Post on 15-Jan-2016
transcript
P1X*Dynamics & Relativity:
Newton & Einstein
Chris ParkesOctober 2005
Special Relativity
Postulates
Time Dilation
Length Contraction
Lorentz Transformation
Relativistic Energy
http://ppewww.ph.gla.ac.uk/~parkes/teaching/DynRel/DynRel.html
Part II - “When a man sits with a pretty girl for an hour, it
seems like a minute. But let him sit on a hot stove for a minute and it's longer than any hour.
That's relativity.”
2005 Centenary
Read the textbook Phenomena close to speed of light, high energy
Michelson Morley Interferometer
Ether was a hypothetical substance through which light travelled•The universe was thought to be a stationary frame of reference full of ether.
So, if the earth is traveling at v with the ether then light should have one velocity c+v (w.r.t. ether).In the perpendicular direction the velocity would just be c
In 1887 Michelson & Morley set up this apparatus in which a light beam is split into two perp. Beams, then
recombined.
By observing interference between the two beams, they showed that the
speed of light was constant
Maxwell’s eqns of electromagnetism (1865)Contain a velocity c for light (emag. waves)
Einstein’s Postulates• First Postulate
– Laws of physics are the same in any inertial frame of reference (principle of relativity)
• See before, co-ordinate axes moving at constant velocity• No preferred frame of reference, no frame is more
“correct”- cannot tell if you are moving at constant velocity
• Second Postulate– The speed of light in vacuum is the same in all inertial
frames• Independent of the motion of the source• at rest I measure speed of light c (~3x108 ms-1)• If it is the headlight of a train moving at u
Newton: speed =c+u Einstein: speed still =c !!!
c
v
Newton’s laws are the low velocity limit of Einstein’s special relativity
Co-ordinate Transformations Revisited
v
vtxx
dtdx
dtdx
'
'
dtdy
dtdy
yy'
'
0
ut
Frame SFrame S’(x’,y’)
x’
yV of S’ wrt S
0 x
uvv xx '
Now, consider point moving, differentiate
Obtain Galilean velocity transformation(as before)
i.e. despite looking convincing this is wrong at high velocities. It gives us the
ucv problem for our train headlight.
We have assumed time is the same for our observer in S (station)and S’ (train) t=t’
No tt’, and the velocities must be defined as:dtdxv '
'' dtdxv
(and similarly with distances…)
Time Intervals: Simultaneous Events• Two events simultaneous in one reference frame are not
simultaneous in any other inertial frame moving relative to the first.
Two bolts seen simultaneously at C
Right bolt seen first at C’
Left bolt seen second at C’
Two lightning bolts strike A,B
S-frame : simultaneous lighting strikesS’-frame : right bolt hit first
Time Dilation• Time interval is shortest in a reference frame where a clock
is at rest. Moving clocks appear slower to an observer.
i.e. an observer looking at a moving clock, measures a longer time on her watch than on the moving clockLets prove this:
v
Stationary observer Moving w.r.t. observer
light
mirror
In both frames (S,S’) observers agree light is travelling at cbut they disagree on the distance (path) the light has travelled
S’S
Frame S: Time takenc
dt
2
speed
distance0
d
This is called the proper time (measured in stationary frame)This is the shortest possible time interval, moving observers measure a longer time
Time Dilation continuedFrame S’ (moving observer):
L L
vt
d
Distance travelled by light =2LBy Pythagoras,
So time taken2
222
222
22
2
2
vtdL
ct
c
Lt
vtdL
Now,
20
222
20
22222
220
2
2
2
2
2
2
2
2
0200
)1(
22
2
2
2
0
tctc
tctvtc
vtctct
ctdd
ct
c
dt
cv
vtctct
So, (By substituting for d in expression above)
Hence, time measured by moving observer2
2
10
cv
tt
Define velocity relative to light speedAnd time “correction” factor
cv
21
1
Hence, can rewrite time as 0tt
Time appears slower
(to stationary observer)
for moving system
t>t0
(So,<1, >1 always)
Time Dilation Example:b-quark decay
• An unstable b quark is produced• The b quark travels ~ 4mm
before decaying• It is travelling at 0.99c (=9)• Hence
4mm
Image reconstructed byDELPHI particle physics
Experiment at CERN
pssms
m
v
st 13103.1
10399.0
004.0 1118
However, the average lifetime (at rest)of a b-quark is 1.5 ps (1 pico-s is 1x10-12 s)
So why did we measure 13ps ? Time dilation
pspstt 5.135.190
Length Contraction• Lengths are longest in a reference frame where the object is at rest. Moving Objects appear shorter to
an observer.
Only applies to lengths in direction of travel
lightmirror
L0
vvt
Stationary observer Moving w.r.t. observerS’SFrame S: Time taken for light to bounce back and forth
c
Lt 0
0
2
speed
distance
L
L0 is the proper length (measured in stationary frame)This is the longest possible length, moving observers measure a shorter length
Frame S’ (moving observer):Time taken for light to travel from source to mirror = t1
Corresponding distance travelledHence,
111 ctvtLd
vcLtLvtct 111
Length Contraction continued
• Length contraction(remember >1)
Time taken for light to travel from source to mirror = t2
Corresponding distance travelledHence,
Frame S’ (moving observer) cont.:
222 ctvtLd
vcLtLvtct 222
So, total time vcL
vcLttt 21
22
)1(
2
2))((
)()(
2
2
22
cL
c
L
vcLc
vcvcvcLvcL
c
v
But,from time dilation cL
cLtt 0222
0
0LL
Moving object appears to shrink by a factor in the direction of travel
L
vt2
Including S’ origin movement
If S’ movement in x direction, then y,z co-ords unaffected
Lorentz Transformations
• Covered special cases– Time dilation,
– length contraction
• General form of how to relate two frames S, S’
vt
Frame S
Frame S’(x’,y’)
x’
y
V of S’ wrt S
0x
•Transform (x,y,z,t) in S to (x’,y’,z’,t’) of S’
A distance x’ in S’ is seen as x’/ w.r.t. S
)(''' vtxxvtxvtx xx
',' zzyy
0
–S’ moving with velocity v along x axis
Spatial Transform
Same as Gallilean transforms
if =1, i.e.=0, low velocity approx.
Lorentz Tranformations : time
Lorentz Transformations Summary:
A distance x in S is seen as x/ w.r.t. S’
Including S origin movement (velocity –v w.r.t. S)
)('
'
vtxx
vtx x
but from spatial transform
)('
'
'
)1('
'
2
2
2
2
21
cxv
cv
cv
x
x
tt
xtt
xvtvt
xvtxvtvt
vtxvt
Hence,
since 2
2
22
2
2211
1
12 11cv
cv
cv
)(' vtxx
)(' 2cxvtt
'
'
zz
yy
Moves from Frame S to a
Frame S’ travelling at velocity v along x axis
Thus,
To move from S’ to S, just reverse sign on ve.g. )'( 2
'c
vxtt )''( vtxx
(tt’for low velocity)
Lorentz Velocity Transform• Particle moving at speed v in S• Frame S’ moving at speed u wrt S’• What speed (v’) is it moving at in S’ ?
Return to our velocity question…
Velocity in frame S dtdxv Velocity in frame S’ '
'' dtdxv
Now, )(' udtdxdx )(' 2c
udxdtdt from Lorentz transforms
Hence,
dtdx
cu
udtdx
dt
dx
21'
'
or
21'
c
xuvx uv
xv
and2'1
'
c
xuvx uv
xv
Is speed of light a constant? Put vx=c (e.g. c for velocity of headlight of train, u is speed of train)
cvcucu
cu
c
uc
cucucx
1
)1(
11 2' i.e. if it moves with c in one frame
it moves with c in all frames second postulate
…and low velocity limit?uvv xx ' back to Gallilean transform
11 2
' c
uv x
So,denominator
Spacetime So, for moving systems as measured by stationary system:Time got longer, distance got shorter.But more than this - from Lorentz transforms we know:Space co-ords in one frame depends on both space and time in other frameAnd time in one frame depends on both space and time in otherSpace & time have become intertwined - ‘joined’ in spacetime.Space and time mix between frames.We have to consider spacetime 4-vector co-ordinates (x,y,z,t)
Spacetime distance is conserved between frames (i.e. a rotation)-the increased time co-ord compensates the reduced spatial co-ord.-Define co-ords as (x,y,z,ct)3D space distance2 is: 4D space-time distance2 is:
22222222 )()( ctzyxictzyx
ict
z
y
x
ict
z
y
x
222 zyx
z
y
x
z
y
x
x
ct
light
Minkowski spacetime
That Equation• Mass is a form of energy
– Can be used to create new particles A->B+C
• Rest mass m0
vmvmp 0
•E = mc2
•m = m0
m0c2
also
420
222
420
220
222
420
420
2420
22 )1(
cmcpE
cmcmvE
cmcmcmE
since 2
2
2
2
2
2
2
2
2
2
2
2
22
11
)1(11
1
1)1(
cv
cv
cv
cv
cv
cv
More useful to express in terms of momentum:
Relativity Top 5• Laws of physics are the same in any inertial frame of
reference• The speed of light in vacuum is the same in all inertial
frames• Moving clocks appear slower to an observer.• Time difference is shortest when the clock is at rest in the
reference frame
• Moving Objects appear shorter to an observer• Object is longest when the object is at rest in the reference
frame
• space and time get mixed spacetime– Transform (x,y,z,t) in S to (x’,y’,z’,t’) of S’
0tt 21
1
where cv
Postulate 1
Postulate 2
TimeDilation
LengthContraction
0LL
LorentzTransformations