Revisiting classical Physics
So we have to start reconsidering old Physics in order to make it
compatible with the 4-dimensional spacetime and special relativity.
For the main part, that means turning vectors in 4_vectors. Which
means at least to find the “time component” of the 4-vector.
Starting, as usual, from the simplest case.
And the simplest is: the “position vector” (x,y,z). In this
case, the answer is obvious, since this is where it all started:
The “position” 4-vector is what we called the “event” 4-vector,
[t,x,y,z]. We know how we found it: it was the Physics that forced it
upon us. In this case, the 4th component is the time.
The next case follows directly from this one: it is the
“displacement”, [δt,δx,δy,δz]. It not used by itself, but as a step in
the calculation of the derivatives, like velocity which is a bit more
complicated; but at least we have a 4-vector to start from. And, even
more complicated, the acceleration.
Another fundamental vector in classical physics is the
Force. In classical physics it is a 3-vector, of experimental origin, and
we do not have any idea on what the 4th term should be.
Although for defining force we may start from the potential. It is
–or so it seems- a scalar. And, as such, it is not dependent on the
reference frame. But, on the other hand, we have one case
(electromagnetism!!) in which we have a scalar potential and a vector
one… It could as well happen that they are components of a sole 4vector!
So we may invert the argument and consider that from a scalar
potential we obtain the gradient and therefore the force. For doing
this, we need to check the transformation properties of that hybrid
thing that the gradient is.
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The 4-vectorization of nabla
Or, the transformation properties of gradients, curls, divergencies
and laplacians in 4-dimensional spacetime.
We have a vector field A whose contravariant components we know.
Since we are dealing with a vector field the components are functions
of the space point and possibly also of time.
Anyway, since we are discussing 4-vectors, we have 4
components, which having a dependence on t,x,y,z will also have
derivatives wrt them, A0 / x i and all others. If we want to change
to a new reference system, we have to change all the components of
our vectors… For every change of IRF exists a matrix of
transformation of coordinates, [Aij] which has an inverse [A’lk], and
which has coefficients x' / x β ; an example of transformation is:


x
'
A'   A
x
For the covariant component a similar equation holds. The
difference with the contravariant one is that in this case the
coefficients are x / x'β , i.e. they are those of the inverse
matrix.

B' 
x
B

x'
And… what about the gradient?
i



 j k
x
y
z
For the initial convention (x=x1, y=x2, z=x3), the 4-gradient
extension of the 3-gradient will have the derivatives wrt
contravariant coordinates.
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Then, well. It turns out that such definition of gradient makes it
defined with covariant components. Because it turns out that, if
we make a change of reference axes from IRF to IRF’ the
coefficients of the transformation matrix will be determined by
the equation


x



x'i x'i x
Thex  x'i are the coefficients of the covariant components!
For further reference then, the components of nabla are:


  

, 1, 2, 3
0
x x x x






,

,

,

x 0 x1 x 2 x 3
[
[
Covariant components
]
]
Contravariant components
The same convention is valid for gradient, divergence and curl.
For the Laplacian, in the first place, it has 4 addends and becomes
the D’Alambertian; in the second place, it being a sort of norm of
the vector, we do not have to choose:
2

[x
0
      
,
,
,
x0 x1 x1 x 2 x2 x 3 x3
Where this new symbol,
2
]
, is called the D’Alambertian.
Note that the D’Alambertian can also be written as
2
2
2



( x 0 ) 2
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Let us now go back to the infinitesimal change of position: the
“displacement”, [δt,δx,δy,δz]. . It is obviously a 4-vector too, in the
Minkowski space. That means that it is an invariant 4-vector,
i.e. that its position with respect to other objects is unchanged
when changing reference frame. Of course, the displacement we
are talking about now is an infinitesimal part of a movement, in
space and time, of an object. It is therefore only timelike, unless
we are talking about light, in which case the LI will be null.
Let us now deal with velocity v. In classical mechanics it is a
relative quantity, i.e. a quantity whose value does depend on the
IRF in which we are studying the system. In ordinary space its
components are (dx/dt, dy/dt, dz/dt) and depend, as well as its
absolute value, on the reference frame chosen. Now, if we
translate literally the 3-vector components in spacetime language
we get (in 3 dimensions):
dx i
dx1 dx 2 dx 3
[ 0 , 0 , 0 ]
0
dx
dx dx dx
At first one would think that the 0-component could be a term like
dt/dt. It does not work, because such 4 terms are NOT a 4vector. For being a 4-vector the 4 terms should vary (for change of
reference frame) as a 4-vector. And in fact the 4 differentials in
the numerator do that. What does not do it is the denominator, i.e.
”dt”. That should have been a scalar, for the 4 terms to behave as
a 4-vector. But it is not a scalar, because under a Lorentz
transformation it varies as the 0-term of a 4-vector. Of course,
dt is a time, and from that point of view it is suitable. But we need
a time which is actually a scalar in spacetime. There are only two
“times” in the market; and the other one, which is the proper time
dτ of that displacement, would do.
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If we divide the 4-vector δx by dτ we obtain a 4-vector. This 4vector is called
0
1
2
3
dx
dx
dx
dx
Four-Velocity  [
,
,
,
]
d d d d
Its norm (i.e. its invariant) is obviously equal to 1, owing to the
invariant of the infinitesimal displacement being equal to (dτ)2.
This is of course a significant change wrt classical physics.
Remark that the norm of the Four-Velocity is 1 only in a system
in which time and space have the same units. Otherwise it norm
is “c”.
Now there is still another step we must make to be done with
four-velocity: make its components intelligible.
Well, we know from the dilatation of time that dt=γdτ. Replacing
this value in the last equation we give the components of the
four-velocity in 3-vector, understandable language.
U  [ ,  x ,  y ,  z ]
In classical mechanics what is conserved in isolated systems
is not the velocity of a system of interacting bodies, but its
total linear momentum. We can extend to 4-vector space the
definition of linear momentum:
P  mU  [m , m x , m y , m z ]
Obviously,
P  P  P 2  m2
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At the limit for β0 we should find back the values of
classical, newtonian physics. The three space components
actually do tend to mv. What is still to be understood is
what does the 0-term (mγ) mean in terms of physics.
In order to understand it, we have to make a connection
with the Classical physics formulae, i.e. the limit for β0.
A hint we already have: The space part of this 4-vector is
the linear momentum, a three component vector which is always
conserved in isolated systems, just the same as a scalar
quantity, the energy.
Now, the quantity under scrutiny is:
0
1
2  2
m
dx
 m  m  (1   )
d
For small values of β2 we approximate the function (1-x)-1/2
with the Taylor formula
Limx0(1-x)-1/2=1+x/2
And therefore:
1
lim
   0
1 
 1
2
lim m  lim
   0
2
   0
2
m
1 
2
 mm
2
2
which is the classical formula for the knetic energy, which is
added to the energy at rest of the particle.
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Four-acceleration, four-force
We have already seen the Velocity 4-vector.
To calculate the four-acceleration we repeat the procedure that
allowed us to calculate without major trouble the four-velocity,
i.e. we derive the 4-velocity wrt dτ.
dU
d d (v)
 c ,
d
d d
v a
and writing D 0   4
D
} {c ddt ,  ( ddt v   ddtv )}
{
We obtain:
c
{
D 4
v a
v a v
v
,  2a  ( 4
)
 D 0 ,  2a  D 0
c
c c
c
} {
In the case of uniform motion (a
should be.
frame):
}
= 0) the 4-acceleration is null, as it
In the case instead of a particle at rest (v=0 in a certain
D  {0, a}
where the time component is null and the space component is the
classical acceleration: the four-acceleration is therefore spacelike.
Moreover, the module (norm) of the four-acceleration (in any IRF,
of course) is the square of the three-dimensional acceleration in
the IRF in which the particle is at rest.
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