Physics 4D Unofficial Study Guide
A list of every formula needed to solve the homework and
quiz problems with explanations.
Compiled by David Eldon.
Physics 4D, Winter 2009
Professor Jenkins
Yes, it's a little long. There's a shortened formula sheet at the end.
I'll update it again if any mistakes are pointed out to me. Otherwise, this is it. I hope it helps.
Optics
Basic Optics
All materials have a property called index of refraction. Light moves slower through materials with a high index, and it also can bend at
the interfaces between materials with different indicies of refraction. The index of refraction is given the letter n.
Snell's law is a relation for the angle of light relative to the surface normal on one side of the surface in terms of the angle from the normal
on the other side.
n1 Sin@Θ1 D = n2 Sin@Θ2 D
Θ1 is the angle between the incident beam of light and the surface normal. n1 is the index of refraction on the side of the interface that the
light comes from.
The index of refraction of vacuum is exactly 1, by definition. The index of refraction of air varies, but is slightly higher than 1, and may
typically be approximated as 1.
The speed of light in a medium is the speed in vacuum divided by the index of refraction:
c
cn =
n
Λn =
Λ
n
The wavelength of light also is decreased in a material with n >1. The frequency of light is unaffected. The subscript n denotes the value
of a variable in a medium with index n.
When light crosses an interface from a region of lower n to a region of higher n, the phase changes by 180°
The intensity of an electromagnetic wave is proportional to the amplitude of the electric field squared, so :
I µ E2
IΘ = I0
EΘ 2
E0 2
Problems using this idea: G 32-43, G 32-58, Q 1-3 (G for Giancoli, your first textbook, and Q for quiz)
2
Lenses
1
1
+
di
m=-
1
=
do
f
di
do
f is positive for converging (convex) lenses, and negative for diverging (concave) lenses. The object distance is positive if the object is on
the same side of the lens as the source of the light, which is by convention the left side of the lens. The image distance is positive if it is on
the opposite side of the lense as the light source. A negative magnification (m) means that the image is inverted.
The image is real if light actually passes through the image. Positive image distances correspond to real images, negative distances
correspond to virtual images.
For a converging lens:
If do > f, ® di > 0, m < 0, image real Hpictured aboveL
If do < f, ® di < 0, m > 0, image virtual
Solving graphically:
Starting from the object, draw a line parallel with the horizontal. When you get to the plane of the lens, stop and draw a new segment from
the intersection of the first line and the lens plane to the focal point on the other side of the lens. This is the red (top) line in the picture.
Draw another line connecting the object and the near focal point (on the same side as the object. Stop at the plane of the lens, and then
draw a horizontal line starting from the intersection of the previous segment and the plane of the lens. This is the grey (bottom) line.
Draw the last line from the object to the center of the lens and continue it on through to the other side. This is the blue (middle) line.
The three lines should intersect at the point on the image that corresponds to the starting point on the object.
This is not as exact as the analytic solution, but it should be close enough to serve as a check on the analytic solution even if you can't draw
very well.
Lens maker's equation:
1
f
= Hn - 1L
1
1
+
R1
R2
This is used to calculate the focal length of a spherical lens. R is positive if the surface curves towards the center of the lens, and negative
if the circle curves away from the center of the lens. That is, for either side of the lens, if the lens is inside the circle, take R to be positive.
Another way of saying this is that convex surfaces are positive and concave surfaces are negative. n is the index of refraction of the
material used. Glass usually has an index close to 1.5. Water is about 1.33.
Problems using this idea: G 33-15, G 33-21, G 33-26, G 33-33, G 33-44, G 33-59, Q 2-1, Q 2-2
3
Mirrors
1
1
+
1
=
di
do
f
Notice that this equation is used for mirrors and lenses, but the signs on the variables mean different things in different situations. Don't
get confused.
The focal length of a spherical mirror is equal to half its radius.
r
f=
2
f is positive for a concave mirror that curves towards the shiny side (like a dish)
f is negative for a convex mirror that curves away from the shiny side (like a ball)
do and di are both positive when they are in front of the mirror. Negative distances refer to things behind the mirror. Notice that a convex
mirror has its focal point behind the mirror, which is consistent with the general rule: In front of miror positive, behind mirror negative.
Problems using this idea: G 32-25, G 32-26, G 32-28, Q 1-1, Q 1-2
Interference
Photons are their own anti particle. They will destroy each other under the right conditions. Because of the wave nature of light, light has
a phase. If two intersecting waves are in phase, the peaks of the waves line up and the amplitude is a sum of the amplitudes of the two
waves. If the waves are out of phase, the amplitude is less. If the phase difference between the waves is half a wavelength, the amplitude
is zero.
Imagine that the surface of a soap bubble is partly reflective so that some light that hits it bounces off, and the rest goes through the first
interface. At the second interface, as the light crosses from the soap-water film to the interior of the bubble (air), some more light bounces
off and the rest goes through. The two reflected beams come back from the bubble and if the thickness is right, they will interfere constructively. The condition is:
1
m+
Λn = 2 t
2
m is a non-negative integer. 2t is twice the thickness of the bubble, which is the difference in distance traveled by the two rays in question.
The 1/2 on the left side is there because of the phase shift as light goes from low n to higher n. If m were 0, the phase shift between the
two reflections would be 360°: 180° due to the extra distance traveled by one of the rays, and 180° due to the first interface.
Λ
Don't forget that the wavelength of light changes inside a medium. It is reduced by a factor of 1/n: Λn = €n
Expanding on this, the phase shift between the two reflections is given by:
t
∆=Π+4Π
Λn
You can recover the condition for maxima by plugging in 2 Π m in for ∆.
The index of light is different for different colors,so a given thickness may produce more constructive interference for one color than for
another, giving soap bubbles their rainbow pattern.
Problems using this idea: G 34-32, G 34-52, Q 3-3
Phasors
Ó Ó
E = E0 CosBk × r - Ω t + ∆F
This is the general form of the electric field in an electromagnetic wave. (The cosine could be a sine, but with a different ∆). Based on
this, you can see that the amplitudes can only be added easily if the waves are in phase. That is, the arguments of the trig functions must
be equal in order for simple adding if amplitudes. But what if things aren't simple? Well, we can handle that. Consider the electric fields
of two waves:
4
Ó
Ó
E1 = E01 CosBk1 ×
Ó
Ó
E2 = E02 CosBk2 ×
`
Ó
r1 - Ω1 t + ∆1 F E1
`
Ó
r2 - Ω2 t + ∆2 F E2
We' ll make this easy by using waves of the same frequency, so let Ω1 = Ω2 ,
È k1 È = È k2 È.
In the double slit case, the two waves really are one wave that split, so they start off in phase H∆1 = ∆2 L, but the distance traveled is different. So define your coordinates or whatever such that ∆1 = ∆2 = 0 and then define the relative phase difference ∆:
2Π
Ó
Ó
Ó
Ó
∆ = k2 × r2 - k1 × r1 = k Hr2 - r1 L = k d Sin@ΘD =
d Sin@ΘD
Λ
Now, we can find the amplitude of the sum of the these two by drawing a phasor diagram. Let's call the sum EΘ . The amplitude of the two
waves together is then E0 Θ . Draw the first amplitude as a vector in phase space, then draw the second vector, starting at the tip of the first
one, with an angle ∆ between the two. Join the tail of the first vector with the tip of the second vector to form a triangle.
This is just a triangle in phase space and you can solve for the length of E0 Θ with some trig. I'll get you started:
E0 Θ x = E01 + E02 Cos@∆D
E0 Θ y = 0 + E02 Sin@∆D
E0 Θ =
E0 Θ x 2 + E0 Θ y 2
These problems can also be solved by just plain adding the waves together. It's helpful to cast the equations for the waves in complex
exponential notation when doing this:
Ó
Ó
Ó
EΘ = E1 + E2
Ó
`
E1 = ReAE01 ã-ä Ω t E E1
Ó
`
E2 = ReAE02 ã-ä HΩ t-∆L E E2
The Re[ ] is an instruction to take the real part of the stuff in the brackets, but it usually is left out because it is understood that the imaginary part of this wave is not physically present. Only the real part is real.
If you're doing slits and stuff, go ahead and just drop the vectors because you can approximate the electric fields as being in the same
direction.
Problems using this idea: G 34-21, G 34-22
Two Slits
Coherent light coming through a pair of two slits (or emitted from two coherent point sources) generates an interference pattern because
light from one source travels a different distance than light from the other source, and if the two waves are not in phase, they will cancel
each other out. The intensity of the resulting pattern is given by:
IΘ = I0 CosB
∆
2
F
2
Where
∆=
2Π
d Sin@ΘD
Λ
d is the distance between the centers of the slits and Λ is the wavelength of the light. Θ is the angle at which the light leaves the slits. Θ=0
is defined as light coming straight out from the slits in a direction normal to the separation of the slits.
5
d is the distance between the centers of the slits and Λ is the wavelength of the light. Θ is the angle at which the light leaves the slits. Θ=0
is defined as light coming straight out from the slits in a direction normal to the separation of the slits.
The positions of the maxima (bright spots) are given by:
d Sin@ΘD = m Λ
m is an integer.
The positions of the minima (dark spots) are given by:
1
d Sin@ΘD = m +
Λ
2
Problems using this idea: G 34-12, G 34-49, Q 3-1, Q 3-2
Single Slit
Light from different parts of a single narrow slit can interfere with light from other parts of the slit, forming a diffraction pattern. The
intensity of this pattern is given by:
IΘ = I0
Β=
Sin@Β  2D
2
2
2Π
D Sin@ΘD
Λ
The approximate positions of the maxima of this pattern are given by (because of the division by Β, the maxima aren't exactly in between
the minima, but close enough):
1
D Sin@ΘD = m +
Λ
2
There is also a maximum in the center at Θ=0. D is the width of the slit.
The exact positions of the minima are given by:
D Sin@ΘD = m Λ
Where m is a NONZERO integer.
Problems using this idea: G 35-13, Q 4-1
Double Diffraction (Single and double slit effects together)
∆=
2Π
d Sin@ΘD
Λ
Β=
2Π
D Sin@ΘD
Λ
IΘ = I0
Sin@Β  2D
2
2
∆
CosB
2
F
2
The Cos[∆/2] part is due to double slit interference. The Part with the Sines and Betas is from the width of the individual slits. You can
see that it will go to 1 as the slit width goes to 0, which is the approximation you used when dealing with double slits earlier.
Problems using this idea : G 35 - 19
Grating
Maxima:
6
Sin@ΘD =
mΛ
d
d is the spacing between slits in the grating, m is an integer.
Resolving Power: If there are two wavelengths, then the resolving power needed to resolve them is defined as:
R=
Λ
=Nm
DΛ
Λ here is some wavelength and DΛ is the smallest difference in wavelength that can be resolved with power R. N is the number of slits in
the diffraction pattern, and m is the integer that determines which maxima you're talking about. According to this formula, it is easier to
resolve higher order peaks than it is to resolve lower order peaks, closer to the central maximum.
This comes from the solution to quiz 4 problem 2, and was not needed for any homework problem.
Problems using this idea: G 35-69, Q 4-2
Resolution/Circular diffraction
Θ is the angle of the edge of the central maximum of the diffraction pattern. It represents the amount of blurring that will occur when light
goes through a circular hole of diameter D.
Θ=
1.22 Λ
D
Problems using this idea: G 35-26, G 35-29
Polaroids
The intensity of light transmitted through a set of two polaroid filters set at angle Θ to one another is given by:
I = I0 HCos@ΘDL2
Here, I0 is the intensity of light coming out of the first filter before it goes into the second one.
The intensity of unpolarized light after it goes through a single polaroid filter is €12 I0 . Therefore, if unpolarized light goes through two
filters, the intensity coming out of the second one is:
I=
I0
2
HCos@ΘDL2
Problems using this idea: G 35-37, G 35-76, Q 4-3
Tidal Accelerations
You may be used to the force of gravity being equal to just mg, but that' s only a special approximation for things that happen close ot the
Earth's surface. Here's the more correct formula:
GMm `
Ó
Fg =
r
r2
Look, see how it depends on distance? Objects that are farther away from the source of gravity (a planet, star, whatever) feel weaker
gravitational forces than objects that are closer. That means that two particles, one above the other, that are falling towards a planet will
seem to drift away from each other (What I mean is that the separation of the particles is parallel to the separation between one particle and
the planet/star/whatever. The three points are colinear).
Now what happens if the two points are at the same distance r from the center of the planet/star/whatever, but still at different positions?
`
Well, that r vector means that they're both accelerating towards the source of gravity, so they'll be drawn together.
Objects in gravitational fields get stretched along the direction from the object to the source of the gravity, and get compressed in the
direction perpendicular to it.
To find the relative acceleration of two particles, just subtract their accelerations. Don't forget about the vectors.
`
Well, that r vector means that they're both accelerating towards the source of gravity, so they'll be drawn together.
7
Objects in gravitational fields get stretched along the direction from the object to the source of the gravity, and get compressed in the
direction perpendicular to it.
To find the relative acceleration of two particles, just subtract their accelerations. Don't forget about the vectors.
Ó GM `
a=
r
r2
GM `
GM `
Ó
Ó
Ó
arelative = a1 - a2 =
r1 r2 = G M
2
r1
r2 2
`
r1
r1 2
-
`
r2
r2 2
=GM
`
`
r2 2 r1 - r1 2 r2
r1 2 r2 2
Problems using this idea: TW 2-9, Q 5-3 (TW for Taylor and Wheeler, your second textbook)
Special Relativity
Concepts and Definitions
We will often talk about two reference frames that are moving at different speeds. They are inertial frames, which means they are not
accelerating. One frame gets a prime and the other doesn't. It is standard to call the frames F and F'. x, y, z, and ct would be coordinates
in frame F. x', y', z', and ct' would be coordinates in F'. Under this convention, it is said that frame F' moves with velocity v relative to F.
v is usually parallel to the x axis. If it is not, it is often possible to rotate the coordinates to make a new, rotated x axis to which v is parallel.
Velocities in the two frames must not be confused with v, the velocity of one frame relative to the other. Subscripts may be used to this
end Hvrel , v1 , v2 , vrocket , etcL, or the velocities of particles in the frames may be represented by the letter u (u' in F').
The speed of light relative to any observer is always exactly 299792458 meters per second. Approximate this as 300 million meters per
second, or 3×108 . This means that velocities do not add normally. Half the speed of light plus half the speed of light does not equal the
speed of light.
Nothing can move faster than light, which is represented with a lowercase c. If you calculate any velocity that is larger than c, you have
made an error, demonstrated that an approximation does not hold under the given circumstances, or proven that a given situation is
impossible.
Events which are simultaneous in one frame are not simultaneous in ANY other frame which moves at a different velocity, unless the
events also occur at the same point in space. Simultaneous events at the same location are simultaneous and at the same location in all
frames (indeed they can be thought of as the same event as they have exactly the same coordinates).
Events do not simply appear to happen at different times because light takes time to bring the information to an observer. They really
happen at different times, even after such concerns are taken into account.
At this point I feel that I should remind you that seeing and observing are different. Seeing is just what you think it means. Observing
means processing what you saw. For example, if you see light arrive at t=1 year from a point that you know is one light year away, you
observe that the light was emitted at t=0. If you see a distant light flash at the same time that you see a nearby light flash, you observe that
the distant light flashed first. Observations take into account things like light's travel time.
Objects aren't really rigid.
It's not just what she said, it's a fundamental property of special relativity. Only point particles are possible. Point particles may form semi
rigid arrangements, but they are always capable of bending under the right circumstances.
Units
It can be helpful to pick a set of units where c=1. In this system, velocities are unitless and are expressed as fractions of the speed of light.
Distance and time have the same units. Suppose you use units of meters for both distance and time. Well, x will be in meters, and t will be
in meters. But ct will also be in meters, because c is just 1. In normal mks units where c=300 million meters per second, ct is also in
meters. Using ct and x instead of t and x is a strategy for avoiding confusion. If you want your distances and times to be in seconds, then
use t and x/c instead of t and x. The quantity v/c is also the same number whether c=1 or not.
Follow this strategy and you may not have to plug in the speed of light at all in some problems. For example, if you say ct=some number
of meters, you don't even have to specify what c is for it to be true!
Other helpful unit related info:
One lightyear is the distance that light travels in a year. One light year divided by the speed of light is equal to one year. A lightmeter is
the time it takes light to travel one meter. A lightmeter times the speed of light is one meter.
Other helpful unit related info:
One lightyear is the distance that light travels in a year. One light year divided by the speed of light is equal to one year. A lightmeter is
the time it takes light to travel one meter. A lightmeter times the speed of light is one meter.
8
Interval
There is a magical thing called the spacetime interval HDsL2 which is constant accross all frames.
HDsL2 = HDxL2 - Hc DtL2
HDs 'L2 = HDsL2
It is common to define the interval as HDsL2 = Hc DtL2 - HDxL2 . However, you'll notice that even if the interval squared gets a sign change, it
is still conserved and the sign will just cancel out. Either convention will yield good results if it is used consistently.
You can see that if two events occur at the same time but at different positions in one frame, the interval will be positive. We call positive
intervals spacelike. If two events occur at the same place but at different times in a frame, then the interval will be negative, and we call
the events timelike. If the interval is zero, we call the separation of the events lightlike, because points along the path traveled by light will
have a zero interval.
Note that if you use the opposite convention for defining the interval, negative intervals are spacelike and positive intervals are timelike.
Just think of what sign you'd get if you had two things happen at the same time but different places, and that's spacelike.
Implications:
You no doubt can see that if the interval is spacelike, then there is one frame where the events are simultaneous. If it is timelike, there is
one frame where the events happen at the same place.
Proper time is time as measured by an observer who is at rest with respect to whatever is being timed. You can see that it's closely related
to the interval, but there is no freedom of convention about its sign!
Hc DΤL2 = Hc DtL2 - HDxL2
Uses:
Suppose you are told that two events happen at two given times and two given locations in frame F. You are also told the time between the
events in frame F', but not the velocity between frames or the seperation between events in F'. You can easily solve this problem using the
fact that the spacetime interval is the same in each frame.
Problems using this idea : Q 5-1, Q 5-2
Lorentz Transformations
Lorentz transformations are a way of finding the coordinates of an event in one frame given the coordinates of the event in another frame.
These are coordinates, as in distances from the origin of the respective frames.
Assuming frame F' moves with velocity v in the positive x direction relative to frame F (Observers in F' see F moving at velocity -v along
their x' axis):
y = y'
z = z'
v
x = Γ x' +
ct '
c
v
ct = Γ
x ' + ct '
c
1
Γ=
1 - Hv  cL2
The above formulas use variables (ct, x) in meters. Want to measure in seconds instead of meters?
x
x'
=Γ
v
+
c
c
t'
c
v x'
t=Γ
+ t'
c
c
Now the only variables that have units are in seconds (t, x/c). v/c has no units.
9
Now the only variables that have units are in seconds (t, x/c). v/c has no units.
Uses: Just about anything. You can derive the other formulas with these.
You can switch x and ct to Dx and cDt or to dx and cdt as long as you switch all of them together:
v
c Dt = Γ
v
Dx ' + cDt '
;
c dt = Γ
c
dx ' + c dt '
c
If the frame velocity is in the y direction instead of x, then just swap x and y in those formulas. If it's at some other angle, rotate your
coordinate system. In order to get equations for the primed variables in terms of the unprimed ones, just swap the positions of the primes.
x' becomes x, x becomes x', etc. Don't forget that v has to switch, too, and v'=-v. Watch what I can do:
x = Γ ' Hx ' - v ' t 'L ;
x ' = Γ Hx - v tL
Look, now each side of each equation is exclusively primed or unprimed, even v. But since v' is always substituted with -v, you never see
it this way. It should be obvious that Γ'=Γ.
Note : Lorentz Transformations are not limited to just the position 4-vector (ct, x, y, z), but can be used on any 4-vector. More on this in
the 4-vector section.
Problems using this idea: TW L-10, TW L-12, TW 5-4, TW 6-5, Q 6-1, Q 6-2
Addition of Velocities
To add two parallel velocity components (assume v is relative velocity of the frames in the x direction):
ux =
ux ' + v
1+
ux ' v
c2
To transform a velocity component that is perpendicular to the relative motion of the reference frames, use:
uy '
uy =
Γ J1 +
ux ' v
c2
N
These formulas can be derived by recognizing that:
dy
dx
ux =
uy =
dt
dt
From there, just transform dx and dt using the Lorentz transformation formulas and replacing the variables (x, t) with their corresponding
differential elements (dx, dt)
Problems using this idea : TW 3 - 11
Angle Transformations
Cos@ΦD =
1 u ' Cos@Φ 'D + v
u
v u' Cos@Φ'D
c2
+1
Φ is the angle that the particle's velocity makes with the x axis. u is the particle's total velocity in F, u' is velocity in F'. Φ' is the angle that
the particle's velocity makes with the x' axis in F'.
This formula is most useful when the velocity in question is c, because then u' and u are both c and you don't have to figure out what u is.
Otherwise, the other formulas for velocity transformations are probably more helpful.
You can derive this the hard way by saying
ux
c
=
dx
,
c dt
then Lorentz Transform dx and dt. Then do a bunch of math about it, or just realize
that it's just the addition of velocity formula where u Cos[Φ] is the x component of velocity that you're adding. Seriously, just plug u
Cos[Φ] in for ux in the parallel velocity addition formula and you have this formula, so don't memorize it, just know how to get it easily.
Problems using this idea : TW L-8, TW L-9
Length Contraction & Time Dialation
10
Length Contraction & Time Dialation
If an object has length L according to an observer at rest with respect to the object, then an observer who sees the object moving at speed
v/c parallel to the object's length will observe a length of:
L
L' =
Γ
Since gamma is always greater than or equal to 1, L' can not be more than L. Objects appear longest to observers that are moving at the
same speed as the object(s) in question. The longest length an object is observed to have is called its proper length.
If the object has width W perpendicular to its direction of motion, then it will appear to have width W'=W. Only the dimension that is
parallel to the motion is contracted. The dimensions that are perpendicular to the motion remain unchanged.
If the time between ticks on a clock is DΤ according to an observer at rest with the clock, then a moving observer will report that the time
between ticks is:
Dt ' = Γ DΤ
DΤ is called the proper time. All measurements made of this time interval from frames moving with respect to the clock will measure
longer times between ticks.
Proper time of an accelerating observer:
What happens if the velocity of something is changing? Well, then Γ becomes a function of t, doesn't it? You have to integrate. Change
the deltas to differentials and go with it:
dt
dΤ =
= dt
1-
Hv@tDL2
Γ@tD
Τ=á
c2
1-
Hv@tDL2
ât
c2
DΤ = à
f
1-
Hv@tDL2
ât
c2
i
If you' re given x as function of t, just change v into dx/dt, take the derivative of the function x[t] that you're given, and plug it in.
DΤ = à
f
i
1
âx
c2
ât
1-
2
ât
Problems using this idea : TW L-12, TW 4-1, TW 5-4, TW 6-5, Q 6-3, Q 7-1, Q 7-2
Momentum and Energy Conservation
Introduce relativistic momentum:
Ó
Ó
p=Γmv
Total (Kinetic + Rest) Relativistic Energy:
E=
m2 c4 + p2 c2 =
m2 c4 + px 2 c2 + p y 2 c2 + pz 2 c2 = Γ m c2
E2 = m2 c4 + p2 c2
Rest Energy (just E when p=0):
E = m c2
Notice that rest energy = mass if c = 1. To get kinetic energy, just take total energy and subtract off the rest energy:
11
m2 c4 + p2 c2 - m c2 = HΓ - 1L m c2
KE =
Notice that there are different ways of expressing all of these things. Which one you want to use depends on how the problem is set up.
Sometimes it's more convenient to have a Γ hanging around, sometimes it's nicer to have the expression with the square root.
When things go REALLY fast (like light) or have no mass (like light):
E>pc
The momentum to use in the above equation is actually the absolute value of momentum (kinetic energy can't be negative).
When things go really slow (like people) or have a lot of mass compared to speed (like people):
E > m c2 +
p2
> m c2 +
m v2
2m
2
This is simply Newton' s formula + rest energy. But when you do classical problems, the rest energy will be on both sides of very equation
and will subtract out, which is why no one noticed it before relativity was discovered.
Problems that use these formulas will usually be blobs colliding and fusing together, gamma rays colliding and making electron-positron
pairs, or possibly some other types of collisions. You could just be asked about two objects bouncing off of each other.
To solve these problems, set up your four equations: conservation of energy, and conservation of momentum in each of the three directions. At least one of the directions will usually be trivial.
In the conservation of energy equation, for example, write the energy of each particle present before the collision on one side of the
equation and set that sum equal to the combined energies of each of the particles present after the collision.
Problems that use this idea : TW 7-9, TW 8-11, TW 8-12, TW 8-15, Q 8-1, Q 8-2, Q 8-3
4- Vectors
When we go from 3 D space to 4 D spacetime, vectors get an extra component. The extra component (usually called the zeroth component
and written at the front) of the position 4-Vector is time (times the speed of light). The extra component of the momentum 4-Vector is
energy (divided by the speed of light).
We denote 4-Vectors by using a greek letter as a superscript or subscript. Here are some relevant 4-Vectors:
rΜ = H ct, x, y, z L = I ct, Ó
rM
Ó
E
E
Μ
p = I €c , px , py , pz M = I €c , p M = H Γ m c, px , py, pz L = H Γ m c, Γ m vx , Γ m vy , Γ m vz L
If the particle has mass, you can write it this way, which is kinda cool:
pΜ = m
d Hc tL
drΜ
=m
dΤ
dΤ
dy
dx
,
,
dΤ
dz
,
dΤ
dΤ
When you multiply two four vectors together, it' s like a normal vector dot product except that the spatial components get negative signs :
pΜ × pΜ =
E
2
- p2
c
Or more generally :
Ó
A Μ × BΜ = A 0 B0 - A × B
The Lorentz Transformation acts on 4-vectors. Take the previously stated formulas and replace ct with the 0th component of the vector, x
with the 1st component, y with the 2nd, and z with the 3rd to get the (more) general form:
A0 = Γ
v
A1 ' + A0 '
c
A1 = Γ A1 ' +
v
A0 '
c
A2 = A2 '
A3 = A3 '
I hope the primes are clear enough. Anyway, this is obviously not the most general form because it still assumes that the velocity is in the
x direction. You can use this on the momentum-energy 4 vector. In the future, you'll learn about 4 vectors for the E-M fields. The
Lorentz transformation works on those, too.
The Lorentz transformation can be represented by a 4x4 matrix called L, which you will no doubt see if you continue in physics.
12
I hope the primes are clear enough. Anyway, this is obviously not the most general form because it still assumes that the velocity is in the
x direction. You can use this on the momentum-energy 4 vector. In the future, you'll learn about 4 vectors for the E-M fields. The
Lorentz transformation works on those, too.
The Lorentz transformation can be represented by a 4x4 matrix called L, which you will no doubt see if you continue in physics.
Problems using this idea : TW 7 - 1
Ÿ Bonus Section (not material you will be tested on, but might make some of the 4 vector stuff clearer)
There are two kinds of 4-Vectors, called covariant and contravariant. When 4-Vectors are multiplied together, a 2nd rank tensor called a
metric is required (It's just a 4x4 identity matrix except the 0,0 component is -1 instead of +1 (and by the sign convention used in this class,
the whole thing is multiplied by -1)). If you don't know about this, some of the negative signs that come out of 4-Vector multiplication
might seem pretty mysterious.
Not because you need to know for this class, but because I think it might make 4 - Vectors less confusing, I' ll write an example of using
the metric to multiply vectors :
â A Μ BΝ ΗΜΝ = A Μ BΝ ΗΜΝ = H A 0 , A 1 , A 2 , A 3 L
ΜΝ
1 0
0
0
0 -1 0
0
0 0 -1 0
0 0
0 -1
B0
B1
B2
Ó
= A 0 B0 - A 1 B1 - A 2 B2 - A 3 B3 = A 0 B0 - A × B
B3
The summation sign tells you to sum over Μ and Ν, but it' s usually left out and it's understood that if an index is repeated (once up, once
down), you sum over it. So since there's a Μ superscript next to a Μ subscript, it's understood that you should sum on Μ. If you see a roman
letter as a superscript or subscript, it means that only the normal spatial components should be used. Greek letters mean that you should
include the 0th component. For example:
â pi = p1 + p2 + p3
;
â pΜ = p0 + p1 + p2 + p3
Μ
i
Light Pressure
Light pushes on things because light carries momentum. Change in momentum is equal to force. Check it out:
E=pc
dp
dE
=
c=Fc
dt
dt
1 dE
F=
Power
=
c dt
c
So, if you' re told how much power something puts out, you can tell how hard it' s pushing on whatever absorbs the power. What if the
power is being directed at a mirror, for example? Well, the force from the light source on the mirror is what we just calculated, but the
mirror reflects the light back as well, so there's another equal reaction force from the light being reflected from the mirror. So the total
force on the mirror is double the force that would act on an object that absorbed all the light that hit it. The other way to think of it is the
change in the light's momentum is twice is great if it turns around than if it stops.
If a fraction f of the total light hitting some object reflects back in the direction it came from, the force on the object is:
F = H1 + fL
Power
c
You can see that if the fraction of light is 1, then we get the doubling of force that we talked about. If the object reflects light at an angle,
you have to use cosines and what have you to figure out what fraction of the light is going back the way it came.
Pressure is just force/area, so if some object (like a star) sends light out in every direction, you can think of the amount of power per unit
area being emitted and use that:
F
Pressure =
A
= H1 + fL
Power 1
Area c
= H1 + fL
F
c
F is the power flux.
You might be asked for the pressure of sunlight on some object in space. You can find the total power radiated by the sun from the
radiative power law from thermo if it isn't given. Then divide the total power by the surface area of a sphere whose radius is equal to your
distance from the sun to get the power/area. Divide by the speed of light for the pressure.
F is the power flux.
13
You might be asked for the pressure of sunlight on some object in space. You can find the total power radiated by the sun from the
radiative power law from thermo if it isn't given. Then divide the total power by the surface area of a sphere whose radius is equal to your
distance from the sun to get the power/area. Divide by the speed of light for the pressure.
Problems using this idea : TW 8 - 3
Mass - Energy equivalence
If something transfers power, it also transfers mass.
E = m c2
dm
1 dE
=
dt
c2 dt
It should be clear that if some mass is transferred in some amount of time over some distance, you have something with the units of
momentum. Transfers of power must then also carry momentum. I guess that means that the powerlines are pushing on your house or
something, but these effects are severely negligible compared to things like gravity, the wind, your neighbor's dog barking, etc.
Problems using this idea : TW 8-5, TW 8 - 40
Formula Sheet
Optics
n1 Sin@Θ1 D = n2 Sin@Θ2 D
c
Λ
cn =
; Λn =
n
n
EΘ 2
IΘ = I0
E0 2
Mirrors and lenses:
1
1
+
1
=
di
do
f
di
m=-
do
f > 0 for converging lens, concave (bowl) mirrior.
f < 0 for diverging lens, convex (ball) mirrior.
do > 0 for object on same side of mirror/lens as incoming light.
di > 0 for image on opposite side of lens as incoming light or in front of mirror
di > 0 means the image is real. di < 0 means that the image is virtual.
1
f
= Hn - 1L
1
1
+
R1
R2
Hfor focal length of spherical lens, convex surfaces have positive RL
Hfor focal lengh of spherical mirrorL
r
f=
2
Constructive interference from reflections off of two parallel surfaces (like the soap bubble):
1
m+
Λn = 2 t
2
Phase shift between reflections from parallel surfaces:
t
∆=Π+4Π
Λn
Phasors:
Draw the amplitude of one electric field as a vector in phase space, then draw the amplitude of the next vector starting from the tip of the
first one. The angle between them should be the phase shift between them, ∆. The length of a vector from the tail of first first field vector
to the tip of the last vector will be the amplitude of the resulting wave.
Phasors:
14
Draw the amplitude of one electric field as a vector in phase space, then draw the amplitude of the next vector starting from the tip of the
first one. The angle between them should be the phase shift between them, ∆. The length of a vector from the tail of first first field vector
to the tip of the last vector will be the amplitude of the resulting wave.
Double slit interference:
IΘ = I0 CosB
∆
2
F
2
, where ∆ =
HMaximaL
d Sin@ΘD = m Λ
1
d Sin@ΘD = m +
Λ
2
2Π
d Sin@ΘD
Λ
HMinimaL
Single slit diffraction:
Sin@Β  2D
2
2
IΘ = I0
Β=
, where
2Π
D Sin@ΘD
Λ
1
D Sin@ΘD = m +
Λ
2
HMaxima HΘ = 0 is the central maximumLL
HMinima, m ¹ 0 !L
D Sin@ΘD = m Λ
Single and double slit effects together:
Sin@Β  2D
2
2
IΘ = I0
∆
CosB
2
F
2
Diffraction grating:
Sin@ΘD =
mΛ
d
R=
Λ
=Nm
DΛ
HResolving powerL
Circular hole diffraction:
Θ=
1.22 Λ
D
Polaroids (aka polarizing filters) (Divide by 2 if light coming into the first filter is unpolarized.):
I = I0 HCos@ΘDL2
Tidal Accelerations
Things get stretched in the direction away from the source of gravity and get compressed in the direction around the source of gravity. To
Ó
`
derive the equation for this, just remember the equation for gravitational acceleration IFg = GrM2 m rM and subtract the two acceleration
vectors.
Ó
Ó
Ó
arelative = a1 - a2 = G M
`
r1
r1 2
-
`
r2
r2 2
15
Relativity
HDsL2 = HDxL2 - Hc DtL2
HDs 'L2 = HDsL2
Hc DΤL2 = Hc DtL2 - HDxL2
Lorentz Transformations (works on general 4-vector if ct,x,y,z replaced with 0,1,2,3 components of 4-vector):
v
ct = Γ
x ' + ct '
c
v
x = Γ x' +
ct '
c
y = y'
z = z'
1
Γ=
1 - Hv  cL2
Addition of velocities (you could replace x with parallel and y with perpendicular for more generality):
ux =
ux ' + v
1+
ux ' v
c2
uy '
uy =
Γ J1 +
N
ux ' v
c2
Length contraction and time dialation:
Proper length is the longest length. Proper time is the shortest time.
L
L' =
Γ
Dt ' = Γ Dt
Accelerating particle (velocity as function of time) :
Τ=á
1-
Hv@tDL2
ât
c2
Momentum and Energy :
p=Γmv
E=
2
m2 c4 + É p È2 c2 =
2
4
2
E =m c +p c
KE =
m2 c4 + px 2 c2 + p y 2 c2 + pz 2 c2 = Γ m c2
2
m2 c4 + p2 c2 - m c2 = HΓ - 1L m c2
4 - Vectors :
rΜ = H ct, x, y, z L = I ct, Ó
rM
Ó
E
E
Μ
p = I €c , px , py , pz M = I €c , p M = H Γ m c, px , py , pz L = H Γ m c, Γ m vx , Γ m vy , Γ m vz L
pΜ × pΜ =
E
2
- p2
c
Light pressure (f is the fraction of light reflected back TOWARDS the source, F is the incoming power flux):
F = H1 + fL
Power
c
Pressure = H1 + fL F
1
c