4 Hamilton's Equations

Consider the Cartesian coordinates

x₁ = x₁(q₁; q₂; q_n; t) ... ... ...

x_n = x_n(q₁; q₂; q_n; t)

where q₁; q_n:generalized coordinates. Also, the generalized momentum

In hamiltonian mechanics, q_j;pb_j and t are chosen to be indepen-dent variables and q__j is a dependent function:

_

q_j = _q_j(q₁; q_n;pb₁; pb_n; t) The Hamiltonian function H is de ned as

(?) H(q₁; q_n;pb₁; pb_n; t) = pb_jq__j L

Introduction to Mathematical Physics 32

"Hamilton equations of motion".

Discussion:

i) What is the physical signi cance of hamiltonian ?

Let's consider the following:

_

Introduction to Mathematical Physics 33 Hence

b

p_kq__k = 2T p_ix_i Plug this into H, we obtain t

H = 2T p_i^x_tⁱ L

= 2T (T V ) p_i^x_tⁱ

= T + V p_i^x_tⁱ

If the transformation between the cartesian and generalized coordinates has no (explicit) time dependence, i. e.

x_i t =

tx_i(q₁; q₂; q_n) = 0 Then the hamiltonian is the total energy of system.

H = T + V

ii) Under which the hamiltonian is a conserved quantity ? Look at the following:

dH

Introduction to Mathematical Physics 34 Hence if the forces are derived from a potential (i. e. no non-conservative force,Q⁰_i = 0) and H has no explicit dependence on time (^H_t = 0), then H is conserved.

Mathematics Department

Introduction to Mathematical Physics 35 Ex. Consider an oscillator, say a 1-D. harmonic oscillator.

The kinetic energy is

T = 1

2m _x²; potential energy is

V = 1

2kx²; and the Lagrangian is

L = T V = 1

2(m _x² kx²) The momentum

p = T

_x = m _x;

and the hamiltonian is H = p _x (1

2 : potential energy:

So the hamiltonian is indeed total energy in this case.

Mathematics Department

Introduction to Mathematical Physics 36 Hamiltonian's equations of motion:

H

p = _2m^2p = _x

H

x = _p + 0 ) p

m = _x; kx = _p ) •x = _p

m = kx m

Mathematics Department

Introduction to Mathematical Physics 37

§Calculus of Variations (Variational Principle)

The problem of the calculus of variation is to minimige (or maximige) functionals.

Consider a function F (x; y; z), and F 2 C^0. We wish to nd a function y(x) s.t.

J (y) =

Z 1 0

F (x; y(x); d

dxy(x))dx has a maximum, t2 P C⁰[0; 1] and y(0) = y₀;y(1) = y₁:

Let's examine the necessary conditions:

Assume that there exists a max, denoted by y₀(x).i. e.

J (y

₀

+ ) J (y

₀

); 8 : scaler

or J (z) J (y₀)f or all admissible z;

z(x) = y₀(x) + (x); y₀(x) = z(x) (x):

Note that 8 2 R; 8 2 P C⁰[0; 1] and (0) = (1) = 0.

Mathematics Department

Introduction to Mathematical Physics 38 In particular, let > 0.

) ¹[J (y₀ + ) J (y₀)] 0;

) lim_!0⁺1

[J (y₀ + ) J (y₀)] =J

(y₀ + )j_!0⁺ 0 (1)

Similarly, let < 0 :

J(y₀ + )j_!0 0 (2) By (1) & (2), we nd

J(y₀ + )j⁼⁰ = 0 (3)

Mathematics Department

Introduction to Mathematical Physics 39 The original integral thus become

J(y₀ + )j⁼⁰ Def. We call the equation

yF (x; y₀; y₀⁰) d

dx y⁰F (x; y₀; y₀⁰) = 0 the "Euler-Lagrange equation" for y₀(x).

Mathematics Department

Introduction to Mathematical Physics 40

Remark: We may solve eq(4) (the Euler-Lagrange equation for y₀(x)) , and we will obtain

max y(x) = y₀(x)

Note: Let's compare the "Euler-Lagrange equation" and the

"Lagrange's equation of motion (1 Dimensional)." If we replace F with L; y with q; y⁰ with q⁰( _q) and x with t; they become identical.

i. e. the later mentioned is the minimum (or maximum) condition for the function L when changing coordinate system.

Ex. Find the shortest distance between two points (x; y) = (0; a) & ex; ey) = (1; b).

Sol. i) D = R1

0 (1 + y⁰²(x))¹²dx;since (ds)² = (dx)² + (dy)²:

ii) To solve the Euler-Lagrange equation : d

Introduction to Mathematical Physics 41 i. e. ¹₂(1 + y⁰²(x)) ¹²y⁰(x) = const:

) y⁰(x) = C; y(x) = cx + a Substitute (0,a) and (1,b) into y(x):

c = b a; ) y = (b a)x + a^#

Remark: We can see the original inequality as K(u + h) k(u); u : extrem:f unction

Then we may write K(u+h) = K(u)+L_uh+R(h);

where L_u : a linear operator, R: remainder, and ^kR(h)k

khk ! 0 as khk ! 0 we call L_uh: Frechet derivative.

Mathematics Department

Introduction to Mathematical Physics 42 Also think about this: By Taylor's Theorem:

f (x₀ + h) = f (x₀) + f⁰(x₀)h + R(x₀; h) where ^kR(x0;h)k

khk !

0 as khk ! 0

Hence we know that

J (y + ) = J (y) + L_y + R( ) L_y

yF (x; y; y⁰) d dx

F

y⁰(x; y; y⁰) = 0 Then we can reach the max.(or min.) where

L_y : Frechet derivative of J at y

More Remarks on Euler-Lagrange's equation:

i) If ^F_y = 0; ^F_y0(x; y; y⁰) : const:

(i. e. _y0F (x; y; y⁰) = _y0F (x; y⁰))

* y⁰ is implicitly a f unction of x:

) We may integrate y' w.r.t then x, y will be solved.

ii) If ^F_y0 0; i: e: ^F_y(x; y; y⁰) = ^F_y (x; y) y(x): implicitly a function of x

iii) If ^F_x 0 : F (x; y; y⁰) = F (y; y⁰) Mathematics Department

Introduction to Mathematical Physics 43 Look at the equation:

d dx

h

F (x; y; y⁰) y⁰_y0F (x; y; y⁰) i

= ^F_yy⁰ + ^F_y0y⁰⁰ y^{00 F}_y0 y^{0 d}_dx^F_y0

= y^{0 F}_{y dx}^{d F}_y0 = 0 i. e.

F (y; y⁰) y⁰

y⁰F (y; y⁰) = const:

Then we have a 1st order O.D.E. y(x):

Mathematics Department

Introduction to Mathematical Physics 44 Review : Extremal R 1

0 F (x; y; y⁰)dx = J (y) where y 2 P C⁰[0; 1]. Hence the necessary condition:

F y

d dx

F

y⁰ = 0 : Euler Lagrange equation

Ex. Find the path y = y(x) which minimige the time of descent for a sliding frictionless bead (mass=1) between y(0) = y₀ and y(x₁) = 0

Introduction to Mathematical Physics 45 ( i. e. conservation of energy)

) T + V = const:

where C₁^{2 k}_2g² By using trignometric substitution

Introduction to Mathematical Physics 46

,which is a cycloid.#

II. Constrained Problems

Ex. Find the curve y=y(x), 0 x 1 s.t. y(0)=y(1)=0 with maximum area R 1

0 y(x)dx with (constraint optimization) xed arc length l 1

Figure.

Constrained maximum:

Mathematics Department

Introduction to Mathematical Physics 47 J (y) = R1

0 F (x; y; y⁰)dx subject to K(y) =

Z 1 0

G(x; y; y⁰)dx = 0 In the above example,

F (x; y; y⁰) = y(x) G(x; y; y⁰) = p

1 + y⁰² l

The point here is to perturb over y₀ + t for which K(y) = 0 Remark: At the required extremal point y₀(x) for any (x) s:t

4K(y⁰) = 0 ) 4J(y⁰) = 0 pf. Assume that 4K(y⁰) = 0 but 4J(y⁰) 6= 0 Let 4J(y⁰) = A 6= 0

) 4J(y⁰)(t ₀) = At 6= 0 if t 6= 0 Since in general, K(y₀ + t ₀) = 0;

) K(y⁰ + t ₀ + s ₁) = 0

Mathematics Department

Introduction to Mathematical Physics 48 i: e: R 1

0 G(x; y₀ + t ₀ + s ₁; y₀⁰ + t ⁰₀ + s ⁰₀)dx R1

0 G(x; y₀; y₀⁰)dx = 0 By Taylor expansion:

R1

0 [G_y(x; y₀; y₀⁰)(t ₀ + s ₁)

+G_y⁰(x; y₀; y₀⁰)(t ⁰₀ + s ⁰₁)] dx = 0 ) tR 1

0 [G_y(x; y₀; y₀⁰) ₀ + G_y⁰(x; y₀; y₀⁰) ⁰₀] dx +sR1

0 [G_y(x; y₀; y₀⁰) ₁ + G_y⁰(x; y₀; y₀⁰) ⁰₁] dx + Higher Order T erms = 0

i: e: 4K(y⁰)(t ₁) + 4K(y⁰)(s ₀) + H:O:T = 0 We may choose ₁ s:t:4K(y⁰)(t ₁) 6= 0 ^{, and}

solve for s = s(t)

Mathematics Department

Introduction to Mathematical Physics 49 Then use the same s(t); t in J:

J (y₀ + t ₀ + s ₁) J (y₀)

= 4J(y⁰)(t ₀ + s ₁) + H:O:T

= t4J(y⁰) ₀ + s4J(y⁰) ₁ + H:O:T

= At + H:O:T (in t)

If y₀ is an extremal, we must have A = 0. (we may verify this by changing sign of t.)

Review: Extremal J (y) = R 1

0 F (x; y; y⁰)dx + B:C: sub-ject to K(y) = R1

0 G(x; y; y⁰)dx = 0 : a constraint: and 4J(y⁰) = R 1

0 (F_y + F_y^{0 0})dx from Taylor series.

Lemma: If the above remark holds, 4J(y⁰) = 4K(y⁰) for some (Lagrange Multiplier).

Mathematics Department

Introduction to Mathematical Physics 50 pf.

Case 1:4K(y⁰) = 0, trivial * 0 = 0; 8 Case 2:4K(y⁰) 6= 0 ^{for some} 0

Let be arbitrary. 4K(y⁰)( t ₀) = 0 for some t (t:

scalar)

) 4K(y⁰) = t4K(y⁰) ₀; t = ^4K(y0)

4K(y⁰) ₀

Now apply the remark, i. e.

4J(y⁰)( t ₀) = 0 Let's plug t into the above equation we nd

4J(y⁰) = 4K(y⁰)

4K(y⁰) ₀4J(y⁰) ₀ (1) So for all

[4J(y⁰) 4k(y⁰)] = 0; = 4J(y⁰) 4J(y⁰) Thus we conclude that at extremum y₀, (1) holds.

Note: It means a necessary condition for constrained entremal problem is (1) holds, .e.

Euler-Lagrange's for the variational problem Mathematics Department

Introduction to Mathematical Physics 51 J (y) K(y) is satis ed. is called the "Lagrange multi-plier".

Sol. The solution is in fact to maximize R1

0 y(x) p

1 + y⁰²(x) l dx By using Euler-Lagrange equation , we have

Introduction to Mathematical Physics 52

From (1) & (2), we obtain x(s) = sin(^s 1) + k₁ y(s) = cos(^s 1) + k₂

where ; k₁ & k₂ are arb. const.

Use boundary conditions

x(0) = y(0) = 0; x(l) = 1; y(l) = 0:

) x(s) = [sin(^{s 2l} ) + sin(₂^l )]

y(s) = [cos(^{s 2l} ) cos(₂^l )]

and Lagrange multiplier 1 = 2 sin(₂^l ) ) s is a circle.

Mathematics Department

Introduction to Mathematical Physics 53 III. Several unknown functions

(1) J = F (x; y₁; y₁⁰; y₂; y₂⁰; y_k; y_k⁰)dx

The extremal y₁; y_k : y_i = y_i (x) +_{i i}(x)

) ^J_i = 0 at y₁; y_k :

The Euler-Lagrange equation:

F y_i

d dx(F_i

y_i⁰) = 0; i = 1; 2; k

Ex. Find the shortest distance arc between 2 pts.

Figure.

(ds)² = (dex)² + (dy)e ² = (dy₁)² + (dy₂)²

d

ds(¹₂p^2y01

y⁰²₁+y⁰²₂) = 0 (1) and _ds^d (¹₂p^2y20

y⁰²₁+y⁰²₂) = 0 (2) since we want to minimize

Mathematics Department

Introduction to Mathematical Physics 54 R L

0 ds = R L 0

p(dy₁)² + (dy₂)^{2 ds}_ds

= R L

0 [(^dy_ds¹)² + (^dy_ds²)²]¹²ds i: e: L = R L

0

py⁰²₁ + y⁰²₂ds F (y₁⁰; y₂⁰) = p

y⁰²₁ + y⁰²₂#

From (1) and (2), y₁⁰ = c₁ & y₂⁰ = c₂ where C₁; C₂ are consts.

* y¹ = x = c₁s + c₃ y₂ = y = c₂s + c₄ i. e. The arc is a straight line.

(2) J = F (x; y; y⁰; ^d_dx^ky_k)dx

Assume y; y⁰; _dx^dkyk speci ed at x₀= x&

x = x₁, and y is the extremal fun, y = y + (x)

Let k=2

0 = J (y + )j⁼⁰

= F (x; y + ; y ⁰ + ⁰; y ⁰⁰ + ⁰⁰)dxj⁼⁰

= (F_y + F_y^{0 0} + F_y^{00 00})dx

Mathematics Department

Introduction to Mathematical Physics 55

= h

(F_{y dx}^d F_y⁰ + _dx^d2₂F_y⁰⁰) dx +F_y⁰ j^xx¹₀ + F_y⁰⁰ j^xx¹₀ d

dxF_y⁰⁰ j^xx¹₀ dx

* (x⁰) = (x₁) = ⁰(x₀) = ⁰(x₁) = 0

Mathematics Department

Introduction to Mathematical Physics 56 ) (F^y + F_y^{0 0} + F_y^{00 00})dx

= (F_{y dx}^d F_y⁰ + _dx^d22F_y⁰⁰) dx By induction, we ned

0 = F_{y dx}^d F_y⁰ + _dx^d22F_y⁰⁰ + + ( 1)^{k d}_dx^kkF_y^[k]:

IV. Several independent variables J (y) = R

Introduction to Mathematical Physics 57

= Z

D x_i( F_zi)dV = Z

D

( _x

iF_zi + F_xizi)dV

) Z

D

(F_{zI xi})dV = Z

D

[(F_zi ) F_xizi ]dV

= R

D(F_z )dV ) R

D(F_zi ) = R

D(F_y F_xizi) dV

= R

D F_zi mds(By divergence theorem) i) If we consider = 0 on D,

Mathematics Department

Introduction to Mathematical Physics 58 Def. Hamilton's Principle states that the trajectories of an object with kinetic energy

T = X

i;k

P_ik(q₁; q₂; q_n; t) _q_iq__k and U = U (q₁; q₂; q_n; t);

render

J =

Z t₁ t0

(T U )dt

stationary. That is, the 1st variation 4J = 0^{. (Note:} q_i are the co-ordinates, and q__i = ^dq_dtⁱ.)

Analysis:

i) note that L = T U is the Lagrangian.

ii) 4J = 0 , _dt^{d L}_{q_i} ^L_qi = 0; i = 1; n:

iii) We always suppose that the position of the object(s) at time t = t₀ and t = t₁ is known. So there will be no problem in the boundary conditions. (Why?)

iv) If T & V are indep. of t, i.e ^T_t = 0 = ^U_t : Mathematics Department

Introduction to Mathematical Physics 59

i L : constant: H is clearly the Hamiltonian.

v) If we make some change of variables, say, q_i; _q_i ! qⁱ; p_i;

(a) & (b) are "Hamilton equations of motion."

Questions :

1) Answer why in analysis iv).

Mathematics Department

Introduction to Mathematical Physics 60 2) Find extremals of the funal.

Z ₂

0

(y⁰² + z⁰² + 2yz)dx subject to y(0) = 0; y(₂) = 1; z(0) = 0 & z(₂) = 1

3) Find extremals of the funal.

Z 1 0

(y⁰⁰² y² + x²)dx subject to y(0) = y(₂) = 0; y⁰(0) = y⁰(₂) = 1

Mathematics Department

Introduction to Mathematical Physics 61

§Special Theory of Relativity

§1. The Galilean Transformation Figure.

Consider the inertial frames << x₁; y₁; z₁ >> and <<

x₂; y₂; z₂ >>, where << x₂; y₂; z₂ >> moves away from

<< x₁; y₁; z₁ >> with velocity v, in the direction of x. Then if O and O⁰ coincide at t=0, we nd

x₁ = x₂ + vt; y₁ = y₂; z₁ = z₂ This is called the "Galilean Transformation"

Mathematics Department

Introduction to Mathematical Physics 62

§2. Breakdown of the Galilean Transformation and of classi-cal Mechanics at high velocities

Ex. Particle velocities never exceed c

It is found experimentally that the velocity of a particle does not increase inde nitely as its energy increases. Instead, the velocity approaches asymptotically the velocity of light

c = 10⁸ m=sec The usual expression

T = m 2 v² therefore is incorrect.

Ex. The addition of velocities

In Galilean Transformation, vector addition applies to ve-locities. But this simple addition fails when the velocity(ies) approach c. The resulting velocity is always never larger than c.

Ex. Time Dilation

The Galilean Transformation assumes that the time t, as assumed before, is the same in

<< x₂; y₂; z₂ >> as in << x₁; y₁; z₁ >>.

It is incorrect when the velocity is high. It has been observed Mathematics Department

Introduction to Mathematical Physics 63 that, for the high energy mesons of the cosmic radiation, time

ows about 9 times more slowly than the laboratory time. The phenomena is called "time dilation"

§3. The Fundamental Postulate of Relativity

? Einstein: It is physically impossible to detect the uniform motion of a frame of reference from observation made entirely within that frame.

Explanation: Any experiment gives precisely the same result, whether it is performed in reference frame 1 or in reference frame 2, as long as it is under no acceleration.

i. e. For example, in frame 1, we nd

F₁ = m₁a₁; in frame 2, we nd F₂ = m₂a₂. There must ex-ist a transformation ,different from Galilean trans., That renders all the laws of nature identical in frames 1 and 2.

§4. The Lorentz Transformation

Mathematics Department

Introduction to Mathematical Physics 64

x₁ = p^x2+vt2

1 (^v_c)² ; x₂ = p^x1+vt1

1 (^v_c)²

y₁ = y₂ ; y₂ = y₁ z₁ = z₂ ; z₂ = z₁

t₁ = ^t2+(

v c2)x₂

p1 (^v_c)² ; t₂ = ^t1+(

v c2)x₁

p1 (^v_c)²

Remarks:

a) The Lorentz Transf. reduces to the Galilean Transf. if we set the velocity of light c equal to in nity.

b) The relative velocity v of the 2 systems cannot be larger than c, for otherwise ,x,y,z;t become imaginary in one system or the other.

c) There are only 4 independent equations, since the 2nd system can be deduced from the 1st system , and vice versa.

d) The transf. equations in frame 1 are identical to those in frame 2, except for the subscripts 1 and 2 interchanged and sign of v switched.

Mathematics Department

Introduction to Mathematical Physics 65

§. Transformation of a length

Consider an observer ₂ in frame 2 xes a ruler of length l₀ on his x-axis, as in the gure below. What will be

Figure.

the length of this ruler for an observer ₁ in frame 1 ? Sol. Let = p ¹

1 (^v_c)²

* The length in frame 1 l₁ can be written as l₁ = (l₀ + vt₂)

And note that on the right end of the ruler, when measured in frame 1, is "simultaneous", i. e. t₁ = 0; and hence

t₂ = (0 ( v

c²l₀)) = (v

c² )l₀ ) l¹ = (l₀ v(v

c² )l₀) = l₀

So we see that a ruler moving in the direction of its length at a velocity v relative to the observer ₂ appears shorter by a factor of ¹ to the observer ₁. ¹ is called the "Lorentz contraction".

Conversely, if the ruler is xed in frame 1, then observer ₂ Mathematics Department

Introduction to Mathematical Physics 66 will see the ruler shortened by a factor of ¹.

Hence this is the general rule: the proper length of an object, measured in a certain direction, is always "Longer" than the same length measured in a frame of reference moving in that particular direction.

Ex. Figure.

Thus we will see as if the cube were rotated through the an-gle = arc sin(^v_c).

§. Transformation of a time interval

Now imagine that observer ₂ measures the duration of a cer-tain phenomenon that is xed w,r,t his frame and that starts at t₂ = 0 and ends at t₂ = T₀. The time T₀ is called "proper time".

What will be the duration of the same phenomenon for observer

1?

Consider the measurement is performed by situating a single clock at the origin O⁰. Observer ₁ uses 2 identical and synchronized clocks, one at O and one at x₁, He chooses x₁ so

Mathematics Department

Introduction to Mathematical Physics 67 that his second clock is next to that of ₂ at the end of the time interval.

Figure.

Thus, for ₁, the time interval begins at t₁ = 0 and ends at T₁ = [T₀ + v

c²x₂] But x₂ = 0. So

T₁ = T₀ = p ^T0

1 (^v_c)²

What if it is ₁ who measures a time interval with a single clock, and ₂ uses 2 clocks?

Then ₂ will nd that ⁰₁s clock runs slow. This phenomenon is called "time dilation".

The general Ruler is: the proper time interval between 2 phe-nomena is always "shorter" than the same time interval measured by a moving observer.

Ex. What if ₁ uses a single clock and look at ⁰₂s clock?

Figure.

Mathematics Department

Introduction to Mathematical Physics 68 t⁰₁ = t₁ + ^vt_c¹ =

q1+(^v_c)²

1 (^v_c)²t₂ > t₂ (Note: this is assuming ₂ moving away from ₁.)

If ₂ moves toward ₁, then

t⁰₁ = t₁ + ^jvt_c^1j = t₁ ^v_ct₁

=

q1 (^v_c)²

1+(^v_c)²t₂ < t₂

In both cases, if ₁ consider the time for light to reach him, then t₁ = t₂ , as always.

§. Simultaneity

If ₁ observes 2 events A & B that, to him, occur at the same x-coordinate x_A1 = x_B1 , and, at the same time, t_A1 = t_B1. He maintains that A and B are simultaneous.

Are they also simultaneous for ₂ ? t_A2 = (t_A1 v

c²x_A1) = (t_A2 v

c²x_A2) = t_B2

But if the events do not occur at the same value of x i. e.

x_A1 6= x^B1 , then

t_B2 t_A2 = (_c^v₂)(x_A1 x_B1) 6= 0 i. e. They are not simultaneous for ₂

Mathematics Department

Introduction to Mathematical Physics 69

§Causality and Maximum Signal Velocity

We may go further in the previous issue: If one event is

"older", say, A, for ₁. Then

t_B2 t_A2 = [(t_B1 t_A1) ( v

c²)(x_A1 x_B1)]

Let us imagine two events A and B. Event A occurs at the origin O, and O⁰, at the moment they coincide, i: e: t_A1 = t_A2 = 0.

Event A is the cause of B, which occurs at x_B1 at a later time t_B1, and in frame 2, at x_B2. According to causality, B cannot oc-cur before A in frame 2.

Event A causes event B through some device, which propagates in some way the signal from A at a velocity v₁ in frame 1. Then

t_B2 = (t_B1 (_c^v2)x_B1) = (t_{B1 c}^v2v₁t_B1)

= t_B1(1 ^v_c¹₂^v)

If causality holds, v₁v c² Since v c, and hence v₁ c, if both v and v₁ > 0.

Mathematics Department

Introduction to Mathematical Physics 70

§Transformation of a Velocity

If ₂ observes that an object has some velocity (v_2x; v_2y; v_2z), what is the velocity of this object according to observer ₁?

Assume that ₁ sees the object with velocity V₁ = (v_2x; v_2y; v_2z). Then

v

_1x

= dx

₁

dt

₁

= dx

₁

dt

₂

dt

₂

dt

₁

Mathematics Department

Introduction to Mathematical Physics 71

* dx₁

dt₂ = d

dt₂[ (x₂ + vt₂)] = (v_2x + v) dt₂

dt₁ = d

dt₁( (t₁ ( v

c²)x₁)) = (1 vv_1x c² )

* dx₁

dt₁ = v_1x = (v_2x + v) (1 vv_1x c² ) ) v^1x = v_2x + v

1 + (^v2x_c2^v) (1) Similarly, we will nd

v_2x = v_1x v

1 (^v1x_c2^v) (2)

If the object moves along the x-axis only, then the above is the velocity transformation.

Mathematics Department

Introduction to Mathematical Physics 72 Ex. This result makes the velocity of light equal to C in both systems ! Consider a photon in frame 2. v_2x = c.

If the object moves rather than x-axis only, we nd

v_1y = dy₁

Introduction to Mathematical Physics 73 Similarly,

v_1z = v_2z

[1 (_c^v₂)v_2x]; and v_2z = v_1z

[1 (_c^v₂)v_1x]

§Transformation of a Acceleration a_1x = dv_1x

dt₁ = dv_1x dt₂

dt₂ dt₁

= d

dt₂[ v_2x + v

1 + (_c^v2)v_2x] (1 v_1xv

c² )= a_2x

3(1 + ^v2x_c2^v)³:

Similarly, we nd

a_2x = a_2x

3(1 + ^v2x_c2^v)³ Mathematics Department

Introduction to Mathematical Physics 74 and

a_1y = 1

2(1 + ^v2x_c2^v)²(a_2y v_2yv

c² + v_2xva_2x);

a_2y = 1

2(1 + ^v1x_c2^v)²(a_1y v_1yv

c² + v_1xva_1x);

a_1z = 1

2(1 + ^v2x_c2^v)²(a_2z v_2zv

c² + v_2xva_2x);

a_2z = 1

2(1 + ^v1x_c2^v)²(a_1z v_1zv

c² + v_1xva_1x)

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Introduction to Mathematical Physics 75

§Relativistic Mass

Observer ₁ has 2 identical masses A and B which slide in opposite directions along the x-axis with zero friction. He sees that their velocities are equal and opposite. The two masses col-lide and rebound with velocities that are again equal and opposite.

We assume that the collision is elastic.

Observer ₂ sees ₁ perform his experiment and measure the velocities of the masses A and B. Let m_A and m_B are the masses in frame 2 before collision. Let us set M to be the total mass in frame 2.

Suppose the total mass of A and B remains constant, i: e:

(1) m_A + m_B = M and

Introduction to Mathematical Physics 76 where m₀ is called the "rest mass". Hence in general, the mass of an object moving with a velocity v w.r.t the observer is

m = m₀

p1 (^v_c)²

The quantity m is called the "relativistic mass".

§Transformation of a Mass For observer ₁,

m₁ = m₀

By rewriting these factors in terms of the factors in frame 2, we nd

Introduction to Mathematical Physics 77

§Relativistic Energy E

E = mc² = m₀c²(1 (^v_c)²) ¹²

= m₀c²

1 2

0 ( (v

c)²)⁰ +

1 2

1 ( (v

c)²)¹+

1 2

2 ( (v

c)²)² + (v c)²

= m₀c²(1 + 1

2m₀v² + 3

8m₀v⁴

c² + (v⁴ c²))

?:¹₂m₀v² is the kinetic energy of classical mechanics.

Mathematics Department

Introduction to Mathematical Physics 78 Def. E mc² is called the "relativistic energy", and

mc² m₀c² is called the "relativistic kinetic energy", where m₀c² is called the "rest energy".

Ex. An excellent demonstration of this existence of "rest en-ergy" is the annihilation of electrons. Positive electrons combine with negative electrons to give 2 gamma rays, each of which

Figure.

has an energy equal to the rest energy of one electron, or 0:511 M ev(= 1:6 10 ¹³joule).

§Some Knowledge Regarding General Relativity I. Light Travels in a force eld

Figure.

light goes "straight" v = 0; a = 0 Figure.

Mathematics Department

Introduction to Mathematical Physics 79

light also goes "straight" v 6= 0; a = 0

Mathematics Department

Introduction to Mathematical Physics 80 Figure.

light travels in a curve !! v 6= 0; a 6= 0

Remark: Time, space and light will be "bent" by a speci c force eld; - gravitational eld, electric eld, magnetic eld, strong-force eld or weak-force eld.

Ex. Correction term of Mercury orbit due to relativistic effect According to Newtonian physics and Kepler's elliptic orbit model, the movements of all the planets can be predicted very well-except for Mercury.

General relativity predicts an r ³ potential correction to Newtonian motion, which solves the error.

II. The Life of a Star-From its Birth to Death i) hydrogen atoms ! ^molecules

ii) Cosmic Clouds

iii) Birth of a Star-possibly a solar system Mathematics Department

Introduction to Mathematical Physics 81 iv) Collapse of a Star-exhaustion of hydrogen gas

v) Red Giant- explosion of a collapsed star vi) White Dwarf-collapse again

vii) Luasar- the "broadcaster"

viii) Black Hole- the nal destiny for great stars Remarks:

i) The black hole is so called because light can never escape from it; also time is "frozen" inside of it, and it is literally a "hole"

in space because it "swallows" all masses close to it.

ii) The idea of a "white hole" arises because of the black holes.

Mathematics Department

Introduction to Mathematical Physics 82

§Prediction of the future?

— The Light Cone Figure.

plane of 3-space

From the gure above, any line on the side of the cone has slope c (light speed). The cone is called the "light cone".

Note that 0 (i: e: (0; 0; 0; 0)) is "present" or "now". Ac-cording to causality, the time-space within the lower cone can in uence the present, but not elsewhere. If we would like to "pre-dict the future", we must know everything within a "future lower cone", yet it is impossible. Figure.

(Impossible since we assume that no signal is to exceed the speed of c and we may only "know" infos. from the 4-dimensional space !!)

Ex. Interesting question

If you want to travel to a star system several hundred of light Mathematics Department

Introduction to Mathematical Physics 83 years away within your life time. How can you do that without do things against the physical laws ?

Sol. There might be several ways to achieve the goal. Let's consider 2 of them:

i) Figure.

Let's say A is the earth and B is the destination . If there is a "higher dimension" than the usual space, we might be able to go from A to B in no time !

ii) Figure.

Similarly to the above- if we only "bend" the space around the spaceship several times and travel in the warped space, we

nd d < D. Then it may shorten the time of travel.

Mathematics Department

在文檔中 Introduction to Mathematical Physics Nai-Sher Yeh (頁 31-83)