In Taylor, read sections 15.1 - 15.13 for this week and 15.13 to 15.18 next week.
This can be represented compactly with matrix notation, introducing the "boost" Λ given by the 4×4 matrix:
and then writing the Lorentz transformation as the matrix expression
Written this way, the Lorentz transformation resembles the rotation matrix about the z-axis in 3-dimensions:
We consider the Lorentz transformation to be an extension of rotations into 4-dimensional space-time. Also, by rotating about the y and z axes, we can produce a Lorentz transformation that boosts in an arbitrary direction from the one given above that boosts along the x axis. You will do this in one of your homework problems.
The "boost" can also be written in "summation notation" as
Finally, the dot product of two 4-vectors is written in this notation as:
where g is the metric of the space. We will always work in flat space where the metric tensor is:
(Note the relation between g and the metric ds² discussed earlier.) With this metric, we see that the scalar product expands to
The usual rules for three-vectors apply to four-vectors, with appropriate modification. Vectors can be added and subtracted, multiplied by a scalar, differentiated, and integrated. Two vectors can by combined to form a scalar product.
Note that this way of writing four-vectors and their scalar product is not unique. Some authors put the time element in position 0. Some authors put the negative sign of the scalar product on the spatial components.
Now we have two kinds of objects in relativity, scalars and four-vectors. The space-time vector for an object is a four-vector and transforms between reference frames according to the Lorentz transformation. The dot product of two four-vectors is a scalar and is the same (invariant) in all reference frames. The mass of an object is also a scalar and is the same in all reference frames. The mass is some times refered to as the "invariant mass", because some authors break the rule that scalars are the same in all reference frames and define a "relativistic mass" that depends on the speed of the object in the frame of reference. This is bad and confusing and we won't use it. Unlearn it if you were taught this.
The position of an object at a particular time is a four-vector, x = (x, ct). As the object moves, the position four-vector changes, tracing out a curve in four-dimensional space. If we consider two nearby points on this curve separated by dx = (dx, cdt) = (vdt, cdt), where v = dx/dt is the usual three-dimensional velocity of the object. We know that this separation has to be time-like (a signal can be transmitted between the two points, or mathematically, dx² = (dx)² - c²dt² < 0). We can find a frame comoving with the object (a frame where the object is at rest) and in that frame, dx0 = (0, cdt0) = (0, cdτ), where t0 = τ is called the proper time. The proper time is simply the time as measured in the rest frame of an object. These two four-vectors respresent the separation between the same space-time points, viewed from two different frames, therefore, their lengths are equal, and we can use this fact to find an expression for the proper time
where γ(v) is the gamma factor for the object's speed in the observer's frame. The proper time is a scalar. All observers will agree on an object's proper time.
While the velocity measured by an observer is still v = dx/dt, this is an awkward object because it is neither a four-vector or a scalar. We wish to have a four-vector version of velocity, four-velocity, for use in relativistic calculations (and it should reduce to regular velocity in the non-relativisitic limit). One way to get such an object is to differentiate the postion four-vector by a scalar, and the scalar can be the proper time.
The four-velocity is a four-vector, and the spatial part reduces to the standard velocity in the non-relativistic limit (the time component reduces to the constant c).
Next we would like a definition for a relativistic momentum. It will need to be a four-vector, and should reduce to the usual momentum, p = mv, in the non-relativisitic limit. It is also useful if relativistic momentum is conserved in the absence of external forces, as it is in the non-relativistic limit. Using the above definition of the four-velocity and the scalar mass, a likely candidate is
In the non-relativisitic limit, the spatial part reduces to the usual three-momentum, p = mv.
What about the fourth component of momentum? In the non-relativistic limit it reduces to the constant mc. But if we expand this and look at higher order terms, we see that
The second term looks like kinetic energy divided by c. And, since we'd like conservation of energy as well as momentum, it is a reasonable step to say that the fourth component of momentum is the relativisitic energy of an object, divided by c.
where p is the relativisitic version of three-momentum, γmv, and E is the relativistic energy. Notice that relativisitic energy is not zero for a particle at rest, rather it is E = mc². For a particle in motion, we say that its relativistic energy is composed of rest energy mc² plus kinetic energy, T = E - mc² = (γ - 1)mc².
There are a number of useful relations that can be found between momentum, energy, and mass. The square of p is an invariant:
It is also given by p² - E²/c². Therefore we have an important relation
Also, since E = γmc², we can show that |p| = γβmc, and β = |p|c/E.© 2007 Robert Harr