Summary on unit 1 (update: 9/11/10)
Constants: c=3 ×108m/s, 1𝑢≈𝑚𝑝≈𝑚𝑛≈1.7×10−27kg, 𝐺=6.7×10−11𝑁𝑚2/𝑘𝑔2.
Sec 1.1-1.9. Coord-vector: r=< 𝑥,𝑦,𝑧 >=𝑟ˆ
r,𝑟=√𝑥2+𝑦2+𝑧2,ˆ
r=<𝑐𝑜𝑠𝜃
𝑥,𝑐𝑜𝑠𝜃
𝑦,𝑐𝑜𝑠𝜃
𝑧>.
For 2 D case, 𝑟=√𝑥2+𝑦2,ˆ
r=<𝑐𝑜𝑠𝜃,𝑠𝑖𝑛𝜃>=<𝑐𝑜𝑠𝜃
𝑥,𝑐𝑜𝑠𝜃
𝑦>.
Vectors (Sec. 1.5): Addition, substraction, magnitude of a vector, unit vector, ⋅⋅⋅.
Displacement: Δr=r𝑓−r𝑖. (Average velocity) = (Total distance traveled)/(travel time), or
v𝑎𝑣𝑔 =Δr/Δ𝑡=(v1Δ𝑡1+v2Δ𝑡2+⋅⋅⋅)/(Δ𝑡1+Δ𝑡2+⋅⋅⋅).
Position up date:r𝑓=r𝑖+v𝑎𝑣𝑔 Δ𝑡.
Instantaneous velocity: v=𝐿𝑖𝑚Δ𝑡→0(Δr/Δ𝑡) . Instantaneous acceleration: a=𝑑v/𝑑𝑡.
For constant acceleration , v𝑎𝑣𝑔 =(v𝑓+v𝑖
2). Otherwise, for sufficiently small Δ𝑡,usee.g. v𝑎𝑣𝑔 ∼v𝑓.
Momentum: p=𝛾𝑚v,𝛾=1/√1−𝛽2,𝛽=∣v∣/𝑐.
∙Nonrelativistic approximation(NR): 𝛾→1, p=𝑚v.
∙Identity: 𝛽=𝛽𝛾/√1+(𝛽𝛾)2,where𝛽𝛾 =𝑝/𝑐𝑚.𝛽=𝑣/𝑐 =(Δ𝑠/𝑐)/Δ𝑡.Δ𝑠=𝛽𝑐Δ𝑡.
The extended Newton’s law of motion: F=Δp/Δ𝑡.
∙If F=0,p=𝑝ˆ
p=𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡, which is Newton’s first law.
∙If F∕= 0, for NR case, it leads to F=Δp/Δ𝑡=𝑚a,wherea=Δv/Δ𝑡is acceleration. This is
Newton’s second law. For the R case, F=Δp/Δ𝑡=(Δ𝛾/Δ𝑡)𝑚v+𝛾𝑚a, with a=Δv/Δ𝑡.
Sec 2.1-2.8, also read 2.9.
Momentum principle:Δp=p𝑓−p𝑖=FΔ𝑡
Momentum Principle is given by a vector equation. The equation is valid for each Cartesian component.
A special case: 1d, NR and F=constant, or 𝑎=𝐹/𝑚 =𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡.
∙From momentum principle: Δp=𝐹𝑛𝑒𝑡Δ𝑡. With 𝑎=𝐹/𝑚,itleadsto𝑣𝑓=𝑣𝑖+𝑎𝑡.
∙For a constant acceleration case, 𝑣𝑎𝑣𝑔 =(𝑣𝑖+𝑣𝑓)/2. (Why?)
∙Position update: Δ𝑠=𝑣𝑎𝑣𝑔 Δ𝑡=(𝑣𝑖+𝑣𝑓)/2Δ𝑡. This leads to Δ𝑠=𝑣𝑖Δ𝑡+(1/2)𝑎Δ𝑡2.(Derive)
Sec 3.1-3.5. Four kinds of forces(or interactions): Gravitational, electromangetic, strong and weak.
Gravitational force: F=−𝐺𝑀𝑚
𝑟2ˆ
r. Near surface of earth (with radius R) 𝐹=𝑚𝑔 =𝑚(𝐺𝑀
(𝑅+ℎ)2)≈𝑚(𝐺𝑀
𝑅2).
Iterative procedure(3d):
∙Begin with the object’s momentum and position (p𝑖,r𝑖), and the force F(r𝑖)at𝑡=𝑡𝑖
∙IL, Iterative Loop: Take time step 𝑡𝑓=𝑡𝑖+Δ𝑡. Apply Momentum principle to update p𝑖to p𝑓.
∙Position update moves the object from r𝑖to r𝑓.
∙Set present (p𝑓,r𝑓), F(r𝑗), 𝑡𝑓to next step (p𝑖,r𝑖), F(r𝑖), 𝑡𝑖Go to IL.
Principle of reciprocity (Newton’s third law): Force on 2 due to 1 and force on 1 due to 2 satisfies the
relationship: F1𝑜𝑛 2=−F2𝑜𝑛 1.
Spring force(1d): Magnitude satisfies the relationship ∣𝐹∣=𝑘∣𝑠∣. With sign: 𝐹=−𝑘𝑠,𝑠=𝐿−𝐿0.The
stretched case has 𝑠>0, leading to attraction. The compressed case has 𝑠<0, leading to repulsion.
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