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Relativistic Invariance (Lorentz invariance)

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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!)

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 dtw.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

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

= -1

= 1

= 0

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

*For more details about contravariant and covariant components see

http://web.mst.edu/~hale/courses/M402/M402_notes/M402-Chapter2/M402-Chapter2.pdf

covariant

components

contravariant

components

covariant

components

contravariant

components

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

covariant

components

contravariant

components

- 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

contravariant

components

covariant

components

- 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

/xµ µ

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

= ( /(ct) , )

4-dimensional

gradient operator

3-dimensional

gradient operator

/xµ µ

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

= ( /(ct) , -)

Note this is not

equal to

They differ by a

minus sign.

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

µµ = µ=0,1,2,3µµ

(repeated index summation )

= 2/(ct)2 - 2

2 = 2/x2 + 2/y2 + 2/z2

Einstein summation notation

Examples:

xµxµ pµxµ

pµpµµµ

pµµ µAµ

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

= L0dx

dx = [ x/x ]dx

= L dx

Summation

over implied

First row of Lorentz

transformation.

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 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.