# Prove rectilinear motinon

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## Main Question or Discussion Point

I've been struggling with this problem from Rindler's "INtroduction to Special Relativity"...

given a particle that has :
$$\frac{dA}{d\tau} = {\alpha}^2 U$$ where A is four-acceleration and U is four velocity... using integration prove that this implies rectilinear motion.

If I integrate both sides and set the constant of integration equal to zero I get:

$$A = {\alpha}^2 x^i$$

I don't know where to go from here. I'd appreciate any help.

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robphy
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Can you proceed knowing that the dot-product of these four-vectors is $$U\cdot A =0$$?

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robphy said:
Can you proceed knowing that the dot-product of these four-vectors is $$U\cdot A =0$$?
Yes, definitely.

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learningphysics said:
I've been struggling with this problem from Rindler's "INtroduction to Special Relativity"...

given a particle that has :
$$\frac{dA}{d\tau} = {\alpha}^2 U$$ where A is four-acceleration and U is four velocity... using integration prove that this implies rectilinear motion.

If I integrate both sides and set the constant of integration equal to zero I get:

$$A = {\alpha}^2 x^i$$

I don't know where to go from here. I'd appreciate any help.
Ok... from here I can say:

$$A \cdot U = (\alpha^2 x^i) \cdot U$$

then
$$0 = \alpha^2 x^i \cdot \frac{dx^i}{d\tau}$$

$$0 = x^i \cdot \frac{dx^i}{d\tau}$$

$$0 = \frac{1}{2} \frac{d}{d\tau}(x^i \cdot x^i)$$

$$0 = \frac{d}{d\tau}(x^i \cdot x^i)$$

$$K = x^i \cdot x^i$$ where K is a constant

taking c=1, and doing the dot product and replacing K with -K:

$$x^2 + y^2 + z^2 - t^2 = K$$

Can the above equation be used to show that the object is moving in a straight line?

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robphy
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Rectilinear motion means that the particle travels along a straight line in space. That is, there is no "turning in space". On a spacetime diagram, this means that the motion must lie on a plane... the plane spanned by the timelike 4-velocity $$U^a$$ and the [necessarily spacelike*] 4-acceleration $$A^a$$. (*Why? Since $$1=U^aU_a$$, then $$0=\displaystyle\frac{d}{d\tau}(U^aU_a)=2U^a\frac{d}{d\tau}U_a=2U^aA_a$$)

So, to ensure that motion lies on this spacetime plane, we should require that
the bivector $$U^{[a}A^{b]}=\frac{1}{2}(U^{a}A^{b}-U^{b}A^{a})$$ (similar to a cross product) is constant during the motion... that is, $$\displaystyle\frac{d}{d\tau}(U^{[a}A^{b]})=0$$.

Let's calculate
\begin{align*} \displaystyle\frac{d}{d\tau}(U^{[a}A^{b]}) &=\displaystyle U^{[a}\left(\frac{d}{d\tau}A^{b]}\right)+ \left(\frac{d}{d\tau}U^{[a}\right)A^{b]}\\ &=\displaystyle U^{[a}\left(\alpha^2 U^{b]}\right)+ \left(A^{[a}\right)A^{b]}\\ &\stackrel{\surd}{=} \alpha^2 0 + 0 \end{align*}

Check for errors.

Oh... one more thing... as a variant of the above calculation
Let's calculate
\begin{align*} \displaystyle\frac{d}{d\tau}(U^{a}A_{a}) &=\displaystyle U^{a}\left(\frac{d}{d\tau}A_{a}\right)+ \left(\frac{d}{d\tau}U^{a}\right)A_{a}\\ \displaystyle\frac{d}{d\tau}(0) &=\displaystyle U^{a}\left(\alpha^2 U_{a}\right)+ \left(A^{a}\right)A_{a}\\ 0&\stackrel{\surd}{=} \alpha^2 (1) + A^aA_a\\ -\alpha^2 &=A^aA_a \end{align*}
which says (again) that $$A^a$$ is spacelike (assuming $$\alpha^2>0$$), but now we learn that $$\alpha$$ is the magnitude of this spacelike $$A^a$$.

I just took a peek at my copy of Rindler, which suggests that $$\alpha$$ must actually be constant, which was not assumed anywhere above. But now, we can prove this starting from the given
\begin{align*} \frac{dA^a}{d\tau} &= {\alpha}^2 U^a\\ A_a \frac{dA^a}{d\tau} &= A_a{\alpha}^2 U^a\\ \frac{1}{2}\frac{d }{d\tau} (A_aA^a) &= {\alpha}^2 (A_aU^a)\\ \frac{d}{d\tau}(A_aA^a) &= 0\\ A_aA^a &\stackrel{\surd}{=} \mbox{constant}\\ \end{align*}

Ok, now it's time to sleep. :zzz:

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Hi robphy. Thanks so much for your help!

In that problem Rindler asks to show that given that given:

$$\frac{dA}{d\tau} = \alpha^2 U$$ ($$\alpha$$ not given as constant), show that the proper acceleration is constant and equal to $$\alpha$$.

I had no problem with those two parts. I believe I did exactly the same thing you did in the second and third parts of your posts.

Right afterwards Rindler says finally show by integration that the equation implies rectilinear motion. So it appears that showing that the magnitude of proper acceleration is constant is not enough to show rectilinear motion (at least in the context of the problem).

I will study the first part of your post which I really don't understand at the moment. I don't know about bivectors, so I can't verify. I really appreciate the time and effort you put into your post and helping me out. Thanks!

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For anyone that's interested here's a different way I came up with to show rectilinear motion. Rectilinear motion means there is some inertial frame where the particle moves in a straight line. This amounts to showing that velocity // acceleration at all times in that frame.

$$A = {\alpha}^2 x^i$$

From here, equating the space components of A and x^i we can say that:

s(t) = k1(t) u(t) + k2(t) a(t) 

s - displacement vector
u - velocity vector
a - acceleration vector

We can prove that if u(t1) // a(t1) at some time t1 then u(t) // a(t) for all t>t1

Assume u(t_k) // a(t_k) 

u(t_k + dt) = u(t_k) + dt a(t_k) 

by  and , u(t_k + dt) // u(t_k) 

by  and , s(t_k) // u(t_k) 

s(t_k + dt) = s(t_k) + dt u(t_k) 

by  and , s(t_k+dt) // s(t_k) 

by , and , s(t_k+dt) // u(t_k+dt) 

and by  and , u(t_k + dt) // a(t_k + dt)

so I believe this pseudo inductive argument shows u(t)//a(t) for t>t1, and we can do the same thing for t<t1.

Finally, we need to see if there is a t1 when u(t1)//a(t1). We can satisfy this condition by picking an inertial frame where u=0 at some point in time. We can always pick such a frame. In any inertial frame where u=0 at some point in time, u//a at all times and so the object moves in a straight line.