Additional 8.033 notes Page 1 of 15
8.033 Relativity ~ Additional Notes
Aberration and the Doppler Effect
If a particle is moving with speed u at an angle q to the horizontal, then, in a
frame moving horizontally at a speed v, the horizontal and vertical velocities
will be
u cos q - v u sin q
ux¢ = uy¢ =
1-
ux v
c2 (
g 1-
ux v
c2 )
And so
u sin q
tan q ¢ =
g (u cos q - v )
When using the aberration and Doppler formulae, the angle is the one
between the direction of motion of the photon and the direction in which the
observer is moving.
Relations for a photon E = pc = w = hc / l
Extra Dynamics Stuff
To find a threshold energy, evaluate E2 – p2c2 in the ZMF after the collision
(in which p is 0)
Equate it to the invariant before the collision in the lab frame.
4-Vectors, Formally
4-vectors are vectors that transform like (c dt, dx, dy, dz).
In deriving them, we use the fact that t is invariant and
dt
dt = dt 2 - dr 2 = dt 1 - r 2 =
g
We also note that
dg d 1 æ dv ö
( ) ( ) ⋅ ççç -2 v ÷÷÷ = g 3vv
-1/2 -3/2
= 1 - v2 = - 1 - v2
dt dt 2 çè dt ø÷
© Daniel Guetta, 2008
, Additional 8.033 notes Page 2 of 15
Examples are
o Velocity 4-vector – obtained by dividing every component above by the
proper time ( dt ). Since dt is invariant, the result is a 4-vector
1 1
U =
dt
( dt, dx , dy, dz ) = g (dt, dx , dy, dz ) = g (1, u )
dt
Note – the g refers to the u, because the dx, etc… are taken in the
frame of the moving object.
o Energy-momentum 4-vector obtained by multiplying U by the
invariant m.
P = mU = g (m, mu ) = (E , p )
o Acceleration 4-vector – obtained by taking the derivative of the
velocity 4-vector with respect to t .
dV æ dg d(gu )÷ö
A= = g ççç , ÷
dt è dt dt ÷÷ø
(
= g g 3uu, g 3uuu + ga )
If u points along the x-direction
(
A = g g 3uxax , g 3ux2ax + gax , gay , gaz )
( 3
( 2
= g g uxax , g g u + 1 ax , gay , gaz 2
x ) )
( 4 4
= g uxa x , g a x , g a y , g a z 2 2
)
o Force 4-vector – obtained by taking the derivative of the momentum 4-
vector with respect to t .
dP æ dE dp ö÷ æ dE ö÷
= = g ççç , ÷= g ççç , F ÷÷
dt çè dt dt ÷ø÷ èç dt ø÷
If the mass is constant, we have = m(dU / dt ) = ma , and we can
write
(
= m g g 3uu, g 3uuu + ga )
Comparing these two expressions
F = m g 3uuu + m ga
If u is in the x-direction, we can use the result above
(
F = m g 3ax , gay , ga z )
A very useful result indeed.
© Daniel Guetta, 2008
8.033 Relativity ~ Additional Notes
Aberration and the Doppler Effect
If a particle is moving with speed u at an angle q to the horizontal, then, in a
frame moving horizontally at a speed v, the horizontal and vertical velocities
will be
u cos q - v u sin q
ux¢ = uy¢ =
1-
ux v
c2 (
g 1-
ux v
c2 )
And so
u sin q
tan q ¢ =
g (u cos q - v )
When using the aberration and Doppler formulae, the angle is the one
between the direction of motion of the photon and the direction in which the
observer is moving.
Relations for a photon E = pc = w = hc / l
Extra Dynamics Stuff
To find a threshold energy, evaluate E2 – p2c2 in the ZMF after the collision
(in which p is 0)
Equate it to the invariant before the collision in the lab frame.
4-Vectors, Formally
4-vectors are vectors that transform like (c dt, dx, dy, dz).
In deriving them, we use the fact that t is invariant and
dt
dt = dt 2 - dr 2 = dt 1 - r 2 =
g
We also note that
dg d 1 æ dv ö
( ) ( ) ⋅ ççç -2 v ÷÷÷ = g 3vv
-1/2 -3/2
= 1 - v2 = - 1 - v2
dt dt 2 çè dt ø÷
© Daniel Guetta, 2008
, Additional 8.033 notes Page 2 of 15
Examples are
o Velocity 4-vector – obtained by dividing every component above by the
proper time ( dt ). Since dt is invariant, the result is a 4-vector
1 1
U =
dt
( dt, dx , dy, dz ) = g (dt, dx , dy, dz ) = g (1, u )
dt
Note – the g refers to the u, because the dx, etc… are taken in the
frame of the moving object.
o Energy-momentum 4-vector obtained by multiplying U by the
invariant m.
P = mU = g (m, mu ) = (E , p )
o Acceleration 4-vector – obtained by taking the derivative of the
velocity 4-vector with respect to t .
dV æ dg d(gu )÷ö
A= = g ççç , ÷
dt è dt dt ÷÷ø
(
= g g 3uu, g 3uuu + ga )
If u points along the x-direction
(
A = g g 3uxax , g 3ux2ax + gax , gay , gaz )
( 3
( 2
= g g uxax , g g u + 1 ax , gay , gaz 2
x ) )
( 4 4
= g uxa x , g a x , g a y , g a z 2 2
)
o Force 4-vector – obtained by taking the derivative of the momentum 4-
vector with respect to t .
dP æ dE dp ö÷ æ dE ö÷
= = g ççç , ÷= g ççç , F ÷÷
dt çè dt dt ÷ø÷ èç dt ø÷
If the mass is constant, we have = m(dU / dt ) = ma , and we can
write
(
= m g g 3uu, g 3uuu + ga )
Comparing these two expressions
F = m g 3uuu + m ga
If u is in the x-direction, we can use the result above
(
F = m g 3ax , gay , ga z )
A very useful result indeed.
© Daniel Guetta, 2008
