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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) 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 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
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
= -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
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 contravariant components covariant components
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 /xµ µ = (/(ct) , /x, /y , /z) = ( /(ct) , ) 4-dimensional gradient operator 3-dimensional gradient operator
partial derivatives /xµ µ = (/(ct) , -/x, - /y , - /z) = ( /(ct) , -) Note this is not equal to They differ by a minus sign.
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 using4-component notation µµ = µ=0,1,2,3µµ (repeated index summation ) = 2/(ct)2 - 2 2 = 2/x2 + 2/y2 + 2/z2 Einstein summation notation
Any four vector dot product has the same valuein all frames moving with constantvelocity w.r.t. each other. 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 • 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.