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Under the Lorentz transformations, quantities are classed as four-vectors, Lorentz scalars etc depending upon how their measurement in one coordinate system transforms as a measurement in another coordinate system.

The proper length and proper time measured in one coordinate system will be a calculated, but not measured, invariant for all other coordinate systems.

So what kind of invariants are proper time and proper length?

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Dear John, it doesn't matter whether a quantity is measured or calculated. In some sense, you may say that the proper time or proper length are measured in the other frames, too. Calculation is sometimes a necessary part of the measurement.

Because the values aren't affected by the transformations, proper lengths and proper times are formally scalars, much like the rest mass which is the proper length of the energy-momentum vector, if you wish.

However, we usually don't use the word "scalar" for such quantities because they're just some "somewhat unnatural" functions of vectors. For example, the proper length is $\sqrt{dx^\mu dx_\mu}$ assuming that $dx^\mu$ (and the convention for the metric) is spacelike. So fundamentally, the proper length is some information about a vector.

Because the proper length is just some "partial information" about a vector, you can't build a "field theory" out of proper lengths themselves. In relativistic field theory, we talk about "scalar fields" and "vector fields" and other "tensor fields", aside from "spinor fields" etc. However, a fundamental theory could have no elementary field associated with the "proper length" even if such constructions made sense: it would have the whole "vector field" while the length of this vector would be a "derived quantity".

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What about the proper four-displacement? It has four components that transform under the LTs, yet are constant when calculated from any other frame. Is it therefore formally an object of four scalar quantities? –  John McVirgo Jan 31 '11 at 17:48
Well, you may call them "scalars" but their definition depends on the reference frame so directly that there's nothing "coordinate-independent" about them. The word "scalar" is usually reserved for something that doesn't depend on any particular objects - such as those you need to specify the four directions of the displacement. –  Luboš Motl Feb 3 '11 at 14:28

allow me a somewhat informal discussion: Pour yourself a cup of coffee, you see that your cup does not move, your can does not move, and that the coffee flowing from the can ends up in your cup. Now, every observer from every other reference frame will agree with you that the coffee ends up in the cup, from his point of view. This is what we call a Lorentz scalar: Every observer agrees upon it, from the point of view of his reference frame.

Proper time and proper length are not Lorentz scalars in this sense. In this case, we first single out a special reference frame R, it is the reference frame where a special object is at rest. Next we have to tell every observer how his reference frame R' is related to R. Every other observer now has to measure time and length in his reference frame R' and is then able to calculate, using the additional information of the relation of R and R', what the corresponding measurements would have told him if he lived in R. If every observer does this calculation correctly, they'll agree upon proper time and proper length :-)

Ergo: Proper time and proper length are simply observables defined with respect to a special reference frame. There is no buzzword for this comparable to Lorentz scalar or Lorentz vector, nor is there the need for one, IMHO.

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I understand your points, but would agree with Lubos that a proper object can still be formally defined. Space and time are the only measurements made in a frame, all other quantities are calculated from these. You can only calculate, rather than measure, something primitive like $dx^\mu$ and then define it as a four-vector based upon how its calculation transforms in other frames. Therefore, the proper $dx^\mu$ for some chosen frame can be calculated for all other frames, and this value can still be classed as an object with 4 scalars - whatever that is. –  John McVirgo Jan 31 '11 at 18:09

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