Do all moving object appear contracted?

In the derivation of Lorentz contraction my prof considered a rod moving away from you with speed $v$. In this case the rod appears shrunk by a factor gamma.
But now let's consider a rod moving towards you. Light from farther end leaves before it does from the closer end; for the measurement to be simultaneous in your frame. Thus rod's length in your frame appears to be greater than proper length. In that case how does the formula L = L0 sqrt(1- (v/c)2) hold.

Your confusion might originate in what it means a simultaneous measurement. The fact that "Light from farther end leaves before it does from the closer end" is irrelevant. That would be true even if the object were at rest.

One way to make a simultaneous measurement to determine the length is to have clocks along the path of the rod, at rest relative to you. Then you can record the time at which the front road is next to a specific clock and then search for a clock that registers the position of the back of the rod next to it, at a time similar than that recorded by the first clock. The length of the rod will be the distance between the two clocks.

All other forms to make a measurement must be equivalent to this method.

• What I mean is that as light leaves the rear end , the front end moves forward by some distance. The light emitted by front end at this instant is what you will see as arriving simultaneously with the light emitted from the rear end at the earlier instant. Thus the rod appears longer. – user264677 Nov 15 '15 at 22:35
• That would not be a relativistic effect, but a visual illusion that would occur in newtonian mechanics too due to the finiteness of the speed of light. To know the actual length you need to correct by this effect. – user83548 Nov 15 '15 at 23:58

The length of the rod will always appear smaller in your frame than in the proper frame, the mathematical relation between both lengths is given by:

$$L=\gamma L_{o} \,\, ,$$ Where $$\gamma=\sqrt{1-\Big(\frac{v}{c}\Big)^{2}} \,\, .$$

I'm not sure if I've understood very well your question, but I think it can answer your question.

• γ<1 and so L<Lo. And so the length should appear contracted. But by the reasoning I gave in the question L should be greater than Lo for a rod moving towards you. – user264677 Nov 15 '15 at 14:36
• Yes, sorry, I've edited correcting this,@user264677, it doesn't change, because the fraction $v/c$ is squared, so even if the result is negative, it will turn to a positive value after the square! – Herr Schrödinger Nov 15 '15 at 14:45
• I don't think you understand it.γ^2 = 1 - (v/c)^2 < 1 and so γ <1 and so L<Lo. – user264677 Nov 15 '15 at 22:24
• @user264677 I did! – Herr Schrödinger Nov 15 '15 at 23:01