# How is Momenergy frame independent?

momenergy = mass * spacetime displacement/proper time for that displacement

What I dont understand is how can the momenergy be frame independent? The unit-4 vector always points in the direction of the worldline?

Agreed, the magnitude of momenergy will be invariant wrt to any frame. But wont the worldline of a particle be different in different frames? So wont the direction of the spacetime interval be frame dependent?

• With spacetime displacement you mean $ds=\sqrt{dx_{\mu}dx^{\mu}}$, right? If so, then I don't see your problem because $ds$ is clearly Lorentz-invariant (and invariant under diffeos in general). This fact is discussed in any literature on special or general relativity.
– psm
Dec 27, 2013 at 11:18

There is some semantic disagreement over the meaning of a vector in physics (although the concepts are not in dispute).

1) Vector as geometric object, or "arrow"

Some would tell you that the momenergy, more commonly called the four-momentum, is like an arrow in space-time with some magnitude and direction (in fact, the magnitude is $mc$ and the direction is tangent to the worldline, as you pointed out). As you change your frame of reference, the components of this four-vector appear to change, as the direction of the arrow relative to you appears to change. But the arrow hasn't changed; it's invariant.

If the four-momentum is really an invariant geometric object, then it must behave like one under coordinate transformations. In particular, recall from linear algebra that when you perform a change of basis, the coordinates of vectors also change; in fact, they change by multiplication by the inverse of the change of basis matrix. This behaviour under change of basis is called contravariance. Contravariant quantities are denoted with superscript indices.

A "frame of reference" in relativity is really a choice of three spacelike directions and a timelike direction to be the basis vectors you use to assign coordinates to points and vectors in spacetime. Change your frame, and your basis changes too. In special relativity, the Lorentz transformation matrix $\Lambda$ gives the change in basis in terms of the rotation angle and boost velocity of your frame change. So the components of the four-momentum must transform with $\Lambda^{-1}$. (*)

If this were not the case, then the transformation properties of the coordinates of the four-momentum would be inconsistent with the picture of the four-momentum as an invariant "arrow" somewhere out there in space-time. But it is the case, so we can treat it in this way. In this sense we may say that the four-momentum is frame-independent.

2) Vector as a sequence of components

Others identify vectors with their components. If I measure the components of a vector quantity to be $(1, 2, 3, 4)$, I will then say that the vector is $(1, 2, 3, 4)$, rather than thinking of the vector as an arrow in space-time whose components just happen to be $(1, 2, 3, 4)$ in whatever my frame of reference happens to be. In this view, the vector actually changes when I change my frame of reference (after all, the vector is the components, and the components change).

If we take this approach, then we impose contravariance as a requirement for a set of four components to denote a meaningful physical quantity in special relativity, rather than deriving it from the invariant nature of a geometric object. Now we cannot say that the four-momentum is really invariant or frame-independent. But it is contravariant, and that's good enough. The calculus of relativity is formulated in terms of contravariant, invariant, and covariant quantities (see below).

(*) Actually, the Lorentz transformation matrix is defined in such a way that the basis vectors transform with $\Lambda^{-1}$. It is defined in this way so that coordinates themselves, which are also contravariant in special relativity, transform with $(\Lambda^{-1})^{-1} = \Lambda$. So in fact the four-momentum also transforms with $\Lambda$. But we can also choose to treat the four-momentum as a covector (with subscript indices), in which case its components transform with the change of basis matrix $\Lambda^{-1}$ itself, not its inverse $\Lambda$.

• Thanks for answer. How would a quantum physicist answer this question then? Because, it is only our observation that forces things into existence, doesnt it? Dec 30, 2013 at 10:06
• Quantum field theory is formulated against the background of special relativity, so the answer is the same. "it is only our observation that forces things into existence" is philosophical nonsense and is not a useful way to think about science. Dec 30, 2013 at 19:09