Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Why is energy-momentum tensor Lorentz invariant?

  1. Jun 8, 2010 #1
    I'm studying General Relativity and facing several problems. We know that energy-momentum must be Lorentz invariant in locally inertial coordinates. I am not sure I understand this point clearly. What is the physics behind?
  2. jcsd
  3. Jun 8, 2010 #2
    It's not. It's covariant.
  4. Jun 8, 2010 #3
    I don't know the answer--I'm just learning too--but I don't think this is it! Covariant is just a more confusing word for the same thing. It's more confusing, because it has two meanings, whereas invariant has one.

    If a tensor is invariant under a specified transformation, its value is unchanged by that transformation (although the components of its coordinate representation may change); so if you apply a Lorentz transformation to your tangent vectors, and the corresponding inverse Lorentz transformation to cotangent vectors, then let a cotangent vector act on a tangent vector to give a real number, it will be the same number as if you hadn't applied the transformations.

    Traditionally, covariant could mean the same as invariant (I think this is the meaning Dickfore has in mind), e.g. covariant derivative. But it can also mean mean, of a tensor, having valence (0,q), where q is some positive integer, e.g. covariant vector = covector, dual vector, cotangent vector, linear functional, linear form, 1-form. The opposite of covariant sense 1 is coordinate dependent / frame dependent. The opposite of covariant sense 2 is contravariant, i.e. having valence (p,0), where p is some positive integer, as a tangent vector does.
  5. Jun 8, 2010 #4
    If we take the well known Energy equation:

    [tex]E = \sqrt{ (m_oc^2)^2 + p^2c^2} = \sqrt{ (m_oc^2)^2 + \frac{(m_o v c)^2}{(1-v^2/c^2)}} [/tex]

    E is the total energy [itex] (m_0c^2)/\sqrt{(1-v^2/c^2)}[/itex],
    [itex]m_oc^2[/itex] is the rest energy ,
    p is the momentum

    and re-arrange it so that:

    [tex]m_oc^2 = \sqrt{E^2 - p^2c^2} [/tex]

    then it is easy to see that the term on the left is invariant because the rest mass and the speed of light are both invariants, so the Energy-Momentum term on the right must be invariant too. I am assuming that the term on the right is what you are calling the energy-momentum.
    Last edited: Jun 8, 2010
  6. Jun 9, 2010 #5
    I don't think that's it. He was probably talking about the stress-energy tensor (also referred to as the energy-momentum tensor (see title)) that appears in the Einstein Field Equations, and covariant means [tex]\nabla_\mu T^{\mu\nu}=0[/tex].

    The short answer is that it is covariant because energy and momentum are conserved quantities. Einstein designed his field equations such that GR obeys this law. See here for a derivation.
    Last edited: Jun 9, 2010
  7. Jun 9, 2010 #6
    What you are mentioning about is the Energy-Momentum four-vector [tex]p^{\mu} (p^0 =m_0 c^2, p^i)[/tex]. Not surprisingly, the four-vector is Lorentz invariant because one always has: [tex]p^{\mu}p_{\mu}=\sqrt{E^2 - p^2c^2}=m_{0}c^2 [/tex]. The physics behind is the validity of the well-known equation [tex]E=mc^2[/tex] in the rest frame leading to the Lorentz invariance of the four-vector.

    What I am talking about is the similar physics behind in the case of Energy-Momentum tensor. :smile:
    Last edited: Jun 9, 2010
  8. Jun 9, 2010 #7
    See post #5. I have given you a link to the derivation.
  9. Jun 9, 2010 #8


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    No, the energy-momentum four-vector is not Lorentz invariant. It transforms like a four-vector. Its *magnitude* is a Lorentz scalar, so its *magnitude* is invariant.

    The subject line of your original post was "Why is energy-momentum tensor Lorentz invariant?" If this refers to the stress-energy tensor, then the answer is that it isn't invariant; it transforms as a rank-2 tensor. If this refers to the energy-momentum four-vector, then the answer is that it isn't invariant; it transforms as a four-vector.

    Are you confusing "invariant" with "conserved?"
  10. Jun 10, 2010 #9
    It's each component of the four-vector that transforms, not the four-vector itself, it must be invariant under a certain transformation. For example, given a vector A, under a certain transformation, one always has:
    [tex]A=A_{\mu}e^\mu = A^{'}_{\nu}e^{\mu^'}[/tex]
    This is exactly the invariance I mean. And so do tensors. The conservation is another story :smile:
  11. Jun 10, 2010 #10
    If you look at the Bianchi identities with the Ricci tensor, you can get an expression like

    R ^{\alpha}_{\ \rho :\alpha}-R_{:\rho}+R^{\nu}_{\ \rho :\nu}=0

    which, because of the symmetry of the Ricci tensor gives you

    2R^{\alpha}_{\ \rho :\alpha}=R_{:\rho}

    Then, raise the suffix rho to get

    g^{\alpha \rho}R^{\alpha}_{\ \rho :\alpha}=\frac{1}{2}g^{\alpha \rho}R_{:\rho}

    Which lets you write

    (R^{\alpha \rho}-\frac{1}{2}g^{\alpha \rho}R)_{:\alpha}=0


    and, since

    (R^{\alpha \rho}-\frac{1}{2}g^{\alpha \rho}R)= T^{\alpha \rho}

    it must be that

    T^{\alpha \rho}_{\ :\alpha}=0

    Which is a form of conservation of energy and momentum.

    *The subscript ":" indicates covariant differentiation.*

    *I don't know why the "R" is missing at the start of the first equation, it's there when I edit my post, but goes away when I click "save."
    Last edited: Jun 10, 2010
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook