4.1 Constant Force and Single Particle

2

Chapter 4:

Classical Mechanics

Gerik Scheuermann University of Kaiserslautern

Germany

3

4.1 Constant Force and Single Particle

A single particle is modeled as an object with position and velocity.

A constant force f interacts with the particle.

Here, m is the mass of the particle and g an acceleration of the par- ticle induced by the force.

r : I→R³ t→r t( ) v : I→R³

t→v t( ) f=mg

4 We are looking for the particle trace.

Integration gives the well-known result:

x : I→R³ x 0( )=x₀ x˙ 0( ) =v 0( ) =v₀

x˙˙ t( )=v˙ t( ) =g x˙ t( ) =v t( ) =gt+v₀ r t( ) x t( )–x₀ 1

2---gt²+v₀t

= =

5 The following figure shows the particle trace.

r v t

gt

v

g v

0

1 -2

0

6 Using the average velocity

we can analyze this also in a v-t diagram, called hodograph.

v r

--t 1 2---gt+v₀

= =

v=r/t gt gt

v

_

v

1-

21

7 We want to compute now the range of a target in direction that has been hit by our particle starting with velocity .We have

rˆ v₀ r

--t 1 2---gt+v₀

= 1 --- rt( ∧r) 1

2--- g( ∧r)t+v₀∧r

= 1

2--- g( ∧r)t=–v₀∧r=r∧v₀ t 2r∧v₀

g∧r ---

=

8 This is independent from the absolute value of r. If the unit direc- tion of our target is , the time needed to hit the target is

The range can now be computed by rˆ t 2r∧v₀

g∧r --- 2rˆ∧v₀

g∧rˆ --- 2v₀

---g rˆ∧vˆ₀ gˆ∧rˆ --- .

= = =

r --t 1

2---gt+v₀

= g∧r=gt∧v₀ r g∧v₀

g∧rˆ

--- t 2 g( ∧v₀)(rˆ∧v₀) g∧rˆ ( )² ---

2 g( ∧v₀)•(v₀∧rˆ) g∧rˆ² ---

= = =

9 The term in the numerator

can be used to maximize the range for a fixed direction and a fixed absolute velocity . The general identity

gives

g 2 gˆ( ∧v₀)•(v₀∧rˆ) rˆ v₀²

2 a( ∧b)•(b∧c) =b²(a•c)–a•(bcb)

2 gˆ( ∧v₀)•(v₀∧rˆ) = v₀²(gˆ•rˆ–gˆ•(vˆ₀rˆvˆ₀))

10 For a variation of , we have to maximize . Since the brackets contain a unit vector like , we have to solve

This means that the angle between and equals the angle between and r.

vˆ₀ –g•(vˆ₀rˆvˆ₀) gˆ

gˆ•(vˆ₀rˆvˆ₀)

– =1

gˆ – =vˆ₀rˆvˆ₀ gˆvˆ₀ – =vˆ₀rˆ .

g v₀

v₀

g

r-g ^{^}

v ^{^} 0

r ^{^}

^

^

11 With more calculations, we get a formula for the maximal range.

This is a paraboloid of revolution.

r_max v₀ ---g 1

1–gˆ•rˆ ---

=

v

2g

12

4.2 Constant Force with Linear Drag

A linear resistance in the direction of the velocity of a particle is also called linear drag. The force on the particle is now

The velocity is given by

This can be solved by using

F=mg–mγv , γ>0 .

v˙=g–γv ⇔ v˙+γv=g .

e^γt(v˙+γv) d dt--- ve ^γt

ge^γt.

= =

13 For a particle with starting velocity , this gives

For the displacement , we integrate to get v₀

v t( )e^{γ( )t} –v₀ ge^γsds 0 t

∫

= =

v t( ) g1–e^–γt ---γ +v₀e^–γt.

= r=x–x₀ r ge^–γt+γt–1

γ²

--- v₀1–e^{– tγ} --- .γ +

=

14 We are now interested in the time a particle needs to hit the line in directionrˆ r.

---r

=

γ > 0

γ = 0

15 We use the formula for the particle trace to obtain the time of flight.

r=gγ^–2e^–γt+γt–1+v₀γ^–11–e^{– tγ} |γ²rˆ∧

γ²rˆ∧r=rˆ∧g e(^–γt+γt–1) γr+ ∧v₀(1–e^–γt) 0=rˆ∧g e( ^–γt+γt–1)+γrˆ∧v₀(1–e^–γt) | : rˆ( ∧g) 0 (e^–γt+γt–1) γrˆ∧v₀

rˆ∧g --- 1 –e^–γt +

=

γt 1 1

2---γT

 – 

 1–e^–γt

, T 2rˆ∧v₀

g∧rˆ ---.

= =

16 T is the time needed in the case without drag. Fort<γ^–1, we have

yt 1 1

2---γT

 + 

  γt 1

2---γ²t²

– 1

6---γ³t³

 + 

  | : 1

2---γ²t 1 1 2---γT

 + 

 

 

 

≈

2γ^–11 1 2---γT

 + 

 ^–1 2

γ---–t 1 3---γt²

≈ +

t

2 1 1

2---γT

 + 

 

γ1 1 2---γT

 + 

 

--- 2 γ1 1

2---γT

 + 

 

---

– 1

3---γt² | t²

+ T²

≈ ≈

t T

1 1

2---γT + --- 1

3---γT²

≈ +

17 Further approximation gives

t T 1 1

2---γT

 – 

  1

3---γT²

≈ +

t T 1 1

6---γT

 – 

 

≈

18 The intersection can now be computed by taking the outer product of the displacement equation with .

With

gˆ r gγ^–2e^–γt+γt–1

v₀γ^–11–e^{– tγ}

| gˆ∧ +

=

gˆ∧r=gˆ∧gγˆ^–2(e^–γt+γt–1)+gˆ∧v₀γ^–1(1–e^–γt) r gˆ∧v₀

gˆ∧rˆ --- 1–e^–γt

---γ

 

 

=

1–e^–γt ---γ t 1

2---γt²

– T 1 1

6---γT

 – 

  1

2---γT²

≈ ≈ – T 1 2

3---γT

 – 

 

=

19 we have

Using the range without drag

we find finally the range as r gˆ∧v₀

gˆ∧rˆ ---T 1 2

3---γT

 – 

 .

≈

R gˆ∧v₀ gˆ∧rˆ ---T ,

=

r R 1 2

3---γT

 – 

 

≈ R 1 4γ

---3rˆ∧v₀ gˆ∧rˆ ---

 – 

 .

=

20

4.3 Constant Magnetic Field

The usual description for the interaction between a charged particle with charge q and mass m is

With , we have

mv˙ q

---vc ×B.

=

ω q

mc---

– 

 B

=

v˙=ω×v .

21 Because of

we get

Rearranging terms, this means

ω×v=–i(ω∧v)=–( )iω •v=v•Ω, Ω=iω

v˙=v•Ω.

v˙ 1 2---Ωv 1

2---vΩ –

+ =0 .

22 We are looking now for an integrating factor R with

This would allow R˙ R1

2---Ω

= R˙^† 1

2---ΩR^† –

=

d

dt--- RvR( ^†) dR ---vRdt ^† Rdv

dt---R^† RvdR^† ---dt

+ +

= =

R1

2---ΩvR^† Rv˙R^† Rv 1 2---ΩR^†

– 

 

+ + =

R 1 2---Ωv v˙ 1

2---vΩ –

 + 

 R^†=0

23 Since we have a constant magnetic field and , we set

and find

Ω=const.

R t( ) e 1 2---Ωt

=

R^†( )t e 1 2--- – Ωt

,

= R 0( ) =R^†( )0 =1,

R^†R=RR^†=1.

24 For the velocity, we have

Separating v into parallel and orthogonal parts with respect to the magnetic field shows

d dt--- RvR( ^†) =0 RvR^†–v₀=0 v=R^†v₀R=e^–0.5Ωtv₀e^0.5Ωt.

v=v_||+v_⊥ Ωv||=v_||Ω Ωv_⊥=–v_⊥Ω R^†v_0||=v_0||R^† R^†v_0⊥=v_0⊥R

25 This results in

This shows that the parallel velocity is constant and that the orthogonal velocity rotates through an angle in time, so v sweeps out a portion of a cone.

v=v_0⊥R²+v_0||=v_0⊥e^Ωt+v_0||.

Ωt

v

0|| v

v

Ω ω

0

0 |_

26 For the trajectory, we get

With

we have

x–x₀ v_0⊥Ω^–1e^Ωt–1 v_0||t . +

=

r=x–x₀+v₀•Ω^–1=x–x₀+v₀×ω^–1

r= (v₀•Ω^–1)e^Ωt+v_0||t= (v₀×ω^–1)e^iωt+(v₀•ω^–1) ωt .

27 Setting

one gets the standard helix

with

a=v₀×ω b=v₀•ω θ = ωt θ= θ ωˆ

r( )θ =ae^iθ+bθ

a•θ=0

θ ae

a x

0 θ i

b