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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.

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relativistic invariance lorentz invariance

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
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’|

slide3

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 dtw.r.t the fixed frame.

slide4

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
How does one “derive” the transformation?Only need two special cases.

Recall the picture

of the two frames

measuring the

speed of the same

light signal.

Eq. 1

Transformation matrix relates differentials

a

b

f

h

a

+ b

Next step: calculate right hand side of Eq. 1 using matrix result for cdt’ and dx’.

f

+ h

slide6

= -1

= 1

= 0

slide10

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.

slide11

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
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
The metric tensor, g, relates covariant and contravariant components

covariant

components

contravariant

components

using indices instead of x y z
Using indices instead of x, y, z

covariant

components

contravariant

components

4 dimensional dot product
4-dimensional dot product

You can think of the 4-vector dot product as follows:

covariant

components

contravariant

components

why all these minus signs
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
Four dimensional gradient operator

contravariant

components

covariant

components

4 dimensional vector component notation
4-dimensional vectorcomponent notation
  • xµ (x0, x1, x2, x3 )µ=0,1,2,3

= ( ct, x, y, z )

= (ct, r)

  • xµ  ( x0 , x1 , x2 , x3 ) µ=0,1,2,3

= ( ct, -x, -y, -z )

= (ct, -r)

contravariant

components

covariant

components

partial derivatives
partial derivatives

/xµ µ

= (/(ct) , /x, /y , /z)

= ( /(ct) , )

4-dimensional

gradient operator

3-dimensional

gradient operator

partial derivatives1
partial derivatives

/xµ  µ

= (/(ct) , -/x, - /y , - /z)

= ( /(ct) , -)

Note this is not

equal to 

They differ by a

minus sign.

invariant dot products using 4 component notation
Invariant dot products using4-component notation

xµxµ = µ=0,1,2,3 xµxµ

(repeated index one up, one down)  summation)

xµxµ = (ct)2 -x2 -y2 -z2

contravariant

covariant

Einstein summation notation

invariant dot products using 4 component notation1
Invariant dot products using4-component notation

µµ = µ=0,1,2,3µµ

(repeated index summation )

= 2/(ct)2 - 2

2 = 2/x2 + 2/y2 + 2/z2

Einstein summation notation

slide23
Any four vector dot product has the same valuein all frames moving with constantvelocity w.r.t. each other.

Examples:

xµxµ pµxµ

pµpµµµ

pµµ µAµ

slide24

For the graduate students:

Considerct =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

= L0dx

dx  = [ x/x ]dx

= L  dx

Summation

over implied

First row of Lorentz

transformation.

4x4 Lorentz

transformation.

slide25

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
  • 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.

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