Michael 發問時間： 科學數學 · 7 年前

# 如何計算其答案和計算過程

In Question 7

the lecture Professor Murayama stated that the weak force acts only on a

billionth of a nanometer, i.e 10^−18 m. This is in contrast to the electromagnetic

force which is long ranged. Mathematically, the potential of such interactions is

given by

V(r)=αW*(e*(−r/a))/r⇐ weak potential V(r)=α EM*1/ r⇐ electromagnetic potential

For the weak potential, the parameter a controls the range of the force. This type of potential is called a Yukawa potential.

Taking a=2×10^−18 m and assuming α W =α EM , what are the ratios of the weak potential to the EM potential at r=2×10^−20 meters? Give your answer to two significant figures.

(continued from Question 7)

Taking a=2×10^−18 m and assuming α W =α EM , what are the ratios of the weak potential to the EM potential at r=2×10^−18 meters? Give your answer to two significant figures.

(continued from Question 7)

Taking a=2×10^−18 m and assuming α W =α EM , what are the ratios of the weak potential to the EM potential at r=2×10^−16 meters? Give your answer to two significant figures.

(continued from Question 7)

(b) Use dimensional analysis with the reduced Plank constant ℏ and the speed of light c to convert the length scale a to an energy.

Give the energy to two significant figures and in units of GeV (giga-electron volt, 1 GeV = 10^9 eV).

__________ GeV

continued from Question 7)

(c) Look up the mass of the W and Z bosons and the recently discovered Higg's boson. Give the difference between the energy scale you found in (Question10) and these masses. Round to the nearest whole number and take the absolute value if the answer is negative (e.g if the difference is −17 enter 17 ).

Difference from W mass = _________ GeV

(continued from Question 11)

Difference from Z boson mass = _________ GeV

(continued from Question 11)

Difference from Higg's boson mass = _________ GeV

### 1 個解答

• 7 年前
最佳解答

You are looking for P = sin(π/7)sin(2π/7)sin(3π/7)

Let s = sin(π/7), c = cos(π/7) so s + c = 1.

From cos(α + β) and sin(α + β) formulas come

cos(2π/7) = c - s = 2c - 1

sin(2π/7) = 2cs

sin(3π/7) = s(2c - 1) + c(2cs) = s(4c - 1), and

sin(4π/7) = 4cs(2c-1)

Since sin(π/7) = sin(4π/7), we have 4c - 1 = 4c(2c - 1) or

(1): 8c = 4c + 4c - 1

This means

However, it will take a bit of work to simplify the still messy expression of

P = 2cs(4c-1)

= s(8c - 2c) and now apply (1) to get

= s(4c + 2c - 1)

Now we take a slight detour:

(2): s(4c + 2c - 1)

= (1-c)(4c + 2c - 1)

= -4c⁴ -2c + 5c + 2c -1 and now apply (1)

= -4c + 3c + (5/2)c - 1 and apply (1) again to get

= c + c -

P = (s(4c + 2c - 1)) apply (2)

= (s(c + c/2 - 1/2))

= s(c + c - - )(c + c - ) apply (2) again

= ((c + c - ) - (1-c)) * (c + c - )

= (4c + c - 3) * (2c + c -1) / 16

= (8c⁴ + 6c - 9c - 4c + 3) / 16 apply (1)

= (10c - 5c - 5c + 3) / 16 apply (1) again

= (-5/4 + 3) / 16

= (7/4) / 16

Hence P = ⅛ * √7

Remark: In the other question referenced, we'd already seen that s(4c +2c -2)= *√7, so we could have avoided this last section of calculation by observing that:

P = s(c + c/2 - 1/2) = *s(4c + 2c - 2) = ⅛ * √7

We've done!You are looking for P = sin(π/7)sin(2π/7)sin(3π/7)

Let s = sin(π/7), c = cos(π/7) so s² + c² = 1.

From cos(α + β) and sin(α + β) formulas come

cos(2π/7) = c² - s² = 2c² - 1

sin(2π/7) = 2cs

sin(3π/7) = s(2c² - 1) + c(2cs) = s(4c² - 1), and

sin(4π/7) = 4cs(2c²-1)

Since sin(3π/7) = sin(4π/7), we have 4c² - 1 = 4c(2c² - 1) or

(1): 8c³ = 4c² + 4c - 1.This means any polynomial in c can be reduced to a quadratic or less.

However, it will take a bit of work to simplify the still messy expression of

P = 2cs³(4c²-1)= s³(8c³ - 2c) and now apply (1) to get

= s³(4c² + 2c - 1)

Now we take a slight detour:

(2

• 登入以對解答發表意見