# Scalar, Invariant, Observer depdanat

1. Dec 4, 2004

### pmb_phy

In the pasxt several people (in anotgher forum) became confused over the terms "Scalar," "Invariant," and "Observer independant."

The term scalar is another name of a tensor of rank 0. The term Invariant is often used to mean the same think in SR/GR. But caution should be exercised when you see these terms so that when you run across something funky on these subtleties when they arise in the physics literature you'll be able to recognize it immediately. To clarify this issue I made an entire web page dedicated to this point.

http://www.geocities.com/physics_world/ma/invariant.htm

I used the widely known example (E.g. See A first course in general relativity, Bernard F. Schutz, page 52. MTW page 65) E = P*Uobs where P is the 4-momentum of a partcile, Uobs is the 4-velocity of an observer and E is the energy as measued by that observer. In this sense the quantity E is a scalar. Many find that fact very hard to deal with since one typically does not think of energy as a scalar

Lest one gets the idea that this is never really found in the physics literature I came across this counter example. From Energy Conservation as the Basis of Relativistic Mechanics II, R. Penrose and W. Rindler, Am. J. Phys., 33, 995-997 (1965). The authors write on page 996
----------------------------------------------------------
We recall that the 3-velocity u is given by U = gamma(u,c); hence the
4-velocity of an inertial observer relative to his own frame is
V=(0,c). Thus

gamma = V*U/c^2

i.e. the Lorentz factor of a particle relative to any inertial observer is
given by the scalar product of the two corresponding 4-velocities divided
c^2. And this product, being invariant, can be evaluted in any inertial
frame of reference.
----------------------------------------------------------
How often have you seen gamma refered to as an invariant?

Other well known examples can be found in the literature. A simialar idea is found in Gravitation by Misner, Thorne and Wheeler, page 65. Eg. if v = (v_x, v_y, v_z) is the velocity of a particle then there is a four vector V whose components in the observers frame of reference are

V = (0, v_x, v_y, v_z)

Same thing for the electric field 4-vector. That 4-vector has components in the observers rest frame of

E = (0, E_x, E_y, E_z)

Notice that the magnitude of V is v and that the magnitude of E is sqrt[ (v_x)^2 + (v_y)^2 + (v_z)^2

As such I'm at a loss to decide on a good definition of "invariant". Should it be "not defined in terms of a basis vector"?

The components of a tensor are

M_ab = M(e_a, e_b)

This is clearly a contraction of M with the two unit vectors e_a and e_b and therefore M_ab is a scalar.

Pete

Last edited: Dec 4, 2004