Mechanics Continuous Expansion, Lecture Notes - Physics, Study notes of Mechanics

Download Mechanics Continuous Expansion, Lecture Notes - Physics and more Study notes Mechanics in PDF only on Docsity!

MechanicsPhysics 151^ Lecture 23Continuous Systems and Fields(Chapter 13)

Where Are We Now? „^ We’ve finished all the essentials^ „^ Final will cover Lectures 1 through 22 „^ Last two lectures: Classical Field Theory^ „^ Start with wave equations, similar to Physics 15c^ „^ Do it with Lagrangian, and maybe with Hamiltonian^ „^ Go into relativistic field theory „^ Not enough time to discuss everything^ „^ Let’s see how much we can do^ „^

And take it easy!

Lagrangian „^ Lagrangian is^ „^ m/ ∆

x^ is the linear density

μ^ (mass/unit length)

„^ k ∆ x^ is the elastic modulus

K^ (force/fractional elongation)

„^ Think about Hooke’s law „ μ^ and^ K^ remain constant as we shrink

∆ x^ Æ^0 2

2 1 2 2

1 1

(^ )

i^ 2 1 2 i^ i i

i^ i i L^ m^ i

k m k^ x^

x x^

x

η^ η^

η η^ η

⎡^

=^

−^ −⎣

⎡^

⎛^

=^

−^ ∆^

⎢^

⎜^

∆^

⎢^

⎣^

∑ ∑

Rearrangea little

It’s^ not^ Young’s modulus LKL L kF

How much the spring is stretchedrelative to its natural length

Continuous Limit „^ Now we have^ „^ Re-label

ηwith the equilibrium position i^

(^2) x 2

1 1 2

i^ i i L^ i

K^

x x η^ η μη^

⎡^

⎛^

=^

−^

⎢^

⎜^

⎝^

⎢^

⎣^

∑

( ) x η η→ i 2

2

2 0

2 1

(^ )^

i x

x^ x^

x

L^

x^ K^

x ηη x dK^ dxdx μη^

η μη ⎡^ ∆ →

+ ∆^ −⎛

=^

−^

⎢^

⎜^

⎢^

⎣^

⎡^

⎯⎯⎯→^

⎢^

⎣^

∑ ∫

Shrink!

Lagrangian per unit length

Lagrange’s Equations „^ First, start from^ „^ Do the usualLagrange’s equations^ „^ That’s wave equation with velocity^ „^ We want to get this from the continuous Lagrangian

2 2

1 1 2

i^ i i L^ i

K^

x x

η^ η

μη^

⎡^

⎛^

=^

−^

⎢^

⎜^

⎝^

⎢^

⎣^

∑ 1

i^ i^

i^ i

d^ L^ L^ i^ i

K^

K^

x

dt^

x^ x^

x^ x

η^ η

η^ η

η^ η

+^

−

⎛^ ⎞^

⎡^

−^

−^ ⎤

∂^ ∂^

⎛^

⎞^ ⎛^

−^ =^

−^

+^

∆^ =

⎜^ ⎟^

⎜^

⎟^ ⎜^

⎢^

∂^ ∂^

∆^ ∆^

∆^ ∆

⎝^

⎠^ ⎝^

⎣^

⎝^ ⎠

(^2) d^ η^0 K^2 dx μη −^

Shrink^ ∆ x

Kv =^ μ

Lagrange’s Equations „^ In the discrete case, we had^ „^ ηbecame i^

η( x )

„^ Simple analogy gives „ But this doesn’t work „^ We must go back to Hamilton’s Principle

i^

i L^ d^

L dt η ⎛^ ⎞ η ∂^

∂^

for each^ i 0 ( )^

L^ d^

L

x^ dt^

x η

⎛^ ⎞ η ∂^

∂^

2

2 1

1

I^ Ldt

dxdt

δ^ δ

=^

=^

∫^

L∫ ∫

Hamilton’s Principle^ „^ Hamilton’s Principle gives

(^

) 2 2 , (^1 12 21 12 21 )

,^ ,^ ,

t^ x^

d^ d dx^ dt t^ x

d^

d

t^ x^

dx^

dt d^

d

t^ x^

dx^

dt

t^ x

d^

d

t^ x^

dx^

dt

dI^ d^

x t dxdt

d^ d

d^

d d^

dxdt d^

d^

d d^

d^

d^ dxdt dx^

dt^

d η^ η η

η η

η η

η

α^ α^

= ⎧^

∂^

∂^

=^

+^

⎨^

∂^

∂^

⎩^

⎧^

⎛^ ⎞^

⎛^ ⎞

∂^

∂^

⎪^

=^

−^

⎨^

⎜^ ⎟^

⎜^ ⎟

∂^

∂^

⎪^

⎝^ ⎠^

⎝^ ⎠

⎩^

L∫ ∫ L^ ∫ ∫ ∫ ∫

L^

L

L^

L^

L

(^2 21 )

( , )^

t^ x

d^

d

t^ x^

dx^

dt

dI^

d^

d^

x t dxdt

d^

dx^ ηη dt α

ζ

α

η

⎧^

⎛^ ⎞^

⎛^ ⎞

∂^

∂^

⎪^

⎛^ ⎞^

=^

−^

−^

⎨^

⎜^ ⎟^

⎜^ ⎟

⎜^ ⎟^

∂^

∂^

⎝^ ⎠^

⎪^

⎝^ ⎠^

⎝^ ⎠

⎩^

L ∫ ∫

L

L

= 0!

Lagrange’s Equation „^ Lagrange’s equation for the 1-dim problem is^ „^ Let’s try it with

d^

d dt^

dx d^

d dt^ ηη^ dx

⎛^ ⎞^

⎛^ ⎞

∂^

∂^ ∂+ −^

⎜^ ⎟^

⎜^ ⎟

∂^

∂^ ∂

⎝^ ⎠^

L^ L^ ⎝^ ⎠

L 2 2

1^ d^^2

dK dt^

dx

⎡^ μ

⎛^ ⎞^

⎛^ ⎞

=^

⎢^

⎜^ ⎟^

⎜^ ⎟

⎝^ ⎠^

⎝^ ⎠

⎢^

⎣^

L

2 2 2

d^ d^

d^ d^

d^ d K^

K

dt^ dt^

dx^ dx

dt^

dx

η

η

η^ η μ

μ ⎛^ ⎞^

⎛^ ⎞− =

−^

⎜^ ⎟^

⎜^ ⎟

⎝^ ⎠^

⎝^ ⎠

Yes, the rightwave equation

Multi-Component Field „^ I defined

η^ as the displacement along

x^ axis

„^ General 3-dim. vibration maybe in any direction^ η^ „^ We are now dealing with 3functions of space and time

(^ ,^ ,^

) η η η x y z

= η

This is getting really tedious ,^ ,^ ,^

,^ ,

,^ ,^ ,^

,^ ,

,^ ,^ ,^

,^ ,

x^ x^ , , , x^ x y y y^ y z z z^ z d^ d^ d^

d x^ dx^ dy^

dz^ dt d d d^ d y^ dx^ dy^

dz^ dt d d d d z^ dx^ dy^

dz^ dt η^ η^ x y z t η^ η η η η^ η η η η^ η ⎛^ η η η

⎜^

⎜^

=^ ⎜^

⎜^

⎜^

⎜^

⎝^

L^ L

Shorthand Notation „^ Let’s use indices

instead of

t ,^ x ,^ y ,^ z

„^ Similar to what we did in relativity „^ We need quantities like „^ Let’s get lazy „ We can write, e.g.

d^ η^ i η i dx^ μ (^2) d η^ i dx dx μ^ ν

d η^ ρη ≡ , ρ μ dx μ

(^2) d η^ ρ , ρ μν dx dx μ^ ν η^ ≡^

and^

etc. d ηη ≡, μ dx μ

(^ ,^ ,^ x ), η^ η=L L^ ρ^ ρ μ^ μ

,

d dx^ μρ μ

ρ

⎛^ ⎞∂^

⎜^ ⎟⎜^ ⎟∂^

L^ ⎝ ⎠

L

Stress-Energy Tensor „^ We got^ „^

NB:^ T is not a tensor in the relativistic sense^ μν^ „^ Suppose

L^ does not depend explicitly on

x^ μ

„^ For^ μ^ = 1, 2, 3, that means no external force „^ For^ μ^ = 0, that means no source/sink of energy

, , d dx^

x μμν νν

μ

⎛^ η^ δ η

∂^

−^

⎜^

⎜^

∂^

⎝^

L^

L

L

Freefield

dT^ μν^0 = dx ν

Stress-energy tensor T ≡^ μν What does this “conservation”condition mean?

Divergence of S-E Tensor „^ The condition

has a form of divergence „^ Integrate over a fixed volume

V^ and use Gauss’s Law

„^ Now we need to know what

T and^ T^ μ^0

are μ dT^ μν^0 = dx ν 0

0

i i dT^ dT

dT^

dT dx^ dt

dx^

dt μνμ

μμ^

μ

=^ ν

+^ =^

+ ∇ ⋅^ = T

d^ T^ dV^^0

dV^

d

μ dt

μ

μ = − ∇ ⋅^

= −^ ⋅

∫^

∫^

∫ T^

T^ S

This vector representsthe “flow”

Total^ T^ μ^0 in the volume

What escapesfrom the surface

Energy Current Density „^ Consider a small piece^ „^ It’s stretched by^ „^ This gives the Hooke’s law force^ „^ The work done by this piece to the next piece is

dx ( ) x η (^

) x dx η +

(^ )^

d ( )

x^ dx^

η x dxdx η

η+ − = d

F^ K dx

η = −^ d

F^ K

η ^ η^ η= − dx

dT K 01 dx

η^ η= −

equals to

Momentum Density „^ First consider^ „^ Again with the 1-dim. elastic rod example^ „^ This isn’t so obvious…

, , T^ μν

∂^ L^ η^ δ≡ −Lμμνη∂ν

(^0) i d i T^

η∂ L=  dx η∂ 2 2 1^ d^ 2

dK dt^

dx

⎡^ μ

⎛^ ⎞^

⎛^ ⎞

=^

⎢^

⎜^ ⎟^

⎜^ ⎟

⎝^ ⎠^

⎝^ ⎠

⎢^

⎣^

L^

d^ 10 d η T dt dx ημ=