Unit symbol

Coordinate: m (x, y, z) Force: N F(f)

Time: s t(T) mass: kg m (m)

Displacement: m r momentum: kg * m/s p(P)

Speed: m/s v(u) Energy: Joule

Acceleration: m/s 2 Impulse: N * s 1

Length: m l(L) kinetic energy: J Ek

Distance: meter second (second) Potential energy: Joule heat

Angular velocity: radians per second ω torque: n * m.

Angular acceleration: rad/s 2 α power: W P.

One:

Newtonian mechanics (preparatory knowledge)

(1): basic formula of particle kinematics: (1)v=dr/dt, r=r0+∫rdt.

(2)a=dv/dt，v=v0+∫adt

(Note: Of the two formulas, the formula on the left is in differential form and the formula on the right is in integral form.)

When v is constant, (1) means uniform linear motion.

When a is constant, (2) indicates uniform linear motion.

As long as we know the motion equation of a particle r=r(t), we can know all its motion laws.

Particle dynamics:

(1) Niu Yi: An unstressed object moves in a straight line at a uniform speed.

(2) Niu Er: The acceleration of an object is directly proportional to the force it receives and inversely proportional to the mass.

F=ma=mdv/dt=dp/dt

(3) Niu San: Action and reaction are on the same straight line, and the direction is opposite to force.

(4) Gravitation: The force between two particles is directly proportional to the product of mass and inversely proportional to the square of distance.

f=gmm/r^2,g=6.67259* 10^(- 1 1)m^3/(kg*s^2)

Momentum theorem: I=∫Fdt=p2-p 1 (the impulse of external force is equal to the change of momentum).

Conservation of momentum: when the external force is zero, the momentum of the system remains unchanged.

Kinetic energy theorem: W=∫Fds=Ek2-Ek 1 (external work equals kinetic energy change).

Conservation of mechanical energy: Ek 1+Ep 1=Ek2+Ep2 when only gravity does work.

(Note: The core of Newtonian mechanics is Niuer: F=ma, which is the bridge between kinematics and dynamics. Our purpose is to know the law of motion of an object, that is, to solve the equation of motion r=r(t). If we know the force, we can get one according to Niu Er, and then we can find it according to the basic formula of kinematics. Similarly, if we know the equation of motion r=r(t), we can find A according to the basic formula of kinematics, and then we can know the force of the object from Niu Er. )

Two:

Special relativity mechanics: (note: γ = 1/sqr (1-u 2/c 2), β = u/c, and u is the velocity of inertial system. )

(1) Basic principle: (1) Principle of relativity: All inertial systems are equivalent.

(2) The principle of invariability of the speed of light: the speed of light in vacuum is a constant that has nothing to do with the inertial system.

(Give the formula first and then give the proof)

(2) Lorentz coordinate transformation:

X=γ(x-ut)

Y=y

Z=z

T=γ(t-ux/c^2)

(3) Speed conversion:

v(x)=(v(x)-u)/( 1-v(x)u/c^2)

V(y)=v(y)/(γ( 1-v(x)u/c^2))

V(z)=v(z)/(γ( 1-v(x)u/c^2))

(d) Proportional effect: △L=△l/γ or dL=dl/γ.

(5) Clock slowness effect: △t=γ△τ or dt=dτ/γ.

(6) Doppler effect of light: ν (a) = sqr ((1-β)/(1+β)) ν (b)

(The light source and detector move in a straight line. )

(7) Momentum expression: P=Mv=γmv, that is, m = γ m 。

(VIII) Basic equation of relativistic mechanics: F=dP/dt

(9) mass-energy equation: E = MC 2

(10) energy momentum relation: e 2 = (E0) 2+p 2c 2.

Note: There are two ways to prove it, one is in three-dimensional space and the other is in four-dimensional space-time. In fact, they are equivalent. )

Three:

Three-dimensional proof:

The axiom summarized by (1) experiment cannot be proved.

(2) Lorentz transformation:

Let the coordinate system (A system) where (x, y, z, t) is located be constant, while the coordinate system (B system) where (x, y, z, t) is located has a speed of U and is positive along the X axis. At the origin of series A, x = 0, and the origin coordinate of series B is X=-uT, that is, X+uT=0. Let x=k(X+uT), (1). Also, because the positions of all points in the inertial system are equivalent, k is a constant related to u (in general relativity, all points are no longer equivalent because of the curvature of space-time, so k is no longer a constant. ) Similarly, there is X=K(x-ut) at the origin of the B system. According to the principle of relativity, the two inertial systems are equivalent, and the two formulas should take the same form, that is, K = K, so there is X=k(x-ut), (2). For y, z, y, z, it has nothing to do with speed, and you can get y. That is t = kt+((kloc-0/-k 2)/(ku)) x, (5). (1) (2) (3) (4) (5) The principle of relativity is satisfied, and the determination of k requires the principle that the speed of light is constant. When the origins of two systems coincide, an optical signal is emitted from the coincidence point, and the two systems have x = CT and x = CT respectively. Substitute them into the formula (1)(2), CT = KT (C+U), CT = KT (C-U). Two formulas are multiplied to eliminate t and t, and K = 65438+.

X=γ(x-ut)

Y=y

Z=z

T=γ(t-ux/c^2)

(3) Speed conversion:

v(x)=dx/dt=γ(dx-ut)/(γ(dt-udx/c^2))

=(dx/dt-u)/( 1-(dx/dt)u/c^2)

=(v(x)-u)/( 1-v(x)u/c^2)

The expressions of v (y) and v (z) can be obtained in the same way.

(4) Scale-down effect:

In system B, there is a thin rod with a length of L parallel to the X axis, then it is obtained from X=γ(x-ut): △X=γ(△x-u△t), and △t=0 (measuring the coordinates at both ends at the same time), then △X=γ△x, that is, △ L = γ△ L, △ L.

(5) the clock slow effect:

According to the inverse transformation of coordinate transformation, t = γ (t+xu/c 2), so △ t = γ (△ t+△ xu/c 2) and △X=0 (the same location to be measured), so △ t = γ△ t 。

(Note: The length, mass and time interval of an object that is relatively stationary with the coordinate system are called intrinsic length, static mass and intrinsic time, which are objective quantities that do not change with the coordinate transformation. )

(6) Doppler effect of light: (Note: Doppler effect of sound is ν(a)=((u+v 1)/(u-v2))ν(b). )

A light source at the origin of system B emits light signals, while the origin of system A has a detector, and the two systems have two clocks respectively. When the origins of the two systems coincide, the calibration clock starts timing. The frequency of light source in system B is ν(b), the wave number is n, and the time measured by system B clock is △t(b). According to the clock slowness effect, the time measured by the clock in system A is △t(a)=γ△t(b), (1). The detector starts receiving at t65438+. (2) Relative motion does not affect the wave number of the optical signal, so the wave number emitted by the light source is the same as that received by the detector, that is, ν(b)△t(b)=ν(a)△t(N), (3). It can be obtained from the above three formulas: ν (a) = sqr ((6544)).

(7) Momentum expression: (Note: dt=γdτ, at this time γ = 1/sqr (1-V 2/C 2) Because the dynamic particle can choose itself as the reference system, β=v/c)

Under galilean transformation, Niu Er kept the same situation, that is, Niu Er was established in any inertial system, but under Lorentz transformation, the original concise form became messy, so Newton's law needs to be revised, and the requirement is to keep the original concise form under coordinate transformation.

In Newtonian mechanics, the forms of V = DR/DT and R are invariant under coordinate transformation ((x, y, z) in the old coordinate system and (x, y, z) in the new coordinate system). As long as the denominator is changed into an invariant (of course, it belongs to dτ when it is not fixed), the concept of speed can be corrected. Let V=dr/dτ=γdr/dt=γv be the relativistic velocity. Newton's momentum is p=mv, and replacing v with v can correct the momentum, that is, p=mV=γmv. Define M=γm (relativistic mass) and then p=Mv. This is the basic quantity of relativistic mechanics: relativistic momentum. (Note: We generally use Newton's velocity instead of relativistic velocity to participate in the calculation)

(8) Basic equations of relativistic mechanics:

From the expression of relativistic momentum, we can know that F=dp/dt, which is the definition of force. Although it is exactly the same as Niuer, it has different connotations. Mass is a variable in the theory of relativity.

(9) Mass-energy equation:

ek =∫Fdr =∫(DP/dt)* dr =∫DP * dr/dt =∫vdp = PV-∫pdv

=mv^2-∫mv/sqr( 1-v^2/c^2)dv=mv^2+mc^2*sqr( 1-v^2/c^2)-mc^2

=mv^2+mc^2( 1-v^2/c^2)-mc^2

=Mc^2-mc^2

That is, e = MC 2 = ek+MC 2.

(10) the relationship between energy and momentum;

E = MC 2, P = MV, γ =1/sqr (1-v 2/c 2), E0 = MC 2, we can get: E2 = (E0) 2+p 2c 2.

Four:

Four-dimensional proof:

The (1) axiom cannot be proved.

(2) Coordinate transformation: dl=cdt, that is, DX 2+DY 2+DZ 2+(ICDT) 2 = 0 holds in any inertial system. DS is defined as a four-dimensional interval, where dS 2 = dx 2+dy 2+dz 2+(icdt) 2, (1), which is constant for optical signal ds, but generally not constant for any two time points. Ds 2 > 0 is called class space division, and ds 2

The mathematical rotation transformation formula is: (keep the Y axis and Z axis fixed, and rotate the X axis and ict axis)

X=xcosφ+(ict)sinφ

icT=-xsinφ+(ict)cosφ

Y=y

Z=z

When X=0 and x=ut, then 0=utcosφ+ictsinφ.

So: tanφ=iu/c, then cos φ = γ and sin φ = iu γ/c are substituted into the above formula:

X=γ(x-ut)

Y=y

Z=z

T=γ(t-ux/c^2)

(3) (4) (5) (6) (8) (10) omitted.

(7) Momentum expression and four vectors: (Note: γ = 1/sqr (1-V 2/C 2), where dt=γdτ).

Let r=(x, y, z, ict) and replace dt in v=dr/dt with dτ, and V=dr/dτ is called four-dimensional velocity.

Then V=(γv, icγ)γv is a three-dimensional component, V is a three-dimensional velocity, and icγ is a four-dimensional component. (The same is true below)

Four-dimensional momentum: P=mV=(γmv, icγm)=(Mv, icM)

Four-dimensional force: f=dP/dτ=γdP/dt=(γF, γicdM/dt)(F is three-dimensional force).

Four-dimensional acceleration: ω =/dτ = (γ 4a, γ4va/c)

Then f=mdV/dτ=mω.

(9) Mass-energy equation:

fV=mωV=m(γ^5va+i^2γ^5va)=0

So four-dimensional force and four-dimensional velocity are always "vertical" (similar to Lorentz magnetic field force)

From Fv = 0: γ 2mfv+γ IC (DM/DT) (IC γ M) = 0 (F, V is a three-dimensional vector, Fv=dEk/dt (power expression)).

So dek/dt = c 2dm/dt means ∫ dek = c 2 ∫ DM, that is, ek = MC 2-MC 2.

So e = MC 2 = ek+MC 2.