Transcript Relativistic mechanics - IIS Severi

Relativistic mechanics
Mass is nothing but
energy.
The body’s inertia depends
on its speed
Let’s consider a body at rest which
absorbs two electromagnetic impulses
According to the momentum conservation law:
After the absorption, the body will remain at
rest.
Where is the energy gone?
Maybe it has increased the temperature of the
body, but …
what can we say if the body is an electron?
Let’s consider the situation by the point of view of an
observer who is moving at speed V towards left.
If the body will remain at rest in the
previous frame,
then it will still move at speed V also with
respect to the new observer.
• But, according to the momentum
conservation law, we expect that the body’s
momentum will be increased:
mafterVafter
E'
 mbeforeVbefore  cos 
c
V must be the same (Vafter = Vbefore)so,
the only possibility is that the mass m must
change!
 E’ = mc2 ;
the energy E’ is transformed in mass.
Let’s go now to check the fundamental law
of dynamics,
that is:
F  ma
or, in a more universal form:
dp
F
dt
F = ma doesn’t work well!
If we apply a constant force – say an electric
force – to an electron:
uniform magnetic field
uniform electric field
–
+
then, accordingly to F = ma ,
dv eE

dt
m

eE
v
t
m
the electron would reach the speed of light! We
only need to wait for time T:
mc
T
eE
Our reference frame
Another frame as appears to us
In special relativity
the fundamental law of dynamics is
dp
F
dt
where the momentum p is defined by the
product mv
So we get:
p = F t
and, finally:
v
F t

2
c
1  F  t 
A massive body can’t reach the speed c
Why must be m = m0 in p = mv
Let’s go back for a while to the relativistic
kinematics.
We saw that x and t aren’t invariant
quantities as regards Lorentz’s Transformation.
If we think to L-T as transformation of
coordinates in a 4-D space, then we have to
find what is the geometrical characteristic
(invariant) in such a 4-D space – the so-called
Minkowsky’s space-time – .
In Euclidean geometry
the distance  between two points is invariant:
2 = (xA – xB)2 + (yA – yB)2 + (zA – zB)2 =
= (x’A – x’B)2 + (y’A – y’B)2 + (z’A – z’B)2
= ’2
according to T-L, the following
expression is invariant:
s2 = c2t2 – x2 = c2t’2 – x’2 = s’2
Where (ct ; x ; y ; z) are the coordinates of the
point (event) – ct is the time component and
the other are the ordinary space coordinates –
This expression can be taught as a way to
calculate the magnitude of a quadrivector
defined in a 4_D geometry.
According to this geometrical
interpretation
we need to find a dynamics quadrivector whose
components are the “classical” momentum :
mv : mvx ; mvy ; mvz 
Because the conservation of momentum is a
consequence of the homogeneity of space,
the component of mv will appear in the
“spatial places” of the dynamical quadrivector.
( ? ; mvx ; mvy ; mvz )
The first component, the time placed one,
has to be related with energy,
because energy conservation is a consequence
of the homogeneity of time.
So the quadrivector is:  E

 ; mv x ; mv y ; mv z 
c

This can be called energy-momentum
quadrivector
The magnitude of the energymomentum must be invariant:
2
2
0
2
E
E
2 2

m
v

2
c
c
where in the right member we have considered
a reference frame in which a body is at rest.
Let m0 be the (rest) mass of the body.
Then:
E0 = m0c2 and m = m0
2 4
m02 2 c 4
m
2 2 2
0c
 m0  v  2
2
c
c

2(c2
–
And, finally:
v2)
=
c2
2
c
2  2 2
c v

 
1
1  2
!