My question concerns the rate of change for vectors. The rate of change is given by:

$\frac{d\vec{G}}{dt}_{space} = \frac{d\vec{G}}{dt}_{body}+\vec{\omega}\times\vec{G}$.

I'm then asked to derive the corresponding relation between:

$\frac{d\vec{G_1}\cdot \vec{G_2}}{dt}_{space}$ and $\frac{d\vec{G_1}\cdot \vec{G_2}}{dt}_{body}$

With $\vec{G_1}$ and $\vec{G_2}$ as two arbitrary vectors.

I'm not entirely sure how to go about this. My reasoning so far is that since $\vec{G_1}\cdot \vec{G_2}$ is a dot product, the derivative has to be 0. I've tried to make sense of this in a physical sense and I suppose that since the dot product describes the "effect" of the first vector on the second (for instance one force on another) this relation is the same between coordinate systems and doesn't really change in time.

However, I'm not sure if this is correct - especially since the rate of change between coordinate systems also involves a cross product and you can't take the cross product of a vector and a scalar?

  • $\begingroup$ Can you clarify what those subscripts are? you state a law earlier for $G$ without a subscript and then start inserting subscripts. Is it true for example that: $\frac{d\vec{G_i}}{dt}_{space} = \frac{d \vec{G_i}}{dt}_{body} + \vec{\omega_i}\times \vec{G_i}?$ for $i=1,2$? $\endgroup$ Oct 16, 2021 at 21:04
  • $\begingroup$ Secondly what does $\frac{d \vec{G_1} \cdot \vec{G_2}}{dt}$ mean? Do you mean to say $ \frac{d}{dt} \left[ \vec{G}_1 \cdot \vec{G}_2 \right] $ or do you mean: $\frac{d \vec{G_1}}{dt} \cdot \vec{G_2}$ ? $\endgroup$ Oct 16, 2021 at 21:06
  • $\begingroup$ G_1 and G_2 are two arbitrary vectors and the first law is a general law for arbitrary vectors. Hope this clarifies :-) $\endgroup$
    – basilikum
    Oct 16, 2021 at 21:07
  • $\begingroup$ yea that answer my first question, please also clarify the second question :) $\endgroup$ Oct 16, 2021 at 21:07
  • $\begingroup$ The problem has d(G_1*G_2)/dt as stated - so inside the derivative $\endgroup$
    – basilikum
    Oct 16, 2021 at 21:07

1 Answer 1


Ok so the crux of this problem is knowing the "dot product rule" for derivatives which states the following:

$$ \frac{d}{dt} \left[ a \cdot b\right] = \frac{da}{dt} \cdot b + a \cdot \frac{db}{dt} $$

This next part might be wrong since I have never seen the $_{space}$ and $_{body}$ notation but if I had to guess I think the problem goes as follows:

Using this we can then evaluate the first part of our answer:

$$ \frac{d}{dt} \left[ \vec{G_1} \cdot \vec{G_2} \right]_{space} = \frac{d\vec{G_1}}{dt}_{space} \cdot G_2 + \vec{G_1} \cdot \frac{d\vec{G_2}}{dt}_{space} $$

Of course this isn't all we can do, we have the general law that

$$ \frac{dG}{dt}_{space} = \frac{dG}{dt}_{body} + \omega\times\vec{G}$$

So we can substitute that in to yield:

$$ \left( \frac{dG_1}{dt}_{body} + \omega\times\vec{G_1} \right) \cdot G_2 + \vec{G_1} \cdot \left( \frac{dG_2}{dt}_{body} + \omega\times\vec{G_2} \right) $$

Now we can expand these products to yield

$$ \frac{dG_1}{dt}_{body} \cdot \vec{G_2} + \left( \omega \times \vec{G_1} \right) \cdot G_2 + \vec{G_1}\cdot \frac{dG_2}{dt}_{body} + \vec{G_1}\cdot \left(\omega \times \vec{G_2} \right)$$

You can now shuffle terms to yield:

$$ \frac{dG_1}{dt}_{body} \cdot \vec{G_2} + \vec{G_1}\cdot \frac{dG_2}{dt}_{body} + \left( \omega \times \vec{G_1} \right) \cdot G_2 + \vec{G_1}\cdot \left(\omega \times \vec{G_2} \right)$$

And now we can "factor" this via the product rule as

$$ \frac{d}{dt} \left[ \vec{G_1} \cdot \vec{G_2} \right]_{body} + \left( \omega \times \vec{G_1} \right) \cdot G_2 + \vec{G_1}\cdot \left(\omega \times \vec{G_2} \right) $$

So we have some kind of relationship now

$$ \frac{d}{dt} \left[ \vec{G_1} \cdot \vec{G_2} \right]_{space} = \frac{d}{dt} \left[ \vec{G_1} \cdot \vec{G_2} \right]_{body} + \left( \omega \times \vec{G_1} \right) \cdot G_2 + \vec{G_1}\cdot \left(\omega \times \vec{G_2} \right) $$

This might be further simplified if you know some rules involving $\omega, G$ terms.

  • $\begingroup$ Thank you so much, this makes a lot of sense! :-) $\endgroup$
    – basilikum
    Oct 16, 2021 at 21:29

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