Special Theory of Relativity(discussion question)

[BatmanACC]

Homework Statement

If you were on a spaceship moving away from Earth at .8C, would you observe any change in your shoe size, mass, or your pulse? Would an observer from Earth note any change in these quantities?

Homework Equations

L = Lo[sqrroot(1-v2/c2)]

M = Mo/ sqrroot(1-v2/c2)

The Attempt at a Solution

Disregarding any logical flaws in the argument (that an observer can't view a pulse, he/she cannot see inside a spaceship unless it is transparent, etc)(although mass/pulse need only be recorded...). I'm not sure as to the answer. Now, shoe size as well as mass are different to both the observer and the captain of the spaceship. This is noted by the 2 given equations. So that is easy to explain.

Note: Although I say its easy to explain I may be wrong and if that is the case correct me :P. But it seems logical that both mass/length are relative to the observer and captain because M=moving mass which is not equal to Mo=rest mass and same for the length contraction. And the captain first observes his foot size and mass at rest (before spaceship takes off). Thus when he is moving the quantities should be different.

As for the pulse I don't have an equation to explain it...I'm going to take a stab in the dark and say there's a reason for that. That reason being that the captains pulse is relevant to his/her excitement level and not on any of the 4 dimensions. As such it should be the same for both the observer and captain.

Thanks for taking a look guys. Just wanted to make sure I got this right as I have a test first period tomorrow :).

Edit - Pulse is contingent on time. That was my mistake. Since the Observer experiences time dilation his view will be different from that of the captain. This correct>?

Edit - Assuming the spaceship is moving via the power of rockets and is simply not gliding in space than the Captain is in an inertial frame of reference. Thus, he experiences no time dilation and therefore notices change in nothing.

Sorry for the edits. Just more clearly thinking it out :P

 

[Chi Meson]

There is no observable change in any measuremnt of an object in your own frame of reference. If there was, it would violate the 2nd posulate, meaning there would be some method of determining an absolute frame of reference.