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Question: The energy of a one-dimensional system is given by $E = \int dx \left[ \left( \frac{d\phi(x)}{dx} \r...

The energy of a one-dimensional system is given by E=∫dx[(dϕ(x)dx)2+ceaϕ(x)+bϕ4(x)]E = \int dx \left[ \left( \frac{d\phi(x)}{dx} \right)^2 + ce^{a\phi(x)} + b\phi^4(x) \right] where a, b and c are some non-zero constants. The dimension of the quantity b/a is given by

A

[ML−1T−2][ML^{-1}T^{-2}]

B

[M−1L−1T2][M^{-1}L^{-1}T^2]

C

[M−1/2L−7/2T][M^{-1/2}L^{-7/2}T]

D

[M−2L−4T4][M^{-2}L^{-4}T^4]

Answer

(C) [M−1/2L−7/2T][M^{-1/2}L^{-7/2}T]

Explanation

Solution

The energy of a one-dimensional system is given by: E=∫dx[(dϕ(x)dx)2+ceaϕ(x)+bϕ4(x)]E = \int dx \left[ \left( \frac{d\phi(x)}{dx} \right)^2 + ce^{a\phi(x)} + b\phi^4(x) \right]

Let's determine the dimensions of each part of the equation. The dimension of energy [E][E] is [ML2T−2][M L^2 T^{-2}]. The dimension of dxdx is [L][L].

For the integral ∫dx[… ]\int dx \left[ \dots \right] to have the dimension of energy, the term inside the square bracket must have the dimension of energy per unit length. So, [(dϕ(x)dx)2+ceaϕ(x)+bϕ4(x)]=[E][L]=[ML2T−2][L]=[MLT−2]\left[ \left( \frac{d\phi(x)}{dx} \right)^2 + ce^{a\phi(x)} + b\phi^4(x) \right] = \frac{[E]}{[L]} = \frac{[M L^2 T^{-2}]}{[L]} = [M L T^{-2}].

According to the principle of dimensional homogeneity, each term inside the square bracket must have the same dimension, which is [MLT−2][M L T^{-2}].

Let [ϕ][\phi] represent the dimension of ϕ(x)\phi(x).

  1. Dimension of the first term: [(dϕ(x)dx)2]=[MLT−2]\left[ \left( \frac{d\phi(x)}{dx} \right)^2 \right] = [M L T^{-2}]

    We know that [dϕ(x)dx]=[ϕ][L]\left[ \frac{d\phi(x)}{dx} \right] = \frac{[\phi]}{[L]}.

    So, ([ϕ][L])2=[MLT−2]\left( \frac{[\phi]}{[L]} \right)^2 = [M L T^{-2}]

    [ϕ]2[L2]=[MLT−2]\frac{[\phi]^2}{[L^2]} = [M L T^{-2}]

    [ϕ]2=[MLT−2][L2]=[ML3T−2][\phi]^2 = [M L T^{-2}][L^2] = [M L^3 T^{-2}]

    [ϕ]=[M1/2L3/2T−1][\phi] = [M^{1/2} L^{3/2} T^{-1}]

  2. Dimension of the second term: [ceaϕ(x)]=[MLT−2]\left[ ce^{a\phi(x)} \right] = [M L T^{-2}]

    For an exponential function eXe^X, the exponent XX must be dimensionless. Therefore, [aϕ(x)]=[1][a\phi(x)] = [1] (dimensionless).

    [a][ϕ]=[1][a][\phi] = [1]

    [a]=[1][ϕ]=[1][M1/2L3/2T−1]=[M−1/2L−3/2T][a] = \frac{[1]}{[\phi]} = \frac{[1]}{[M^{1/2} L^{3/2} T^{-1}]} = [M^{-1/2} L^{-3/2} T]

    Since eaϕ(x)e^{a\phi(x)} is dimensionless, the dimension of cc must be [MLT−2][M L T^{-2}].

  3. Dimension of the third term: [bϕ4(x)]=[MLT−2]\left[ b\phi^4(x) \right] = [M L T^{-2}]

    [b][ϕ]4=[MLT−2][b][\phi]^4 = [M L T^{-2}]

    [b]=[MLT−2][ϕ]4[b] = \frac{[M L T^{-2}]}{[\phi]^4}

    Substitute the dimension of [ϕ][\phi]:

    [b]=[MLT−2]([M1/2L3/2T−1])4[b] = \frac{[M L T^{-2}]}{([M^{1/2} L^{3/2} T^{-1}])^4}

    [b]=[MLT−2][M(1/2)×4L(3/2)×4T−1×4][b] = \frac{[M L T^{-2}]}{[M^{(1/2) \times 4} L^{(3/2) \times 4} T^{-1 \times 4}]}

    [b]=[MLT−2][M2L6T−4][b] = \frac{[M L T^{-2}]}{[M^2 L^6 T^{-4}]}

    [b]=[M1−2L1−6T−2−(−4)][b] = [M^{1-2} L^{1-6} T^{-2 - (-4)}]

    [b]=[M−1L−5T2][b] = [M^{-1} L^{-5} T^2]

Now, we need to find the dimension of the quantity b/ab/a:

[ba]=[b][a]\left[ \frac{b}{a} \right] = \frac{[b]}{[a]}

[ba]=[M−1L−5T2][M−1/2L−3/2T]\left[ \frac{b}{a} \right] = \frac{[M^{-1} L^{-5} T^2]}{[M^{-1/2} L^{-3/2} T]}

[ba]=[M−1−(−1/2)L−5−(−3/2)T2−1]\left[ \frac{b}{a} \right] = [M^{-1 - (-1/2)} L^{-5 - (-3/2)} T^{2 - 1}]

[ba]=[M−1+1/2L−5+3/2T1]\left[ \frac{b}{a} \right] = [M^{-1 + 1/2} L^{-5 + 3/2} T^1]

[ba]=[M−1/2L−10/2+3/2T]\left[ \frac{b}{a} \right] = [M^{-1/2} L^{-10/2 + 3/2} T]

[ba]=[M−1/2L−7/2T]\left[ \frac{b}{a} \right] = [M^{-1/2} L^{-7/2} T]

Comparing this result with the given options, the calculated dimension matches option (C).