is defined.
Ok, ##T_{00}## indeed i wrote the inverse of the answer, sorry.
##T_{03} = dp_{3}/dV = \frac{F_3 dt}{dV} = \frac{F_3 dt}{A dz} = \frac{F_3}{A v} = \frac{F}{A \sqrt{F / \mu}}##
And ##T_{33}##? Pressure ##F/A## at z direction, no?
 
  • #4
Orodruin
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Why do you think ##p_3## is non-zero? Is the rod moving?
 
  • #5
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Why do you think ##p_3## is non-zero? Is the rod moving?
No, but the waves are propagating, no?
 
  • #6
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No, but the waves are propagating, no?
Is there more to the problem than what you wrote in the OP? The OP says nothing about propagating waves.
 
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  • #7
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Is there more to the problem than what you wrote in the OP? The OP says nothing about propagating waves.
No. But since this problem is from a special relativity book, i thought that rigid objects would be meaningless here, and we should consider the deformation, and so the wave.
But i think i got it, i am overthinking the problem, maybe?
 
  • #8
Orodruin
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Well, nothing says that the rod is rigid to deformation by new external forces. There just are no such forces mentioned. All I that is mentioned is a rod under tension. Presumably after any motion arising from the application of the forces providing the tension has dissipated and the rod has settled.
 
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