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Dimension of the stress energy tensor

  1. Jun 26, 2015 #1
    The coefficient of the stress energy tesor in the GR equation reduces to 8π/Ν, where N = {"(Kg)m/s^2.} Is it correct to conclude that all the elements of the stress energy tensor must have the dimension of N = (Kg)m/s^2 since the curvature and metric tensors on the other side of the equation are dimensionless?
     
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  3. Jun 26, 2015 #2

    Orodruin

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    No. This depends completely on the choice of metric. The metric tensor does not need to be dimensionless, it can even have different dimensions in different components depending on the coordinates (c.f., the metric in polar coordinates on R^2 is diag(1,r^2)).
     
  4. Jun 26, 2015 #3

    bcrowell

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    No, I don't think this is right.

    First off, nobody actually does calculations in GR using SI units. The normal way you would work would be to use geometrized units, in which G=c=1, finish the calculation, and then at the end if you had a result where you needed to plug in numerical data, reinsert the factors of G and c.

    The next thing to realize is that it's often natural to use coordinate systems where different coordinates have different units, e.g., spherical coordinates in which r and t have units of distance, but the the angles are unitless. When you write out a tensor in terms of its components, in this type of mixed-unit coordinate system, the different components of the tensor can all have different units. Therefore it doesn't always make sense to talk about what units a tensor has.

    But suppose we're using a coordinate system such that all the coordinates have units of distance. Then the metric is unitless. The Christoffel symbols are derivatives of the metric, so they have units of distance^-1. The Riemann tensor can be expressed in terms of derivatives of the Christoffel symbols, so it has units of distance^-2. So does the Ricci tensor. Therefore in geometrized units, the stress-energy tensor also has units of distance^-2. This makes sense because the stress-energy tensor can be thought of in the Newtonian limit as a mass density, and mass has units of distance in geometrized units, so mass density is distance/distance^3=distance^-2.

    This means that the units of the stress-energy tensor in SI can't be newtons, because newtons are unitless in geometrized units.
     
  5. Jun 26, 2015 #4
    Thanks for your response. Newtons^-1 is the dimension of the coefficient of T. If the Ricci tensor has dimension of distance^2, would it not follow that T has dimension N(distance^2)?
     
  6. Jun 27, 2015 #5

    pervect

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    My $.02. If you use geometric units, all the coefficients of the stress energy-tensor do have the same dimension if you use a geometric units as defined in "Gravitation" by Misner, Thorne, Wheeler, henceforth MTW. In this setup you measure time in cm, distances in cm, masses in cm. This is so much easier than dealing with different units in the different components of the tensor that it's the way I think about things. BTW, rho would have units of energy / unit volume, which is mass*acceleration*distance/volume, which is cm*(1/cm)*cm / cm^3 = 1 / cm^2, different than your definition which is cm*cm / cm^2. If you need conventional units with the geometric approach, you just stick back the needed values of c and G in your geometric unit result based on a dimensional analysis. Not every author uses this approach though, nothing forces one to do things this way, and any particular text may not adopt this approach.

    There's also the issue of whether you are working in a coordinate basis or a "frame-field" non-coordinate basis. The approach I suggest makes more sense in a frame-field basis.
     
  7. Jun 27, 2015 #6

    bcrowell

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    We're really talking about two different issues here: (1) geometrized units versus SI, and (2) units of tensors and their components. Your first two sentences are about #1, your last sentence about #2. They're not related, and really Orodruin and I have just hijacked the thread by having a side discussion about #2.
     
  8. Jun 27, 2015 #7

    Orodruin

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    I have moved the side discussion to a separate thread. Let us continue there and keep this thread on topic.
     
  9. Jun 28, 2015 #8
    For clarificatiion:
    The stress energy tensor has a coefficient with specific dimensions. In order for the equation to be consistent, T, R and g must have dimensions to make the equation consistent. My understanding of the metric and curvature tensors is that they are geometrical and either have no dimension or (as has been pointed out above) the dimension of distance. In either case, T must have dimensions to make the equation a proper one. N (perhaps with dimensions of distance) seems to have the dimensions to acheive that.
    Is this not true?
     
  10. Jun 28, 2015 #9

    Orodruin

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    The dimension of a tensor component (in coordinate basis) is dependent on the dimensions of the coordinate. For example, take a vector V in Cartesian coordinates in two dimensions and transform it to polar coordinates. The components ##V^r## and ##V^\phi## will now have different dimensions. However, if you get the dimensions correct for one component, they will be correct for the rest as well.

    How to assign dimensions to a tensor as a whole object is discussed in the thread that was split from this one (see above).
     
  11. Jun 28, 2015 #10

    samalkhaiat

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    In any units system, the energy-momentum tensor has the unit of energy density, i.e. pressure. So (in 4-dimensional space-time) in powers of Mass, Length and Time [tex][\mathbb{T}] = M L^{-1} T^{-2} .[/tex] You can check that using the definition of the tensor in any field theory [tex]\mathbb{T} \sim \frac{\partial \mathcal{L}}{\partial (\partial \varphi)} \ \partial \varphi - \eta \ \mathcal{L} , \ \Rightarrow \ [\mathbb{T}] = [\mathcal{L}] .[/tex] Or, as you would like to do, from the Einstein field equations [tex]R \sim \frac{G_{N}}{c^{4}} \ \mathbb{T} .[/tex] Notice that [tex]\Gamma \sim g \ \partial g , \ \ \ \ R \sim \partial \Gamma + \Gamma^{2} .[/tex] Since [tex][g] = L^{0} , \ \ \ [\partial] = L^{- 1} ,[/tex] it follows that [tex][\Gamma] = L^{-1} , \ \ \ [R] = L^{-2} .[/tex] You also know the units of Newton’s constant [tex][G_{N}] = M^{-1}L^{3}T^{-2} , \ \ [c]^{4} = L^{4} T^{-4} .[/tex] So [tex][\mathbb{T}] = \frac{[c]^{4} [R]}{[G_{N}]} = M L^{-1}T^{-2} .[/tex] So (in 4-dimensional space-time), if you want to use the natural units where [itex]c = \hbar = 1[/itex], i.e. [itex]L = T = M^{-1}[/itex], you get [itex][\mathbb{T}] = L^{-4} = M^{4}[/itex]. And in the co-called geometric units (again in 4-dimensional space-time): [itex]c = G_{N} = 1[/itex], i.e. [itex]L = T = M[/itex], the energy-momentum tensor will have the same units as the Ricci tensor, [itex][\mathbb{T}] = [R] = L^{-2}= M^{-2}[/itex].
     
  12. Jun 28, 2015 #11

    bcrowell

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    No, see #3.
     
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