Tuesday, October 29, 2019

special relativity - Is acceleration relative?


A while back in my Dynamics & Relativity lectures my lecturer mentioned that an object need not be accelerating relative to anything - he said it makes sense for an object to just be accelerating. Now, to me (and to my supervisor for this course), this sounds a little weird. An object's velocity is relative to the frame you're observing it in/from (right?), so where does this 'relativeness' go when we differentiate?


I am pretty sure that I'm just confused here or that I've somehow misheard/misunderstood the lecturer, so can someone please explain this to me.



Answer



I find the phrase "acceleration need not be relative anything" to be awkward, but I can see where it comes from.


For the moment restrict our consideration the Galilean relativity (just to keep the math simple). Consider two frames of reference one ($S$) in which the body is at rest and another ($S'$) in which it moves with velocity $\vec{v'_i} = \vec{u} = u \hat{z}$.


So we have the initial velocity of the body in frame $S$ as $v_i = 0$, and $v' = v + u \hat{z}$


Now assume that the the body accelerates from time $t$ at acceleration $\vec{a} = a \hat{Z}$ resulting in a velocity in frame $S$ of $\vec{v_f} = a t \hat{z}$.


Compute the final velocity in frame $S'$ as $v'_f = v_f + u \hat{z} = (u + a t)\hat{z}$, and from that the acceleration in the primed frame as $a' = a$.


So the acceleration is the same in all frames (you can check the cases for $a \not\parallel u$ yourself), and it is reasonable to say that accelerations are not relative anything.



All of this is a consequence of the simple form of the transformation between frames:


$$ \vec{x'} = \vec{x_0} + \vec{u} t $$ $$ t' = t $$


So what about Einsteinian relativity?


Here the transformation between frames is more complicated, and the math is much more complicated resulting in observers in different frames seeing different accelerations, but they will all agree on the acceleration as measured in the body's own frame. In my opinion "the acceleration need not be relative" risks causing unnecessary confusion on these points. The magnitude and direction measured will measure in depend on the frame of the observer, which is often what is meant when people say "it's relative".


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