/rel517 spring2002 began 1/23we 11:30a Rm226
last cycle: /rel517atsok
text: Rindler's Relativity: Special, General, Cosmological Oxford 2001
LECTURE 1 a novel start
23jan02we initial comment: Even Galilean x' = x - vt mixes time into space!
HANDOUTS: /drawing.rel (and /rel517. Updates of /rel517 will not be listed)
HOMEWORK #1: Execute the drawing suggested in /drawing.rel (which is
also on my web page)
The course's key transformation will be
the Lorentz transformation, (set down 1904 by Hendrik Anton Lorentz)
t' = (t - vx)/sqrt{1-vv},
x' = (x - vt)/sqrt{1-vv},
from time and space coordinates t,x of an inertial observer Unprime ( ),
to time and space coordinates t',x' of another inertial observer Prime ('):
we have started with a railroad track---just one spatial dimension.
Prime moves at velocity v as seen by Unprime. My units have v as the
fraction that the speed is, of the speed of light, i.e. v/c -> v.
The Lorentz transformation can be described (if you already know it all!) as a
``shear in spacetime'', so why not talk first about an ordinary shear?
Weeks later, ``tensors''. History: Tensors were invented to study
stresses and strains, the key word being tension! (So that is another reason
favoring a quick look at ordinary shear in statics.)
-------> -------> ------->
____________ ____________ ____________
| | | | | | | /|\
| | | | | | | |
| | | | | | | |
| | | | \|/ | | |
____________ ____________ ____________
<------- <-------
crummy text better text good text also
the experiment balances forces balances torques
.-- --.
| . --------> . |
octagonally punched sample: . /-----\ . squeeze out the
[lousy ascii art] | / \ /|\ 4 original faces
| | | | and see just a stretch
| | | | and a squeeze, at
\|/ \ / | mutual right angles:
. \-----/ . i.e., the dotted arrows
| . <------- . | running ``at 45deg''.
.-- -- .
LECTURE 2 25jan02fr the word `tensor'
HANDOUT: /tiltedrectangle.rel
to cover a bit on that, and maybe the few extra words needed
to cement the tx Lorentz transformation, intended to break the ice on
2d spacetime diagrams---and to dispel any mystery about /drawing.rel
at the risk of detracting from
project ``Elektrodynamik bewegter K\"orper''
HOMEWORK #2,3,4,5,6: For those who clear your drawing, the tasks specified in..
HANDOUT: /drawingfollowup.rel which is also on my web page.
project ``Elektrodynamik bewegter K\"orper'':
We prepare with a geometrically-focused review of e&m
The Maxwell equations as simplified by Lorentz, in Heaviside-Lorentz (HL)
units show how Gauss, Ampere, Faraday (also Henry), then Maxwell, harbor a
constant (c) with the units of speed: H-L units are clearer than SI units!
(SI is polluted by fusspot mu0=4pi*10^{-7}N/A^2 and epsilon0=1/(cc*mu0), which
we happily avoid. Amazingly, even Feynman the great radical, felt forced
to teach in SI to conform; of course he knew better, as do most physicists.)
Maxwell's equations, in words:
Electric flux comes out of electric charge, | Gauss, electric
and leftcirculates around magnetic-pole current.| Faraday, Henry
Magnetic flux comes out of magnetic pole, | Gauss, magnetic
and rightcirculates around electric current. | Oersted, Ampere
``Current'' includes rate of change of flux. | Maxwell's ``displacement''
in symbols:
oriented length, area, volume. How units force the (1/c)'s: the Tube.
Phi_E = Q_E Phi_B = Q_B Gauss, electric then magnetic
dots denote d/dt . . dots denote d/dt
Bmf = (1/c)(I_E + Phi_E) -Emf = (1/c)(I_B + Phi_B)
Oersted,Ampere Maxwell `Maxwell',Faraday,Henry
conservation of charge conservation of pole
LECTURE 3 Invariance of speed 1 suggests x'=x-vt needs matching t'=t-vx and
28jan02mo same det for both ways, forces det=1 hence puts in the 1/sqrt{1-vv}.
conservation of charge conservation of pole
How I put poles into Maxwell's equations and into the Lorentz force:
conservation of pole is parallel to Maxwell's displacement-current
support of conservation of charge: i.e. the I+Phidot combination is desired.
The link between Emf and PhiBdot, and the q_E (v/c)^B force got by
considering a current loop moving through a B field, is imitated as
link between Bmf and PhiEdot, for a
loop carrying a current of poles moving through an E field.
HANDOUT: Bill Watterson's Calvin & Hobbes cartoon on time dilation.
LECTURE 4 Why v is the velocity: contains hint for your /drawingfollowup.rel
1/30we work on velocity: v means x-jump divided by t-jump.
Jump is between two same-Prime-place events.
But now we leave spacetime, including c-moving waves, to do it differently.
Einstein's 1905 title, ``Zur Elektrodynamik bewegter K\"orper''
``On the Electrodynamics of Moving Bodies'',
suggests something else:
Rindler quotes Einstein1952 in text, p.38: ``What led me more or less directly
to the special theory of relativity was the conviction that the electromagnetic
force acting on a [charged] body in motion in a magnetic field was nothing else
than but an electric field [in the body's rest frame].''
This quote confirms my conviction based on Einstein's 1905 title,
and affirms my intention to get the Lorentz transformation of the electric and
magnetic fields, directly from the Lorentz force law---indeed pretend I hadn't
told you yet about the spacetime Lorentz transformation at all!
Then I will struggle to get [back] to spacetime by looking at how the
magnetic field around a current-carrying wire turns partly electric
when I run along the wire (``charge without charge!'') indeed, that will give
me the Lorentz-Fitzgerald contraction, for my spacetime foot-in-the-door.
[My endgame, to the spacetime Lorentz transformation, will be lifted
from a chapter on Relativity in an old elementary text of F.W.Sears.]
So wipe the spacetime slate clean, to go straight to
E and B to wrench the Lorentz transformation `from electrical engineering'!
Lorentz force -> Lorentz transformation lesson. Use c=1.
Result is E|| and B|| preserved, E'perp = (Eperp + v ^ B)/sqrt{1-vv}
B'perp = (Bperp - v ^ E)/sqrt{1-vv}.
The part hardest to explain quickly is the sqrt{1-vv} denominators.
(The quick hint on this is that the 2x2 matrix parts, in needing to have
the same determinant for +v as for -v, must have unit determinant, and
omitting the sqrt{1-vv} denominators instead yields a bad 1-vv determinant.)
[Must use invariance of charge, and guess same for pole.]
Dissect components. det=1 excuses sqrt{1-vv} denominator inserts.
Finishes E&B Lorentz transformation! On to Lorentz-Fitzgerald contraction:
The 1/sqrt{1-vv} concentration of transverse field directly
informs us of the sqrt{1-vv} longitudinal contraction, provided we
know that the out-of-sight sources are still the same number of Tubes.
The lack of change of longitudinal field similarly tells us of
the absence of any transverse contraction or expansion---this is the e&m
version of Einstein's pipes.
[So how did we know the invariance of charge? Neutrality of atoms,
confirmed with great precision by Millikan's 1909 oil-drop experiment.
But Einstein 1905 could not have used that---also Rutherford's nuclear
atom arrives 1908 (Rutherford 1911 useful for us to appreciate that neutral
lead at rest harbors lots of very fast electrons, but slow nuclei).
But Thomson 1898 had got the electron. That fast electrons are charged
the same as slow electrons could then have looked right already by 1905
---at least, maybe!] Harvey Fletcher helped Millikan: Phy.Today v35June1982p43
So far, c is just the speed needed in Oersted magneto-electric crossover,
if we are honest about sidelining Maxwell's 1865 c-moving E,B waves.
And for convenience we put c=1.
Q: So how do we learn about 1 being a
limiting speed in some sense? A: The sqrt{1-vv} denominators blow up!
The transverse fields go infinite! The Lorentz-Fitzgerald contraction
squashes the 3d world into a badly overprinted 2d pancake! At v=1 you go an
infinitesimal distance to what used to be far away, in no time at all;
so a 1-traveler is seen by nontravelers to be carrying a stopped clock.
(The same thought for other v would say, a sqrt{1-vv}-factor-slowed clock.)
Leave the Maxwell-eq. shock waves for next time---to tell us that
indeed ``c is a law'' as vacuum-light's only allowed speed, for all observers,
and also to give us Maxwell-equation geometric practice.
HOMEWORK#7 Get sum w of velocities u and v (all in one
direction: i.e. no y, z to worry about)
from the fact that Doppler factors multiply, explicitly, from
sqrt{1+w / 1-w} = sqrt{1+v / 1-v} sqrt{1+u / 1-u} .
Hints: squaring gets rid of the sqrt's, then solve for w. Ans.: w={u+v / 1+uv}
LECTURE 5 Einstein's pipes
2/1fr Maxwell shock wave -> ``c is a law'' done right
HANDOUTS: ``The Tube and the Step'', AJP Letter (page proofs)
/relatube (i.e. on alpha system as ~eli/relatube)
Relativistic Mechanics chapter (43) from Sears and Zemansky
Next, an aside on ``c is a law'' to exercise the Maxwell equations:
I was tempted to say ``Maxwell's equations contain a constant c ,
hence c is itself a law of nature. If we assume that Maxwell's equations
work in any inertial frame of reference (we do!), then
motion at c in one inertial frame must convert to motion at c in any
inertial frame.'' Unconvincing! Counterexample: c/2, c/17, &c are
each a universal constant, yet these are *not* invariant speeds---indeed
this list if completed would falsely claim all speeds to be invariant!
But the following *does* do the invariant-speed-c job *right*:
Maxwell ~1865 found electromagnetic wave solutions moving at speed c and
noticed that c matched the already measured speed of light in vacuum.
Of course this was no coincidence: E&M had swallowed optics!
Maxwell might have himself developed relativity if he had not died young.
I will find c-moving solutions (in class) by working out a shock wave---
easier!: no need to mess with sines and cosines. More fun, too: geometry.
These shocks *do* prove c to be an invariant speed: Prime, Unprime
both see any shock move at c ---so when they examine the same shock,
whose Primed speed must be the transform of the Unprimed, that still
is true---hence the result of transforming speed c must come out c .
It is easy to now use this c-invariance to get the
spacetime Lorentz transformation: t+x and t-x reciprocal eigenstretches!
LECTURE 6 Fitzgerald contraction -> Lorentz transformation [after Sears]
2/4mo charge without charge
---a COPPER wire caricatured---:
Bridge to the Lorentz contraction by looking at the magnetic field
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
around an electric current!:
^^^^^^^^^^^^^^^^^^^^^^^^^^^
Current I positive upwards, produces field B around (up->around via
rthand rule)
at radius r out: B*2pir = I (c=1) but (neglecting resistance) E=0.
B = I/2pir around .
Let the current come from L-separated negative charges -q moving v down ,
through a line of +q charges at rest also L-separated.
Current = (charge per unit length)*speed
I = (q/L)v so B = qv/2pirL around.
Now go to system S' moving downwards with the negative charges hence v down .
No parallel fields or Eperp. E'perp = (0 + v down ^ B around)/sqrt{1-vv}
B'perp = (Bperp=B around)/sqrt{1-vv}
E' = E'perp = out vqv/(2pirL*sqrt{1-vv}) because down^around = out
B' = B'perp = B/sqrt{1-vv} around = qv/(2pirL*sqrt{1-vv}) around
The simplest thing here is the enhancement B->B' by factor 1/sqrt{1-vv}.
But also B' = I'/2pir around (I put r'=r by using e.g. Einstein's pipes).
^^^^^^^^^^^^^^^^
So I' is also enhanced similarly, i.e., I' = I/sqrt{1-vv}. But now the
negative charges aren't moving, rather, it is the positive charges
that are moving +v. Were they still L apart, there would be no enhancement.
So they must be closer together by factor sqrt{1-vv}. <-******LORENTZ CONTR.!
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ******This is the
Lorentz-Fitzgerald contraction, derived through electromagnetic reasoning!*****
A perhaps more dramatic manifestation of the Lorentz contraction
is the nonzero value of E' = E'perp = vvq/(2pirL*sqrt{1-vv}) out,
positive outwards. The wire is no longer electrically neutral, it is now
electrically positive---``charge without charge''! This, partly because
^^^^^^^^^^^^^^^^^^^^^
we saw the positive charges to be now more *closely* spaced. But the
negative charges, in being stopped, have now *lost* their Lorentz contraction,
so the negative charges are more *widely* spaced, and that contributes
some more pos-neg imbalance. So do both respacings quantitatively
account for our already known E' = vvq/(2pirL*sqrt{1-vv}) out ? Yes:
The electric field out of a wire of (charge per unit length) = k'
is easily had from applying Gauss's law to a meter length of cylinder
around the wire, to give charge = k'*meter = E'2pir*meter = flux; so
E' = k'/2pir = vvq/(2pirL*sqrt{1-vv}) hence
k' = vvq/(L*sqrt{1-vv}), has been got from Gauss's law. <------------<----\
We now will see that this k' is exactly accounted for |
by the Lorentz contraction of the positives, and the |
Lorentz expansion of the negatives: |
How much ( k'+ ) comes from the + charges? k'+ = (q/L)/sqrt{1-vv}. |
How much ( k'- ) comes from the - charges? The former -q/L which applied |
to charges moving v down has got reduced by a losing of a sqrt{1-vv} |
Lorentz contraction to spacing enhanced by factor 1/sqrt{1-vv}, so is |
now -q/(L/sqrt{1-vv}) = -(q/L)*sqrt{1-vv}. The former +q/L and -q/L |
canceled to zero, so we can leave off both these former values and just add /|\
the new values, getting k' = (k'+) + (k'-) |
= (q/L)/sqrt{1-vv} - (q/L)*sqrt{1-vv} |
k' = (q/L)*( 1/sqrt{1-vv} - sqrt{1-vv} ) |
= (q/L)sqrt{1-vv}*( 1/(1-vv) - 1 ) |
= (q/L)sqrt{1-vv}*( 1 - (1-vv) ) / (1-vv) |
= (q/L)sqrt{1-vv}*(vv)/(1-vv) so |
k' = vv(q/L)/sqrt{1-vv} which, here got from reckoning the ---------> |
Lorentz-contraction-expansion details, checks with what was got |
directly from Gauss's law about 18 lines above.----------------------->-----/
[ Also, collect answers this way: current I, charge per length k=0,
pattern 0 we now know transforms to k' = Iv/sqrt{1-vv}
I I' = I/sqrt{1-vv}
for COPPER. You expect to see current = charge density * v, and this
seems puzzlingly reversed! The units at least are ok: c=1 has v dimensionless.
We will see that this is general when we learn 4-vectors;
you are catching a glimpse of ``a spacelike 4-vector''. ]
Summary. The largest spacing between the hearses of a funeral
march is seen when comoving with the march---that is, perceiving the march
as at rest. To those not moving with the march, the hearses are more
closely spaced. The squeezer is sqrt{1-vv} .
If we see *two* comparable funeral marches moving along the same
road but differently, so that they sift through each other, each
march is separately contracted with respect to us the observer,
according as to how discrepant its motion is with ours. If both marches
bear different markings, and have vastly different velocities so that by
comoving with one of them we drive the speed of the other up to a
large fraction of the speed of light, then the markings of the rapid
movers will grossly dominate in density over the markings of those we joined.
If we switch from meaningless markings to electric charge,
and if one is +
and the other is - , then we can adopt a state of motion where the two will
cancel to neutrality, or we can have either the + charges appear more
densely than the - , or vice versa, by varying choices of our own state
of motion. (``Appear'' is ``to be''! I don't wish you to think of
this as you being fooled by a trick.) In particular, a net nonzero
charge of either sign can be evoked from a net neutrality---given a
net current---by changing our observing state of motion.
Current density without charge density, in one frame of reference,
thus evokes nonzero charge density in another frame of reference:
``charge without charge''.
^^^^^^^^^^^^^^^^^^^^^
* denotes an extra-credit assignment (usually difficult). Such
* credits count in your grade's numerator, but not in the denominator.
*HOMEWORK*8 (easy, but perhaps tedious, hence the asterisks): ANTICOPPER
Change the above ``COPPER wire'' example to have the fun of getting the same
conclusions using different details. My suggestion is that you get
your original upwelling current I also without net charge, by using a
+v flow of positive charges +q, and at-rest negative charges -q,
taking both kinds to be spaced L apart as seen by you, so the only
difference from mine will be the signs of the charges.
To best benefit from this, you must redo anything you need that I
worked out in my example, because you wish to pluck the Lorentz contraction
right out of *your* example, not cite it out of mine!
But you *do* use the law of transformation of electric and magnetic fields,
just the same as I did---that I did *not* derive within the example.
*HOMEWORK*9 (harder, and independent credit): What I used to like better
was to use +q/2 v-moving charges and -q/2 -v-moving charges. But that is
harder! Harder because there are then two at first unknown Lorentz
contractions to juggle (though equal so this trouble melts away).
Worse, you have to use relativistic addition of velocities---spoiling
the game of inventing relativity! So why do I list this as a problem at all?
Because connecting inputted addition of velocities to an outputted
Lorentz contraction, is something, after all. [That COPPER is
cleaner than this was unknown to me until I made up the COPPER problem
merely to caricature the actual situation for a usual metal!]
HOMEWORK*10* DARK LIGHT (easy!) Show that changing the minus sign to a + ,
in the Maxwell equation representing Faraday's and Henry's discoveries of
induction, (i.e. change -Emf = ... to +Emf = ...)---show that that changes
the c-speed shockwave finding into a proof that for the thus
DAMAGED MAXWELL EQUATIONS,
no shockwave solution exists. [This assignment is
strongly recommended, because it forces you to run over the
orientation of every vector in my shock-wave argument, hence will
dispel vagueness about those. Easy, because otherwise, it is
straight imitation of what I did 1feb02fr on the board. Should I
write that out in an as yet unwritten open /shock.rel file?]
HOMEWORK*11* ROTATION Show that the transformation of E,B which leaves the
DAMAGED MAXWELL EQUATIONS of the previous assignment invariant, is no
longer a shear (cosh and sinh), but is rather plain rotation (cos and sin).
(Don't try *11* until I repeat the E,B Lorentz transformation
``using compound interest'' instead of sneaking in the det=1 condition.)
LECTURE 7 light-bounce clock easy! (compare Sears!) via diagonal=1sec
2/6we discussion of why e&m field isn't regarded as having a velocity
HANDOUT: /doubleDoppler.dvi
LECTURE 8 distinct simultaneities from countercurrents in parallel wires,
2/8fr use clockface tagging---watch out for charge `on' an open system!
compounding (i.e. exponentiating) an infinitesimal rotation
compounding Lorentz transformation properly, rapidity
time dilation from the lightbounce clock (needs a 3rd dimension!)
HANDOUT: pp1-4notes,``Quick Derivation of the Lorentz Transformation'' with
sheet ``Quicker Derivation of the Lorentz Transformation'' attached
[My charge w/o charge isn't Wheeler&Misner's wormhole charge without charge!]
HOMEWORK #12 purely geometric proof of the (1-v)/(1+v) doubleDoppler
frequency reduction factor, coincides with Exercise dD1 in the handout.
Hint: I will sketch the key drawing---ask me to if I forget! c=1 helps!
LECTURE 9 review the clockface way to relativity of simultaneity via a circuit
2/11mo time dilation and relativistic Doppler effect, as
geometric mean of the two nonrelativistic Doppler effects (/doubleDoppler.dvi)
HOMEWORK*13* OIL-DROP (easy but not a one-liner) Estimate the accuracy of
charge invariance which follows from feasibility of the Millikan oil-drop
experiment. [Harvey Fletcher helped Millikan: Physics Today,v35 June1982p43]
E.g., assume that q' = (1 + K vv) q and argue a lower bound for coefficient
K on the basis of the usual student lab experiment, no need to search for
the state of the art; of course, v is the speed as fraction of the
speed of light. How big are the drops? Assume that 1000 volts/cm
supports a singly charged drop---1 electron. (If you wish to use
mksHL units, you will wish to first convert volts/cm to Tubes/meter^2.)
http://www.snow.edu/~larrys/PHSX222L/PHSX222LFletcher
Also try http://www.snow.edu/~larrys/PHSX222L/PHSX222LLabs/Millikan.html
and perhaps http://www.snow.edu/~larrys/PHSX222L/PHSX222LOilDrop/
though it is a bit silly.
From bschwarz@aip.org Mon Mar 11 16:45:08 2002
Subject: Re: query for Bertram Schwarzschild re 10^{-21}
> about his PhysToday mention that neutron charge < 10^-21 e
> My guess is that this comes from the ``Millikan'' oildrop experiment.
I got that limit out of the year 2000 Particle Data Group's compilation
Euro. Physical J. C vol 15, pp1-878 (2000)
The PDG resides at LBL and CERN
Bert Schwarzschild
HOMEWORK*14* RELATIVITY OF SIMULTANEITY FROM A CIRCUIT Let 2N minus
numbered charges (``electrons'') circulate around two close parallel vertical
`COPPER' wires, fastened together on top and on bottom, at speed v .
As in COPPER, the electrons descend in the left wire, but rise on the right.
Number the charges 1, 2,... , 2N. When charge 1 crosses on top, charge N+1
crosses on bottom. The charges are again spaced L apart (making the setup
NL tall). So far, Unprime is speaking. But our COPPER discussion
(above) shows that Prime who moves with speed v down . sees fewer
electrons on the left owing to their having lost their v-related Lorentz
contraction. Hence when electron 1 crosses on top, it is not
electron N+1 which crosses on bottom---the bottom-crossing event
Prime-simultaneous to that 1-top-crossing event is different.
[a] Which numbered electron is that (N is large)? From this, determine
[b] what Unprime time-interval delta_t separates two events simultaneous
for Prime, expressed in terms of the speed v (as usual a fraction of
the speed of light), and the Unprime distance delta_x between them.
[c] Finally, confirm by sticking delta_t' = 0 in the (t,x)->(t',x') Lorentz
transformation written out with deltas---differences between events.
[d*---5* more extra yet for this] Prime's shortfall of electrons on the left
must be extra electrons on the right. But also Prime's measure for height
of the whole setup is now contracted. From both these things get
how closely spaced Prime sees those electrons on the right.
[e*5*] Answer d changed Unprime's v-induced Lorentz contraction to a larger
Lorentz contraction. Equate this larger contraction for a new
doublevelocity w to the formula for plain w-Lorentz-contraction to
determine how doublevelocity w depends on v. Check that this agrees
with what we have already learned about additon of velocities v and v.
(One v is for Unprime's velocity of rising of those right electrons.
The other v is for Prime seeing Unprime's self and circuit moving v upwards.
But please use this only for a check! Otherwise you will have
missed getting a limited version of addition of velocities entirely
from figuring out things about a circuit: Elektrodynamik bewegter
Elektronen! Had you used a new speed u for Prime's descent, you
could have fought to adding u and v rather than just v and v .)
HOMEWORK*15* Fix up 6->4 clockface figure's overall scale. This means
that on the left, 6-division 12top 1 2 3 4 5 6bottom spaced L apart
goes to Prime's 4-division 12top 1 2 3 4bottom, and overall scale means to
get the Lorentz-contraction factor for the whole circuit's Prime diagram.
You might wish to sketch, and say how many L's go into each key spacing.
HANDOUT: pp5-8 notes, ``The Vector Idea in Relativity''
LECTURE 10 4vector guess for completing 3-momentum (as in pp5-8notes, handout)
2/13we time `dilation' from the measured mile, fast Christmas,
i.e. the Lorentz *contraction* of a stick-in-time
INVARIANTS: (t+x)(t-x) area hence tt-xx and EE-pp = mm hence mass m
I.e., ``E=mc^2'' makes us think Energy and mass are the same,
so two different words are undesirable prolixity. No. Mass is
the *length* of the 4vector whose *components* are Energy and px py pz.
perhaps fast-relativity interlude: d'Inverno's radaric Ltz.tf.quickie
LECTURE 11 the 4vector of current density is THE geometrically obvious 4vector
2/15fr d'Inverno's radaric bounce off event crossing ( ) and (') clocks
HOMEWORK #16 Complete d'Inverno to k-involving Lorentz transformation, &
find v[k], then k[v], where v is the speed, k is a Doppler factor,
t1' = k t1, t2 = k t2',
t = (t2+t1)/2, x = (t2-t1)/2, t' = (t2'+t1')/2, x' = (t2'-t1')/2.
HANDOUT: pp9-12 notes, ``Making (Rest) Mass Out of Energy''
LECTURE 12 e&m field's energy density, EE/2 + BB/2 , from capacitors,
2/18mo charged up and poled up
reviewed 4vector-guess approach to energy and momentum
LECTURE 13 shock hints E^B is a good flux vector for energy---it is!(on friday)
2/20we You do a similar exercise in HW#20 below.
mass without mass---the above handout `Making Mass Out of Energy'
LECTURE 14 -d/dt of (EE+BB)/2 =?= div(E^B) for sourceless Maxwell---it is!
2/22fr Faraday's muscular tubes of flux fetch us the 3x3 stress tensor.
sourcefree Maxwell's diff.equ'ns divE=0, divB=0, curlB=Edot, -curlE=Bdot
HANDOUT: Notes, pp9-14, repeats Making Rest Mass out of Energy, plus pp13,14,
``Rapidity from Multiplying Doppler Factors'' containing easy Exercises!
HOMEWORK #17 Exercise from ``Notes'' p3margin, invariance to show EE-pp=mm.
I.e., The square of the time component of a 4vector minus the square
of the space component, is INVARIANT. Use THIS PRINCIPLE to show that
E^2 - p^2 = m^2 where m is the rest mass, E is the energy, p is the momentum.
HOMEWORK #18 Exercise from Notes, p11 bottom, mass of the Lambda. I.e.,
A moving Lambda decays into a proton (positive) and a pi-minus (negative pion)
and both make tracks in a suitable medium, and so can be studied.
The energy of the pion (including its rest energy) is 226 MeV,
pion's x-component of momentum is 143 MeV/c,
pion's y-component of momentum is 107 MeV/c,
pion's z-component of momentum is zero.
Proton's energy is 1209 MeV (including its rest energy),
proton's x-momentum is 755 MeV/c,
proton's y-momentum is -107 MeV/c, (Yes, I was lazy: see next problem.)
proton's z-momentum is zero.
FIND the restmass of the Lambda (which you may wish to cite as a rest energy).
HOMEWORK*19* Apply a random real orthogonal 3by3 matrix to both
3vectors in the Lambda problem (#18) to produce ``unlazy data'' for
the same question---and rework the problem using your new data.
That random matrix is got from almost any three messy-looking real orthonormal
3vectors, easily generated from two random 3vectors a, b by normalizing
a to get vec1 , getting vec2 by normalizing a^b ,
then set vec3 = vec1^vec2 . I have been typing ^ for crossproduct.
The matrix uses vec1 vec2 vec3 for its three columns.
HOMEWORK #20 Show that (E^B)z checks out as density for flux: EE/2
uniform pressure plus EE electric-oriented tension, with likewise
magnetic terms, explicitly, ExBy-EyBx checks out as density for flux:
z-momentum flux = ( -ExEz-BxBz, -EyEz-ByBz, (EE+BB)/2-EzEz-BzBz ).
*EXTRA* Reason from the words, `` EE/2 uniform pressure plus EE
electric-oriented tension, with analogous magnetic terms '' ,
to the explicit algebraic expression given for z-momentum flux.
Below is the array of four densities on top and their fluxes beneath:
of energy of x-momentum of y-momentum of z-momentum
(EE+BB)/2 EyBz-EzBy EzBx-ExBz ExBy-EyBx
EyBz-EzBy (EE+BB)/2-ExEx-BxBx -ExEy-BxBy -ExEz-BxBz
EzBx-ExBz -EyEx-ByBx (EE+BB)/2-EyEy-ByBy -EyEz-ByBz
ExBy-EyBx -EzEx-BzBx -EzEy-BzBy (EE+BB)/2-EzEz-BzBz
It shows pressure contributing positively, tension contributing negatively.
LECTURE 15 How I know that above Tij array is T^ij not mixed, not T_ij:
2/25mo Symmetry vetoes mixed. 4divergencelessness needs an ^ to latch onto.
Tracelessness in mixed version eta_ij=diag(+1 -1 -1 -1) raise,lower
LECTURE 16 2/27we 4vector guess for energy,momentum checks on 2 optic packets!
HANDOUT: /optic4momentum.rel which assigns...\/ (TeX'ed version next time)
HOMEWORK #21 I did the energy ( open alpha file
~eli/optic4momentum.rel ), you do the 3momentum, for those 2 packets.
LECTURE 17 conservation of charge expressed in indices
3/1fr j^i_,i = 0 to cons. of Q = \int dxdydz j^0 for j> small far away
HANDOUTS: /optic4momentum.rel.dvi and /optic4momentum.rel.tex :TeX `lesson'
LECTURE 18 invariance of body's charge captured in lopsided 4-tube integral
3/4mo 4vector momentum checked by lopsided \int dtdxdydz u_i (T^ij_,j=0)
|factorization used and explained
Maxwell's differential equations with sources, cut into components
support currentdensity contra4vector j^i and give us F^ij_,j=j^i [&\/ dual]
charge conservation three ways: Millikan ji,i Fij,ij
INVARIANTS: (t+x)(t-x) area hence
mass, mm = EE - pp . Our ``mass'' is restmass, an invariant.
dx^i dx_i = dt^2 - dx^2 - dy^2 - dz^2 = ds^2, relativity's arclength, and
dtdxdydz 4-volume
charge of an isolated body, magnetic pole of an isolated body
^^^^^^^^^^^^^
LECTURE 19 dual---the pole-sourced Maxwell equations: eps_ijkl eps^ijkl
3/6we co and contra: L^i_j and its contragredient, L^i_k L_j^k = 1^i_j
which tells the easy way to invert a Lrntz-tf: Linv =(eta L eta)transpose
(E,B)->(B,-E) same as G^ij = eps^ijab F_ab / 2!
HOMEWORK #22 Express [a] F_ij F^ij and [b] G_ij G^ij and [c] F_ij G^ij
which are obviously invariant (why?) in terms of E> and B> to learn some
combinations of E> and B> that are thus proven invariant with little trouble.
LECTURE 20 rotational ghost of last time's Lorentz-transformation story
3/8fr j^i = eps^iabc j_abc / 3!
Maxwell ``Grassmanned'' to df=j dg=k g=*f
HOMEWORK*23* You check the signs of df=j and dg=k if j^123 is (+)charge density
and k^123 is (north)pole density, and G^ij = eps^ijab F_ab / 2! More below.
Attraction: To make it more fun, I am assigning it before checking it myself.
The following, updated in /indices.rel517.tex for today's handout
/indices.rel517.dvi , will both be handed out improved on Monday.
FARADAIC SLOGANS:
The flux of energy is momentum. The word
The flux of momentum is stress. ``density'' is omitted.
The flux of moment of momentum is torque. Is word ``twiststress'' better?
conservation of position
F^01_,1 + F^02_,2 + F^03_,3 = j^0 = j_123 = f_23,1 + f_31,2 + f_12,3 hence
f_ij = eps_ijab F^ab / 2! F^ij = eps^ijab f_ab / 2!
F^ij_,j = j^i j_ijk = f_jk,i + f_ki,j + f_ij,k df = j
Again j^m = eps^mijk j_ijk / 3! = eps^mijk (f_jk,i + f_ki,j + f_ij,k)/3!
= eps^mijk f_ij,k / 2! since the eps symbol does enough antisymmetrizing
= eps^mijk (eps_ijab F^ab / 2!)_,k / 2!
= eps^ijmk eps_ijab F^ab_,k / 4 moving m involves two canceling minuses
= 1^ijmk_ijab F^ab_,k / 4 1^ijmk_pqrs explained below
= 1^mk_ab F^ab_,k / 2
= (1^m_a 1^k_b - 1^m_b 1^k_a) F^ab_,k / 2
= F^mk_,k / 2 - F^km_,k / 2
= F^mk_,k winds us back to F^mk_,k = j^m for a check.
1^ijmk_pqrs is 1 if upper and lower index strings are an even permutation
of each other, is -1 if odd, is 0 otherwise. See O.Veblen,
``Invariants of Quadratic Differential Forms'', a Cambridge U. Press Tract,
where instead of 1 Veblen uses \delta . The Kronecker delta or unit
matrix is the one of these everyone already knows.
Note that 1^ijmk_0123 = eps^ijmk and 1^0123_pqrs = eps_pqrs so that
both the Kronecker delta and the Levi-Civita epsilon specialize the
Veblen delta or 1 symbol.
0 F^23 F^31 F^12 0 Bx By Bz interchanges E and B
f__ = F^32 0 F^03 F^20 = -Bx 0 Ez -Ey in describing
F^13 F^30 0 F^01 -By -Ez 0 Ex display F^^ -> display f__.
F^21 F^02 F^10 0 -Bz Ey -Ex 0 Duality further wishes to
change the sign of the E entries, and duality gets from F^^ to G^^.
Indeed f__ to G^^ should require two eta^^ raisings. But those doubly
change the E signs which cancels out, but also singly changes the B signs,
thus doing the exact opposite of our defined duality. So we must
throw in yet an overall minus. So roughly
G^^ = - eta^^ eta^^ f__ = - eta^^ eta^^ eps____ F^^ and, better,
G^ij = - eta^ia eta^jb f_ab = - eta^ia eta^jb eps_abcd F^cd / 2!
And G^ij_,j = k^i are the pole-sourced Maxwell equations.
g_ij = eps_ijab G^ab / 2! should almost complete the story.
But then is the analogue of df = j indeed dg = k ? Exploring
that would complete *HOMEWORK*23* above.
LECTURE 21 invariance of EdotB and EE-BB done by 3vectors, parallels HW#23
3/11mo v-boost along E^B needs v half as rapid as u = E^B/.5(EE+BB)
HANDOUT: /indices.rel517.dvi and /indices.rel517.tex bound together
LECTURE 22 double:u->v is u=2v/(1+vv) which *was* Monday's quadratic equation.
3/13we square F and detrace -> e&m energy-momentum-stress T^_ via T00 check
HOMEWORK #24 a Select a SINGLE (') CLOCK. WHAT LINE represents it?
b MARK a series of events at THAT one (') clock. THESE WILL
REPRESENT CROSSINGS OF STEEPLES. LIST the t and t' values of those
steeple-crossing events.
c Refer to these noted values, to TELL how your selected (') clock is
seen to be going SLOW by the ( ) confederation of steeple managers.
d Refer to these same values to TELL how the steeple time (t values)
seen by (') runs FAST compared to (') `wristwatch' time.
e WHAT LINES on the figure represent YOUR SELECTED STEEPLES? IF
YOUR CHOICE OF EVENTS IN b WERE OFF-x-CONTOURS, YOU WILL HAVE TO DRAW
MORE x-CONTOUR LINES.
HOMEWORK #25 ...is a continuation:
f Select a second (') clock, and make another list of t and t' values
for that (') clock AT THOSE EVENTS WHERE THIS SECOND (') CLOCK PASSES
THE EARLIER CHOSEN STEEPLES.
g Use these t and t' values for (at least 2 of) those events, to SHOW
that the (') confederation finds a steeple clock (an ( ) clock) to be
going SLOW.
h Cite the factor by which either example of moving clock is slowed,
and explain how it has come FROM READING TIMES OF EVENTS MARKED ON
YOUR FIGURE, either in part b or in part f.
i COMPARE WITH sqrt(1 - v^2). This means that your part h was NOT
worked out as that: you did subtractions and division in h, but never
extracted a square root, for example.
*j* philosophy about comparing clocks analogous to HW*6* about comparing sticks
RINDLER HOMEWORK p157 each # symbol marks ten points
# Ex 7.1 If T^ij is a 4-tensor, show that under a spatial rotation,
[a] T^00 remains unchanged
[b] T^0i and T^i0 go as 3-vectors
[c] T^ij for i,j restricted to 1,2,3 goes as a 3-tensor
Exercise 7.2 teaches factorization:
# Ex 7.2i Given E_ij U^j is a covariant 4-vector for all timelike U^j.
Prove E_ij is a covariant 4-tensor. [Hint given in Rindler]
# Ex 7.2ii Given c_ij for every coord.system, and c_ij A^ij is
invariant for any contravar.tensor A^ij. Show c_ij is a covar.tensor.
# Ex 7.2iii Given g_ij B^i B^j invariant, for any vector B^i. Given g_ij=g_ji.
Show g_ij a covar.tensor
# Ex 7.3 Given a_ij constant and symmetric.
Prove (a_ij A^i A^j)_,k = 2 a_ij A^i A^j_,k [Hint given in Rindler]
# Ex 7.4 symmetric and antisymmetric: Split tensor A_ij into
symmetric A_(ij) = (1/2)(A_ij + A_ji)
and antisymmetric A_[ij] = (1/2)(A_ij - A_ji) parts.
[a] Prove A_ij = A_(ij) + A_[ij]
[b] If B^ij = B^ji prove A_ij B^ij = A_(ij) B^ij (so if A antisym, zero)
[c] If C^ij = -C^ji prove A_ij C^ij = A_[ij] C^ij (so if A symm, zero)
*Ex*7.4*[a* EXTRA]*Prove that the only split [a] possible into symm,antisymm
parts is the one given. [Hint: Work back from [a].]
LECTURE 23 raising eps_ or lowering eps^, both weighted, needs det of metric
3/15fr angular momentum and center of mass : M^ijk = x^i T^jk - x^j T^ik
hence conservation of position! The 10 mechanical conservation laws.
The const. u_..-indexed currents are j^k = u_ij M^ijk/2! which Grassmanns to
j_abc = j^k eps_kabc = u_ij M^ijk eps_kabc/2! so the Grassmann 3-forms are
j[u] = u_ij M^ijk eps_kabc dx^a dx^b dx^c / 2!3! Conservation is ``dj=0''.
And spin is in: conservation finds nothing missing and e&m polarization is in.
virtues of Grassmann forms: integrals use no normal, no Jacobian, no weight
If time: parking a long car in a short garage---which was indeed stated
HOMEWORK #26 GRASSMANN ONLY---``JACOBIANS WITHOUT JACOBIANS'':
Compute the volume of a sphere of radius R by employing the *following*
*strategy* or the strategy on pp1,2 of ~eli/circleGrassmann.tex
also ~eli/circleGrassmann.dvi
area of a circle auf Grassmann---on board 3/15fr I was off by factor 1/2
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ I fix that in /circleGrassmann.dvi pp1,2
but here give the .tex of the easier /circleGrassmann.dvi page 3:
\centerline{\bf Faster and without formal transforming of a tensor:}\bigskip
\centerline{area of a circle}
$$= \int_{\theta=0}^{2\pi}\int_{r=0}^R dx dy$$
$$= \int_{\theta=0}^{2\pi}\int_{r=0}^R d(r\ \cos\theta) d(r\ \sin\theta)$$
$$= \int_{\theta=0}^{2\pi}\int_{r=0}^R (dr\ \cos\theta - r\ \sin\theta d\theta)
(dr\ \sin\theta + r\ \cos\theta d\theta)$$
$$= \int_{\theta=0}^{2\pi}\int_{r=0}^R (dr\ \cos\theta\ r\ \cos\theta d\theta)
- (r\ \sin\theta d\theta dr\ \sin\theta)$$
$$= \int_{\theta=0}^{2\pi}\int_{r=0}^R
(r\ \cos^2\theta\ dr\ d\theta - r\ \sin^2\theta\ d\theta\ dr)$$
\line{The anticommutation of differentials is here invoked.\hfil}
$$= \int_{\theta=0}^{2\pi}\int_{r=0}^R
(r\ \cos^2\theta\ dr\ d\theta + r\ \sin^2\theta\ dr\ d\theta)$$
$$= \int_{\theta=0}^{2\pi}\int_{r=0}^R r\ dr\ d\theta$$
$$= {RR\over2} \int_{\theta=0}^{2\pi} d\theta$$
$$= {RR\over2} 2\pi$$
$$= \pi RR.$$\bigskip
The only step outside 1st-year calculus is at the single intercalated comment.
\medskip
Why keep the longer version which transforms a tensor? That points up
the connection between Grassmann forms and antisymmetric covariant tensors.
People used to tensors might even see the longer version as more transparent.
\bye
LECTURE 24 circle by simple substitution car and garage and key events
3/25mo introduce the rotating barrel, noninvariance of C.M. trajectory
HANDOUT: /circleGrassmann.dvi pp1,2 improve 3/15 lecture, p3 is faster
HANDOUT: /circleGrassmann.tex in your mailboxes
HOMEWORK #27 Make a drawing like the one you have made
of t , x , t' , x' contours,
that will AUTOMATICALLY show BOTH POINTS OF VIEW of the car and garage:
Let the car's rest frame be ('), the garage's be ( ). Constant-t slices
should show a foreshortened car parking nicely in a longer garage [this if you
do not push it as hard as in *28*]; constant-t' slices should
without! further! ado! show the other point of view. The speed v and choices
of lengths must of course be adjusted, in order for this other point of
view to see the car-at-rest as being actually longer than the garage.
Another interesting thing to mark on this picture is the world-line
segment of the closed garage door, and also, the first x-rays' world-line
segment, to see the x-rays stopped from leaving by the closed door.
THERE WILL BE LITTLE IN THE PICTURE, so how well you grasp this will
be shown by your writing about what you see in the picture.
Please---in all such commentary---refer to DEFINITE EVENTS.
SUGGESTED EXTRA DETAIL: Let the garage be yellow, the car front-to-back
as given by the Lorentz contraction, be red, and since even more can
get in (see next problem), attach a TRAILER, colored green.
*HOMEWORK*28* What rest length L0 does the longest thin car have,
that will fit into a thin garage of rest length G0 , if it `parks' at
speed v ? RULES: When the front end of the car hits the back wall of
the garage, light & x-rays will appear, and I count that radiation as
[a new] part of the car. So if any of that stuff gets outside the
[fast] door, you have NOT succeeded in `parking inside the garage'.
>>>>Hint<<<<: You may be surprised to find that, even in the garage's rest
frame, the (Fitzgerald-contracted) length L of that longest car is longer
than G0 , i.e. that L > G0 , even though the original `paradox' created
itself by thinking about parking for the less severely counterintuitive
case where L is < G0 , with only rest length L0 being > G0.
LECTURE 25 Doppler as compared lengths of a null interval in t,x relativity
3/27we review of center of mass X made vivid by a rotating barrel
stealing from electromagnetic E,B 6-vector to learn X and J
F^ij goes like M^ij
HANDOUT: /rotatingbarrel
HOMEWORK*29* CAR AND GARAGE AS DOPPLER EFFECT; KEY EVENTS: just 1 equation!
The key events are front of car
meets back of garage : boom!, and back of car enters front of garage : latch!.
^^^^^ ^^^^^^
For the *longest* v-moving car, boom! and latch! are connected by the
first crash light. HOW does that make it *obvious* that the rest lengths
L0 of car and G0 of garage (these symbols being just names, no equation!)
must be related by the optic Doppler factor (the 1 equation)?
``Obvious'' should be a careful verbal explanation, not yet quantitative,
which culminates in the stated insight. Then that insight writes the
answer as the sole quantitative relationship, one first and last equation.
[This is my way of sharing the insight---that this Dopplerian connection
is, seen rightly, indeed quite direct---by the usual focus on homework.]
Once the Lorentz-Dopplerian answer is thus swallowed in, as it were, a single
bite, the sense in which, basing the parking of a long car in a short garage
on the Lorentz-Fitzgerald contraction, is wrong,
also becomes clear.
The clarity involves resolving the
paradox that
Doppler relates how two observers see a length of light,
whereas here we relate L0 and G0, neither being light, so
though their ratio turns out to be the optic Doppler factor,
that must be an accident as it cannot be an actual Doppler effect.
But it is no accident! And indeed periods, wavelengths of light
cannot be intrinsic times or lengths of the light itself, as a
photon's clock is stopped, its ruler is scrunched.
Hence car, garage,
and the crash light join in a lesson on the meaning of the Doppler
effect itself: Of the three distances between boom! and
latch!, that measured by the light, by the car's passengers, by the
garagekeeper, the first is of course zero so must lie outside of what
Doppler deals with. So it must be the other two, that
Doppler deals with. Always.
All your Doppler-effect wavelength pairs
are longest cars matching garages!
Needle light in itself has no period, no wavelength, keeps no record,
has no mind, as it is too fast. Mass makes mind.
[Nadelstrahlung (Einstein) is to keep clear of ``mass without mass''.]
Car and garage has just taught me, and now you too, that the null interval
Doppler deals with, need not be parallelled by other light.
HOMEWORK #30 Bring in faster-than-light (ftl) phenomena (legal as not every-
thing need be a signal). If you can't think of any, peek at p.56, Rindler.
LECTURE 26 how to build a time machine given 2 ftl guns: vetos ftl
3/29fr COLLECTed ftl nonsignal example: learning other's colored bead
HOMEWORK #31 TIME MACHINE It is 2002. A secret project is revealed:
A gun has been developed that shoots pulses which travel at four
times the speed of light, i. e., at speed 4. It shoots when
you press the trigger, not otherwise. It reloads rapidly.
The pulses can be detected as far down the road as you like. A railroad
car has been developed which goes at half the speed of light, i. e.,
at 1/2. Copies of both have already been set up at the Centaurus system,
4 [`light'] years `to the right' of us and at rest with respect to us.
Using the relaying scheme explained in class, show that a signal
started here 1 January, 2006 will arrive here in September 2003.
HOMEWORK*32* Show that, to turn two guns which shoot at speed u > 1
into a time machine, with the help of two railroad cars at ordinary
speed v < 1, speed v must surpass the value 1/u. [No signs given in
these speeds; `right or left' is left to you.]
HANDOUTS: /stresstensor.dvi /constacc.rel
LECTURE 27 reviewed TIME MACHINE
4/1mo s^ijk = M^ij P^k + M^jk P^i + M^ki P^j (Pauli Lubanski) is a tensor
(because M^ij is: tricky but was done). s^ijk is mass*spin at rest.
But s^ijk is also slide-invariant (unlike M^ij): ^^^^
The slide invariance via typical component:
int means \int dxdydz |the cancellations emphasized
s^120 = int (x^1-a^1)T^20-(x^2-a^2)T^10 P^0 | (-a^1 P^2 + a^2 P^1)P^0
+ int (x^2-a^2)T^00-(x^0-a^0)T^20 P^1 | (-a^2 P^0 + a^0 P^2)P^1
+ int (x^0-a^0)T^10-(x^1-a^1)T^00 P^2 | (-a^0 P^1 + a^1 P^0)P^2
*******************************************************************************
4vec s_a (which +--- squares to -mmss) = (J*P, (J - X^P)sqrt{mm + P*P}). Hence
mmss = (mm + PP)(J-X^P)*(J-X^P) - ((J-X^P)*P)^2 | J,P,X are vec3s
= (mm + PP)(J-X^P)*(J-X^P) - (J*P)^2 |* is dotprod, ^ is crossprod
*******************************************************************************
The epsilon lowering from s^ijk to s_a is nonmetric duality,
s_a = eps_aijk s^ijk / 3! which is also the most obvious 3->1 relabeling
hence no time vs. space sign flip happens. Unwise to write or discuss s^a.
LECTURE 28 polar coordinates and constant acceleration (asymptotes assumed)
5/3we M&M's -> Einstein-Podolsky-Rosen, relativity of reality!
HOMEWORK #33 Show that the following three on-board methods of defining
distance between back r1 and front r2 (with r1 < r2 and ship
accelerating forwards, passengers tend to fall backwards, clarifying what
I mean by ``back'' and ``front'') give three different answers, by finding
the three answers: [a] Slowly pace off the distance.
[b] Use a radar at r1 against a reflector at r2.
[c] Use a radar at r2 against a reflector at r1.
Note that the 1st method for on-board distance accesses r as a distance.
The inertial observer also uses r , i.e., paces the no-brainer r2-r1 .
HOMEWORK #34 You are conveniently seated, in this ship, at Earth's
gravitational acceleration g = 9.8 meter/secsec .
[a] (not really an essential step, and trivial, but worth noting):
Convert units of g ``to c=1'' in the sense that ``meter'' is
replaced by the (light)second, i.e. sec, as unit of length.
[b] How far back is the horizon? (where the acceleration gets infinite)
[c] Compare the seat-at-9.8m/ss gradient of acceleration on the constantly
accelerating ship, with the gradient of acceleration (just above
ground, as this is not intended to be a problem in geology!)---at
the earth's surface. Hint: vastly different
HOMEWORK*35* By bouncing radar repeatedly between back end at r1 and
front end at r2 of the finitized-segment version of the long ship of
HW#34, clarify the sense in which the clock at r1 ``runs slower'' than
the clock at r2 . As they are identically built clocks, intrinsically
running at the same rate, care in specifying procedures is needed to
explain ``runs slower''.
*10*MORE EXTRA POINTS: Check that the answer conforms quantitatively with
the well-known slowing down of clocks downhill in a gravitational potential.
HOMEWORK*36* Integrate d(rapidity)/ds = a in damaged relativity
where ds^2 = dt^2 + dx^2 instead of minus, parallelling the integrations
I will have done (on the board---this is an advance notice so I'm not sure)
for undamaged relativity to get a hyperbola. You will instead get a circle.
HOMEWORK*37* on HANDOUT: /fts which is also /paradoxes.FOUR Debug
the time machine built out of acoustic rigged rulers and clocks with a
flashlight providing the ``faster than sound'' gun. [Welcome to try
other of the /paradoxes of course, those on spec.rel, all for extra credits.]
LECTURE 29 *the* long rocketship
4/5fr HANDOUTS: /biot /relfromem and new version of /constacc.rel
LECTURE 30 4/8mo ship's universal time alpha, radar, began d\alpha/ds = a
LECTURE 31 YOU do HW*36* ABOVE on damaged relativity :calculus exercise:
4/10we ME d(rapidity)/ds = a in undamaged|relativity (over)confirms asymptotes
HANDOUT: /constaccrel.calculus.dvi and for the TeX /constaccrel.calculus.tex
LECTURE 32 Lorentz force, force density: rho Ek to j^a F_a^k to ``force''
4/12fr Does our ` Unprime =~= Prime Lorentz force ' start, still make sense?
HOMEWORK*38* [a] Write out forcedensity^k = j^a F_a^k to show how the k=1,2,3
part does come out the Lorentz forcedensity> = rho E> + j> cross B>, and
[b], how the k=0 part relates to power.
LECTURE 33 applications: Biot-Savart Ampere from Gauss (/biot /relfromem)
4/15mo historic application: de Broglie's wavelengths
HANDOUT: de Broglie's `` lambda=h/p '' Two Ways mainly for METHOD 2
HANDOUT: /force.rel.dvi
LECTURE 34 E and B of a moving charge (E by Lorentz contraction but not B!)
4/17we Compton effect. spatial curvature on merry-go-round in flat spacetime
HOMEWORK*39* Show that Lorentz contraction of the space bearing Faraday's
lines of force gives the E field (but not B) of a moving charge.
HOMEWORK*40* MERRY-GO-ROUND (mgr) of angular velocity \omega: [b] Find
radius R where a careless observer on mgr adding up what he sees as
``contracted'' meter sticks nearby but off mgr, add up to a [c] c[R]
FALSE restframe maximum circumference, using the FALSE-cirmcumf. [a] c[r]
function. What is the on-mgr tile-floor (paced) [d] OnC[r]?
circumference, and its value at R [e] OnC[R]?
and the true off-mgr circumference [f] OffC[r]?
and [g] OffC[R]?
(``FALSE'' as the off observers don't see their sticks contracted.)
Does the false c[r] write us a metrical 2sphere (is doublextra) [*h*] ?
LECTURE 35 crossspeed projective action on 3velocities:sky boosts conformally
4/19fr A velocity is a cosine: Bradley 1729, the aberration of light
HOMEWORK*41*:THE ABERRATION OF LIGHT Ignore my ``velocity is a cosine'' trick
and instead show directly that a v-boost towards ``the north star''
changes stars' colatitudes according to
$$\cos\theta' = {\cos\theta + v\over1 + v \cos\theta}$$ ---by boosting an
appropriately labeled lightlike 4vector (e.g. ``the photon's 4-momentum'').
LECTURE 36 the vt, hypotenuse ct right triangle seals the velocity=cosine fact
4/22mo (same as light-bounce clock) ...but see next lecture!
Seeing Lorentz transformations in the planetarium sky---axis, coaxis
in and out of Mercator's projection for a boost
LECTURE 37 ``A velocity is a cosine'': I have been missing the obvious! The
4/24we planetarium *itself* shows this immediately! Origin is zero velocity.
Pick another point, and that *is* a nonzero velocity. Boost with it and
the equatorial stars indeed slide to the cos(colatitude) = v parallel.
The photon sliding through the moving 2-slit Tycho-Brahe telescope is
only for people not oriented to the planetarium as reversed 3-velocities,
and the funny thing is that till now I have been one of those people!
Mercator1568, and generalized Mercator projections take twist-and-boost
field on the sky to a constant vector field on a plane or cylinder.
LECTURE 38 L^i_j = exp F^i_j = exp(F^ik eta_kj) has F^ik + F^ki = 0:
4/26fr turns on electromagnetic-field F^ik speak for Lorentz transformations!
transparency lecture run-through (begun Wed) completed
e homerapidity, b hometwist: EdotB = eb, EdotE-BdotB = ee-bb.
Mercatorlatitude = homerapidity = argtanh v = argtanh sin latitude or
application of relativity to geography: tanh Mercatorlatitude = sin latitude
LECTURE 39 offsky motion : homothetic lemma
4/29mo crossspeed C and the Poyntingnull way home---we do it *from* home
To avoid log of a matrix, can use ev, ew. e and b as log ew's.
The ev's rays (3velocities) are the fixed stars.
dipolars: their E> and B> mimic our optic shock: unit crossspeed,
homeless, em invariants 0 & 0, emfield nilpotent
LECTURE 40 application of relativity to geometry: 1-d projectivity is a boost!
5/1we projective geom. preserves lines. 8-dimensionality for plane proven
HANDOUT: /crossratio.rel AC:CD::AB:BD = doubleDoppler = {AC*BD\over AB*CD}
HANDOUT: /rel.projective mainly for its attachment /rel.lines both now open
-1 u w 1
LECTURE 41 crossratio is u->w boost's doubleDopplerfactor (1+v)/(1-v)=wDF/uDF
5/3fr ln crossratio = 2*rapidity, additive eigenvalue
crossratio = exp 2*rapidity, multiplicative like an ew (Eigenwert)
dipolars | C=1 | homeless | nullEMinvariants | nilpotent | KineticEnergy=EE/2
If try to chase dipolar E>,B> down, they redshift to approach nothing.
HANDOUT: /crossratio.rel update shows`jingle' form (AC BD)/(AB CD)=crossratio
ayceeBEEdee OVER aybeeCEEdee
HANDOUT: /spininsky.rel for next week
LECTURE 42 spin in the planetarium sky---at least as spin rays
5/6mo
from /spininsky.rel :
Spin begins by instead merely turning stereo's complex plane to complexphere.
Then the sky->sky Lorentz transformations become complexphere->complexphere
w->w'={aw+b\over cw+d} maps (because everywhere analytic is that limited).
*HOMEWORK*42* Show that these a b turn out to compose under matrix
c d multiplication.
Hence have a 2by2 representation of the Lorentz transformations.
*HOMEWORK*43* conventionalize the Pauli matrices from what I present
in /spininsky.rel :
That is, go back to when w is presented as a w0/w1 ratio and exchange
the 0 1 labels so call that a w1/w0 ratio instead. Then where I first
put in i you put in -i , and take it from there.
LECTURE 43 spinors w^a w^a. w_a w_a. lifted from spinrays w = {x+iy\over t+z}
5/8we LAST v^k = \sigma^k_ab. w^a w^b. sky ``sky without sky''(fake miracle)
general v^k = \sigma^k_ab. h^ab. h hermitean limits to real 4vectors
*HOMEWORK*44* So L^k_m v^m = \sigma^m_ab. S^a_c S^b._d. h^cd. which
you are to factorize to get L in terms of S
HANDOUT: /spininsky.rel :updated
homework checklist *asterisks* frame ``extra-credit'' items
HOMEWORK #1 drawing (/drawing.rel)
HOMEWORK #2 contours all in /drawingfollowup.rel
#3 speed
#4 color
#5 contraction
*6*philosophy
HOMEWORK #7 addition of velocities from multiplying Doppler factors
HOMEWORK*8* ANTICOPPER
HOMEWORK*9* + and - charges contributing equally to current I
HOMEWORK*10* DARK LIGHT no light without the - sign in -Emf = (1/c)Phi_B_dot
HOMEWORK*11* if not for that - sign, Lorentz transformation would be a rotation
HOMEWORK #12 purely geometric proof of the (1-v)/(1+v) doubleDoppler factor
HOMEWORK*13* q'=(1+Kvv)q, estimate Millikan-oildrop bound on K, `1000 volts'
HOMEWORK*14* [a b c] RELATIVITY OF SIMULTANEITY FROM A CIRCUIT
HOMEWORK*14* [d e] special-case addition of velocities FROM THAT CIRCUIT
HOMEWORK*15* Fix up 6->4 clockface figure's overall scale.
HOMEWORK #16 Complete d'Inverno to k-involving LorentzTransf; v[k]; k[v]
HOMEWORK #17 Exercise from ``Notes'' p3margin, *invariance* to show EE-pp=mm
HOMEWORK #18 Exercise from Notes, p11 bottom, mass of the Lambda
HOMEWORK*19* Random-rotate my lazy-man's 3vec data in HW17 and redo.
HOMEWORK #20 Show that (E^B)_z checks out as density for given 3momentum flux.
HOMEWORK #21 See ~eli/optic4momentum.rel :you get 3momentum of optic packets
HOMEWORK #22 E,B-reduce [a] F_ij F^ij and [b] G_ij G^ij and [c] F_ij G^ij
HOMEWORK*23* Check signs of df=j and dg=k
HOMEWORK #24 a Select a SINGLE (') CLOCK. What line?
b MARK a series of CROSSINGS OF STEEPLES. LIST the t and t'
c From these TELL how your selected (') clock is seen SLOW by () steeples
d From same TELL how steeple time runs FAST compared to `wristwatch' time
e WHAT LINES on the figure represent YOUR SELECTED STEEPLES?
HOMEWORK #25 ...is a continuation:
f Select second (') clock. List t and t' when THIS PASSES THOSE STEEPLES.
g Use f DATA to SHOW that (') gang finds a steeple clock (an ( ) clock) SLOW.
h Get the movingclockslowing factor FROM b or f DATA, then...
i ...COMPARE WITH sqrt(1 - v^2).
*j* philosophy about comparing clocks analogous to HW*6* about comparing sticks
RINDLER HOMEWORK p157 each # symbol marks ten base points
# Ex 7.1 T^ij under 3rotation into scalar T00 3vecs T0i Ti0 3tensor Tij
# Ex 7.2i E_ij U^j a vector, all timelike U^j, shows E_ij a tensor
# Ex 7.2ii c_ij A^ij scalar all A^ij tensors shows c_ij a tensor
# Ex 7.2iii g_(ij) B^i B^j scalar with any vector B^i shows g_(ij) a tensor
# Ex 7.3 a_ij constant & symm, then (a_ij A^i A^j)_,k = 2 a_ij A^i A^j_,k
# Ex 7.4 A_ij B^(ij) = A_(ij) B^(ij) AND A_ij C^[ij] = A_[ij] C^[ij]
*Ex*7.4a*[10EXTRA]* Prove the split into sym, asym given by Rindler is unique
HOMEWORK #26 GRASSMANN ONLY---JACOBIANS without JACOBIANS: volume of a sphere
HOMEWORK #27 CAR AND GARAGE through drawing
HOMEWORK*28* CAR AND GARAGE analytic
HOMEWORK*29* Why Doppler does car and garage, many words but just 1 equation!
HOMEWORK #30 Bring in faster-than-light
HOMEWORK #31 TIME MACHINE 4years, speeds 4 and 1/2
HOMEWORK*32* TIME MACHINE, speeds u and v
HOMEWORK #33 lengths [a] pace [b] r1 r2 r1 radar [c] r2 r1 r2 radar
HOMEWORK #34 seat@9.8m/ss [a]c=1 g [b] horizon? [c]vgl Erd g-grad
HOMEWORK*35* bouncing repeatedly clarifies r1 clock slower than r2 clock
*10*MORE EXTRA POINTS: vgl slowing of downhill clocks
HOMEWORK*36* d(rapidity)/ds = a if dsds=dtdt+dxdx. You will get a circle
HOMEWORK*37* /paradox.FOUR , signals not faster than sound<-(flashlight exists)
*[Welcome to try others of the /paradoxes ---of course, those on
spec.rel---all for extra credits. See also scattered Exercises in handouts]*
HOMEWORK*38* [a] recognize j^a F_a^k k=1,2,3 [b] k=0 and power
HOMEWORK*39* cartoon of flux of a moving charge
HOMEWORK*40* MERRY-GO-ROUND abcdefg,*h*
HOMEWORK*41* ABERRATION OF LIGHT by transforming a lightlike 4vector
HOMEWORK*42* Show linfrac compose as matrices: v. /spininsky.rel
HOMEWORK*43* conventionalize Pauli: v. /spininsky.rel
HOMEWORK*44* L^k_m in terms of spinor S^a_b