Twiglet wrote:Farsight and Mistermack:
Please answer the following basic question using special relativity.
Two objects, each one metre long, are moving away from each other with a speed of 2.7x10^8m/s
How long does object B appear as observed from object A?
That's .9c, so using applying the Lorentz factor √(1-v²/c²) and natural units where c=1:
length = √(1-.9²/1²)
length = √(1-.81)
length = √(.19)
length = .435 metres
Twiglet wrote:Next, assume Object A is at rest, and object B is moving away from it at 2.7x10^8m/s
Object B carries a clock and waits for 1 second to elapse.
How long has elapsed on Object As clock.
Assume c=3x10^8m/s
Please show working.
Thankyou.
I presume you're asking what B would observe if he could see A's clock, and that you aware that the situation is symmetrical. The working is as above, but we're dealing with time
dilation rather than length
contraction, so take a reciprocal. The answer is 1/.435 seconds or:
2.3 seconds.
Note that arithmetical exercises doesn't get to the bottom of
why we observe length contractiobn and time dilation, or indeed why c is the limit. For this you have to understand the underlying Pythagoras' Theorem and E=mc². And to understand the latter you have to understand the terms. You have to understand E, m, and c, which means you have to understand energy mass and time. Only then do you fully understand special relativity.
Edit:
Twiglet wrote:The distinction is basic. Relativity makes assumptions.
Only then do you get past the assumptions and get to the bottom of it. Then you appreciate that mistermack, in a roundabout way, is right.