Question about length contraction

[Chenkel]

Hello everyone!

I was researching about time dilation and I feel I have a pretty good understanding of it its very cool and fascinating, but my understanding of length contraction is fuzzy. It's not so clear why to me length contraction only happens in the direction of motion.

If there is a stationary observer and another observer (in motion) that has a clock with a longer period than the stationary clock because of time dilation then I would expect from the perspective of the stationary frame that the moving frame would be contracted in all directions (not just the direction of motion), because if you shine a light in any direction for the clock with the longer period the light should travel less far (in all directions) than for the clock with the shorter period, wouldn't this indicate a shrunken reference frame in all directions?

I feel I'm missing something pretty basic and I feel it might be confusion about one of the postulates of special relativity, hopefully I've explained enough of my confusion that I can get an answer that makes some sense.

Let me know what you think, thank you, any shed of light will be appreciated

 

[PeroK]

Hello everyone!

I was researching about time dilation and I feel I have a pretty good understanding of it its very cool and fascinating, but my understanding of length contraction is fuzzy. It's not so clear why to me length contraction only happens in the direction of motion.

If there is a stationary observer and another observer (in motion) that has a clock with a longer period than the stationary clock because of time dilation then I would expect from the perspective of the stationary frame that the moving frame would be contracted in all directions (not just the direction of motion), because if you shine a light in any direction for the clock with the longer period the light should travel less far (in all directions) than for the clock with the shorter period, wouldn't this indicate a shrunken reference frame in all directions?

I feel I'm missing something pretty basic and I feel it might be confusion about one of the postulates of special relativity, hopefully I've explained enough of my confusion that I can get an answer that makes some sense.

Let me know what you think, thank you, any shed of light will be appreciated

Time dilation and length contraction (in the direction of motion) are symmetric. Each observer measures the other's clock to run slow and lengths to be contracted (in the directtion of motion).

Consider two observers moving towards each other, each with a metre stick oriented at right angles to the direction of motion. As they pass each other, whose metre stick would be shorter?

Note: remember that all inertial motion is relative.

 

[Ibix]

You're missing the relativity of simultaneity, which is the third difference between inertial frames along with time dilation and length contraction. You won't be able to make any sense of relativity without it.

The easiest thing to do is to use the Lorentz transforms, which are the general case from which you can derive length contraction and time dilation. If you have a light bulb at rest at the origin in some frame and it emits a spherical light pulse at time zero then at time $t$ the light satisfies $c^2t^2=x^2+y^2+z^2$. You can use the Lorentz transforms to confirm that you also have a sphere in another frame in relative motion.

You might also like to consider what happens to the little bits of light that go straight along the $x$ and $y$ axes ($ct=\pm x$, $ct=\pm y$). Again, apply the Lorentz transforms. Don't try to do it by guess and by God with the time dilation and length contraction formulae.

 

[Orodruin]

If there is a stationary observer and another observer (in motion) that has a clock with a longer period than the stationary clock because of time dilation then I would expect from the perspective of the stationary frame that the moving frame would be contracted in all directions (not just the direction of motion), because if you shine a light in any direction for the clock with the longer period the light should travel less far (in all directions) than for the clock with the shorter period, wouldn't this indicate a shrunken reference frame in all directions?

What you describe here is completely unrelated to length contraction and it is not how length contraction works. When we talk about length contraction and time dilation it is always about effects observed after the light travel time has been accounted for so light travel time is irrelevant to the effects.

You're missing the relativity of simultaneity, which is the third difference between inertial frames along with time dilation and length contraction. You won't be able to make any sense of relativity without it.

I’d rather say that the relativity of simultaneity is the basic concept and results from the Lorentz transformations. Time dilation and length contraction are not so fundamental as to call them out as two out of three things different between frames.

 

[Hill]

Here is a way that has helped me to visualize length contraction. A "slogan" to remember is, Leading Clock Lags. If there is a rod moving along its length with clocks on each end, then when the trailing clock shows 9:00, the leading clock shows, say, 8:55.
As the rod flies by me, I mark the positions of its two ends at some moment and then measure a distance between the marks. But for a guy sitting on the rod, I've marked the position of the leading end at 8:55 and the position of the trailing end 5 minutes later, after the rod has moved ahead somewhat. Of course, the distance between my marks is shorter than the length of his rod.

 

[Sagittarius A-Star]

If there is a stationary observer and another observer (in motion) that has a clock with a longer period than the stationary clock because of time dilation then I would expect from the perspective of the stationary frame that the moving frame would be contracted in all directions (not just the direction of motion), because if you shine a light in any direction for the clock with the longer period the light should travel less far (in all directions) than for the clock with the shorter period, wouldn't this indicate a shrunken reference frame in all directions?

No, length contraction exists only in direction of motion, i.e. x-direction.

Call the "stationary" frame $S$ and the "moving" frame $S'$.

Frame $S'$:
Consider the following L-shape light clock at rest in $S'$: An object sends simultaneously two light-pulses, one in x'-direction and one in y'-direction to mirrors, each in a distance of $L_0$ and receives both reflections in after
$\Delta t' = \frac{2L_0}{c}$.

Frame $S$:

For the moving longitudinal light clock, the "tick" period is (with Lorentz contraction $L=L_0/\gamma$):
$\Delta t = \frac{L}{c-v} + \frac{L}{c+v} = L (\frac{c+v}{c^2-v^2} + \frac{c-v}{c^2-v^2}) = \frac{2L}{c} \frac{1}{1 -v^2/c^2} = \frac{2L}{c} \gamma^2 = \frac{2L_0}{c} \gamma = \Delta t' * \gamma$.

The moving transversal light clock is time-dilated because of aberration of the angle. The light pulse does not move in y-direction, but has ...

• component in x-direction: $v$
• component in y-direction: $c\sqrt{1-v^2/c^2}$ .

Therefore:
$\Delta t = \frac{2L_0}{c\sqrt{1-v^2/c^2}} = \Delta t' * \gamma$.

 

[PeroK]

Here's the symmetry argument/thought experiment. Imagine a bullet train moving through a tunnel, which is only a bit larger in cross section than the train. If the tunnel's cross section were contracted in the train frame, then the train would not fit in the tunnel beyond a certain speed. Likewise, if the tunnel's cross section were dilated in the train frame, then the train's cross section would be dilated in the tunnel frame and again the train would not fit. The only physically viable conclusion is that there is no length contraction or dilation in the directions perpendicular to the motion.

It's important, of course, to understand why length contraction in the direction of motion does not lead to any similar physical contradictions.

 

[Orodruin]

It's important, of course, to understand why length contraction in the direction of motion does not lead to any similar physical contradictions.

Is it relativity of simultaneity? I bet it is relativity of simultaneity!

 

[PeroK]

Is it relativity of simultaneity? I bet it is relativity of simultaneity!

Is simultaneity even a thing?

 

[Orodruin]

Is simultaneity even a thing?

Not if you want to discuss invariant statements …

 

[Dragon27]

If there is a stationary observer and another observer (in motion) that has a clock with a longer period than the stationary clock because of time dilation then I would expect from the perspective of the stationary frame that the moving frame would be contracted in all directions (not just the direction of motion), because if you shine a light in any direction for the clock with the longer period the light should travel less far (in all directions) than for the clock with the shorter period, wouldn't this indicate a shrunken reference frame in all directions?

Why would it? The light will travel less far with respect to the moving clock (i.e. the vector movement of the light minus the vector movement of the clock), which indicates that the moving light clock is ticking slower. If the clock was shrunken in perpendicular direction to compensate for that, that would mean that the light will reach the extremity of the moving clock at the same time as the light in the stationary clock, which means there's no time dilation. It would make no sense.

 

[Chenkel]

So I had a thought experiment and it goes something like this, if I have a rod in a rest frame and it has a proper length of $L_0$, and it moves its length in $t_0$ seconds and then there is a moving reference frame S' moving in the direction of the rod then from the moving reference frame S' the time for the rod to move its length should be $$t_0\sqrt{1 - \frac {v^2} {c^2}}$$

The rod moving its length should happen at different times depending on which reference frame you use, from the moving reference frame S' the rod moves its full length in less time than from the stationary reference frame S, this would imply length contraction in the direction of motion.

Hopefully I made things clear enough to get the idea across, what we think is simultaneous (the rod moving its length) is actually only relatively simultaneous because it moves its length in different time intervals depending on which reference frame you use.

 

[Ibix]

So I had a thought experiment and it goes something like this, if I have a rod in a rest frame and it has a proper length of $L_0$, and it moves its length in $t_0$ seconds

If it moves at all, it can't be in its rest frame. So this thought experiment doesn't make sense.

You could say something like, in frame S the rod is at rest and has length $L_0$. In a frame S', moving with speed $v$ parallel to the length of the rod, the rod has length $L'=L_0/\gamma$ and moves its own length in time $L_0/(\gamma v)$.

 

[Chenkel]

If it moves at all, it can't be in its rest frame. So this thought experiment doesn't make sense.

You could say something like, in frame S the rod is at rest and has length $L_0$. In a frame S', moving with speed $v$ parallel to the length of the rod, the rod has length $L'=L_0/\gamma$ and moves its own length in time $L_0/(\gamma v)$.

In your equations is the following true? $\gamma = \frac {1} {\sqrt{1 - \frac {v^2}{c^2}}}$

 

[Nugatory]

In your equations is the following true? $\gamma = \frac {1} {\sqrt{1 - \frac {v^2}{c^2}}}$

That is the standard convention.
You will also see $\gamma = \frac {1} {\sqrt{1-v^2}}$, which is the same thing written using units in which $c=1$, for example measuring time in seconds and distances in light-seconds. This convention unclutters the equations without losing any of the essential physics.
Another notational improvement is to use $\beta=v/c$ which achieves the same uncluttering without committing our choice of units.

 

[Janus]

Consider the light clock thought experiment. But add a second light pulse going back and forth in line with the motion.
If there was no length contraction, you would get this result:

Note that the clock "in motion" ticks half as fast when we compare the vertically moving light pulses( the white dots).
But, the horizontal pulse hasn't even reached the mirror before the vertical pulse has made a round trip. If you were riding along with this clock, you would measure the pulses as staying in sync. This must remain true for our "stationary" observer also, or all kinds of contradictions can arise.
But if you apply length contraction in the the direction of motion(and only in the direction of motion),you get this:

where the round trip times for both vertical and horizontal pulses match according to both the stationary and moving observer.

 

[Ibix]

But if you apply length contraction in the the direction of motion(and only in the direction of motion),you get this:

where the round trip times for both vertical and horizontal pulses match according to both the stationary and moving observer.

I "know" Janus well enough to believe that this diagram is accurate, but I really struggle to see the left and right going pulses in the moving light clock as travelling at the same speed. It's really easy to eyeball the pulse speed with respect to the mirrors in the Galilean sense and see the left-going pulse as super fast while the right going pulse crawls along.

 

[Janus]

I "know" Janus well enough to believe that this diagram is accurate, but I really struggle to see the left and right going pulses in the moving light clock as travelling at the same speed. It's really easy to eyeball the pulse speed with respect to the mirrors in the Galilean sense and see the left-going pulse as super fast while the right going pulse crawls along.

It might have helped if I had put grid lines as an reference, but that might have also cluttered the image and made it more confusing.

 

[Ibix]

It might have helped if I had put grid lines as an reference, but that might have also cluttered the image and made it more confusing.

I don't think it does help actually, because I've seen similar diagrams with grid lines and I get the same visual effect.

I wasn't complaining. It was more a comment on how I can't see the correct behaviour, even though I know what ought to be happening and I have no doubt you've drawn that correctly.

 

[PeroK]

All these diagrams and animations. Best to stick to algebra!

 

[Ibix]

All these diagrams and animations. Best to stick to algebra!

I like a mix, actually. You need to be able to do the algebra, but I like diagrams for the intuition they build.

 

[Chenkel]

If it moves at all, it can't be in its rest frame. So this thought experiment doesn't make sense.

You could say something like, in frame S the rod is at rest and has length $L_0$. In a frame S', moving with speed $v$ parallel to the length of the rod, the rod has length $L'=L_0/\gamma$ and moves its own length in time $L_0/(\gamma v)$.

So you are saying that the moving reference frame perceives the rod in the stationary reference frame as having shortened length of $L'=L_0/\gamma$ and if we divided this value by the velocity we should get the time it takes the moving reference frame to cover the length of the rod?

Either reference frame can be considered stationary when you have two inertial reference frames moving in relative motion because there's no privileged reference frame?

So whatever is moving relative to the "stationary reference frame" is contracted along its direction of motion by the Lorentz factor from the perspective of the stationary reference frame?

 

[Ibix]

So you are saying that the moving reference frame perceives the rod in the stationary reference frame as having shortened length of $L'=L_0/\gamma$ and if we divided this value by the velocity we should get the time it takes the moving reference frame to cover the length of the rod?

I can't work that out.

I strongly recommend against calling frames 'moving' and 'stationary'. Einstein did it in his 1905 paper and it still confuses people. Either label frames (S, S', S'' or whatever) or name them after things that are at rest in them (the Earth's rest frame, the rocket's rest frame, etc).

In any frame where the rod is moving parallel to its length it takes time $L/v$ to travel its own length, where $L$ is the length of the rod as measured in that frame and $v$ is the velocity of the rod in that frame. Hopefully that is obvious. It also means that $-v$ is the velocity of the frame with respect to the rod's rest frame, and therefore $L=L_0\sqrt{1-(-v)^2/c^2}$ if $L_0$ is the rest length of the rod.

So whatever is moving relative to the "stationary reference frame" is contracted along its direction of motion by the Lorentz factor from the perspective of the stationary reference frame?

Anything that is moving in an inertial frame is, if measured using that frame, contracted in its direction of motion by a factor of $1/\gamma$. If you designate one frame "the stationary frame" things that are moving in that frame will be contracted if measured using that frame, just like any other, if that's what you are trying to ask.

 

[Orodruin]

I like a mix, actually. You need to be able to do the algebra, but I like diagrams for the intuition they build.

For me, algebra builds the intuition …

 

[Mister T]

Hello everyone!

I was researching about time dilation and I feel I have a pretty good understanding of it

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?

 

[Chenkel]

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?

Yes, I understand that part, I'm just a little fuzzy on length contraction however but I appreciate all the help on this forum.

 

[Chenkel]

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?

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, however 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.

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.

 

[Mister T]

Yes, I understand that part, I'm just a little fuzzy on length contraction however but I appreciate all the help on this forum.

Let's say that I send out a flash of light at the origin of my rest frame. Then, as measured in my rest frame, I send out another flash of light at my origin 3 seconds later. Your rest frame moves past my origin at a fast enough speed that you observe 5 seconds of elapsed time between the flashes. We explain this using time dilation. You observe my clocks running slow so that only 3 seconds elapse on my clocks while 5 seconds elapse on your clocks.

Now, given the above observations, how can it be that I observe your clocks running slow?

Don't neglect the above observations in your explanation.

 

[Ibix]

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

No! In the reference frame of the rod it is at rest, so it goes nowhere. In the frame where it is moving at 10m/s it is length contracted by a tiny amount ($\approx 5\times 10^{-15}\mathrm{m}$ if my mental arithmetic is to be trusted) and therefore it travels its length in very slightly less than 1s.

You keep mixing up measurements in different reference frames, which is part of why you are confused.

 

[PeterDonis]

You observe my clocks running slow so that 5 seconds elapse on my clocks while only 3 seconds elapse on your clocks.

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.

 

[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.

 

[Mister T]

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.