### Lecture3

```Relativistic Invariance
(Lorentz invariance)
The laws of physics are invariant under
a transformation between two
coordinate frames moving at constant
velocity w.r.t. each other.
(The world is not invariant, but the laws
of physics are!)
Review: Special Relativity
Einstein’s assumption: the speed of light is independent of the (constant )
velocity, v, of the observer. It forms the basis for special relativity.
Speed of light = C = |r2 – r1| / (t2 –t1) = |r2’ – r1’ | / (t2’ –t1‘)
= |dr/dt| = |dr’/dt’|
C2 = |dr|2/dt2 = |dr’|2 /dt’ 2
Both measure the same speed!
This can be rewritten:
d(Ct)2 - |dr|2 = d(Ct’)2 - |dr’|2 = 0
d(Ct)2 - dx2 - dy2 - dz2 = d(Ct’)2 – dx’2 – dy’2 – dz’2
d(Ct)2 - dx2 - dy2 - dz2
is an invariant!
It has the same value in all frames ( = 0 ).
|dr| is the distance light moves in dt w.r.t the fixed frame.
A Lorentz transformation relates position and time in the two frames.
Sometimes it is called a “boost” .
•
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/ltrans.html#c2
How does one “derive” the transformation?
Only need two special cases.
Eq. 1
Transformation matrix relates differentials
Recall the picture
of the two frames
measuring the
speed of the same
light signal.
a b
f h
a
+b
f
+h
Next step: calculate right hand side of Eq. 1
using matrix result for cdt’ and dx’.
=1
=0
= -1
c[- + v/c] dt =0
 = v/c
But, we are not going to need the transformation matrix!
We only need to form quantities which are invariant under
the (Lorentz) transformation matrix.
Recall that (cdt)2 – (dx)2– (dy)2 – (dz)2 is an invariant.
It has the same value in all frames ( = 0 ).
This is a special invariant, however.
Suppose we consider the four-vector:
(E/c, px , py , pz )
(E/c)2 – (px)2– (py)2 – (pz)2 is also invariant.
In the center of mass of a particle this is equal to
(mc2 /c)2 – (0)2– (0)2 – (0)2 = m2 c2
So, for a particle (in any frame)
(E/c)2 – (px)2– (py)2 – (pz)2 = m2 c2
covariant and contravariant
components*
*For more details about contravariant and covariant components see
http://web.mst.edu/~hale/courses/M402/M402_notes/M402-Chapter2/M402-Chapter2.pdf
The metric tensor, g, relates
covariant and contravariant components
covariant
components
contravariant
components
Using indices instead of x, y, z
covariant
components
contravariant
components
4-dimensional dot product
You can think of the 4-vector dot product as follows:
covariant
components
contravariant
components
Why all these minus signs?
• Einstein’s assumption (all frames measure the
same speed of light) gives :
d(Ct)2 - dx2 - dy2 – dz2 = 0
From this one obtains
the speed of light.
It must be positive!
C
= [dx2 + dy2 + dz2]1/2 /dt
Four dimensional gradient operator
covariant
components
contravariant
components
4-dimensional vector
component notation
•
xµ
contravariant
components
•
xµ
covariant
components
 ( x0, x1 , x2, x3 )
= ( ct, x, y, z )
= (ct, r)
µ=0,1,2,3
 ( x0 , x1 , x2 , x3 )
= ( ct, -x, -y, -z )
= (ct, -r)
µ=0,1,2,3
partial derivatives
/xµ
=
 µ
4-dimensional
(/(ct) , /x , /y , /z)
= ( /(ct) , )
3-dimensional
partial derivatives
/xµ 
µ
Note this is not
equal to 
= (/(ct) , -/x , - /y , - /z)
= ( /(ct) , -)
They differ by a
minus sign.
Invariant dot products using
4-component notation
contravariant
covariant
xµ xµ = µ=0,1,2,3 xµ xµ
Einstein summation notation
(repeated index one up, one down)  summation)
xµ xµ
= (ct)2 -x2 -y2 -z2
Invariant dot products using
4-component notation
µµ = µ=0,1,2,3 µµ
Einstein summation notation
(repeated index summation )
= 2/(ct)2 - 2
2
= 2/x2
+ 2/y2 + 2/z2
Any four vector dot product has the same value
in all frames moving with constant
velocity w.r.t. each other.
Examples:
xµxµ
pµpµ
pµxµ
pµµ
 µAµ
 µ µ
For the graduate students:
Consider ct = f(ct,x,y,z)
Using the chain rule:
d(ct) = [f/(ct)]d(ct) + [f/(x)]dx
+ [f/(y)]dy + [f/(z)]dz
d(ct) = [ (ct)/x ] dx
= L 0 dx
Summation
over  implied
First row of Lorentz
transformation.
dx  = [ x/x ] dx
= L  dx 4x4 Lorentz
transformation.
For the graduate students:
dx  = [ x/x ] dx
= L  dx
/x  = [ x/x ] /x
= L  /x
Invariance:
dx  dx  = L  dx L  dx
= (x/x)( x/x)dx dx
= [x/x ] dx dx
=  dx dx
= dx dx
Lorentz Invariance
• Lorentz invariance of the laws of physics
is satisfied if the laws are cast in
terms of four-vector dot products!
• Four vector dot products are said to be
“Lorentz scalars”.
• In the relativistic field theories, we must
use “Lorentz scalars” to express the
interactions.
```