# Resulting Dot and Cross Products in Relativistic Speeds

Imagine the following problem:

Person 1 travels with velocity $$v$$, and person 2 has a velocity of $$u$$ according to the rest frame. They both travel in straight line, with an angle $$\theta$$ between their trajectories. Find the speed of Person 2 in Person 1's frame:

The solution of this problem is starts with realizing that one of the person's can be taken to be traveling on the x-axis of the rest frame, making
$$\mathbf{v} = \langle v,0 \rangle$$ and $$\mathbf{u} = \langle u\cos\theta,u\sin\theta \rangle$$ Now by the relativistic addition of velocities we realize that, the y-component of the Person's 2 in velocity in Person's 1 frame ($$u_y'$$) and the x-component ($$u_x'$$) are: $$u_x'=\frac{u\cos\theta-v}{1-\frac{uv\cos\theta}{c^2}}$$ $$u_y'=\frac{u\sin\theta}{\gamma \left( 1-\frac{uv\cos\theta}{c^2} \right)}$$ where: $$\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$

we can now find the speed ($$\| \mathbf{u'} \|$$) as: $$\| \mathbf{u'} \|=\sqrt{u_x'^2+u_y'^2}$$

Now if you go through the process of simplifying it you will find out that: $$\| \mathbf{u'} \| = \frac{1}{1-\left (\frac{\mathbf{v}\cdot\mathbf{u}}{c^2}\right )}\sqrt{\left (\mathbf{u}-\mathbf{v}\right )^2 - \frac{(\mathbf{v}\times\mathbf{u})^2}{c^2}}$$

I am curious why such equation is true and why do each of the products (dot and cross, especially the cross) appear in this equation?? Thank you in advance!

• You could eliminate the cross product and write it in terms of only the dot product (or vice versa). – G. Smith Dec 11 '20 at 22:00

If you calculated $$\boxed{||{\vec{u'}}||^2 = {u'}_x^2+ {u'}_y^2}$$ you would obtain $${u'}^2 = \frac{1}{1-\dfrac{uv\cos\theta}{c}} \left(u^2+v^2-2uv\cos\theta-\frac{u^2v^2\sin^2\theta}{c^2}\right)$$ Then one can notice that $$uv\cos\theta = \vec{u}\cdot\vec{v}$$ using the definition of the scalar product and $$uv\sin\theta = |\vec{u}\times\vec{v}|$$ from the definition of the cross product. Finally $$(\vec{u}-\vec{v})^2 = (\vec{u}-\vec{v}) \cdot (\vec{u}-\vec{v}) = \vec{u} \cdot \vec{u} + \vec{v} \cdot \vec{v} - 2 \vec{u} \cdot \vec{v} = u^2 + v^2 - 2 u v\cos\theta$$