I am looking at a derivation of the parallel axis theorem in a book. The photo is below

enter image description here

I'm not sure why but I just cannot figure out how they got from line 2 to line 3. I have thought through all of the properties I know (I have expanded and used the vector triple product, the cyclic permutations of scalar triple product etc) but I cannot seem to figure out that step.

Would very much appreciate if someone could enlighten me as to what property/rule was used here.

  • 2
    $\begingroup$ lagrange's identity $\endgroup$ – juan arroyo Jan 5 '17 at 19:25

$$ (a \times b) \cdot (c \times d) = \varepsilon_{kij} a_i b_j \varepsilon_{klm} c_l d_m = (\delta_{il} \delta_{jm}-\delta_{im}\delta_{kl}) a_i b_j c_l d_m = (a \cdot c)(b \cdot d) - (a \cdot d)(b \cdot c), $$ so in this case, $$ \lVert \hat{n} \times (r-a) \rVert^2 = \lVert \hat{n} \rVert^2 \lVert r-a \rVert^2 - (\hat{n} \cdot (r-a))^2. $$

  • $\begingroup$ Thank you for your reply! I thought I would be OK with the rest of the proof but I have also gotten stuck on the penultimate line. I have read the note below the proof several times but I don't see how it applies as you do not have an independent vector r in the integral that is the third from last line. How can the result of integrating the vector r with respect to the mass be used when inyegrating the magnitude of r or when integrating the dot product of a with r? $\endgroup$ – Meep Jan 5 '17 at 20:16
  • $\begingroup$ If it's the same as this edition of the solution manual to Riley, Hobson and Bence, the definition of $I_0$ becomes $ \int (r^2-(n \cdot r)^2) \, dm $, which deals with the terms with two $r$s, and linearity of the integral gives $\int a \cdot r \, dm = a \cdot \int r \, dm$, which is what happens to the linear terms. $\endgroup$ – Chappers Jan 5 '17 at 20:36
  • $\begingroup$ Thank you for your reply. I still have questions about this so I feel I should make a new post, but thank you for all of your insights! $\endgroup$ – Meep Jan 6 '17 at 12:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.