Question about length contraction

[Chenkel]

The simple explanation is that if the speed of light is the same in two different frames (one where the rod is at rest and one where it moves with speed $v$), then the rod cannot have the same measured length in both frames.

You can see this by having a light signal travel from one end of the rod to the other and bounce back again. If the rod has rest length $L_0$, then the time for this round trip in the rod's frame is $\Delta t' = \frac{2L_0}{c}$. And, if we assume we have already derived the formula for time dilation, the round trip takes a time of $\Delta t = \gamma \Delta t' = \frac{2\gamma L_0}{c}$ in the frame in which the rod is moving.

However, if $L$ is the length of the rod in this frame, then the round trip time is given by:
$$\Delta t = \frac{L}{c-v} + \frac{L}{c+v} = \frac{2Lc}{c^2 - v^2} = \frac{2Lc}{c^2(1 - v^2/c^2)} = \frac{2\gamma^2L}{c}$$And by equating these two ways to calculate $\Delta t$ we see that:
$$L = \frac{L_0}{\gamma}$$Which is length contraction.

I'm a little confused, what does $\frac{L}{c-v}$ and $\frac{L}{c+v}$ mean? It looks the denominator is some relative velocity but I don't understand relative velocities when the speed of light is invoked.

 

[Nugatory]

I'm a little confused, what does $\frac{L}{c-v}$ and $\frac{L}{c+v}$ mean? It looks the denominator is some relative velocity but I don't understand relative velocities when the speed of light is invoked.

We have a flash of light emitted at one end of the rod, travelling to a mirror at the other end, and being reflected back to the source. What is the round trip time?

Using the frame in which the rod is at rest, the distance covered is $L_0$ in both directions, total travel time is $L_0/c$.

Watching the exact same situation using a frame in which the rod is moving at speed $v$: The light is moving at speed $c$ and the far end of the rod is moving away from it at speed $v$. Thus the light is closing the gap between emission event and the mirror at speed $c-v$ instead of $c$. On the return leg the light is still moving at speed $c$, but the near end of the rod is moving towards the light so the gap is being closed more quickly, speed $c+v$.

It may be easier to see this if you look at the distance the flash of light travels on the two legs - it has farther to go on the outbound leg than the return leg.

 

[Sagittarius A-Star]

If there is a rod with proper length (L sub zero) of 10 and it is to go at 10 meters per second along its length

Calculation of the length contraction in scenario from posting #33:

Consider two events $C$ and $D$ at both ends of the rod. The spatial distance between those events in frame $S$ is
$\Delta x = L_0$.

Assume, that those events happen at the same time with reference to frame $S'$. Then
$\Delta t' = 0$
$L' = \Delta x'$.
The events $C$ and $D$ could be i.e. checking the $x'$ coordinates of the front end and the back end of the moving rod.

Inverse Lorentz transformation of $\Delta x'$:
$L_0 = \Delta x = \gamma (\Delta x' + v \Delta t') = \gamma L'$
$\Rightarrow$
$L' = L_0 / \gamma$.

 

[PeroK]

I'm a little confused, what does $\frac{L}{c-v}$ and $\frac{L}{c+v}$ mean? It looks the denominator is some relative velocity but I don't understand relative velocities when the speed of light is invoked.

The kinematics of light are no different in SR than Newtoinian physics, as long as you use a single inertial reference frame. Light moves with speed $c$ and objects with mass move with speeds less than $c$. You can do kinematics the way you always did.

It's only when you transform to a second inertial reference frame that SR differs from Newtonian physics. In Newtonian physics, the speed of a light signal would change from frame to frame. In SR, of course, the speed of light is invariant.

 

[Orodruin]

It's only when you transform to a second inertial reference frame that SR differs from Newtonian physics. In Newtonian physics, the speed of a light signal would change from frame to frame. In SR, of course, the speed of light is invariant.

I feel the qualifier “for the purposes of kinematics” is necessary here. Of course, once you bring dynamics into the game Newtonian mechanics differ from SR even if you keep to one frame.

 

[Sagittarius A-Star]

I'm a little confused, what does $\frac{L}{c-v}$ and $\frac{L}{c+v}$ mean? It looks the denominator is some relative velocity but I don't understand relative velocities when the speed of light is invoked.

You will find this also explained in chapter "1.3.3 Length contraction", page 24, before equation (1.17) in Morin's book about relativity:
https://scholar.harvard.edu/files/david-morin/files/relativity_chap_1.pdf

 

[Chenkel]

You will find this also explained in chapter "1.3.3 Length contraction", page 24, before equation (1.17) in Morin's book about relativity:
https://scholar.harvard.edu/files/david-morin/files/relativity_chap_1.pdf

I just read that section and it makes a lot sense. Thank you.

 

[cianfa72]

Do you understand the symmetry of time dilation? That is, if I observe time dilation in your rest frame you will observe time dilation in my rest frame. We will each observe the clocks in the other's rest frame running slow relative to the clocks in our own rest frame. Note that that is required by the first postulate. How is such a thing possible?

Just to highlight that the process to measure time dilation involves one moving clock (that reads just its proper time $\tau$) and two clocks at rest in the chosen inertial frame synchronized according it (by definition these synchronized clocks read the frame coordinate time $t$). In this sense this process is asymmetric.

 

[Nugatory]

Just to highlight that the process to measure time dilation involves one moving clock (that reads just its proper time $\tau$) and two clocks at rest in the chosen inertial frame synchronized according it (by definition these synchronized clocks read the frame coordinate time $t$). In this sense this process is asymmetric.

We can do it with one clock at rest and one moving clock: the moving clock transmits current readings by light signal to the stationary clock; the stationary-frame coordinate time of the emission is determined by subtracting the light travel time from the reception time. Of course this is exactly equivalent to your two synchronized clocks at rest - but has the advantage of not obscuring the symmetry between the two frames.

 

[cianfa72]

We can do it with one clock at rest and one moving clock: the moving clock transmits current readings by light signal to the stationary clock; the stationary-frame coordinate time of the emission is determined by subtracting the light travel time from the reception time.

Therefore the moving clock encodes in the light signal it transmits the current reading of its own clock and its current position w.r.t. the stationary frame.

 

[Nugatory]

Therefore the moving clock encodes in the light signal it transmits the current reading of its own clock and its current position w.r.t. the stationary frame.

Don’t even need the position, we can calculate the change in position from $v\Delta t$ where $\Delta t$ is the elapsed time between two receptions at the stationary clock. (And we’re allowed two receptions - they’re equivalent to your setup with one reading at each of two clocks)