Coordinate (x, y, z): m Force f (f): n
Time t (t): s mass m(M): kg
Displacement r: m momentum p: kg*m/s
Speed v (u): m/s energy E: J
Acceleration A: m/sec 2 Impulse: N * sec
Length l (l): m kinetic energy ek: j
Distance s (s): m potential energy EP: j
Angular velocity ω: radians/second torque: N*m
Angular acceleration: rad/s 2 α power p: w
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: Without force, all objects are always in a state of static or uniform linear motion.
(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: Two forces acting on the same object are balanced if they act on the same straight line in opposite directions.
(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. )
Special relativistic mechanics
(Note: "γ" is a relativistic factor, 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))
4. Scaling 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
8. Basic equation of relativistic mechanics: F=dP/dt.
9. mass-energy equation: E = MC 2
10. Relationship between energy and momentum: 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-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.
Ke Ling
x=k(X+uT) ( 1)。
Moreover, because the positions of each point in the inertial system are equivalent, K is a constant related to U (in general relativity, due to the curvature of space-time, each point is no longer equivalent, so K is no longer a constant. ) Similarly, there is X=K(x-ut) at the origin of system B. 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, but the speeds are opposite.
So there is
X=k(x-ut) (2)。
For y, z, y, z, regardless of speed, you can get.
Y=y (3)。
Z=z (4)。
Substituting (2) into (1) gives: x = k 2 (x-ut)+Kut, i.e.
T=kt+(( 1-k^2)/(ku))x (5)。
(1)(2)(3)(4)(5) The principle of relativity is satisfied, and the principle of invariability of the speed of light is needed to determine k. When the origins of the two systems coincide, an optical signal is emitted from the coincidence point, and there are x=ct and X=cT for the two systems respectively.
Substitute into the formula (1)(2): ct=kT(c+u), cT=kt(c-u). Multiply two formulas to eliminate t and t:
K = 1/sqr ( 1-u 2/c 2) = γ。 Substituting γ into the coordinate transformation of (2) and (5):
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. Scaling 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. 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 slow effect of the clock, the time delta measured by the clock in system A is
△t(a)=γ△t(b) ( 1)。
The detector starts receiving at t 1+x/c and ends at t2+(x+v△t(a))/c, and then
△t(N)=( 1+β)△t(a) (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)。
Can be obtained from the above three formulas:
ν(a)= sqr(( 1-β)/( 1+β))ν(b)。
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)
Newton's second law remains unchanged under galilean transformation, that is, Newton's second law holds in any inertial system, but under Lorentz transformation, the original concise form becomes 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;
According to the expression of relativistic momentum, the definition of force F=dp/dt is exactly the same as Newton's second law, but the connotation is different. 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. Energy momentum relation:
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-dimensional proof
The axiom 1. cannot be proved.
2. Coordinate transformation: dl=cdt, that is, DX 2+DY 2+DZ 2+(ICDT) 2 = 0 holds in any inertial system. Define dS as a four-dimensional interval,
ds^2=dx^2+dy^2+dz^2+(icdt)^2( 1)。
Then the optical signal dS is always equal to 0, and the dS at any two time and space points is generally not 0. ds^2>； 0 is called class space partition, 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. Omit.
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.
8. Omit.
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.