Is the four current in Relativity an invariant quantity?

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Is the four current in relativity an invariant quantity? I know the divergence is zero for the four gradient, i.e. the continuity equation. But is the four current a vector in the sense that it has invariant properties?
 

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  • #2
Orodruin
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The 4-current is a 4-vector.
 
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The 4-current is a 4-vector.
So it is an invariant? I was under the impression that the four vector referred to the four momentum and the four current was apart of the energy tensor and measured fluxes and therefore densities as well. The four T^ob components of the energy tensor form an invariant?
 
  • #4
PeterDonis
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I was under the impression that the four vector referred to the four momentum and the four current was apart of the energy tensor and measured fluxes and therefore densities as well.
What "four-current" are you referring to? Do you have a reference?
 
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What "four-current" are you referring to? Do you have a reference?
The four current of momentum.
 
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Orodruin
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The four current of momentum.
Please provide the reference Peter asked for.
 
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Please provide the reference Peter asked for.
I've seen this in textbooks before but here is Professor Susskind talking about the momentum current. The four components of the fourth component. The T^oa components it stops at about 1:38:50
 
  • #8
vanhees71
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Physicists are often sloppy in their language and confuse students, who are not familiar with it. Vectors or more generally tensors are all invariant objects. The vector/tensor components depend of course on the basis and cobasis they refer to, and these components transform in a specific way under change of the basis. Objects with an upper index transform contravariantly to the basis vectors, which thus carry a lower index and transform covariantly. Any tensor component with a lower index also transforms covariantly and with an upper index contravariantly.
 
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Orodruin
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I've seen this in textbooks before but here is Professor Susskind talking about the momentum current. The four components of the fourth component.
Momentum current is not the same thing as 4-current. What is typically intended when you just say "4-current" without further specification is the electromagnetic 4-current ##J^\mu = (\rho,\vec j)##, where ##\rho## is the charge density and ##\vec j## the (spatial) current density, which is a 4-vector.

The components of the stress energy tensor do not transform like the components of a 4-vector because they are the components of a second rank tensor. Of course, as pointed out in the previous post, each index by itself transforms in the appropriate manner and the stress energy tensor itself is an invariant object. However, your time direction is not invariant under Lorentz transformations and therefore (in general) ##T'^{0\nu} \neq \Lambda^\nu_{\phantom\nu\mu} T^{0\mu}##.
 
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Momentum current is not the same thing as 4-current. What is typically intended when you just say "4-current" without further specification is the electromagnetic 4-current ##J^\mu = (\rho,\vec j)##, where ##\rho## is the charge density and ##\vec j## the (spatial) current density, which is a 4-vector.

The components of the stress energy tensor do not transform like the components of a 4-vector because they are the components of a second rank tensor. Of course, as pointed out in the previous post, each index by itself transforms in the appropriate manner and the stress energy tensor itself is an invariant object. However, your time direction is not invariant under Lorentz transformations and therefore (in general) ##T'^{0\nu} \neq \Lambda^\nu_{\phantom\nu\mu} T^{0\mu}##.
The divergence of ##T^{0\nu}## is an invariant though, correct?. Maybe I'm confusing concepts but I thought the collection of the four components that make up ##T^{0\nu}## represents measured energy flux across all surfaces, and therefore an invariant.
 
  • #11
vanhees71
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It's important to keep in mind that ##\partial_{\nu} T^{0 \nu}## is an invariant if and only if ##\partial_{\nu} T^{0 \nu}=0##. That's why you get a four-vector by integration over ##\mathrm{d}^3 \vec{x}## in some reference frame if and only if energy and momentum are conserved, i.e., if ##T^{\mu \nu}## is the total energy-momentum tensor of a closed system, where ##\partial_{\mu} T^{\mu \nu}=0##.
 
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Orodruin
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To expand a little bit on #11, generally ##\kappa^\nu = \partial_\mu T^{\nu\mu}## is a 4-vector. Thus the 0-th component of that 4-vector is invariant only if the 4-vector itself is the zero vector.
 
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To expand a little bit on #11, generally ##\kappa^\nu = \partial_\mu T^{\nu\mu}## is a 4-vector. Thus the 0-th component of that 4-vector is invariant only if the 4-vector itself is the zero vector.
Kappa could be considered an outside four vector acting on some dust cloud type of matter described by the energy tensor?
 
  • #14
vanhees71
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To be very pedantic ##\kappa^{\nu}## are four-vector components (or even more pedantic, the components of a vector field).
 
  • #15
vanhees71
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##T^{\mu \nu}(x)## are 2nd-rank-tensor-field components. That's why ##\partial_{\mu} T^{\mu \nu}## are vector-field components.
 
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  • #16
haushofer
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##T^{\mu \nu}(x)## are 2nd-rank-tensor-field components. That's why ##\partial_{\mu} T^{\mu \nu}## are vector-field components.
Well, to be pedantic: if it supposed to be vector components under general coordinate transformations, the partial derivative should be a covariant one :P
 
  • #17
vanhees71
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Sure, but I thought we discuss SR in Minkowski coordinates and keep the things simple first ;-).
 

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