Please explain how length contraction and time dilation result in a constant c

[Amin2014]

TL;DR

moving car, light speed

if light pulse is to the right, but moving car's velocity is to the left, we would expect c-v, but we get c instead which is greater.
but if light pulse is to the right, and moving car's velocity is also to the right, we would expect c+v, but we get c instead which is smaller.
how could the same length contraction and time dilation explain both larger and smaller than expected results?

 

[Ibix]

You have forgotten the relativity of simultaneity, which is less popularised than length contraction and time dilation, but probably more important. And, critically, the sign of the synchronization difference depends on direction, which is how you get different answers for things travelling in the $\pm x$ directions.

 

[robphy]

Restricting to relative-motion in 1 spatial dimension…
light pulses involve the Doppler effect, which are direction-dependent (since the Doppler factor k depends on the sign of relative-velocity),
whereas length-contraction and time-dilation do not depend on direction since \gamma is an even-function of relative-velocity.

Using light-pulses to explain length-contraction and time-dilation isn’t as simple as you how describe. You need radar (light pulses, there and back).

 

[Ibix]

As a general point, when starting out with relativity I'd advise you to forget about length contraction and time dilation. They are special cases of the more general Lorentz transforms, and it's very easy to try to understand things in terms of them when it is inappropriate to do so. The general recipe for relativity problems is to write down the coordinates of one or more events in one frame, then transform to the other frame. Then try to understand the other frame's description.

So, for the sake of argument, suppose we emit two light pulses in opposite directions from $x=0$ at time $t=0$ and they hit targets, stationary in the lab frame, one light second away one second later. We have three events of interest - $(x,t)=(0,0)$, $(1,1)$ and $(-1,1)$. What does this look like in a frame moving at speed $v$ in the $+x$ direction? Plug the coordinates into the Lorentz transforms$$\begin{eqnarray*}
x'&=&\gamma(x-vt)\\
t'&=&\gamma\left(t-\frac{v}{c^2}x\right)\\
\gamma&=&\frac{1}{\sqrt{1-v^2/c^2}}
\end{eqnarray*}$$and you will find that the coordinates in this frame are $(x',t')=(0,0)$, $(\gamma(1-\beta),\gamma(1-\beta))$, and $(\gamma(1+\beta),-\gamma(1+\beta))$ where $\beta=v/c$, and $c=1$ in these units.

Notice that both light pulses must have travelled at $c$ for the distance travelled ($\gamma(1\pm\beta)$) to equal the time taken in these $c=1$ units. But note that the two light pulses didn't move the same distance as each other (because the targets are moving in the $-x$ direction in this frame) and didn't take the same time as each other - this is the relativity of simultaneity at work.

 

[Nugatory]

You might want to look at the relativistic velocity addition formula, which combines the effects of relativity of simultaneity, length contraction, and dilation for this sort of problem. This formula tells us the speed $w$ of something relative to you if it is moving at speed $u$ relative to me and I am moving at speed $v$ relative to you: $w=\frac{u+v}{1+uv/c^2}$ and replaces the classical $w=u+v$.

Using a frame in which I am rest, we have a flash of light moving to the left with speed $-c$ and another one moving to the right at speed $+c$. You are moving at speed $v$ to the right relative to me. The velocity addition formula tells us that the speed of the right-moving flash relative to you is $\frac{c+v}{1+vc/c^2}=c$ and similarly the speed of the left-moving flash relative to you is $-c$.

However, the best way of understanding this stuff is to go back to the Lorentz transformations which underlie all of the other formulas. Here what you will do is…. Read @Ibix’s post above which landed while I was still writing this.