Take on Length Contraction at relativistic speeds

[A.T.]

No I get it, it's not an actual physical length change but a measurement of length in a different system of coordinates.

In these discussions people often argue whether something is "physical" or "real", but often mean different things by these terms or don't even have a precise definition of them in their own mind.

Consider a circular train track and an unlimited supply of train engines, entering that circle from an tangential side track. The number of engines you can fit on that circle, depends on their contracted length. The faster they go, the more of them you can fit in.

Whether you consider that "real" or "physical" is semantics.

 

[David Lewis]

The number of engines you can fit on that circle, depends on their contracted length. The faster they go, the more of them you can fit in.

Will the track be length contracted in the reference frame of the engine?

 

[Ibix]

Will the track be length contracted in the reference frame of the engine?

The engine is not inertial, so length contraction does not apply to its frame. It's an acceptable approximation to consider any short section of a train as inertial, and deduce that it must be length contracted, but you cannot think of the whole track this way from the engine frame.

 

[A.T.]

Will the track be length contracted in the reference frame of the engine?

As @Ibix wrote, you can consider an instantaneous inertial co-moving frame, where the usual Lorentz transformation applies, but an engine is only at rest for an instant.

Alternatively you can consider a rotating frame, where all the engines filling the circle are permanently stationary. So they are non-contracted and thus have a total length greater than 2πr, but still fit onto a circumference of radius r. You can interpret this as an non-Euclidean spatial geometry in the rotating frame. But note that the usual simultaneity convention doesn't work continuously along the circumference, so you cannot use simple Lorentz transformations to analyse this globally.

The point I was making with this example is that for storing moving stuff in a loop, the contracted length is the relevant one. So regardless if you call length contraction "real" or "physical", you have to account for it because it has concrete practical consequences.

 

[Ebeb]

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He didn't say 'a shadow'; he said 'shadow-play'.

That's even worse.

You are taking his words too literally.

I thought the goal of communication in science was to be as specific as possible?

The 'projection' and 'shadow' wording follows from looking at a stick from a non-90 degree angle, resulting in a shorter view of the stick. This misleading analogy blocks the interested layman's mind from really understanding how it works .

Consider two trains moving relative to each other. The dots represent wagons of the train. In a train frame the wagons are simultaneously of the same color. But every split second the wagons change color, simultaneousely in their respective frame.
A contracted train is a collection of wagons of a different color: a 3D section through a collection of (in time) successive 3D trains of the 4D train object.

The contracted train is not a projection or shadow of a length of the proper train. It's the actual length of the moving 3D train (made of wagons of a different color and changing colors in time), present in the observers 3D space/world.

I leave it to the reader to philosophy about what is phyically real or not, but keeping the following in mind: if an observer considers a train at rest relative to him being a "physically real" object present in a "physically really existing 3D space/world", then the contracted train is also a "physically real" object present in that 3D space/world.

 

[DaveC426913]

A contracted train is a collection of wagons of a different color

I haven't processed your post fully yet, but I have to say, you really missed an opportunity here: instead of wagons, you really should have used horses.

 

[Viopia]

One never experiences length contraction in one's own reference frame.
That goes for anyone else in the same reference frame (i.e. stationary with respect to you).

So, observer A observes no length contraction in object O.

Hi. I am not a physicist but I watched an interesting video on YouTube regarding time dilation and length contraction. If the formula for length contraction shown in the video is correct, something traveling at the speed of light, ie: a photon, would experience an infinite length contraction. This would mean that the photon would pass the start and finish line of any distance instantaneously. Would this also mean that an entangled photon would be at the same point in spacetime as its entangled partner and so there would be no "spooky action" at a distance from the photons frame of reference?

 

[jbriggs444]

the photons frame of reference?

There is no such thing as the rest frame for a photon.

It would be self-contradictory to have an inertial rest frame for light in a theory where light moves at c in all inertial frames.

 

[1977ub]

For some interesting imagery regarding null intervals you might enjoy working your way through this:

https://www.mathpages.com/rr/s9-09/9-09.htm

 

[Janus]

I watched an interesting video on YouTube regarding time dilation and length contraction.

At the start of the video he quotes the expert as saying "The photon experiences no time", he then goes on to say that zero time exists for the photon between the moment it is emitted and received. However, this second statement doesn't really follow the first according to Relativity. To see why, you have to look at the actual time dilation equation:
$$ T = \frac{t}{\sqrt{1-\frac{v^2}{c^2}}}$$
here T is time as measured by our observer, and t is time as measured by the clock that is moving relative to the observer.
So for example is v=.0866c, then T= t/0.5) = 2t, which equates to our observer measuring every tick of the moving clock as equally 2 ticks of his own clock.
But what happens if we make v=c? then we get T= t/0. But division by zero is undefined and has no answer. So for a photon, time is undefined. It has no physical meaning. It does not "experience time" because the whole concept has no meaning for it. This is not the same as it experiencing zero time. Zero time, even if it has no duration is still a defined duration, and there is no defined duration of time that light measures.
And thus if you go on to say that " the photon doesn't experience length either" this means that length is equally undefined for a photon. This is just another way of saying that a photon has no valid reference frame from which to measure time or space.

 

[PeterDonis]

This thread has run its course and is now closed.

Edit: A thread hijack has also been cleaned up.