I Question about length contraction

[Nugatory]

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 then in one second it will travel its length from the reference frame of the rod

This makes no sense, because “the reference frame of the rod” is the frame in which the rod is at rest. Let’s try describing your setup again:

We have a rod, and using that frame in which it at rest its length is ten meters; its proper length $L_0$ is ten meters. Of course using this frame it is not moving, but we can say that a bird flying alongside the rod at ten meters per second will take one second to travel the length of the rod.

Or we can use the frame in which the bird at rest. Using this frame the rod is moving past the stationary bird at ten meters per second and its length is $L_0/\gamma$. At that speed it takes $1/\gamma$ seconds for the length of the rod to pass the bird.

It would be a good (as in will clear things up for you more than anything else) exercise to draw a Minkowski spacetime diagram showing the paths through spacetime of the bird and both ends of the rod.

 

[PeroK]

This is my elementary way of understanding it and like I said I'm still a bit fuzzy on it but it seems to me that the relativity of a certain event might explain length contraction.

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.

 

[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

meaning it will cover its length in gamma seconds in the stationary reference frame, because it takes longer to cover its length in the stationary reference frame it means its length is contracted.

No, as others mentioned. Assume the rod is at rest in frame $S$ and a bird at rest in frame $S'$.

Define event $A$: the front end of the rod meets the bird.
Define event $B$: the back end of the rod meets the bird.

Define $\Delta x$: spatial distance between event $A$ and event $B$ in frame $S$.
Define $\Delta t$: temporal interval between event $A$ and event $B$ in frame $S$.

Define $\Delta x'$: spatial distance between event $A$ and event $B$ in frame $S'$.
Define $\Delta t'$: temporal interval between event $A$ and event $B$ in frame $S'$.

From you problem description:
$\Delta x = 10m$,
$\Delta t = 1s$,
$v = 10\frac{m}{s}$.

Lorentz transformation of $\Delta x$:
$\Delta x' = \gamma (\Delta x -v \Delta t ) = 0$

Inverse Lorentz transformation of $\Delta t'$:
$\Delta t = \gamma (\Delta t' +v \Delta x'/c^2 ) = \gamma \Delta t'$
$\Rightarrow$
$\Delta t' = \frac{1}{\gamma} \Delta t = \frac{1}{\gamma}s$.

 

[Sagittarius A-Star]

the "traveling its length" event could happen in gamma seconds relative to a stationary reference frame meaning it will cover its length in gamma seconds in the stationary reference frame, because it takes longer to cover its length in the stationary reference frame it means its length is contracted.

I think you misunderstand time dilation.
Time dilation always refers to two tick-events of the moving clock.
Assume a clock is at rest in frame $S$ at location $x=0$.
Then the spatial distance between two tick events in this frame is $\Delta x = 0$.
Lorentz transformation of $\Delta t$:
$\Delta t' = \gamma (\Delta t -v \Delta x / c^2) = \gamma \Delta t$.

The Lorentz transformation only reduces to the time-dilation formula in the special case of $\Delta x=0$.
In other cases you must use the Lorentz transformation instead of the time-dilation formula.

 

[Herman Trivilino]

No, this is wrong. You specified that 3 seconds elapse on your (Mister T's) clock between the flashes. Now you're saying that 5 seconds elapse on your clock between flashes. You're contradicting yourself. The correct statement is the other way around.

Oops. Let me go back and fix it. Thanks for catching that.

 

[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)