it’s relatively simple…
DESCRIPTION
It’s Relatively Simple…. Nick Bremer, Erik Cox, Scott McKinney, Mike Miller, Logan Petersen, AJ Schmucker, Nick Thull. IN REVIEW. Einstein’s Postulates Lorentz Frames Minkowski Inner Product. Einstein’s Postulates. There exists a Lorentz frame for Spacetime. - PowerPoint PPT PresentationTRANSCRIPT
It’s Relatively Simple…
Nick Bremer, Erik Cox, Scott McKinney, Mike Miller, Logan Petersen, AJ Schmucker, Nick
Thull
IN REVIEW
• Einstein’s Postulates• Lorentz Frames• Minkowski Inner Product
Einstein’s Postulates
1. There exists a Lorentz frame for Spacetime.
2. A Lorentz transformation of a Lorentz frame gives a Lorentz frame.
1st Lorentz Postulate
– For stationary events, Physical Clock Time and Coordinate Time should agree.
– That is, we assume that stationary standard clocks measure coordinate time.
2nd Lorentz Postulate
• The velocity of light called c = 1.• Light always moves in straight lines with
unit velocity in a vacuum.• , time and spatial position• Note: Think of the light pulse as a
moving particle.
),( 0rvTTT
Minkowski Space(Geometry of Spacetime)
• The symmetric, non-degenerate bilinear form of the inner product has the properties
• <x,y>=<y,x>• <x1 + x2, y> = <x1, y> + <x2, y>• <cx,y> = c<x,y>• The inner product does not have to be positive definite, which
means the product of it with itself could be negative.• Non-degenerate meaning only the zero vector is orthogonal
to all other vectors• Spacetime has it’s own geometry described by the
Minkowski Inner Product.
Minkowski Inner Product• Defined on R4:• u = (u0,u1,u2,u3)• v = (v0,v1,v2,v3)• <u,v>:=u0v0- u1v1- u2v2- u3v3
• <•,•> also called the Lorentz Metric, the Minkowski metric, and the Metric Tensor
• M = R4 with Minkowski Inner Product• “•” represents the usual inner product (dot product)
in R3.• In this case you have an inner product that allows
negative length.
WorldLines
Vector Position Functions andWorldlines
• In Newtonian physics/calculus, moving particles are described by functions t r(t)
• r(t) = ( x(t) , y(t) , z(t) )
This curve gives the “history” of the particle.
Vector position functions andWorldlines cont…
• View this from R4 perspective t ( t, r(t) )
• In the above ‘t’ represents time and ‘r(t)’ represents the position.
• This can be thought of as a “curve in R4”, called the Worldline of the particle.
• A worldline (at +b) is given a non-Euclidian “geometry” in M by the Minkowski Metric.
A Brief Description of Relativistic Time Dilation
The Einstein – Langevin Clock
• Time is measured where the period between light emission and return is regarded as one unit.
Light source
Mirror
LCt 2 Where L = length of tube, and c = speed of light.
The Relativistic Time Dilation Factor
Let t’ = time of ½ pendulumLight Source
Mirror
L L = ct’
Consider a spaceship with an Einstein-Langevin Clock onboard. We will look at what happens after time t.
Time Relative to a Stationary Lorentz Frame
Light Source
Mirror
L
Ship has moved D = vt.
vt
ct
O
Z
Y
X
O’
Z’
Y’
X’
“e”
Observe:
(ct)2 = L2 + (vt)2
We’ll let t be the measure of ½ of the light pendulum according to the clock of the stationary Lorentz frame.
Relativistic Dilation Factor
• Solve for
22
22
22222
'
vc
ctt
vc
Lt
tvLtc
(ct)2 = L2 + (vt)2
We call
the relativistic dilation factor. This shows up in many equations in relativity theory. Too make things easier, we often set c = 1.
22)(
vc
cv
And recall: L = ct’
What Is Proper Time?Proper time is the elapsed time measured by a moving object.
However, this formula only applies for constant velocities.
Recall:
)1
1)( : Note(
')(1
'
2
2
vv
tvtv
tt
where represents the elapsed time of the moving object in proper time and t is the elapsed time of the moving object in coordinate time (using as a conversion factor).)(v
t
Solution:Take small time intervals on the worldline, each with approximately constant velocity.
)( ttx
)(tx vdtxd
ttxttx
)()(
t
Average Velocity
for small
2
1
2
1
22
011lim
t
t
t
ttdtvtv
1t
2ttvt 21'
With constant velocity, we can use:
As , summing the intervals together yields the following:
0t
Proper Time!
Proper Time
t
a
dttxt tv )(12
)(,
where t-a is the elapsed time measured on the stationary clock and the resulting integral is the elapsed time measured on the moving clock.
Note: t
a
dtat tv )(12
Note: T is Tau
Parameterize the Worldline Using Coordinate Time
Recall: Spacetime = E = the set of all possible events e
where E is modeled by ,,4M
4)(,)( txttz Parameterize using coordinate time:
)(,1,1 tvdtxd
dtdz
Differentiate:
Worldline: Eses )(
where and )(),()( sxstse s
Proper time represents a non-Euclidian arc-length. Proof: )(,1 tv
dtdz
Fact About Proper Time
)(,1,)(,1, tvtvdtdz
dtdz
)()(1 tvtv
2)(1 tv
Notice the similarities between the common arc-length formula and the expression above:
b
a
b
a
dtdtdz
dtdzdt
dtdy
dtdx ,
22
)(,, )(12
txtdtdtdtzd
dtzdt
a
t
a
tv
2)(1, tvdtdz
dtdz dtdt
dtzd
dtzd tv )(1
2,
This shows that Proper time is a non-euclidean arc-length.
t = b
t = a
x = x(t), y = y(t)
z(t) = ( x(t),y(t) )
For any curve, we can make a change of parameter using a function to get
)(tet )(tss ))(( tses
Since Proper Time is similar to arc-length, can the worldline be parameterized by proper time? Yes, provided the quantity called proper time is “independent of parameter”.
dtdz
dtdz
dsdt
dsdt
dtdz
dsdt
dtdz
dsdz
dsdz ,,,
2
dtdz
dtdz
dsdt
dsdz
dsdz ,,
dsdt
dtdz
dsdz
By Chain Rule
Differentiate z with respect to s:
Change parameter of the worldline:
dsdsdt
dtdz
dtdzds
dsdz
dsdz s
b
s
b
,,
Pick b so t(b) = a (i.e. if s = b, then t = a)
Make a substitution: )(stt dsdsdtdt
)))((),((,,)(
)(
stxstdtdtdz
dtdzdt
dtdz
dtdz t
a
st
bt
Thus showing proper time is independent of parameter.
Four-velocity
What is four-velocity?
4-dimensional, relativistic analog of traditional three-
velocity, represented by: ord
dz )(U
Four-velocity and Three-velocity
How is four-velocity related to traditional three-velocity?
Recall that we can parameterize worldlines by proper time:
)))((),(()( txtzz
)(v
dxd
ddt
ddzU
,)(
ddt
dtxdv
),(
)(),( v
dtxdv
Factor out
vv ,1)(
Note: shown later.)(vddt
Important Fact About Four-velocity
Four-velocity is a timelike unit vector
Proof: Show 1, d
dzddz
t
a
dttxt tv )(12
))(,(
Recall:
Differentiate using Fund. Thm. Calculus: )(1)(1
22
tvtvt
a
dtdtd
dtd
Using Chain Rule and inverses, we get:
)(,
11
)(12
v
dtdz
dtdzd
dt
tv
dtxd
dtdz
,1Recall:
dtdz
dtdzdt
xdddt
dtdz
ddz
,
1,1
dtdz
dtdzdt
xd
dtdz
dtdzdt
xdddz
ddz
,
1,1,,
1,1,
dtxd
dtxd
dtdz
dtdz
,1,,1
,
11
,1,,1
,1,,1
dtxd
dtxd
dtxd
dtxd
Thus showing and proving four-velocity
is a timelike unit vector.
1, d
dzddz
Classical Momentum
Conservation of Momentum
Recall the Classical Law of Conservation of Momentum
22112211 vmvmvmvm
It can be shown, however, that this Law is NOT Lorentz invariant for inelastic collisions.
Due to the principle of covariance, the laws of physics should hold for any Lorentz Frame.
Something must not be quite right!
So what do we do?
Answer: Would a 4-dimensional analog work?
Question: If the Classical Law isn’t quite right, how do we fix it?
What if we use 4-velocity[ ]in place of standard velocity?
)(U
Gearing up
Note that for small , || v
d
xddtxd
Recall: and
(space component of )
),()(
d
xdddtU
2||1 vdtxd
)(U
Relativistic Momentum
Our first analog of the Classical Law will look like:
22112211 UmUmUmUm
We’ll call this equation *. Let’s consider the time and space components of *.
Time Component
The time component of U(T) is simply: )(vddt
Which gives:
)()()()( 22112211 vmvmvmvm
As you may have noticed, this seems trivial since for small all it says is that . However, we will see that this will be a new law analogous to the Classical Law of Conservation of Energy.
v
2121 mmmm
Space Component
Now let’s examine the space part of *. Recall that:
dxdv
dtxd
21
dxd
dtxd
v
21
1
a.k.a. vv )(
21 vdtxd
(space component of ))(U
Space ComponentNow we can write the space component of * as:
222111222111 )()()()( vvmvvmvvmvvm
Notice that when is small,
and we get the Classical Law of Conservation of Momentum.
11
1)(2
v
v v
Philosophy
It behooves us to stop and think here for a moment. It seems that the space component of * is very close to the Classical Law of Conservation of Momentum. Does this make sense?
It does. When moving from 3 dimensions to 4, we added time. If we look at the space part of our 4-dimensional analog, it seems reasonable to see things that were developed in 3-dimensional space.
So now what?
is very close to the Classical Law of Conservation of Momentum. If we could somehow redefine mass as some sort of rest mass times , then this equation would match the Classical Law.
222111222111 )()()()( vvmvvmvvmvvm
)(v
MassCan Mass Change?
If so, what happens if Mass is not constant?
How to Define Rest Mass
•Fix a Lorentz coordinatization
•Arbitrarily choose a particle at rest
•Define the rest mass of that particle to be one
In other words,
Equals the post collision mass and velocities
The initial mass and velocity
Than by solving for m1 and than looking as v1 approaches zero
as just the mass of an object having velocity zero
Rest mass is than defined
Relativistic Momentum also called 4- Momentum
or energy momentum
p = (Rest mass)(4-velocity)
The break down of p
The time component in P is related to the Newtonian concept of Kinetic energy.
While the space component in P is related to the Classical Law of Conservation of Momentum.
Substitution and Distribution
Recall,
Inspired by the tendency of Newtonian concepts to appear in spatial components of
relativistic momentum. We determine relativistic mass to equal
M=(rest mass)(time dilation factor)
Hence,
p can now be defined as a vector with relation to M.
Observations on Kinetic Energy
Background on Kinetic Energy
• Kinetic energy is the energy of motion
• Kinetic energy is a scalar quantity; it does not have a direction.
• The Kinetic energy of an object is completely described by magnitude alone.
221 vmKE
v = speed of the object
= mass of the objectm
Setup for Conversion to Newton
Newtonian Perspective versus the classical definition of momentum
vmp
dtpdF
:Newton’s 2nd Law of Motion Explains how an object will change velocity if it is pushed or pulled upon
Information for the Proof
The Kinetic Energy’s rate of change is particularly interesting:
vFvdtpdv
dtvdmvm
dtd
)21( 2
Claim:
Conversion to Newtonian KE
)21()21( 2 vvmdtdvm
dtd
Breakup the using the product rule for dot-products2v
)()21()21( vvdtdmvvm
dtd
Pullout the constant m21
Conversion (cont.)
)()21()()21(dtvdvv
dtvdmvv
dtdm
Distribute the dot-product
))(2)(21()()21(dtvdvm
dtvdvv
dtvdm
Combine the terms
dtvdmvm
dtvdv
)21)((2
Reduce
ConclusionRecall that , so we know that F
dtvdm
Fv
dtvdmv
vFKEdtd
)(
So…
Before we move on, this equation can be used to define Relativistic Energy as:
vFdtdE
Relativistic Energy
E = mc2
Working out the equation…• Let m = rest mass
UUm ,2
pp,
),(),,( pppp oo
UmmU ,
122 mmRecall: unit vector of four velocity: <U, U>
Recall: Four momentum p = mU
Continuing on…
pppo
2
),(),,( pppp oo
From the previous slide we had:
Recall: Relativistic Mass M = po
ppM 2
This is also the Minkowski Inner Product definition
Which gives us the following equation:
ppMm 22
Differentiating both sides with respect to …
ppMdtd
20
dtpdp
dtdMM
220
dtpdvM
dtdMM
220
vFMdt
dMM
220
By Product Rule for Dot Product
Recall: vMp
Recall: by Newton’s II
Fdtpd
t
Continuing on…
vFdt
dM 0
dtdE
dtdM
0
dtdM
dtdE
dtdt
dMdtdtdE
vFMdt
dMM
220
From the previous slide we had…
By factoring out 2M we get…
Recall: Relativistic Energy
vFdtdE
Almost there…
CME
dtdt
dMdtdtdE
From the previous slide…
If we consider smaller and smaller masses that get closer and closer to zero, the energy will also get smaller and smaller; therefore not allowing for any remainder C. So we will claim C is zero.
Which brings us to…
E = M!!!
proving the equation E = mc2 where
c (speed of light) is set to 1
Special Thanks
Thanks to Dr. Deckelman for his guidance on the creation of this presentation.