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

Physical meaning of tensor contraction

  1. Dec 14, 2011 #1


    User Avatar
    Gold Member

    I have seen PeterDonis state that predictions of GR are expressed as contracted tensors i.e. scalars. And they are coordinate independent.
    So I have a question what these scalars represent physically?
    Could it be that they represent measurements of physical quantities (with tensors dual representing measurement device in certain state of motion)? Because measurements of physical quantities are coordinate independent while physical quantities are coordinate dependent.
  2. jcsd
  3. Dec 15, 2011 #2


    Staff: Mentor

    Basically this is the idea, yes (note that the term "tensors" includes vectors, which are just (1, 0) tensors, and their duals, which are (0, 1) tensors). For example, say I measure the frequency of a light beam. The light beam is described (in the simplest case) by a 4-momentum vector [itex]k^{a}[/itex], and the frequency I measure is given by the contraction of that 4-momentum with my 4-velocity [itex]u^{b}[/itex], thus: [itex]\hbar \omega = g_{ab} k^{a} u^{b}[/itex]. (More precisely, the 4-velocity is that of the device that measures the frequency; I call it "my" 4-velocity because I'm assuming the measuring device is at rest relative to me.) The components of the 4-vectors are frame-dependent, but the measured frequency is a frame-invariant scalar.

    The term "coordinate-dependent" or "frame-dependent" is often used in a way that tends to confuse the issue; for example, people say that the frequency of a light beam is frame-dependent. What they really mean is that, as is obvious from the above, the measured frequency depends on the 4-velocity of the measuring device. But given the 4-velocity of the measuring device, the measured frequency is an invariant; even if the device is moving relative to me, I can still use the components of its 4-velocity in my frame, along with the components of the photon's 4-momentum in my frame, to calculate the [itex]\omega[/itex] measured by the device and get the same answer.

    In fact, even those "components" of vectors can be specified in frame-invariant terms. In GR, the tools used to do this are called "frame fields" and are well worth studying; for example, see here:


    Basically, a "frame" is a set of one timelike unit vector (which can be thought of as the 4-velocity of a measuring device) and three spacelike unit vectors (which can be thought of as three spatial axes defined by the measuring device and carried along with it). The "components" of any other vector (or tensor) in the "frame" are just the contractions of that vector with the unit vectors of the frame. So in the above example, if I defined a frame using the 4-velocity [itex]u^{b}[/itex] as the timelike unit vector, then the measurement result [itex]\hbar \omega[/itex] is the "time component" of the photon's 4-momentum [itex]k^{a}[/itex] in that frame. Similarly, contracting [itex]k^{a}[/itex] with the three spacelike unit vectors of the frame gives the components of the photon's ordinary spatial momentum.

    So really, *any* number we use in relativity physics can be specified as a scalar invariant formed by contracting the appropriate vectors and tensors; and anything that is "frame-dependent" is really just dependent on which particular vectors and tensors we choose to contract it with.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook