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Prévia do material em texto

About the 4−acceleration
Author: Mário Raia Neto
Email: mraianeto@gmail.com
Instituition: UFSCar
Degree: Undergraduate
Date: 2019-2
Contents
1 The 4-acceleration vector 3
1.1 Some other mathematical "mambo-jambo" . . . . . . . . . . . . . . . . . 7
2 Norm of 4−acceleration vector 8
3 Some physical models and their 4-acceleration 11
3.1 Motion with constant speed . . . . . . . . . . . . . . . . . . . . . . . . . 11
3.2 Acceleration in 1+1 dimension: . . . . . . . . . . . . . . . . . . . . . . 12
4 The 4−acceleration in non-inertial coordinate systems 14
References 16
1 The 4-acceleration vector
Definition 1.1. The 4−acceleration vector is defined as the second derivative of the
position in spacetime and the first derivative of 4−velocity :
A =
d2
dτ2
[X] =
d
dτ
[U] (1)
In component form we have then,
A =
d2
dτ2
[X] =
d
dτ
[U] =⇒
=⇒ Aµ = dU
µ
dτ
=
d2xµ
dτ2
=
d
dt
(
dxµ
dt
dt
dτ
)
dt
dτ
=
dt
dτ
[
d
dt
(
dxµ
dt
dt
dτ
)]
=
= γv
[
d2xµ
dt2
γv +
dγv
dt
dxµ
dt
]
=⇒
Aµ = γv
[
d2xµ
dt2
γv +
dγv
dt
dxµ
dt
]
(2)
Expliciting the basis vectors we have then,
A =
3
∑
µ=0
Aµ
∂X
∂xµ
≡ Aµ ∂X
∂xµ
=
{
γv
[
d2xµ
dt2
γv +
dγv
dt
dxµ
dt
]}
∂X
∂xµ
(3)
Remark 1.
dt
dτ
=:
1√
1− ||v||
2
c2
≡ γ||v||2 ≡ γ(|| v ||2)≡ γv
Where || v || is the norm of the relative 3− velocity (i.e. of the spatial components of
the four-velocity ) between the frame S and S′.
Now, suppose the cartesian basis, then (3) becomes:
A =
3
∑
µ=0
Aµ
∂X
∂xµ
= A0
∂X
∂x0
+A1
∂X
∂x1
+A2
∂X
∂x2
+A3
∂X
∂x3
⇐⇒
⇐⇒ A = A0 ∂X
∂x0
+A1
∂X
∂x1
+A2
∂X
∂x2
+A3
∂X
∂x3
= A0
∂
∂x0
(x0,x1,x2,x3)+
3
+A1
∂
∂x1
(x0,x1,x2,x3)+A2
∂
∂x2
(x0,x1,x2,x3)+A3
∂
∂x3
(x0,x1,x2,x3) ⇐⇒
⇐⇒ A = A0
(
∂x0
∂x0
,
∂x1
∂x0
,
∂x2
∂x0
,
∂x3
∂x0
)
+A1
(
∂x0
∂x1
,
∂x1
∂x1
,
∂x2
∂x1
,
∂x3
∂x1
)
+
+A2
(
∂x0
∂x2
,
∂x1
∂x2
,
∂x2
∂x2
,
∂x3
∂x2
)
+A3
(
∂x0
∂x3
,
∂x1
∂x3
,
∂x2
∂x3
,
∂x3
∂x3
)
⇐⇒
A = A0(1,0,0,0)+A1(0,1,0,0)+A2(0,0,1,0)+A3(0,0,0,1) (4)
Lemma 1.0.1. The derivative of the gamma factor is given by:
dγv
dt
=
γ3v
c2
dxµ
dt
d2xµ
dt2
≡ γ
3
v
c2
v`v̇` (5)
Proof. First of all, it’s reasonable the existence of a gamma factor derivative because
now we must to consider variable relative velocities between the frames of reference.
In other words, we have relative motion considering now accelerated particles, and
then the relative velocity is changing pointwise, in particular it’s value || v || is chang-
ing . Hence, the gamma factor is a dependent quantity, with respect to coordinate
time t. Then,
dγv
dt
=
d
dt
(
1√
1− ‖v‖
2
c2
)
=−1
2
(
1− ‖v‖
2
c2
)− 32 d
dt
(
1− ‖v‖
2
c2
)
=
=− 1
2
√√√√(1− ‖v‖2c2
)3
[
d
dt
(1)− d
dt
(
‖v‖2
c2
)]
=
=−1
2
[(
1− ‖v‖
2
c2
)− 12]3[
0−
(
d
dt
(c−2)‖v‖2 + d
dt
(‖v‖2)c−2
)]
=
=
γ3v
2
[(
d
dt
(‖v‖2)c−2
)]
=
γ3v
2
[(
d
dt
(〈v,v〉)c−2
)]
=
=
γ3v
2c2
[(
〈 d
dt
v,v〉+ 〈v, d
dt
v〉
)]
=
γ3v
2c2
[(
〈v̇,v〉+ 〈v, v̇〉
)]
=⇒
4
dγv
dt
=
γ3v
2c2
[(
〈v̇,v〉+ 〈v, v̇〉
)]
(6)
Now, with inner product properties we can rewrite equation (6) as,
dγv
dt
=
γ3v
2c2
[(
〈v̇,v〉+ 〈v, v̇〉
)]
=
γ3v
2c2
[(
〈v, v̇〉+ 〈v, v̇〉
)]
=
γ3v
2c2
[
2〈v, v̇〉
]
=
=
γ3v
c2
[
〈v, v̇〉
]
=⇒
dγv
dt
=
γ3v
c2
〈v, v̇〉 ≡ γ
3
v
c2
〈v̇,v〉 ≡ γ
3
v
c2
〈a,v〉 ≡ γ
3
v
c2
〈v,a〉 (7)
Now, it should be stressed that equation (7) is also written as,
dγv
dt
=
γ3v
c2
〈v,a〉 ≡ γ
3
v
c2
ηk`vka` (8)
Where the metric ηk` ≡ η
(3)
µν =:
1 0 00 1 0
0 0 1
 is the spatial part of Minkowski metric
ηµν. The metric η
(3)
µν is needed to calculate the norm of the 3-vectors separately from
the covariant 4-vectors, because in this derivation a 3-vector appears naturally. Then,
〈v,a〉 ≡ 〈v,a〉(3) (9)
Where the inner product 〈·, ·〉(3) is the Inner Product of the vector space E3 which
is the "space part" of Minkowski spacetime. The inner product〈·, ·〉(3) is needed to
calculate the inner product of the 3-vectors separately from the covariant 4-vectors,
because in this derivation 3-vectors appears naturally. Now, the space part of space-
time is a Euclidean Space just by convention and simplicity. In fact the space part of
Minkowski spacetime is just a 3-dimensional vector space.
In component form then we have,
γ3v
c2
ηk`vka` =
γ3v
c2
v`a` =⇒
dγv
dt
=
γ3v
c2
v`v̇` ≡
γ3v
c2
〈a,v〉(3) (10)
�
Now, using lemma 1.0.1 and the components of equation 3 we can rewrite the 4-acceleration
5
in different forms:
1. Direct Substituition:
A =
3
∑
µ=0
Aµ
∂X
∂xµ
=
{
γv
[
d2xµ
dt2
γv +
dγv
dt
dxµ
dt
]}
∂X
∂xµ
=
=
{
γv
[
d2xµ
dt2
γv +
γ3v
c2
aivi
dxµ
dt
]}
∂X
∂xµ
(11)
A =
3
∑
µ=0
Aµ
∂X
∂xµ
=
{
γv
[
d2xµ
dt2
γv +
dγv
dt
dxµ
dt
]}
∂X
∂xµ
=
{
γv
[
d2xµ
dt2
γv +
γ3v
c2
v̇ivi
dxµ
dt
]}
∂X
∂xµ
(12)
2. "With 4-velocity":
Remembering that
dxµ
dτ
=: γv
dxµ
dt
then,
A =
{
γv
[
d2xµ
dt2
γv +
dγv
dt
dxµ
dt
]}
∂X
∂xµ
=
{
γv
[
d2xµ
dt2
γv +
γ3v
c2
aivi
dxµ
dt
]}
∂X
∂xµ
=
=
{
γv
[
d2xµ
dt2
γv +
γ2v
c2
aiviγv
dxµ
dt
]}
∂X
∂xµ
=⇒
A =
{
γv
[
d2xµ
dt2
γv +
γ2v
c2
aivi
dxµ
dτ
]}
∂X
∂xµ
=
{
γv
[
d2xµ
dt2
γv +
γ2v
c2
aiviuµ
]}
∂X
∂xµ
(13)
But perhaps the best way to write these components are:
A =
{
γv
[
d2xµ
dt2
γv +
γ3v
c2
〈a,v〉(3)
dxµ
dt
]}
∂X
∂xµ
=
{
γv
[
d2xµ
dt2
γv +
γ2v
c2
〈a,v〉(3)uµ
]}
∂X
∂xµ
(14)
Now, using 14 we can calculate the components:
6
• The Temporal Component:
A0 = γv
[
d2x0
dt2
γv +
γ3v
c2
〈a,v〉(3)
dx0
dt
]
=
d2x0
dt2
γ
2
v +
γ4v
c2
〈a,v〉(3)
dx0
dt
=
=
d2(ct)
dt2
γ
2
v +
γ4v
c2
〈a,v〉(3)
d(ct)
dt
= 0γ2v +
γ4v
c2
〈a,v〉(3)c =
γ4v〈a,v〉(3)
c
=⇒
A0 =
γ4v
c
〈a,v〉(3) (15)
• The Spatial Components:
Ai = γv
[
d2xi
dt2
γv +
γ3v
c2
〈a,v〉(3)
dxi
dt
]
= aiγ2v +
γ4v
c2
〈a,v〉(3)vi =⇒
Ai = aiγ2v +
γ4v
c2
〈a,v〉(3)vi =⇒

A1 = a1γ2v +
γ4v
c2 〈a,v〉(3)v
1
A2 = a2γ2v +
γ4v
c2 〈a,v〉(3)v
2
A3 = a3γ2v +
γ4v
c2 〈a,v〉(3)v
3
(16)
With (15) and (16) we can write the 4-acceleration in a "Rn−vector" form:
A = γv
(
γ3v
c
〈a,v〉(3),aiγv+
γ3v
c2
〈a,v〉(3)vi
)
=
(
γ4v
c
〈a,v〉(3),aiγ2v +
γ4v
c2
〈a,v〉(3)vi
)
(17)
1.1 Some other mathematical "mambo-jambo"
We obtained the equation (17) using just derivative calculus. Moreover we can rewrite it
as:
A =
(
γ4v
c
〈a,v〉(3),aiγ2v +
γ4v
c2
〈a,v〉(3)vi
)
=
(
γ4v
c
〈a,v〉(3),aiγ2v
)
�
(
0,
γ4v
c2
〈a,v〉(3)vi
)
=
=
(
γ4v
c
〈a,v〉(3),0
)
�
(
0,aiγ2v
)
�
(
0,
γ4v
c2
〈a,v〉(3)vi
)
=
(
γ4v
c
〈a,v〉(3),
γ4v
c2
〈a,v〉(3)vi
)
�
(
0,aiγ2v
)
=
=
γ4v
c2
〈a,v〉(3)
(
c,vi
)
� γ2v
(
0,ai
)
=⇒
7
A =
γ4v
c2
〈a,v〉(3)
(
c,vi
)
� γ2v
(
0,ai
)
(18)
2 Norm of 4−acceleration vector
It’s a mathematical fact that given a inner product we have in general a norm associ-
ated with it. The Minkowski inner product is not a usual metric in the sense of positive
definiteness. But still we can use as a inner product. Then,
‖A‖=:
√
〈A,A〉 (19)
But, since 19 is a scalar, then we are able square the two sides:
(
‖A‖
)2
=
(√
〈A,A〉
)2
= 〈A,A〉
Which defines a new quantity:
‖A‖2 = 〈A,A〉 (20)
In component form 20 is written as:
‖A‖2 ≡ ds2 =: ηµνAµAν (21)
Or,
‖A‖=
√
ηµνAµAν ≡
(
ηµνAµAν
) 1
2 (22)
Now, given the Minkowski metric ηµν =:

−1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
, the equation 21 becomes,
ηµνAµAν = η00A0A0 +η01A0A1 +η02A0A2 +η03A0A3+
+η10A1A0 +η11A1A1 +η12A1A2 +η13A1A3+
+η20A2A0 +η21A2A1 +η22A2A2 +η23A2A3+
+η30A3A0 +η31A3A1 +η32A3A2 +η33A3A3 ⇐⇒
⇐⇒ ηµνAµAν = (−1)A0A0 +(0)A0A1 +(0)A0A2 +(0)A0A3+
8
+(0)A1A0 +(1)A1A1 +(0)A1A2 +(0)A1A3+
+(0)A2A0 +(0)A2A1 +(1)A2A2 +(0)A2A3+
+(0)A3A0 +(0)A3A1 +(0)A3A2 +(1)A3A3 ⇐⇒
ηµνAµAν =−(A0)2 +(A1)2 +(A2)2 +(A3)2 = AµAµ (23)
Agora, por (13) e (17) tem-se que:
AµAµ =−(A0)2+(A1)2+(A2)2+(A3)2 =−
(
γ4v
c
〈a,v〉(3)
)2
+
[(
a1γ2v+
γ4v
c2
〈a,v〉(3)v1
)2
+
+
(
a2γ2v +
γ4v
c2
〈a,v〉(3)v2
)2
+
(
a3γ2v +
γ4v
c2
〈a,v〉(3)v3
)2]
=
=−
(
γ8v
c2
〈a,v〉2(3)
)
+
[(
(a1)2γ4v +
2γ2vγ
4
v
c2
〈a,v〉(3)a1v1 +
γ8v
c4
〈a,v〉2(3)(v
1)2
)
+
+
(
(a2)2γ4v+
2γ2vγ
4
v
c2
〈a,v〉(3)a2v2+
γ8v
c4
〈a,v〉2(3)(v
2)2
)
+
(
(a3)2γ4v+
2γ2vγ
4
v
c2
〈a,v〉(3)a3v3+γ8v
c4
〈a,v〉2(3)(v
3)2
)]
⇐⇒
Ordering the terms we have,
AµAµ =−
(
γ8v
c2
〈a,v〉2(3)
)
+
(
(a1)2 +(a2)2 +(a3)2
)
γ
4
v+
2γ6v〈a,v〉(3)
c2
(
(a1v1)+(a2v2)+(a3v3)
)
+
γ8v
c4
〈a,v〉2(3)
(
(v1)2 +(v2)2 +(v3)2
)
=⇒
AµAµ =−
(
γ8v
c2
〈a,v〉2(3)
)
+
+
(
(a1)2 +(a2)2 +(a3)2
)
γ
4
v +
2γ6v〈a,v〉(3)
c2
(
(a1v1)+(a2v2)+(a3v3)
)
+
9
+
γ8v
c4
〈a,v〉2(3)
(
(v1)2 +(v2)2 +(v3)2
)
(24)
Note that the terms
(
(a1)2 +(a2)2 +(a3)2
)
,
(
(a1v1)+ (a2v2)+ (a3v3)
)
and
(
(v1)2 +
(v2)2 +(v3)2
)
are, in fact, the summations of the inner product between the 3−velocities
and 3− accelerations:

(a1)2 +(a2)2 +(a3)2 = δi jaiv j = ‖a‖2
(a1v1)+(a2v2)+(a3v3) = ηi jaiv j = 〈a,v〉(3)
(v1)2 +(v2)2 +(v3)2 = δi jviv j = ‖v‖2
And then the equation (19) can be written as:
AµAµ =−
γ8v
c2
〈a,v〉2(3)+‖a‖
2
γ
4
v +
2γ6v
c2
〈a,v〉2(3)+
γ8v
c4
〈a,v〉2(3)‖v‖
2 (25)
Is it possible then simplify the expression (20) like:
AµAµ =
[
−γ
8
v
c2
〈a,v〉2(3)+
2γ6v
c2
〈a,v〉2(3)+
γ8v
c4
〈a,v〉2(3)‖v‖
2
]
+‖a‖2γ4v =
=
[
γ6v
c2
〈a,v〉2(3)
(
− γ2v +2+
γ2v
c2
‖v‖2
)]
+‖a‖2γ4v =
=
{
γ6v
c2
〈a,v〉2(3)
[
2− γ2v
(
1− ‖v‖
2
c2
)]}
+‖a‖2γ4v ⇐⇒
AµAµ =
{
γ6v
c2
〈a,v〉2(3)
[
2− γ2v
(
1− ‖v‖
2
c2
)]}
+‖a‖2γ4v (26)
Remark 2. γv =: 1√
1− ‖v‖
2
c2
and γ2v =:
1(
1− ‖v‖
2
c2
) . Then,
γ
−2
v = 1−
‖v‖2
c2
(27)
Using equation (22) on (21) we have then,
AµAµ =
{
γ6v
c2
〈a,v〉2(3)
[
2− γ2v
(
1− ‖v‖
2
c2
)]}
+‖a‖2γ4v =
10
=
{
γ6v
c2
〈a,v〉2(3)
[
2− γ2vγ−2v
]}
+‖a‖2γ4v =
{
γ6v
c2
〈a,v〉2(3)
[
2−1
]}
+‖a‖2γ4v =
=
γ6v
c2
〈a,v〉2(3)+‖a‖
2
γ
4
v ⇐⇒
‖A‖2 ≡ AµAµ =
γ6v
c2
〈a,v〉2(3)+ γ
4
v‖a‖2 (28)
With equation (23), (16) and (17) we define the norm of the 4−acceleration:
‖A‖ ≡
(
ηµνAµAν
) 1
2
=
√
γ6v
c2
〈a,v〉2(3)+ γ4v‖a‖2 (29)
3 Some physical models and their 4-acceleration
Knowing that we have derived the most general expression for the accelerated motion
based just on the derivatives of a absolute spacetime quantity which is the 4-position X,
is easy now to reduce the norm for some restricted physical situations.
3.1 Motion with constant speed
In total analogy with a Newtonian case (and based on a elementary derivative calculus
fact), constant speed means that the 3−acceleration is inexistent, i.e.
a = v̇ = 0 (30)
Then, due to this fact we have then that the norm of four acceleration becomes,
‖A‖=
√
γ6v
c2
〈a,v〉2(3)+ γ4v‖a‖2 =
=
√
γ6v
c2
〈0,v〉2(3)+ γ4v‖0‖2 = 0
But also, we have the situation where the norm of 3−velocity (the speed) remains un-
changed, rather it’s direction do not. This is the case of an orthogonal nature between
3−velocity and 3−acceleration:
a⊥ v =⇒ 〈a,v〉(3) = ‖a‖‖v‖cos
(
π
2
)
= 0 (31)
11
‖A‖=
√
γ6v
c2
〈a,v〉2(3)+ γ4v‖a‖2 =
=
√
0+ γ4v‖a‖2 =
√
γ4v‖a‖2 = γ2v‖a‖
Then, in this situation the norm of 4−acceleration is given by:
‖A‖= γ2v‖a‖ ⇐⇒ α = γ2a (32)
Furthermore, the form of the 4−acceleration vector is then (using equation (17) and for
the case of a⊥ v),
A =
(
γ4v
c
〈a,v〉(3),aiγ2v +
γ4v
c2
〈a,v〉(3)vi
)
=
(
γ4v
c
0,aiγ2v +
γ4v
c2
0vi
)
=
(
0,aiγ2v
)
=⇒
A =
(
0,aiγ2v
)
⇐⇒
{
A0 = 0
Ai = γ2va
i (33)
3.2 Acceleration in 1+1 dimension:
This situation is chracterized then by the spatial motion of the particle along a line, there-
fore, is a one dimensional spatial motion and then the 3−velocity and 3−acceleration are
colinear. This fact implies that the angle between the 3−vectors are zero,
〈a,v〉= ‖a‖‖v‖cos(0) = ‖a‖‖v‖ (34)
and then the norm of four-acceleration reduces to:
‖A‖=
√
γ6v
c2
〈a,v〉2(3)+ γ4v‖a‖2 =
=
√
γ6v
c2
(‖a‖‖v‖)2 + γ4v‖a‖2 =
√
γ6v
c2
‖a‖2‖v‖2 + γ4v‖a‖2 ⇐⇒
⇐⇒ ‖A‖2 =
(
γ6v
c2
‖v‖2 + γ4v
)
‖a‖2 =
=
[
γ
4
v
(
γ
2
v
‖v‖2
c2
+1
)]
‖a‖2 =
{
γ
4
v
[(
1
1− ‖v‖
2
c2
)
‖v‖2
c2
+1
]}
‖a‖2 =
{[
‖v‖2
c2−‖v‖2
]}
γ
4
v‖a‖2 =
12
=
{[
‖v‖2− (c2−‖v‖2)
c2−‖v‖2
]}
γ
4
v‖a‖2 =
{[
c2
c2−‖v‖2
]}
γ
4
v‖a‖2 =
{[
c2
c2
(
1−‖v‖
2
c2
)−1]}
γ
4
v‖a‖2
=
{
1
1− ‖v‖
2
c2
}
γ
4
v‖a‖2 = γ2vγ4v‖a‖2 = γ6v‖a‖2 ⇐⇒
⇐⇒ ‖A‖2 = γ6v‖a‖2 ⇐⇒ ‖A‖=
√
γ6v‖a‖2 =
√
(γ3v‖a‖)2 = γ3v‖a‖
Then, in this situation the norm of 4−acceleration is given by:
‖A‖= γ3v‖a‖ ⇐⇒ α = γ3a (35)
Furthermore, the form of the 4−acceleration vector is then (using equation (17)),
A =
(
γ4v
c
〈a,v〉(3),aiγ2v +
γ4v
c2
〈a,v〉(3)vi
)
=
(
γ4v
c
‖a‖‖v‖,aiγ2v +
γ4v
c2
‖a‖‖v‖vi
)
=
=
(
γ4v
c
‖a‖‖v‖,axγ2v +
γ4v
c2
‖a‖(vx)2
)
=
(
γ4v
c
axvx,γ2va
x
[
1+
γ2v
c2
(vx)2
])
=
=
(
γ4v
c
axvx,γ2va
x
[
1+
(
1
1− (v
x)2
c2
)
1
c2
(vx)2
])
=
(
γ4v
c
axvx,γ2va
x
[
1+
(
(vx)2
c2
1
1− (v
x)2
c2
)])
=
=
(
γ4v
c
axvx,γ2va
x
[
1+
(
(vx)2
c2− (vx)2
)])
=
(
γ4v
c
axvx,γ2va
x
[(
(c2− (vx)2)+(vx)2
c2− (vx)2
)])
=
=
(
γ4v
c
axvx,
γ2va
xc2
c2− (vx)2
)
=
(
γ4v
c
axvx,
γ2va
xc2
c2
(
1− (v
x)2
c2
)
)
=
(
γ4v
c
axvx,γ2va
x 1
1− (v
x)2
c2
)
=
(
γ4v
c
axvx,γ2va
x
γ
2
v
)
=⇒
A =
(
γ4v
c
axvx,γ4va
x
)
⇐⇒
{
A0 = γ
4
v
c a
xvx
Ax = γ4va
x
(36)
13
4 The 4−acceleration in non-inertial coordinate systems
Given then the four acceleration,
A =
{
γv
[
d2xµ
dt2
γv +
γ3v
c2
〈a,v〉(3)
dxµ
dt
]}
∂X
∂xµ
=
[
d2xµ
dt2
γ
2
v +
γ4v
c2
〈a,v〉(3)
dxµ
dt
]
∂X
∂xµ
(37)
It’s necessary to say that this vector have this particular form with respect a particular
reference frame named inertial. In non-inertial systems, we have therefore non-inertial
coordinates, and then the four acceleration vector is described by an (absolute) covariant
derivative along a line, parametrized by proper time:
A = Aµ
∂
∂xµ
=
{
Duµ
Dτ
}
∂
∂xµ
=
{
uν∇νuµ
}
∂
∂xµ
=
{
dxν
dτ
∇ν
(
dxµ
dτ
)}
∂
∂xµ
=
=
{
dxν
dτ
[
∂ν
(
dxµ
dτ
)
+Γµ να
dxα
dτ
]}
∂
∂xµ
=
{
∂ν
(
dxµ
dτ
)
dxν
dτ
+Γµ αν
dxα
dτ
dxν
dτ
}
∂
∂xµ
=
Remark 3. It’s extremely important to show the conceptual difference between the
covariant detivative and the absolute covariant derivative. The difference isn’t about
the derivative operator, rather, it’s in the type of vector fields. It’s possible to differen-
tiate a vector field in a domain D4 ⊂M and along a line D⊂M . The 4−acceleration
and 4−velocity are vector fields along a line.
Hence, the symbol
{
∂ν
(dxµ
dτ
)dxν
dτ
}
∂
∂xµ is a total derivative of 4−velocity with respect
of proper time:
{
∂ν
(
dxµ
dτ
)
dxν
dτ
}
∂
∂xµ
≡
{
∂
∂xν
(
dxµ
dτ
)
dxν
dτ
}
∂
∂xµ
≡
{
∂
(dxµ
dτ
)
∂xν
dxν
dτ
}
∂
∂xµ
≡
≡
{
∂uµ
∂xν
dxν
dτ
}
∂
∂xµ
=
{
duµ
dτ
}
∂
∂xµ
=⇒
d
dτ
[U] =
{
duµ
dτ
}
∂
∂xµ
(38)
Which agrees with the definition 1, up to the Christoffel terms.
14
=
{
duµ
dτ
+Γµ αν
dxα
dτ
dxν
dτ
}
∂
∂xµ
=⇒
A = Aµ
∂
∂xµ
=
{
Duµ
Dτ
}
∂
∂xµ
=
{
duµ
dτ
+Γµ αν
dxα
dτ
dxν
dτ
}
∂
∂xµ
≡
≡
{
aµ +Γµ ανuαuν
}
∂
∂xµ
(39)
Remark 4. It’s interresting to note that if we change the notation, for the four ve-
locity components, we can rewrite the equation (39) to another one, which carries a
deep geometrical meaning:
A =
{
duµ
dτ
+Γµ αν
dxα
dτ
dxν
dτ
}
∂
∂xµ
=
{
d
dτ
(
dxµ
dτ
)
+Γµ αν
dxα
dτ
dxν
dτ
}
∂
∂xµ
We reach out precisely the form of the geodesic equation:
A =
{
d2xµ
dτ2
+Γµ αν
dxα
dτ
dxν
dτ
}
∂
∂xµ
(40)
And then the condition A = 0, which is a system submited of the action of inertial
forces (or Constrain Forces), constrain precisely a geodesic motion of a particle:
A =
{
d2xµ
dτ2
+Γµ αν
dxα
dτ
dxν
dτ
}
∂
∂xµ
= 0 ⇐⇒ d
2xµ
dτ2
+Γµ αν
dxα
dτ
dxν
dτ
= 0 (41)
Acctually, what I’ve said about ficticious forces are valid just in Speciatal Relativity.
This is, in fact, the picture of Special Relativity in non-inertial reference frames,
pretty analogous to the Newtonian counter part. But in General Relativity this way to
describe the world is changed by the fact that all reference frames are equivalent; the
geodesic equation then, will describe the motion of a free particle in a gravitational
field (a.k.a. curvature of spacetime).
15
References
[1] RINDLER, W. Introduction to Special Relativity, 2ed, Clarendon Press, Oxford,
1991.
[2] FARAONI, V. Special Relativity, 1ed, Springer, 2013.
16
	The 4-acceleration vectorSome other mathematical "mambo-jambo"
	Norm of 4-acceleration vector
	Some physical models and their 4-acceleration
	Motion with constant speed
	Acceleration in 1+1 dimension:
	The 4-acceleration in non-inertial coordinate systems
	References