- #1

therealDR.DOG

- 4

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**Dimensional analysis...**

I am having a real hard time wrapping my head around dimensional analysis. Can someone please explain it in english? Thanks.

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- Thread starter therealDR.DOG
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- #1

therealDR.DOG

- 4

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I am having a real hard time wrapping my head around dimensional analysis. Can someone please explain it in english? Thanks.

- #2

Dickfore

- 2,988

- 5

Every physical quantity

- time T
- length L
- mass M
- electric current I
- thermodynamic temperature [itex]\mathrm{\Theta}[/itex]
- amount of substance N
- luminous intensity J

Luminous intensity is a

Amount of substance is directly proportional to the number of elementary entities in the ensemble. The unit is called a mole. The actual number of particles present in a mole of substance is considered a physical constant, although I would think that it makes more sense to call it a conversion factor. It is the

[tex]

N_A = 6.022 \times 10^{23} \, \mathrm{mol}^{-1}

[/tex]

Similarly, due to developments in Statistical Physics, it had been shown that temperature is directly proportional to some energy ("thermal energy"). The unit of thermodynamic temperature is

[tex]

k_B = 1.3806 \times 10^{-23} \, \frac{\mathrm{J}}{\mathrm{K}}

[/tex]

The above two constants combine to give another famous constant - the

[tex]

R = k_B \, N_A = 8.314 \, \frac{\mathrm{J}}{\mathrm{K} \cdot \mathrm{mol}}

[/tex]

The unit of electric current is the

[tex]

\mu_0 = 4\pi \times 10^{-7} \, \frac{\mathrm{N}}{\mathrm{A}^2}

[/tex]

Thus, in principle, all quantities have dimension only w.r.t. to the three mechanical base units:

[tex]

[f] = \mathrm{T}^{\tau} \, \mathrm{L}^{\lambda} \, \mathrm{M}^{\mu}

[/tex]

The exponents (the

Some rules for dimensions are:

- A dimensionless quantity has
**zero**dimension w.r.t. each base physical quantity [itex][f] = \mathrm{T}^0 \, \mathrm{L}^0 \, \mathrm{M}^0 = 1[/itex]. - Both sides of an equality have the same dimension, i.e. [itex]f = g \Rightarrow [f] = [g][/itex]
- You may only add (or subtract) quantities with the same dimension, i.e. [itex]h = f \pm g \Leftrightarrow [f] = [g] = [h][/itex].
- The dimension of a product (quotient) of two physical quantites is he product of their respective dimensions, i.e. [itex][f \cdot g] = [f] \cdot [g][/itex] (the exponents for each base quantity add (subtract).
- A mathematical function (exp, log, sin, cos, etc.) accepts a dimensionless argument and returns a dimensionless result, i.e. [itex]y = f(x), \Rightarrow [x] = [y] = 1[/itex].

Last edited:

- #3

Jakeus314

- 47

- 0

- #4

Dickfore

- 2,988

- 5

Example:

Let us look at the following formula:

[tex]

v = v_0 + a \, t

[/tex]

according to the summation rule, we know that we must have:

[tex]

[v] = [v_0] = [a \, t]

[/tex]

Further, according to the product rule, we also have:

[tex]

[a \, t] = [a] \cdot [t]

[/tex]

Thus, if we want to express the dimension of

[tex]

[a] = \frac{[v]}{[t]}

[/tex]

- #5

therealDR.DOG

- 4

- 0

its just weird because for kinematics I am having a way easier time then the majority of the homework in chapter one which involves a lot more than just changing units.

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