Revisions to Energy-Momentum Tensor under Lorentz Transformation

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edited Sep 6, 2014 at 22:36

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In relativity, the symmetric energy-momentum tensor is given by $$ T^{ij}, $$ where $T^{00}$ is the energy density and $\frac{1}{c}T^{10}$ is the momentum density. Thus: $$ \left(\frac{1}{c}T^{00}dV, \frac{1}{c}T^{10}dV\right)^{T}$$ is the 4-momentum. Under a Lorentz transformation, this should transform like 4-vectors where $$ \frac{1}{c}T^{00}dV= \left[\frac{1}{c}T'^{00}dV'+\frac{v}{c^2}T'^{10}dV'\right] \left( 1-\frac{v^2}{c^2}\right)^{-1/2}\\dV=dV'\sqrt{1-\frac{v^2}{c^2}}.$$ After simplifications, we have: $$ T^{00}= \left[T'^{00}+\frac{v}{c}T'^{10} \right] \left( 1-\frac{v^2}{c^2}\right)^{-1}$$ But if we apply the Lorentz transformation to the tensor directly we get $$ T^{00}= \left[T'^{00}+\frac{v}{c}T'^{10}+\frac{v^2}{c^2}T\ ^{11} \right]\left( 1-\frac{v^2}{c^2}\right)^{-1}$$ What accounts for the difference? I think the first is wrong but have no idea why. Thanks.

In relativity, the symmetric energy-momentum tensor is given by $$ T^{ij}, $$ where $T^{00}$ is the energy density and $\frac{1}{c}T^{10}$ is the momentum density. Thus: $$ \left(\frac{1}{c}T^{00}dV, \frac{1}{c}T^{10}dV\right)^{T}$$ is the 4-momentum. Under a Lorentz transformation, this should transform like 4-vectors where $$ \frac{1}{c}T^{00}dV= \left[\frac{1}{c}T'^{00}dV'+\frac{v}{c^2}T'^{10}dV'\right] \left( 1-\frac{v^2}{c^2}\right)^{-1/2}\\dV=dV'\sqrt{1-\frac{v^2}{c^2}}.$$ After simplifications, we have: $$ T^{00}= \left[T'^{00}+\frac{v}{c}T'^{10} \right] \left( 1-\frac{v^2}{c^2}\right)^{-1}$$ But if we apply the Lorentz transformation to the tensor directly we get $$ T^{00}= \left[T'^{00}+\frac{v}{c}T'^{10}+\frac{v^2}{c^2}T\ ^{11} \right]\left( 1-\frac{v^2}{c^2}\right)^{-1}$$ What accounts for the difference? I think the first is wrong but have no idea why.

Some minor grammar corrections.

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edited Apr 19, 2014 at 16:19

Ali

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In relativity, the symmetric energy-momentum tensor is given by $$ T^{ij}, $$ where $T^{00}$ is the energy density and $\frac{1}{c}T^{10}$ is the momentum density. Thus: $$ \left(\frac{1}{c}T^{00}dV, \frac{1}{c}T^{10}dV\right)^{T}$$ is the 4-momentum. Under a Lorentz transformation, this should transform like 4-vectors where $$ \frac{1}{c}T^{00}dV= \left[\frac{1}{c}T'^{00}dV'+\frac{v}{c^2}T'^{10}dV'\right] \left( 1-\frac{v^2}{c^2}\right)^{-1/2}\\dV=dV'\sqrt{1-\frac{v^2}{c^2}}.$$ After simplifications, we have: $$ T^{00}= \left[T'^{00}+\frac{v}{c}T'^{10} \right] \left( 1-\frac{v^2}{c^2}\right)^{-1}$$ But if we apply the Lorentz transformation to the tensor directly we get $$ T^{00}= \left[T'^{00}+\frac{v}{c}T'^{10}+\frac{v^2}{c^2}T\ ^{11} \right]\left( 1-\frac{v^2}{c^2}\right)^{-1}$$ What accounts forthefor the difference? I think the first is wrong but have no idea why. Thanks.

In relativity, the symmetric energy-momentum tensor is given by $$ T^{ij}, $$ where $T^{00}$ is the energy density and $\frac{1}{c}T^{10}$ is the momentum density. Thus: $$ \left(\frac{1}{c}T^{00}dV, \frac{1}{c}T^{10}dV\right)^{T}$$ is the 4-momentum. Under a Lorentz transformation, this should transform like 4-vectors where $$ \frac{1}{c}T^{00}dV= \left[\frac{1}{c}T'^{00}dV'+\frac{v}{c^2}T'^{10}dV'\right] \left( 1-\frac{v^2}{c^2}\right)^{-1/2}\\dV=dV'\sqrt{1-\frac{v^2}{c^2}}.$$ After simplifications, we have: $$ T^{00}= \left[T'^{00}+\frac{v}{c}T'^{10} \right] \left( 1-\frac{v^2}{c^2}\right)^{-1}$$ But if we apply the Lorentz transformation to the tensor directly we get $$ T^{00}= \left[T'^{00}+\frac{v}{c}T'^{10}+\frac{v^2}{c^2}T\ ^{11} \right]\left( 1-\frac{v^2}{c^2}\right)^{-1}$$ What accounts forthe difference? I think the first is wrong but have no idea why. Thanks.

In relativity, the symmetric energy-momentum tensor is given by $$ T^{ij}, $$ where $T^{00}$ is the energy density and $\frac{1}{c}T^{10}$ is the momentum density. Thus: $$ \left(\frac{1}{c}T^{00}dV, \frac{1}{c}T^{10}dV\right)^{T}$$ is the 4-momentum. Under a Lorentz transformation, this should transform like 4-vectors where $$ \frac{1}{c}T^{00}dV= \left[\frac{1}{c}T'^{00}dV'+\frac{v}{c^2}T'^{10}dV'\right] \left( 1-\frac{v^2}{c^2}\right)^{-1/2}\\dV=dV'\sqrt{1-\frac{v^2}{c^2}}.$$ After simplifications, we have: $$ T^{00}= \left[T'^{00}+\frac{v}{c}T'^{10} \right] \left( 1-\frac{v^2}{c^2}\right)^{-1}$$ But if we apply the Lorentz transformation to the tensor directly we get $$ T^{00}= \left[T'^{00}+\frac{v}{c}T'^{10}+\frac{v^2}{c^2}T\ ^{11} \right]\left( 1-\frac{v^2}{c^2}\right)^{-1}$$ What accounts for the difference? I think the first is wrong but have no idea why. Thanks.

Some minor grammar corrections.

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edit approved Apr 19, 2014 at 16:19

Drarp

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In relativity, the symmetric energy-momentum tensor is given by $$ T^{ij} $$$$ T^{ij}, $$ where $T^{00}$ is the energy density and $\frac{1}{c}T^{10}$ is the momentum density. Thus: $$ \left(\frac{1}{c}T^{00}dV, \frac{1}{c}T^{10}dV\right)^{T}$$ is fourthe 4-momentum. Under a Lorentz transformation, this should transform like a four4-vectors where $$ \frac{1}{c}T^{00}dV= \left[\frac{1}{c}T'^{00}dV'+\frac{v}{c^2}T'^{10}dV'\right] \left( 1-\frac{v^2}{c^2}\right)^{-1/2}\\dV=dV'\sqrt{1-\frac{v^2}{c^2}}$$$$ \frac{1}{c}T^{00}dV= \left[\frac{1}{c}T'^{00}dV'+\frac{v}{c^2}T'^{10}dV'\right] \left( 1-\frac{v^2}{c^2}\right)^{-1/2}\\dV=dV'\sqrt{1-\frac{v^2}{c^2}}.$$ After simplifications, we have: $$ T^{00}= \left[T'^{00}+\frac{v}{c}T'^{10} \right] \left( 1-\frac{v^2}{c^2}\right)^{-1}$$ But if we apply the Lorentz transformation to the tensor directly we get $$ T^{00}= \left[T'^{00}+\frac{v}{c}T'^{10}+\frac{v^2}{c^2}T\ ^{11} \right]\left( 1-\frac{v^2}{c^2}\right)^{-1}$$ What makes theaccounts forthe difference? I think the first is wrong but have no idea why. Thanks.

In relativity, the symmetric energy-momentum tensor is given by $$ T^{ij} $$ where $T^{00}$ is the energy density and $\frac{1}{c}T^{10}$ is the momentum density. Thus $$ \left(\frac{1}{c}T^{00}dV, \frac{1}{c}T^{10}dV\right)^{T}$$ is four-momentum. Under a Lorentz transformation, this should transform like a four-vectors where $$ \frac{1}{c}T^{00}dV= \left[\frac{1}{c}T'^{00}dV'+\frac{v}{c^2}T'^{10}dV'\right] \left( 1-\frac{v^2}{c^2}\right)^{-1/2}\\dV=dV'\sqrt{1-\frac{v^2}{c^2}}$$ After simplifications, we have $$ T^{00}= \left[T'^{00}+\frac{v}{c}T'^{10} \right] \left( 1-\frac{v^2}{c^2}\right)^{-1}$$ But if we apply the transformation to the tensor directly we get $$ T^{00}= \left[T'^{00}+\frac{v}{c}T'^{10}+\frac{v^2}{c^2}T\ ^{11} \right]\left( 1-\frac{v^2}{c^2}\right)^{-1}$$ What makes the difference? I think the first is wrong but have no idea why. Thanks.

In relativity, the symmetric energy-momentum tensor is given by $$ T^{ij}, $$ where $T^{00}$ is the energy density and $\frac{1}{c}T^{10}$ is the momentum density. Thus: $$ \left(\frac{1}{c}T^{00}dV, \frac{1}{c}T^{10}dV\right)^{T}$$ is the 4-momentum. Under a Lorentz transformation, this should transform like 4-vectors where $$ \frac{1}{c}T^{00}dV= \left[\frac{1}{c}T'^{00}dV'+\frac{v}{c^2}T'^{10}dV'\right] \left( 1-\frac{v^2}{c^2}\right)^{-1/2}\\dV=dV'\sqrt{1-\frac{v^2}{c^2}}.$$ After simplifications, we have: $$ T^{00}= \left[T'^{00}+\frac{v}{c}T'^{10} \right] \left( 1-\frac{v^2}{c^2}\right)^{-1}$$ But if we apply the Lorentz transformation to the tensor directly we get $$ T^{00}= \left[T'^{00}+\frac{v}{c}T'^{10}+\frac{v^2}{c^2}T\ ^{11} \right]\left( 1-\frac{v^2}{c^2}\right)^{-1}$$ What accounts forthe difference? I think the first is wrong but have no idea why. Thanks.