# Differential form of the velocity equation in a non-standard configuration

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## Main Question or Discussion Point

I'm reading a text on special relativity (Core Principles of Special and General Relativity), in which we start with the equation for composition of velocities in non-standard configuration. Frame ##S'## velocity w.r.t. ##S## is ##\vec v##, and the velocity of some particle in ##S'## is ##\vec u##. Then the particle's velocity in ##S'## is:

$$\vec u'=\frac{\vec u-\vec v}{1-\vec u.\vec v\ /\ c^2}+\frac{\gamma}{c^2(1+\gamma)}\frac{\vec v\times(\vec v\times\vec u)}{(1-\vec u.\vec v\ /\ c^2)}$$

where ##\gamma=\frac{1}{\sqrt{1-v^2/c^2}}## and ##\vec v## and ##c## are constants. Then the text states that "differentiating" the above equation gives us
$$d\vec u'=\frac{1}{\gamma(1-\vec u.\vec v\ /\ c^2)^2}\bigg[d\vec u-\frac{\gamma}{c^2(1+\gamma)}(\vec v.d\vec u)\ \vec v+\frac{1}{c^2}\vec v\times\vec u\times d\vec u\bigg]$$

I'm struggling with proving this. Just to reduce some of the notational headache, if we denote ##f(\vec u)=\frac{1}{1-\vec u.\vec v\ /\ c^2}##, then
$$df(\vec u)=\frac{f(\vec u)^2\vec v.d\vec u}{c^2}$$

Also let ##K\equiv \frac{\gamma}{c^2(1+\gamma)}##. Then the original equation is:
$$\vec u'=f(\vec u)(\vec u-\vec v)+Kf(\vec u)\ (\vec v\times (\vec v\times\vec u))$$

Differentiating (writing ##f## without its argument for convenience),
$$d\vec u'=(\vec u-\vec v)\ df+fd\vec u + K\ df\ (\vec v\times (\vec v\times\vec u)) + Kf\vec v\ (\vec v.d\vec u)-Kfv^2d\vec u$$
$$=\frac{f^2(\vec u-\vec v)\vec v.d\vec u}{c^2}+fd\vec u + K\ \frac{f^2\vec v.d\vec u}{c^2}\ (\vec v\times (\vec v\times\vec u)) + Kf\vec v\ (\vec v.d\vec u)-Kfv^2d\vec u$$
$$=f^2\bigg[\frac{(\vec u-\vec v)\vec v.d\vec u}{c^2}+\frac{d\vec u}{f} + K\ \frac{\vec v.d\vec u}{c^2}\ (\vec v\times (\vec v\times\vec u)) + \frac{K}{f}\vec v\ (\vec v.d\vec u)-\frac{K}{f}v^2d\vec u\bigg]$$

Beyond this, I'm really not able to get to the final result despite trying a bunch of times. Not sure if I'm overcomplicating things or missing some magical identity that simplifies everything. Would appreciate any help. Thanks!

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anuttarasammyak
Gold Member
I'm reading a text on special relativity (Core Principles of Special and General Relativity), in which we start with the equation for composition of velocities in non-standard configuration. Frame S′ velocity w.r.t. S is v→, and the velocity of some particle in S′ is u→. Then the particle's velocity in S′ is:
I read your "and the velocity of some particle in S′ is u→. " as "and the velocity of some particle in S(NO DASH) is u→. ". Just a confirmation.

Shirish
I read your "and the velocity of some particle in S′ is u→. " as "and the velocity of some particle in S(NO DASH) is u→. ". Just a confirmation.
Yes that's correct, there shouldn't be a prime. I'll edit the question.

anuttarasammyak
PeroK
Homework Helper
Gold Member
I'm reading a text on special relativity (Core Principles of Special and General Relativity), in which we start with the equation for composition of velocities in non-standard configuration. Frame ##S'## velocity w.r.t. ##S## is ##\vec v##, and the velocity of some particle in ##S'## is ##\vec u##. Then the particle's velocity in ##S'## is:

$$\vec u'=\frac{\vec u-\vec v}{1-\vec u.\vec v\ /\ c^2}+\frac{\gamma}{c^2(1+\gamma)}\frac{\vec v\times(\vec v\times\vec u)}{(1-\vec u.\vec v\ /\ c^2)}$$

where ##\gamma=\frac{1}{\sqrt{1-v^2/c^2}}## and ##\vec v## and ##c## are constants. Then the text states that "differentiating" the above equation gives us
$$d\vec u'=\frac{1}{\gamma(1-\vec u.\vec v\ /\ c^2)^2}\bigg[d\vec u-\frac{\gamma}{c^2(1+\gamma)}(\vec v.d\vec u)\ \vec v+\frac{1}{c^2}\vec v\times\vec u\times d\vec u\bigg]$$

I'm struggling with proving this. Just to reduce some of the notational headache, if we denote ##f(\vec u)=\frac{1}{1-\vec u.\vec v\ /\ c^2}##, then
$$df(\vec u)=\frac{f(\vec u)^2\vec v.d\vec u}{c^2}$$

Also let ##K\equiv \frac{\gamma}{c^2(1+\gamma)}##. Then the original equation is:
$$\vec u'=f(\vec u)(\vec u-\vec v)+Kf(\vec u)\ (\vec v\times (\vec v\times\vec u))$$

Differentiating (writing ##f## without its argument for convenience),
$$d\vec u'=(\vec u-\vec v)\ df+fd\vec u + K\ df\ (\vec v\times (\vec v\times\vec u)) + Kf\vec v\ (\vec v.d\vec u)-Kfv^2d\vec u$$
$$=\frac{f^2(\vec u-\vec v)\vec v.d\vec u}{c^2}+fd\vec u + K\ \frac{f^2\vec v.d\vec u}{c^2}\ (\vec v\times (\vec v\times\vec u)) + Kf\vec v\ (\vec v.d\vec u)-Kfv^2d\vec u$$
$$=f^2\bigg[\frac{(\vec u-\vec v)\vec v.d\vec u}{c^2}+\frac{d\vec u}{f} + K\ \frac{\vec v.d\vec u}{c^2}\ (\vec v\times (\vec v\times\vec u)) + \frac{K}{f}\vec v\ (\vec v.d\vec u)-\frac{K}{f}v^2d\vec u\bigg]$$

Beyond this, I'm really not able to get to the final result despite trying a bunch of times. Not sure if I'm overcomplicating things or missing some magical identity that simplifies everything. Would appreciate any help. Thanks!
As that's a vector equation, you could try doing it for one component at a time. Start with ##du_x##. The other two should be the same.

anuttarasammyak
Gold Member
I am an old student of vector mathematics. Vector product
$$\mathbf{v}\times\mathbf{u}\times\mathbf{du}$$
in the second formula of OP seems ambiguous. Is it
$$(\mathbf{v}\times\mathbf{u})\times\mathbf{du},$$
$$\mathbf{v}\times(\mathbf{u}\times\mathbf{du})$$
or other?

I am an old student of vector mathematics. Vector product
$$\mathbf{v}\times\mathbf{u}\times\mathbf{du}$$
in the second formula of OP seems ambiguous. Is it
$$(\mathbf{v}\times\mathbf{u})\times\mathbf{du},$$
$$\mathbf{v}\times(\mathbf{u}\times\mathbf{du})$$
or other?
That's how it's written in the book, unfortunately. I'm thinking it should be ##(\mathbf{v}\times\mathbf{u})\times d\mathbf{u}##, since ##\mathbf{v}\times(\mathbf{u}\times d\mathbf{u})## is zero as ##\mathbf{u}## and ##d\mathbf{u}## are parallel.

anuttarasammyak
Gold Member
We can accelerate a moving body in any direction, so not always zero. It seems that formula in the textbook has something wrong. In the way of OP you may have a chance to get the right result.

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