How do I "pull out" a vector in this operation?

Question

I have the expression:

$$\vec{a}-(\vec{v}\cdot \vec{a})\vec{v}$$

Is it possible to do this:

$$\vec{a}(1-\vec{v}\cdot\vec{v})$$

I doubt it because when I continue my problem I don't result in the right answer. I can provide the full context if needed.

The full context:

The relativistic expression for the momentum of a massive particle is

$$\vec{p}=\gamma m \vec{v}$$

and I need to derive an expression for the 3-force in terms of the 3-acceleration $\vec{a}$ and $\vec{\beta}=\frac{\vec{v}}{c}$, then use that expression to show that:

$$\gamma m \vec{a}=\vec{F}-\vec{\beta}(\vec{\beta}\cdot \vec{F})$$

So for the first part, I found the expression like this:

$$\vec{F}=\frac{d\vec{p}}{dt}=\gamma m\vec{a}+(-\frac{1}{2}(1-\frac{v^2}{c^2})^{-3/2}(\frac{-2\vec{v}}{c}\frac{\vec{a}}{c}))m\vec{v}$$

$$=\gamma m \vec{a}(1+\beta ^2 \gamma ^2)$$

For the second part, I just went step by step from the inside to the outside of $\vec{F}-\vec{\beta}(\vec{\beta}\cdot \vec{F})$, which resulted in:

$$\vec{F}-\vec{\beta}(\vec{\beta}\cdot \vec{F})=\gamma m \vec{a}(1+\beta^2\gamma^2)-\frac{\gamma m}{c}(1+\beta^2\gamma^2)(\vec{v}\cdot\vec{a})\vec{v}$$

It was at this point that I wanted to "pull out" the $\vec{a}$ to continue the problem.

Answer 1

Define the arrays

$$ v = \left(\begin{array}{c}v_1 \\ v_2 \\ \vdots \end{array}\right) ~~~\mbox{and}~~~ v\cdot = (v_1, v_2, \cdots) $$

note that $x\cdot y$ is the usual inner product. In this case

$$ a - (v\cdot a)a = a - v(v\cdot a) = a - vv\cdot a = (\Bbb{1} - vv\cdot)a $$

Answer 2

You can “pull out” the vector $\vec a$, but you’re going to end up with a matrix. The dot product $\vec x\cdot\vec y$ can be written as $\vec x^T\vec y$, so the expression that you’re starting with is $\vec a-(\vec v^T\vec a)\vec v$. The parenthesized quantity is a scalar, so taking a small liberty with scalar multiplication we can move it to the right and regroup: $$\vec a-(\vec v\vec v^T)\vec a=(I-\vec v\vec v^T)\vec a.$$ The product $\vec v\vec v^T$ is a matrix (the outer or tensor product of $\vec v$ with itself.)

If $\vec v$ is a unit vector, this expression is, of course, the orthogonal rejection of $\vec a$ from $\vec v$.

Answer 3

Your expression for $F$ is incorrect, it shouldn't be parallel to $a$ (that's one of the many weird things about SR). $$F=\frac{d}{dt} \gamma m v = m\left(\gamma \frac{dv}{dt}+v\frac{d\gamma }{dt} \right) = m\left(\gamma a + v\frac{v\cdot a}{c^2}\gamma^3 \right)$$ since $\frac{d}{dt} v^2 = 2v\cdot a$ (verify this by writing out $v= (v_1,v_2, v_3)$). Now, we want to show that $$F - \gamma m a = m\gamma^3 \frac{v\cdot a}{c^2}v = (\beta\cdot F) \beta$$ So $$\beta\cdot F = \frac{1}{c}v\cdot F = \frac{m}{c} \left(\gamma v\cdot a +v^2 \frac{v\cdot a}{c^2}\gamma^3\right) = v\cdot a \frac{m}{c}\gamma\left(1+\frac{v^2}{c^2}\gamma^2 \right)$$ $$ = v\cdot a\frac{m}{c}\gamma(\gamma^2)$$ and so $$(\beta\cdot F)\beta = \frac{v\cdot a}{c^2}m\gamma^3 v$$ as desired.