# Speed of Light and Special Relativity

If light is travelling vertically upwards and I am travelling horizontally and perpendicular to light, the velocity of light in the vertical direction will get reduced. How does this happen? How does motion in x-direction cause a reduction in speed of light in y-direction ?

Also, in Lorentz Transformation, suppose two observers measure a rod, can one of the observers see that the other person who is moving w.r.t. him, measure the rod to be less than his own measurement ? In other words, can the metre sticks of a moving observer get bigger than the observer who views the moving observer and sees himself to be at rest ?

• What is your level of familiarity with these topics? Are you comfortable with applying lorentz transformations?
– user12029
Dec 26, 2013 at 6:53

You're asking so many questions that the only way to answer satisfactorily would be to completely rejustify special relativity. I suggest you take a look at a book like Special Relativity which does such a thing!

How does this happen?

If a photon is at position $(0,0,ct)$ at time $t$, that is, it has a spacetime event at time t of $e=(ct,0,0,ct)$, one has to apply the Lorentz Transform, which would yield $e'=(ct',x',y',z')=(\gamma c t,-\beta\gamma ct,0,ct)$. So $ct=ct'/\gamma$, and therefore:

$$e'=(c t',-\beta c t',0,ct'/\gamma)$$

Now, if we want to find the speed of the photon in this frame, that will be the magnitude of the spatial portion of the event $e'$, differentiated by $t'$ (and in this case, since everything is linear, that has the same effect as dividing by $t'$):

$$\|(-\beta c,0,c/\gamma)\|=\sqrt{c^2(\beta^2+(1-\beta^2))}=c$$

It moves at the speed of light, still. The reduction in the y-direction, in this frame of reference, is a consequence of the Lorentz transformation, which can be viewed as a consequence of the constancy of the speed of light. Of course you can notice that if the motion in the y direction didn't decrease, motion in the $x$ and $z$ directions would have to stay the same for the speed of light to stay constant, but this doesn't make sense, because it implies the beam of light is being dragged along with your reference frame.

Also, in Lorentz Transformation, suppose two observers measure a rod, can one of the observers see that the other person who is moving w.r.t. him, measure the rod to be less than his own measurement ? In other words, can the metre sticks of a moving observer get bigger than the observer who views the moving observer and sees himself to be at rest ?

I'll limit my reply to this. Try to work it out from the definition of a Lorentz transformation. The answer to "can one of the observers see that the other person who is moving w.r.t. him, measure the rod to be less than his own measurement" is yes. In your next sentence, if you mean to imply that some observer measures a meter stick at longer than a meter, the answer is no. A meter stick at rest is a meter, and a meter stick flying by (along its axis) in either direction is slightly shorter than a meter, by a factor of $1/\gamma$.

• Can $\gamma$ be $<1$ Dec 26, 2013 at 7:29
• @Iota: $\gamma = 1/\sqrt{1 - v^2/c^2}$ so it cannot be less than one unless the speed $v$ is imaginary (so $v^2$ is negative). Dec 26, 2013 at 7:53