today in physics 218: lorentz invariantsdmw/phy218/lectures/lect_77b.pdf · microsoft powerpoint -...

14 April 2004 Physics 218, Spring 2004 1

Today in Physics 218: Lorentz invariants

The Einstein summation conventionThe Minkowski invariant intervalProper time and four-velocityFour-momentum and the relativistic energy

Movie made from VLBI radio observations of the quasar 3C273, showing the ejection of material from the active nucleus at apparent speed 7c (!) transverse to the line of sight (Tim Pearson, Caltech). See problem 12.6, on the next homework, to understand why the ejecta look like they move faster than light.


Scalar products of four-vectors, and the summation convention

Last time we introduced a bookkeeping device for scalar products, the contravariant and covariant forms of four-vectors:

The scalar product can be obtained either by multiplication of the vectors that represent the four-vectors:

( )

0

10 1 2 3

2

3

, .

a

aa a a a a a

a

a

µµ

⎛ ⎞⎜ ⎟⎜ ⎟

= = −⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠


Scalar products of four-vectors, and the summation convention (continued)

or more compactly as

( )

0

10 1 2 3 0 0 1 1 2 2 3 3

2

3

0 0 ,

b

ba a a a a b a b a b a b

b

b

a b

⎛ ⎞⎜ ⎟⎜ ⎟

− = − + + +⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

= − + a b◊

30 0

0.a b a bµµ

µ== − +∑ a b◊


Scalar products of four-vectors, and the summation convention (continued)

Or even more compactly. When Einstein started using four-vectors in relativity, he quickly got tired of writing all the sums, and began using the following convention: for an index that is repeated, once covariant and once contravariant, one assumes that the term in which it appears is to be summed over that index from zero to 3:

We will use Einstein’s summation convention henceforth.

3

0

no sum there must be some mistake...

a b a b

a ba b

µ µµ µ

µν

µ

µ µ

=⇔

⇒

⇒

∑


Useful scalar products: the invariant interval

Consider two events A and B, that can be seen by observers in two different inertial reference frames. The observers would give different coordinates for the events,but the same value of

This is called the invariant interval.

y

,x x

vy

zz

A

, , , and ,B BA Ax x x xµ µ µ µ

B

( )( )22 2 2 22 .

A B BAI x x x x x x

c tc t d d

µ µ µµ µ µ= − − = ∆ ∆

= − ∆∆ += − +

, ,BAx x Iµ µ , ,BAx x I

µ µ


Useful scalar products: the invariant interval (continued)

If I < 0, the interval is dominated by the time-difference part, and is described as timelike as a result.

Since I < 0, there are no two frames of reference for whichbut one can have , so that A

and B can appear to one observer to happen at the same spatial point.The I < 0 case describes all pairs of events that are connected by cause and effect, since in this case the time order is always preserved: if I < 0 and A occurs before B in one reference frame, if occurs before B in all inertial reference frames.

20 (because 0),t d∆ = ≥ 2 0d =



If I > 0, on the other hand, one can always find a frame in which A and B are simultaneous as well as pairs of frames in which the events occur in different order. Such an interval is called spacelike.Example (Griffit$hs prob$lem 12.2$1):The coordinates of events A and B areAssuming the interval between them is spacelike, find the velocity of the reference frame in which they are simultaneous. Solution: in the new frame,

( )0 ,t∆ =

( ) ( ), ,0,0 , , ,0,0 .A A B Bt x t x

( )( ) ( ) ( )

0

.A B

A A B B

c t c t t

t x t x c t xγ β γ β γ β

= ∆ = −

= − − − = ∆ − ∆



Thus,

Since for spacelike separations one can find pairs of frames in which the two events occur in opposite orders, these sorts of separation cannot apply to events that are related by cause and effect.

2

, or

.

t xc

tv c cx

β

β

∆ = ∆

∆= =

∆



The case I = 0 is referred to as a lightlike separation, because

in all reference frames. Two events separated by a lightlikeinterval can be causally related only if the “cause” travels at the speed of light.

( )

22

2d ct

=∆


Proper time and four-velocity

The proper time, dτ, is a time interval defined in a clock’s rest frame, from the dilated interval observed in another frame:

It’s invariant because there’s only one frame at rest with respect to your clock. Yourspeed, according to me, is

y

,x x

uy

zz

!dt !dτ

2

21 .ud dtc

τ = −

You

Me, measured

in my framettd

d=

⎧⎨⎩

u


Proper time and four-velocity (continued)

According to your time, though, your speed is

It is useful to construct a four-vector velocity from this new object

2

2

measured in my fra.

1

me, in yours.u

dd u

c

τ τγ= = =

−

⎧⎨⎩

u uη

:η0

2 2

2 2

, ,

1 1u u

d dt cc cd du u

c c

γ η γτ τ

= = = = = =

− −

x u uη

Four-veloci. tydxd

µµη

τ=


Proper time and four-velocity (continued)

It’s a four-vector, and thus Lorentz-transforms like one:

because is a four-vector, and the derivative with respect to proper time is as invariant as proper time.

Contrast this with u: this velocity is not part of a four-vector, because both need to be transformed. This is why the Einstein velocity addition rule you learned in freshman physics is more complicated than the Lorentztransformation.

,µ µ ννη η= Λ

xµ

and t


Flashback: Einstein velocity addition rule

y

,x x

vy

zz

( )

2

2

2

2

1

1

1

1

1

1

1

xx

x

yy

x

zz

x

u vdxu vudtc

uu vu

cuu vu

c

v c

γ

γ

γ

−= =

−

=−

=−

=−

u


Four-momentum and relativistic energy

From the four-velocity, one can define the four-momentum,or “energy-momentum four-vector,” as

where m is the rest mass of the moving body – also an invariant, because nothing has more than one rest frame.

The invariant that can be constructed from the scalar product of the four-momentum with itself expresses relativistic energy and momentum conservation (note: the following is G$riffiths probl$em 12.2$6, more or less):

( )( )20 0 ,

,

uu

mc Ep m m cc c

m

γη γ= = = =

=p η


Four-momentum and relativistic energy (continued)

Since the right-hand side is constant, or equivalently since

( ) ( )

( )

( )

2 22 02

2 2 220 22 2

2 2

22 2 2 2

2 2 2 2

22 2 2 2

1 1

11 1

.

Ep p p pc

m c mm mu uc c

um c m cu c u c

E p c mc

µµ

η

= − + = − +

= − + = − +

− −

⎛ ⎞= − + = −⎜ ⎟⎜ ⎟− −⎝ ⎠

⇒ − =

p p

u uη η

◊

◊◊

2 2 2 2 2 2, .p p p p E p c E p cµ µµ µ= − = −



The energy-momentum invariant, and the Lorentztransformation of the four-momentum, are very useful in solving problems on the kinematics of relativistic particles. Example (Griffit$hs probl$em 12.3$0): Suppose you have a collection of particles, all moving in the x direction, with energies Find the velocity of the center of momentum frame, in which the total momentum is zero.Solution:

1 2 1 2, ,... and momenta , ,...E E p p

1 2 1 2 Let ... and ... Then,T TE E E p p p= + + = + +2

0 , or .TTT TT

c pEp p v cc E

γ β β⎛ ⎞= − = = =⎜ ⎟⎝ ⎠



Example (Griffit$hs probl$em 12.3$3): A neutral pion, with rest mass m and relativistic momentum decays into two photons. One of the photons is emitted in the same direction as the original pion, and the other in the opposite direction. Find the energy of each photon. Solution: The total energy of the pion is given by

and energy and momentum are conserved, so the energy and momentum of the photons (1 and 2) are given by

3 4 ,p mc=

( ) ( ) ( )22 2 22 2 2 2 2 2

2

3 251 , or4 16

5 ,4

E p c mc mc mc

E mc

⎛ ⎞⎛ ⎞⎜ ⎟= + = + =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

=



But photons have zero rest mass , so

where we’ve take photon 1 to travel along the pion’sdirection. Add these to get subtract them to get

21 2 1 2

5 3, .4 4

E E mc p p mc+ = − =

( )E pc⇒ =2

1 2

21 2

5 ,43 ,4

E E mc

E E mc

+ =

− =

1 ,E 2 :E

2 21 2

1, .4

E mc E mc= =

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