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I'm very new to physic. I do understand the basic concept of proportional.

it says if $$L \alpha S$$ then $$L= kS$$

I understand this part. but my professor says that

$$L \alpha S$$

and $$M \alpha S^3$$

then:

$$M \alpha S^3\alpha L^3$$

i really don't get the last line

I do understand that $$M= j(S^3)$$

but i don't understand the connection between the three. why this statement is valid?

$$M \alpha S^3\alpha L^3$$

Can someone show me the relationship between the three in basic elementary algebra?

user3270418
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2 Answers2

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Formally, if you have $L \propto S$ then there is a non-zero factor $k$ such that $L = kS$. Take the cubic power of both sides: $L^3 = k^3S^3$ and now divide by $k^3$ (you can do it since $k$ is non-zero): $$S^3 = \frac{1}{k^3}L^3.$$ This means $S^3 \propto L^3$. If further $M \propto S^3$ then you can write $M \propto S^3 \propto L^3$.

Gibbs
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  • can you please also show $M \propto S^3 \propto L^3$ in the equality form? – user3270418 Nov 05 '17 at 16:27
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    There is no equality of the three terms, there are just two different equalities, i.e. $M = jS^3$ and $S^3 = \frac{1}{k^3}L^3$. This is equivalent to saying that $M \propto S^3 \propto L^3$. – Gibbs Nov 05 '17 at 16:30
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$L \propto S \implies L = k_1S$ for some $k_1 \in \mathbb{R}_{>0}$
Cubing this equation we get, $$L^3 = (k_1^3) S^3$$ Rearranging give us the following, $S^3 = \frac{L^3}{k_1^3}$.
$M \propto S^3 \implies M = k_2S^3$ for some $k_2 \in \mathbb{R}_{>0}$.
Now substituting in the equality that we had earlier got $M = k_2\cdot\left(\frac{L^3}{k_1^3}\right) = \left( \frac{k_2}{k_1^3}\right)L^3$
$$\therefore M \propto L^3$$

Harsh Parekh
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