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Tesnsors and Units Consistancy

  1. Jun 28, 2009 #1
    Tensors and Units Consistency

    Where do the unit labels go on tensors?

    This discussion is continued from this thread https://www.physicsforums.com/showthread.php?t=321298"

    Greg, I've been trying to make sense of things in mixed units. Taking the dot product as you've done is a good test, Dale.

    x·x = xμ xμ

    If xμ has units (T, L, L, L) then xμ has units (1/T, 1/L, 1/L, 1/L)

    The scalar product would be unitless.

    The metric has interesting units. So that xν = ημν xμ , works out, the metric would have units

    \left[ \begin {array}{c}
    {\frac{1}{T^2} \; \frac{1}{LT} \; \frac{1}{LT} \; \frac{1}{LT}} \\
    {\frac{1}{LT} \; \frac{1}{L^2} \; \frac{1}{LT} \; \frac{1}{LT}} \\
    {\frac{1}{LT} \; \frac{1}{LT} \; \frac{1}{L^2} \; \frac{1}{LT}} \\
    {\frac{1}{LT} \; \frac{1}{LT} \; \frac{1}{LT} \; \frac{1}{L^2}}
    \end{array} \right]

    I don't know if it makes sense in terms of the full tensor.

    [tex] \eta = \eta_{\mu \nu}\: dx^{\mu} \otimes dx^{\nu} [/tex]

    α β γ δ ε ζ η θ ι κ λ μ ν ξ ο π ρ ς σ τ υ φ χ ψ ω . . . . . Γ Δ Θ Λ Ξ Π Σ Φ Ψ Ω
    ∂ ∫ ∏ ∑ . . . . . ← → ↓ ↑ ↔ . . . . . ± − · × ÷ √ . . . . . ¼ ½ ¾ ⅛ ⅜ ⅝ ⅞
    ∞ ° ² ³ ⁿ Å . . . . . ~ ≈ ≠ ≡ ≤ ≥ « » . . . . . † ‼
    Last edited by a moderator: Apr 24, 2017
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  3. Jun 28, 2009 #2


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    It depends on what convention for the metric you are using.

    If you are taking your coordinates to be (t, x, y, z) (and not (ct, x, y, z)) and your metric has units of square-distance:

    [tex]ds^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2[/tex]​

    and so

    [tex]g_{\mu\nu} =
    \left[ \begin {array}{cccc}
    c^2 & 0 & 0 & 0 \\
    0 & -1 & 0 & 0 \\
    0 & 0 & -1 & 0 \\
    0 & 0 & 0 & -1
    \end{array} \right]

    That means that if [itex]X^{\mu}[/itex] has components of dimension (T, L, L, L) then [itex]X_{\mu} = g_{\mu\nu}X^{\nu}[/itex] has components of dimension (L2/T, L, L, L); [itex]X_{\mu}X^{\mu}[/itex] will thus have dimension L2. Because of that, I would think of [itex]X^{\mu}[/itex] as "having dimension L" (i.e [itex]\sqrt{|X^{\mu}X_{\mu}|}[/itex]), but I don't know if that's the official way of looking at it.

    With non-Cartesian coordinates it gets even worse because you could have (in spherical polar coordinates)

    [tex]X^{\mu} = (t, r, \phi, \theta)[/tex]
    [tex]ds^2 = c^2 dt^2 - dr^2 - r^2(d\theta^2 + \sin^2 \theta d\phi^2)[/tex]​
  4. Jun 28, 2009 #3


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    Hi Phrak! :smile:
    No, you have a fundamental misconception about co- and contra-vectors …

    xμ and xμ have the same units,

    and the scalar product has those units squared.

    (and the problem with L/T disappears if you regard L as have the same dimensions as T, or cT)
  5. Jun 28, 2009 #4


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    True, if you are using a coordinate system in which all four components have the same units...
    ...but the whole point of this thread is how to cope when the components don't have the same units. This problem can (but needn't) be avoided in SR with Minkowski coordinates but it can't be avoided in GR.
  6. Jun 28, 2009 #5
    Re: Tensors and Units Consistency

    Thank you DrGreg. That all made a great deal of sense. And thanks for helping Tiny. This should prove interesting, where these things are masked when working with symbols, and/or where c is set equal to one. It has been for me!

    What seems to be important is consistency; if the units of the covariant and contravarant vectors are chosen, for instance, they dictate the units of the metric.

    Could there be some 'natural' units inherited from the manifold, specifically (T, L, L, L) for a contravariant vector in cartesian coordinates?

    I went back to the definition of a contravariant vector,
    [tex]V = \frac{dx^\mu}{d \lambda} \partial _{\mu} [/tex]
    or even
    [tex]V = \frac{dx^\mu}{d \tau} \partial _{\mu} \ .[/tex]
    The bases have 'natural' units of 1/T and 1/L, but I can't make anything of the units[itex]d/d\lambda[/itex] should posses.
    Last edited: Jun 28, 2009
  7. Jun 29, 2009 #6


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    The components of the metric are dimensionless, or the line element would have the wrong dimensions.

    [tex]ds^2 = g_{jk} \, dx^j \, dx^k[/tex]
  8. Jun 29, 2009 #7


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    That's true enough in coordinates where dx0, dx1, dx2, dx3 and ds are all distances, e.g. standard (ct, x, y, z) Minkowski coordinates, but it's not true in (t, x, y, z) coordinates or [itex](t, r, \phi, \theta)[/itex] coordinates.
  9. Jun 29, 2009 #8


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    Re: Tensors and Units Consistency

    To my way of thinking, it's the other way round.
    1. Decide whether your metric is in distance-units (e.g. [itex]ds^2 = c^2dt^2 - dx^2...[/itex]) or time-units (e.g. [itex]ds^2 = dt^2 - dx^2/c^2...[/itex])
    2. Decide on a (+---) or a (-+++) signature (e.g. [itex]ds^2 = c^2dt^2 - dx^2...[/itex] or [itex]ds^2 = -c^2dt^2 + dx^2...[/itex])
    3. Decide on your coordinate system (e.g. (ct, x, y, z) or (t, x, y, z) or [itex](t, r, \phi, \theta)[/itex])

    Those 3 decisions are independent of each other, but once made, the 16 individual components of the metric tensor are completely determined (for a given metric; assume the flat metric for this discussion, but the same holds in any other spacetime).

    Then, for any given contravariant vector, the components of the covariant covector are determined from the original vector and the metric.

    Let's consider an example: 4-momentum. To keep the equations simple I'll stick to one spatial dimension but it works with all three. My choices are

    [tex]ds^2 = c^2dt^2 - dx^2[/tex]

    [tex]X^{\mu} = (t, x)[/tex]​

    (But you could make different choices, and different equations would follow.)

    Define contravariant 4-momentum of a mass m > 0 as

    [tex]P^{\mu}=m\frac{dX^{\mu}}{d\tau} = \left(m\frac{dt}{d\tau}, m\frac{dx}{d\tau}\right) = (E/c^2, p)[/tex]​

    Bearing in mind what the metric components are (see post #2), the covariant 4-momentum is then

    [tex]P_{\mu} = (E, -p)[/tex]​

    Note you can read the energy and momentum off this directly without rescaling (apart from the -1 factor). The norm is

    [tex]\sqrt{|P_{\mu}P^{\mu}|} = \sqrt{|E^2/c^2 - p^2|} = mc[/tex]​

    So I would say this tensor as a whole has dimensions of momentum.
    Last edited: Jun 29, 2009
  10. Jun 29, 2009 #9


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    Yes, I said a dumb thing. Must stop coffee break posting.
  11. Jul 20, 2009 #10
    Thanks DrGreg. That was very well done and enlightening. I haven't been able to find any good response other than to one. It seems the underlying manifold could be something such as energy and momentum, say, rather than time and space and generate the same metric. That would be very odd. I'm not sure what an energy-momentum manifold would be about.
  12. Jul 21, 2009 #11


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    Re: Tensors and Units Consistency

    The tensor experts in this forum might correct me, but I think it is correct to say that there is only one manifold, and certainly only one metric, regardless of whether the vector or covector you are considering is 4-velocity, 4-momentum, 4-force, 4-current, 4-potential, 4-frequency or whatever. 4-vectors don't actually reside in the manifold itself but in a tangent space (or cotangent space).

    Actually you raise an interesting point. At a given event on the manifold, is the space of all possible 4-momenta considered to be different to the space of all possible 4-forces (for example)? The two spaces have different physical interpretations, but are they mathematically distinct? They're both "the tangent space".
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