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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!

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  • $\begingroup$ You could eliminate the cross product and write it in terms of only the dot product (or vice versa). $\endgroup$ – G. Smith Dec 11 '20 at 22:00
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Dot and cross products appear in the final answer because (naively) dot products is a tool to tell you how much of the two vectors are in the same direction (parallel) and cross product is a tool to tell you how much of the two vectors are perpendicular to each other.

Because the parallel and perpendicular components of velocity contract diffently, you should expect to see the dot and the cross products.

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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 $$ (\vec{u}-\vec{v})^2 = (\vec{u}-\vec{v}) \cdot (\vec{u}-\vec{v}) = \vec{u} \cdot \vec{u} + \vec{v} \cdot \vec{v} - 2 \vec{u} \cdot \vec{v} = u^2 + v^2 - 2 u v\cos\theta $$

If You wonder, why there is scalar and cross product in the formula, notice that the scalar product is just a number and the cross product is squared, so it is just the squared length of product vector.

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