The question is: Say a rod is travelling towards an observer O at a velocity v. The front of the rod is point A and the back is point B. First it wants me to calculate the apparent velocity of the rod to the observer.(adsbygoogle = window.adsbygoogle || []).push({});

So lets say at [tex]t_0=0[/tex], A is at O (ie x=0)

then when A is a distance d from O (x=-d), the time is [tex]t_2=\frac{-d}{v}[/tex]

The time at which the observer sees A at a distance d is [tex]t_1=t_2+\frac{d}{c}=d\left(\frac{1}{c}-\frac{1}{v}\right)[/tex]. (ie the time at which A is actually at d plus the time it takes the light to travel from x=-d to x=0)

therefore

When A gets to 0, the observer sees the object at the same time as it is actually there, [tex]t_0=0[/tex]

therefore the difference in the time between when he sees it at x=d and x=0 is[tex]{\Delta}t=t_0-t_1=0-d\left(\frac{1}{c}-\frac{1}{v}\right)=d\left(\frac{1}{v}-\frac{1}{c}\right)[/tex]

so the apparent velocity [tex]v_{app}[/tex] is then

[tex]v_{app}=\frac{{\Delta}d}{{\Delta}t}=\frac{0-(-d)}{d\left(\frac{1}{v}-\frac{1}{c}\right)}=v\left(\frac{c}{c-v}\right)[/tex]

What i'm confused about is whether I should this have a [tex]\gamma[/tex] term in it to account for special relativity?

The reason I think this is, is that if the rods system is O', then [tex]t_2[/tex] and consequenly [tex]t_1[/tex] should be instead [tex]t_2'[/tex] and [tex]t_1'[/tex] respectively, which would then make [tex]{\Delta}t[/tex] instead be [tex]{\Delta}t'[/tex]

so that [tex]{\Delta}t'=\frac{{\Delta}t}{\gamma}[/tex] which would then multiply my final solution by a factor of [tex]\gamma[/tex]. (ie. [tex]v_{app}=v\gamma\left(\frac{c}{c-v}\right)[/tex])

which method is correct...if either

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# Homework Help: Special Relativity Question

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