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Space-Time Symmetry
Properties of Space
•
Three Dimensionality:
If we consider any arbitrary point in space then maximum three perpendicular lines can be
drawn from this point. These mutually perpendicular lines are called three axes of the
coordinate system. Therefore position of an arbitrary point in space can be defined using
three coordinates (x, y, z). So space is three dimensional.
•
Flatness:
Space is Flat. This means
(i) In a right angle triangle,
(Hypotenuse)2 = (Base)2 + (Normal)2
(ii) The sum of three angles of a triangle is equal to π radians.
(iii) The shortest distance between two points in the space is a straight line.
For most of Classical Mechanics Problems, the space is assumed to be completely flat.
•
Isotropy:
Isotropy of space means uniformity of direction. All the directions of space are equally
preferred.
E.g.: if we have a source of light in space, then it will send light rays in all the directions with
equal speed.
Properties of Space
• Homogeneity:
Homogeneity of space means space is alike everywhere. If an
experiment is performed in space at one place then the identical
experiment performed anywhere in space will give the identical
results.
Also If F=ma is valid in one coordinate system then it will also be
valid in another coordinate system.
• Space Reflection:
Space reflection means transformation of coordinates under
which coordinates change sign.

 
For example if r  xi  yj  zk in right handed
coordinate system
  
then in left handed coordinate system  xi  yj  zk
However laws of physics (e.g. F=ma) remains invariant.
Properties of Time
• One Dimensionality
Time is specified by only one coordinate ‘t’. Hence time is one
dimensional.
• Homogeneity
(i) The laws of physics does not change with time.
For example: If we perform an experiment today and repeat
it after one month, the result of the experiment will be the
same.
(ii) interval of time are not affected by origin of time.
One minute of today will be exactly equal to one minute of
yesterday or tomorrow.
• Isotropy
The laws of physics remains invariant by changing t to –t.
Conclusion
• The properties of space do not change with time.
• Time interval has same value for all times i.e., time
flows uniformly.
• Free space is homogenous.
• Free space is isotropic.
• Free space has property of reflection.
These symmetry properties of space & time are
called space-time invariance principle.
According to these, laws of nature are same at all
points in space & for all times.
Taylor’s Theorem
 
If r1 & r2 are two variables , having infinitesi mal change


dr1 & dr2 respective ly, then according to Taylor' s
theorem in vector form




 
 
f( r1  dr1,r2  dr2 )  f(r1,r2 )  dr1.1 f  dr2 . 2 f

U ˆ U ˆ U ˆ
where 1 f 
i
j
k
x1
y1
z1

U ˆ U ˆ U ˆ
& 2 f 
i
j
k
x2
y2
z 2
Frame of Reference
This train platform serves as the “rest” frame for Observers A and B, but it
moves with a speed v towards Observer C, who stands on the roadside.
A
B
C
v
Homogeneity of Space & Newton’s Third law of motion
• Consider two particles A & B interacting with each
other.
 
Let r1 & r2 be the position vectors of the particles in
coordinate system S.
 
U
(
r
The P.E. of interaction of two particles will be
1 , r2 )
• Consider another coordinate system S’ which is
displaced infinitesimally through a displacement dr.
The position vectors of the particles in S’ will be

 

r1  dr & r2  dr
Then P.E. of interaction of two particles will be

 

U (r1  dr , r2  dr )
Z’
S’
Z
S
m1

dr
A

r1
O
Y

r2
X
B
m2
X’
Y’
O’
Homogeneity of Space & Newton’s Third law of motion
• According to Principle of Homogeneity




U(r1,r2 )  U( r1  dr ,r2  dr ) ...(1)
From Taylor' s theorem in vector form






f( r1  dr ,r2  dr )  f(r1,r2 )  dr .1 f  dr . 2 f

U ˆ U ˆ U ˆ
where 1 f 
i
j
k
x1
y1
z1

U ˆ U ˆ U ˆ
& 2 f 
i
j
k
x2
y2
z 2
 eq.(1) 




U(r1,r2 )  U(r1,r2 )  dr .1U  dr . 2U


 dr .1U  . 2U  0

 
 dr .(1U   2U )  0

since dr  0


 (1U   2U )  0 ...(2)
but for conservati ve forces


F  U


 F12  1U


& F21   2U
 eq. (2) becomes


 F12  F21  0


 F12   F21 ...(3)
which is Newton' s Third Law.
Homogeneity of Space & Law of Conservation of Linear Momentum
• If m1 & m2 be
 the masses of the particles A & B which are moving with
velocities v1 & v2 respectively at any time then according to Newton’s second
law of motion



dv1 
m1a1  F12 or m1
 F12 ...(4)
dt



dv2 
& m2 a2  F21 or m2
 F21 ...(5)
dt
adding (4) & (5)



dv1
dv2 
m1
 m2
 F12  F21
dt
dt
according to Newton' s third law ( eq. (3) )


F12   F21




dv1
dv2
d (m1v1 ) d (m2 v2 )
 m1
 m2
0


0
dt
dt
dt
dt


d
 (m1v1  m2 v2 )  0
dt


 (m1v1  m2 v2 )  constant
which is law of conservati on of linear momentum.
Rotational Invariance of Space & Law of Conservation of
Angular Momentum
• Consider two particles A & B interacting with each
 
other. Let r & r be the position vectors of the
particles in coordinate system S. Distance between
 
two particles will be given by r1  r2
1
2
• The P.E. of interaction between the particles
depends upon the scalar distance between two
particles i.e.,
 
 
U (r1 , r2 )  U ( r1  r2 )
Z’
Z
A
S
S’

r1

r2
O
X
 
r2  r1
B
Y’
Y
X’
Fig. 1
Suppose the system is
rotated through a small angle
d then the position
vectors will get changed to

 

r1  dr1 & r2  dr2 but according
to isotropy of space, the
distance between two
particles will remain same.
Continued…
 
• Therefore P.E. U (r1 , r2 ) before rotation is equal to

 

P.E.U (r1  dr1, r2  dr2 ) after rotation
 

 

U (r1 , r2 )  U (r1  dr1 , r2  dr2 ) ...(1)
• Applying Taylor’s theorem

 


 
 
U (r1  dr1 , r2  dr2 )  U(r1,r2 )  dr1.1U  dr2 . 2U ...(2)
• Substituting the value to eq.
(1),
we
get


 



U (r1 , r2 )  U(r1,r2 )  dr1.1U  dr2 . 2U
 
 
 dr1.1U  dr2 . 2U  0 ...(3)
Continued…
• We know


F  U




 F12  1U & F21   2U
• Hence equation
(3)
becomes




 dr1.F12  dr2 .F21  0
 
 
 dr1.F12  dr2 .F21  0 ...(4)
• We know, when coordinate
system is

through a vector angle d then,



r  d  dr ...(5)
Z
X
where dr is an arc which subtends a vector angle d
at the origin.



r1  d  dr1 ...(6)



r2  d  dr2 ...(7)

rotated r  dr dr
d 
r
O
Y
from Newton' s Second law




dp1
dp2
F12 
& F21 
...(8)
dt
dt
Substituti ng eq. 6,7 & 8 in eq. (4)





dp1 
dp 2
(r1  d).
 (r2  d).
0
dt
dt
Interchang ing Dot & Cross product
  dp1
  dp 2
d.r1 
 d.r2 
0
dt
dt
   dp1  dp 2 
 d. r1 
 r2 
0
dt
dt 


since d  0


  dp1  dp2 
  r1 
 r2 
  0 ....(9)
dt
dt 

 


d (r1  p1 ) dr1   dp1
Now

 p1  r1 
dt
dt
dt


  dp1
 
 v1  mv1  r1 
but v1  v1  0
dt
 

d (r1  p1 )  dp1

 r1 
...(10)
dt
dt
 

d (r2  p2 )  dp2
Similarly
 r2 
...(11)
dt
dt
hence eq.(9) becomes
 
 
 d (r1  p1 ) d (r2  p2 ) 


0
dt
dt


 
d  
 (r1  p1 )  (r2  p2 )   0
dt
but angular momentum is given by
  
Lrp
 
d  

L1  L2  0  L1  L2  constant
dt




Implicit & Explicit Time Dependence
• Suppose we are looking at the movement of a classical particle. The
relevant variables here are position x(t) and momentum p(t).
• For example, angular momentum L⃗=x⃗×p⃗ .
Since x and p depend on the time, L also depends on time, but in this case
it does so only because x and p depend on time.
We have basically a function L=L(x,p) which then becomes L(x(t),p(t)).
This is because in the definition of L , the time does not play a role.
Therefore, we say that this quantity has only an implicit time dependence.
In particular, ∂L/∂t=0 .
• If, however, derived quantity f is defined such that the time occurs explicitly
in the definition,
for example a=∂v/∂t
then a=a(v,t)
there is direct dependence (explicit time dependence) on time
& ∂a/∂t will not be equal to zero.
Homogeneity of flow of Time & Law of Conservation
of Energy

We know, force is related to P.E. by
F 

U
...(1)
r
We know gravitational force between two
masses m1 & m2 is given by

Gm1m2 
F 
r
2
r

It is clear from the expression that force
does not depend on time explicitly
(directly). Therefore P.E. will also not
depend on time explicitly.
U
 0 ....(2)
t
Continued…
• Also K.E. will also not depend on time explicitly
but depends on time implicitly
K.E. T 

1 2
mv
2
T
 0 ....(3)
t
• Total energy of system
E=T+U where E is function of r & t.
i.e. E=E(r,t)
Continued…
as E  E (r , t )
E
E
 dE 
dr 
dt
r
t
 (T  U )
 (T  U )

dr 
dt
r
t
U   T
U 
 T

dr 
dr   
dt 
dt 
r
t 
 r
  t
using eq. (2) & (3),
U 
 T
dE  
dr 
dr 
r
 r

dE  T U  dr



...(4)

dt  r r  dt
U
but
  F ....(5)
r
T   1 2  m v 2
&
  mv  
r r  2
 2 r
m v
v
 2v  mv
2 r
r
T
r v

m
r
t r
T
v

 m.
r
t
T

 ma ....(6)
r
substituti ng eq. (5) & (6) in eq. (4)
dE
dr
 ma  F 
dt
dt
but F  ma
dE

0
dt
or E  constant
Frames of Reference
Inertial & Non-Inertial Frames
• Inertial Reference Frame:
 Any frame in which Newton’s Laws are valid!
 Any reference frame moving with uniform motion (nonaccelerated) with respect to an “absolute” frame “fixed” with
respect to the stars.
• Non-Inertial Reference Frame:
 Any frame in which Newton’s Laws are not valid!
 Any reference frame moving with non-uniform motion
(accelerated) with respect to an “absolute” frame “fixed” with
respect to the stars.
• The most common example of a non-inertial frame = Earth’s
surface!
• We usually assume Earth’s surface is inertial, when it is not!
– A coord system fixed on the Earth is accelerating (Earth’s rotation
+ orbital motion) & is thus non-inertial!
– For many problems, this is not important. For some, we cannot
ignore it!
RELATIVITY
Problems with Classical Physics


Classical mechanics are valid at low speeds  c
But are invalid at speeds close to the speed of light
Relativity


a special case of the general theory of relativity for
measurements in reference frames moving at
constant velocity.
predicts how measurements in one inertial frame
appear in another inertial frame. How they move
wrt to each other.
Reference Frames

The problems described will be done using reference
frames which are just a set of space time coordinates
describing a measurement. eg.

r  r x, y, z, t 
z
t
y
x
Galilean-Newtonian Relativity


According to the principle of Newtonian Relativity,
the laws of mechanics are the same in all inertial
frames of reference.
i.e. someone in a lab and observed by someone
running.
Galilean-Newtonian Relativity
Galilean Transformations
Galilean Transformations





We will consider effect of uniform motion on
different quantities & laws of physics.
We will establish a relationship between the space
& time coordinates in two inertial frames of
reference.
The basic relations were obtained by Galileo & are
known as Galilean Transformation Equations.
allow us to determine how an event in one inertial
frame will look in another inertial frame.
assume that time is absolute.
Galilean Transformations


When second frame is moving relative to first along positive
direction of X-axis.
In S an event is described by (x,y,z,t). How does it look in S′?
z
z′
S
v
S’
vt
x′
t
P(x,y,z.t)
(x’,y’z’.t’)
t′
x
O
O′
x′
Y
Y’
Galilean Transformations


We will assume that time of occurrence is same in both the
frames
t = t′
From the diagram, x  x'  vt
 x'  x  vt

And there is no relative
motion in Y & Z-directions
z
v
z′
vt
x′
y'  y
t
z'  z
t′
O
Y
O′
Y’
x
x′
• So the Galilean transformations are
x'  x  ut
y'  y
z'  z
t'  t
• Inverse Galilean transformations are
x  x'ut
y  y'
z  z'
t  t'
Galilean Transformations for Velocity


Velocities can also be transformed.
Using the previous equations,
dx' d ( x  vt) dx
v' x 


v
dt
dt
dt
dy ' dy
v' y 

 vy
dt dt
dz ' dz
v' z 

 vz
dt dt
 v' x  v x  v
addition law for velocities
Galilean Transformations for Acceleration
dv' x d (vx  v) dvx dv
a' x 



dt
dt
dt dt
dv
since v is constant

0
dt
dvx
 a' x 
 ax
dt
& a' y  a y , a' z  a z
Hence accelerati on remains invariant
Galilean Transformations for Force
• The mass is a constant quantity in Newtonian Mechanics and
does not depend on state of rest or motion of the object.
• According to Newton’s second law of motion
F  ma
• w.r.t frame of reference S & S’
F  ma (in S)
F '  ma ' (in S')
• But acceleration is invariant under Galilean Transformations
a  a'
 ma  ma '
or F  F '
Galilean Transformations
Transforming Lengths
Galilean Transformations


How do lengths transforms transform under a
Galilean transform?
Note: to measure a length two points must be
marked simultaneously.
Galilean Transformations
S
S′
XA


v
XB
Consider the truck moving to the right with a velocity v.
Two observers, one in S and the other S′ measure the length
of the truck.
Galilean Transformations
S
v
S′
XA


XB
In the S frame, an observer measures the length
= XB-XA
In the S′ frame, an observer measures the length
= X′B-X′A
Galilean Transformations
S
XA


v
S′
XB
We know x’=x-vt
each point is transformed as follows:
x 'B  xB  vtB
x ' A  xA  vt A
Galilean Transformations
x' B  xB  ut B
S
U
S′
0
x' A  x A  ut A
Therefore we find that
0
XA
XB
x' B  x' A  xB  x A  u(t B  t A )
if we measure poeition simulatenously then
tB  t A

x' B  x' A  xB  x A
Hence for a Galilean transform, lengths are
invariant for inertial reference frames.
Important consequence of a Galilean
Transformations

All the laws of mechanics are invariant under a
Galilean transform.
Galilean Transformations


When second frame is moving relative to first along a straight line
along any direction
In S an event is described by (x,y,z,t). How does it look in S′?
Z’
P(x,y,z.t)
(x’,y’z’.t’)
S’
Z

r'

r
S
Y’
vt
O
Y
X’
X
O’
• S & S’ are two frames of references.
• S’ is moving w.r.t. S with velocity ‘v’
• The coordinates of point P are (x,y,z,t) & (x’,y’,z’,t’)
relative to S & S’ respectively.
• Time is measured from the instant at which the
two origins O & O’ coincide
• Then after time t, O’ is separated from origin O by
displacement vt
• Then OO '  vt
& let OO '  r , O ' P  r '
• Applying the triangle law of vectors in triangle
OO’P, we get
OP  OO ' O ' P
 r  r ' vt
or r '  r  vt ...(1)
• Now
r  xi  yj  zk
r '  x 'i  y ' j  z 'k
& v  vx i  v y j  vz k
• Therefore eq. 1 becomes
( x ' i  y ' j  z ' k )  ( xi  yj  zk )  (vx i  v y j  vz k )t
comparing coefficients of i , j & k , we get
x '  x  vx t
y '  y  vyt
z '  z  vz t
Invariance of Law of Conservation of
Linear momentum
• Under Galilean Transformations, velocity is not
invariant, therefore linear momentum is also not
invariant.
• We have to check that whether law of
conservation of linear momentum holds or not
• According to whether law of conservation of
linear momentum “in the absence of external
force, the total momentum of the system remains
unchanged or conserved”
• Consider two particles
m1u1  m2u2  m1v1  m2v2
u1  u1'  v
u2  u 2'  v
v1  v1'  v
v2  v 2'  v
m1 (u1'  v)  m2 (u 2'  v)  m1 (v1'  v)  m2 (v 2'  v)
 m1u1'  m2u 2'  m1v1'  m2 v 2'
Invariance of Law of Conservation of
Kinetic Energy
• In S
1
1
1
1
m1u12  m2u22  m1v12  m2 v22
2
2
2
2
1
1
1
1
 m1 (u1.u1 )  m2 (u2 .u2 )  m1 (v1.v1 )  m2 (v2 .v2 )
2
2
2
2
1
1
1
1
 m1 (u1'  v).(u1'  v)  m2 (u 2'  v).(u 2'  v)  m1 (v1'  v).(v1'  v)  m2 (v 2'  v).(v 2'  v )
2
2
2
2
u1  u1'  v
u2  u 2'  v
v1  v1'  v
v2  v 2'  v

1
1
1
1
m1u1'2  m2u2'2  m1v1'2  m2v2'2
2
2
2
2

Are all the laws of Physics invariant in all inertial
reference frames?
Problems with Newtonian- Galilean
Transformation


Are all the laws of Physics invariant in all inertial
reference frames?
For example, are the laws of electricity and
magnetism the same?
Problems with Newtonian- Galilean
Transformation



Are all the laws of Physics invariant in all inertial
reference frames?
For example, are the laws of electricity and
magnetism the same?
For this to be true Maxwell's equations must be
invariant.
Problems with Newtonian- Galilean
Transformation
Problems with Newtonian- Galilean
Transformation

From electromagnetism we know that,
c2 

1
 0 0
Since and
constants then the speed of light
are
0
0
is constant
.c  3  108 m / s
Problems with Newtonian- Galilean
Transformation

From electromagnetism we know that,
c2 


1
 0 0
Since and
constants then the speed of light
are
0
0
is constant
.c  3  108 m / s
However from the addition law for velocities
cv
Problems with Newtonian- Galilean
Transformation



Therefore we have a contradiction!
Either the additive law for velocities and hence
absolute time is wrong
Or the laws of electricity and magnetism are not
invariant in all frames.
Fictitious Forces
• Some problems (like rigid body rotation!): Using an inertial frame is
difficult or complex. Sometimes its easier to use a non-inertial frame!
• “Fictitious Forces”: If we are careful, we can the treat dynamics of
particles in non-inertial frames.
– Start in inertial frame, use Newton’s Laws, & make the coordinate
transformation to a non-inertial frame.
– Suppose, in doing this, we insist that our equations look like Newton’s
Laws (look like they are in an inertial frame).
 Coordinate transformation introduces terms on the “ma” side of F =
mainertial. If we want eqtns in the non-inertial frame to look like Newton’s
Laws, these terms are moved to the “F” side & we get: “F” = manoninertial.
where “F” = F + terms from coord transformation  “Fictitious Forces”
!