Why does length contraction occur and how does it relate to time dilation?

[syfry]

Ok I'll make note of your points and will check out the Spacetime Physics book.

Thanks all who chimed in and contributed to helping me understand this better!

 

[PeroK]

Fair enough. "Learn about (something)" isn't the same as saying "learn the (something)". Had meant in a general way learn the gist of its possibilities, what things mean. But even that was too fast. I'd have to write down what they're saying word for word to slowly read at my own pace, that's all.

Try this:

https://scholar.harvard.edu/david-morin/special-relativity

The first chapter is free.

It takes more than a minute to learn SR!

 

[syfry]

I recommend Spacetime Physics by Taylor and Wheeler, 1st Edition if you can find it.

Over $200 for 1st edition

(found a beat up one for $12 -ish)

 

[Vanadium 50]

The 2nd edition is free to download, but I think the 1st edition is better. But not $200 better.

 

[PeroK]

Over $200 for 1st edition

(found a beat up one for $12 -ish)

Are you a student or a collector?

 

[Herman Trivilino]

Yeah, had added on 'long' for a better chance that my question wasn't misinterpreted. But then after using the word 'pole' for the traveler and 'object' for the target, I still messed up anyway and referred to traveler as an object.

You misunderstand. What I mean is that you have two poles, a long one and a short one. When you look at the short one do you wonder if it's shorter because one end was moved, or both ends were moved, or if it has moved all along its length to make it short.

 

[Demystifier]

If a long pole at relativistic speed would strike an object, would we calculate from the front of that object as if the rest of its length had shrunk toward the front, or, would we calculate as if the front had shrunk back toward the middle of pole?

This question only makes sense if the pole is accelerating. It is answered in my https://arxiv.org/abs/physics/9810017 . A short answer is that it depends on the application point of the force.

 

[syfry]

This question only makes sense if the pole is accelerating. It is answered in my https://arxiv.org/abs/physics/9810017 . A short answer is that it depends on the application point of the force.

Interesting that accelerating it by push vs pull makes a difference in contraction. Nice thinking!

Didn't find in which part of the doc my question is answered, will look again later.

Realizing now that a visual example of the pole in barn paradox might've shown whether both ends of a pole contract inward or if its inward contracting happens all from only one end. (would've been a better way to have phrased my original question!)

Gonna go look for a visual example now.

 

[Demystifier]

Interesting that accelerating it by push vs pull makes a difference in contraction. Nice thinking!

Didn't find in which part of the doc my question is answered, will look again later.

Realizing now that a visual example of the pole in barn paradox might've shown whether both ends of a pole contract inward or if its inward contracting happens all from only one end. (would've been a better way to have phrased my original question!)

Gonna go look for a visual example now.

If the rod is pulled, i.e. the force is applied to the front, then the front moves as if all mass was concentrated there, which means that the rod contracts from the back. Is that an answer to your question? If not, then I haven't understood the question.

 

[syfry]

If the rod is pulled, i.e. the force is applied to the front, then the front moves as if all mass was concentrated there, which means that the rod contracts from the back. Is that an answer to your question? If not, then I haven't understood the question.

That seems like it would answer my question from a case of accelerated motion like you had said. Very interesting!

I'm instead interested in another aspect that shows up in steady motion (and might still show up in your accelerated scenario).

Might be able to rephrase the question better:

For a rod moving at a steady relativistic speed toward a target, in any frame where either the rod contracts or the target plus the distance to it contracts, does that contraction then add in a tiny amount of distance between both objects because one of their fronts would shrink away toward its midsection?

Let's say the target is 5 meters deep and the rod is 100 meters long. If the speed would contract the target so it's now only 1 meter deep, that's 4 meters of contraction: 2 from its front and 2 from its rear. Now the rod has to travel 2 extra meters to hit.

In another frame, if the math is the same so the contraction is 80%, then the rod will instead contract into 20 meters long, which is 80 meters of contraction: 40 from its front and 40 from its rear. Now the rod has to travel 40 extra meters to hit.

It's quite likely I'm erring mathematically.

But if everything is correct, then the only thing that seemsto logically make sense is for the front to firmly stay put without contracting, so only the rear is contracting... like in your example with acceleration by pulling. (but in my case using only steady or inertial motion)

 

[jbriggs444]

For a rod moving at a steady relativistic speed toward a target, in any frame where either the rod contracts or the target plus the distance to it contracts, does that contraction then add in a tiny amount of distance between both objects because one of their fronts would shrink away toward its midsection?

No. That description is not correct. You've forgotten about the relativity of simultaneity.

Write down the starting coordinates for the front and back of the pole in the original rest frame.
Write down the starting coordinates for the front and back of the 5 meter barn in the original rest frame.

Let's say the target is 5 meters deep and the rod is 100 meters long.

So, for instance using (x,t) coordinates.

Front end of pole: (100, 0)
Back end of pole: (0, 0)
Front door of barn: (105, 0) -- I put the barn 5 meters in front of the pole
Back door of barn: (110, 0)

Now transform to a coordinate system moving at 90% of the speed of light.

If you actually do the work, you'll notice that the scenario is not beginning all at the same time any more. None of your reasoning accounts for that.

 

[PeterDonis]

If the rod is pulled, i.e. the force is applied to the front, then the front moves as if all mass was concentrated there

This is only an approximation. But in this approximation, the rod's motion can be taken to be Born rigid and there is no "contraction" at all in the rod's rest frame. In the original frame (in which the rod was at rest before it started moving), the contraction is just the kinematics of the Rindler congruence. All parts of the rod move as if all the mass was concentrated there.

which means that the rod contracts from the back.

Not in the approximation described above. In that approximation the contraction is pure kinematics and you cannot assign it to any particular part of the rod; all parts of the rod move in a Born rigid manner and the rod's length in its instantaneous rest frame does not contract at all.

If you want to go beyond this approximation, and analyze the motion of the rod as a collection of independent atoms connected by inter-atomic forces, then it is no longer true that the front moves as if all the mass was concentrated there. The front will accelerate more than that initially, and then will get pulled back by the part of the rod just in back of it, as a wave of disturbance passes along the rod from front to back. The result will be that the rod's individual parts oscillate about an "equilibrium" motion that is the Born rigid motion described above; but depending on the rod's material properties, it might take quite sometime for such an equilibrium to be reached. But even in this more complicated analysis, it is still not the case that the rod contracts from the back. The average contraction is still purely kinematic, and the oscillations about that average occur in all parts of the rod equally.

It is possible that you are thinking of "contraction" in the sense of the rod's length in its instantaneous rest frame becoming smaller due to the effects of proper acceleration, which causes the rod's material to compress. This is valid, but it is not the same as the length contraction the OP is asking about. Nor can this compression be assigned to a particular part of the rod in any case.

 

[syfry]

No. That description is not correct. You've forgotten about the relativity of simultaneity.

Write down the starting coordinates for the front and back of the pole in the original rest frame.
Write down the starting coordinates for the front and back of the 5 meter barn in the original rest frame.So, for instance using (x,t) coordinates.

Front end of pole: (100, 0)
Back end of pole: (0, 0)
Front door of barn: (105, 0) -- I put the barn 5 meters in front of the pole
Back door of barn: (110, 0)

Now transform to a coordinate system moving at 90% of the speed of light.

If you actually do the work, you'll notice that the scenario is not beginning all at the same time any more. None of your reasoning accounts for that.

Thanks, that likely answers the question but wanna confirm a small difference wouldn't change anything.

In my scenario, the rear of either object is irrelevant because only their fronts would ever touch (so whichever order the front and rear would shrink inward is perhaps irrelevant to this thought exercise too). In the scenario where the pole enters the barn, then of course the rear of each would matter.

What I've instead been proposing (a bit unclearly) would be like the barn's door hadn't ever opened and the pole would splat against the door to the barn's entrance, without ever entering.

But, the splat was delayed a moment as the barn's front had been length contracted a number of meters away from the incoming pole (which let's say had been a light hour of distance away, and traveling at 90% the speed of light, so its total journey even after the length contraction would've added back in a wee amount of travel time... not even a split second, but still non zero)

I had previously mentioned looking up the pole in barn as an analogy, but had changed my mind after failing to connect the setup with my scenario, so the pole in barn doesn't seem appropriate.

 

[Herman Trivilino]

Let's say the target is 5 meters deep and the rod is 100 meters long. If the speed would contract the target so it's now only 1 meter deep, that's 4 meters of contraction: 2 from its front and 2 from its rear. Now the rod has to travel 2 extra meters to hit.

Two extra meters compared to what? The way you've described the scenario, ALL 100 meter rods that move at that speed contract by that amount.

 

[Demystifier]

This is only an approximation. But in this approximation, the rod's motion can be taken to be Born rigid and there is no "contraction" at all in the rod's rest frame.

Agreed. I was talking about contraction in the inertial frame in which the rod was at rest before it started moving. And BTW, in the paper I discuss the conditions under which the Born rigid approximation is a good approximation.

In the original frame (in which the rod was at rest before it started moving), the contraction is just the kinematics of the Rindler congruence.

Agreed.

All parts of the rod move as if all the mass was concentrated there.

Now I'm confused, which frame are you talking about? Clearly it is not so in the inertial frame, because the accelerated rod Lorentz contracts, so at each inertial time the back end is faster than the front end. In the non-inertial frame of the rod no part of the rod moves at all, so your sentence above is true in a trivial sense. But note that observers sitting on different parts of the rod feel different proper accelerations. If the force is applied to the front end, then only the observer on the front end feels proper acceleration as if the whole mass of the rod was concentrated in that point.

 

[PeroK]

Thanks, that likely answers the question but wanna confirm a small difference wouldn't change anything.

In my scenario, the rear of either object is irrelevant because only their fronts would ever touch (so whichever order the front and rear would shrink inward is perhaps irrelevant to this thought exercise too). In the scenario where the pole enters the barn, then of course the rear of each would matter.

What I've instead been proposing (a bit unclearly) would be like the barn's door hadn't ever opened and the pole would splat against the door to the barn's entrance, without ever entering.

But, the splat was delayed a moment as the barn's front had been length contracted a number of meters away from the incoming pole (which let's say had been a light hour of distance away, and traveling at 90% the speed of light, so its total journey even after the length contraction would've added back in a wee amount of travel time... not even a split second, but still non zero)

I had previously mentioned looking up the pole in barn as an analogy, but had changed my mind after failing to connect the setup with my scenario, so the pole in barn doesn't seem appropriate.

My advice is to forget these subsidiary questions and focus on the basics for now. If you understand length contraction, then you'll be able to answer the questions for yourself.

Long, equationless posts are not a good sign, IMO, that you have a good approach to learning physics.

The case of "uniform" acceleration on SR is a tricky subject, which is best left until you have masteted the conceptual and mathematical basics.

Finally, the real thing to learn is what precisely is meant by a measurement of length for a moving object?

 

[syfry]

Two extra meters compared to what? The way you've described the scenario, ALL 100 meter rods that move at that speed contract by that amount.

Please rephrase that? I'm not sure what you're asking.

 

[syfry]

My advice is to forget these subsidiary questions and focus on the basics for now. If you understand length contraction, then you'll be able to answer the questions for yourself.

That's all I can do, right? Little choice. I'll do so with the question in mind.

It's merely surprising that no one's ever brought up that if an object contracts, now that implies slightly more added spacing in between it and its destination.

Thanks to all who chimed in and who suggested good learning material.

 

[Ibix]

It's merely surprising that no one's ever brought up that if an object contracts, now that implies slightly more added spacing in between it and its destination.

It's not that nobody brings it up - Bell's spaceships paradox is an example of studying the effects of a changing length contraction, IMO. It's just that (as in this thread) it usually devolves into minutiae about how you are accelerating the object, where you apply the force, whether it stops afterwards, and worrying about whether the finite speed of mechanical waves in the object can be neglected, and probably more.

Once you've specified all that, answering the question can involve some messy book-keeping which isn't particularly exciting and usually leaves people none the wiser.

 

[syfry]

It's just that (as in this thread) it usually devolves into minutiae about how you are accelerating the object, where you apply the force, whether it stops afterwards, and worrying about whether the finite speed of mechanical waves in the object can be neglected, and probably more.

Once you've specified all that, answering the question can involve some messy book-keeping which isn't particularly exciting and usually leaves people none the wiser.

In that case, physics obviously works and such measurements are more pain than practical, so I'll assume time dilation will automatically fill any discrepancy such as extra spacing between objects. (that aren't accelerating)

Discussion was fruitful because of the good recommendations and because Demystifier introduced us to their work on length contracted differences in between ends of an accelerated object. Maybe that'll lead to a discovery!

Also jbriggs444 had personally helped with better understanding how to approach the concept of reference frames.

 

[Herman Trivilino]

Please rephrase that? I'm not sure what you're asking.

You're saying the rod contracts by 2 meters. But all rods of that length moving at that speed have contracted by the same amount. So what are you comparing? Two rods in the same rest frame that have contracted by the same amount or one rod in motion relative to the other?

Obviously, the latter. But since they are in motion relative to each other how shall you compare their lengths? Will you do it while you're at rest relative to the uncontracted rod while the contracted rod comes whizzing past. Do you make the comparison when the leading ends of the rod are aligned, when their centers are aligned, or when their tail ends are aligned?

And does it matter?

 

[Ibix]

In that case, physics obviously works and such measurements are more pain than practical, so I'll assume time dilation will automatically fill any discrepancy such as extra spacing between objects. (that aren't accelerating)

So here's an example of what I was saying. What are you actually measuring that length contraction of the rod is relevant? If I measure the time it takes something to get from A to B I usually mean the time between when one point on the object passes A and when the same point passes B. Then length contraction is irrelevant to one frame and only the length contraction of the distance between A and B matters in any other.

 

[syfry]

You're saying the rod contracts by 2 meters. But all rods of that length moving at that speed have contracted by the same amount. So what are you comparing? Two rods in the same rest frame that have contracted by the same amount or one rod in motion relative to the other?

Obviously, the latter. But since they are in motion relative to each other how shall you compare their lengths? Will you do it while you're at rest relative to the uncontracted rod while the contracted rod comes whizzing past. Do you make the comparison when the leading ends of the rod are aligned, when their centers are aligned, or when their tail ends are aligned?

And does it matter?

I'll try to close any holes in my logic.

First, it's only one rod, and one target. The target can be any rectangular object, a stone, a slab of wood, anything solid.

So let's say we had already previously held the rod in our reference frame. And we had visited the target in our reference frame. We had measured each in our own reference frame.

Now, sometime later, for whatever reason, the objects are in outer space and the rod is on the way to its target.

I'm not sure how to select reference frames that wouldn't cause issues, but let's try this: the target is orbiting the Milky Way galaxy, so I'd suppose its rest frame is the galaxy. The rod is approaching from a direction, and you know which reference frame the rod is in. (one that works and doesn't cause extra issues)

Now, knowing what we know about their previously measured size at rest in our own reference frame (they're no longer in it), what amount of extra spacing will appear between the rod and its target at the rod's current relativistic speed?

I forget what even started that line of thought, to be honest.

 

[syfry]

I've encountered a difficult to analyze part in my reading up on relativity, it's actually an old personal difficulty. In the tidbit below, which seems to potentially be the birth of length contraction:

"The most successful explanation was the Lorentz–FitzGerald contraction. These two physicists indepen-dently proposed that lengths are contracted in the direction of the motion by precisely the right factor, namely√1 v2=c2, to make the travel times in the two arms of the Michelson–Morley setup equal, thus yielding the null result. This explanation was essentially correct, although the reason why it was correct wasn’t known until Einstein came along"

It's from the footnotes at bottom of page 8 in this pdf:

https://scholar.harvard.edu/files/david-morin/files/relativity_chap_1.pdf

What I'm trying to wrap my head around is, what are they supposing was length contracted in their setup? The light itself? Can someone shed light on how they interpreted the situation? (in plain words please)

 

[PeterDonis]

observers sitting on different parts of the rod feel different proper accelerations. If the force is applied to the front end, then only the observer on the front end feels proper acceleration as if the whole mass of the rod was concentrated in that point.

Now I'm the one who is confused. Please give the mathematical expression for "proper acceleration as if the whole mass of the rod was concentrated in that point".

 

[Demystifier]

Now I'm the one who is confused. Please give the mathematical expression for "proper acceleration as if the whole mass of the rod was concentrated in that point".

Take a pen and make fixed marks on the rod, the marks are $n=1,2,...,100$. The mark $n=1$ denotes the front end, while $n=100$ denotes the back end. Now apply the force on the front end of the rode. The whole rod will move and any part of the rod with mark $n$ will have some trajectory $x_n(t)$, where $x$ and $t$ are position and time in the inertial laboratory frame.

Now take another, more compact body, shaped like a small ball, rather than a rod. Let this ball have the same mass as the rod, and apply the same force on it. It will have some trajectory $x_{\rm ball}(t)$.

Now, since the force is applied to the front end of the rod, in the approximation of Born rigidity I claim that
$$x_1(t)=x_{\rm ball}(t)$$
For any trajectory $x(t)$ I'm sure you know how to calculate the proper acceleration $a(t)$, so the equality above implies
$$a_1(t)=a_{\rm ball}(t)$$
In fact, if the force is constant, the proper accelerations $a_n(t)$ and $a_{\rm ball}(t)$ do not depend on $t$. But $a_n$ depends on $n$, that's what I mean by the statement that different parts of the rod have different proper accelerations.

Is it clear now?

 

[Ibix]

What I'm trying to wrap my head around is, what are they supposing was length contracted in their setup? The light itself? Can someone shed light on how they interpreted the situation? (in plain words please)

Anything that can be described as at rest when you use one frame of reference will be length contracted when using any other. Anything else (a beam of light, or the distance between a pair of objects in motion with respect to one another) will do something similar, but it won't obey the same formula because other factors come into play.

So Lorentz and Fitzgerald were proposing that the interferometer was contracted. I'm not sure what they thought would happen to the light itself - it doesn't matter for this experiment. You can work out what happens to it in a full relativistic model easily enough if you want.

Note that if you idealise the interferometer to just a few mirrors floating in space, not physically connected but at rest with respect to one another, the distance between the components will also length contract as observed using a frame where they are in motion. This must be the case because I could lay a ruler between them, and if the ruler touches both mirrors all frames must describe this - so if a ruler just touches both mirrors and you determine that the ruler is length contracted then you must also determine that the gap is length contracted or the ruler would not fit exactly.

 

[robphy]

I've encountered a difficult to analyze part in my reading up on relativity, it's actually an old personal difficulty. In the tidbit below, which seems to potentially be the birth of length contraction:
It's from the footnotes at bottom of page 8 in this pdf:

https://scholar.harvard.edu/files/david-morin/files/relativity_chap_1.pdf

What I'm trying to wrap my head around is, what are they supposing was length contracted in their setup? The light itself? Can someone shed light on how they interpreted the situation? (in plain words please)

These visualizations might help:

Here is a (2+1)-d spacetime diagram (a position-vs-time graph) of the Michelson-Morley experiment.
All light signals are sloped at 45-degrees.

• https://www.geogebra.org/m/XFXzXGTq
  Relativity-LightClock-MichelsonMorley-2018 (robphy)

Without length-contraction (in the first diagram),​

light signals at the origin event emitted along the two arms and reflected by the mirrors​

are received at distinct events TY, then TX.​

The time-difference depends on the velocity of apparatus, as they expected,​

in violation of the principle of relativity.​

​

With length-contraction (in the second diagram)​

[the separation of the worldlines marking the ends of the X-arm (along the relative-velocity axis) is shorter],​

the signal along the relative-velocity axis has a shorter round-trip time so that​

the reception events TY and TX are now coincident, as experimentally observed,​

in accord with the principle of relativity.​

​

​

​

​

 

[PeterDonis]

since the force is applied to the front end of the rod, in the approximation of Born rigidity I claim that
$$x_1(t)=x_{\rm ball}(t)$$

For the case of the rod this can only be true, if it's true at all, in the equilibrium state after the rod has stopped oscillating. But that's not how you're asserting it. Are you claiming the rod does not oscillate? That as soon as a force is applied to the front end, the entire rod immediately assumes Born rigid motion? You know that's impossible, right? To make the motion Born rigid from the outset, you would need to apply a precisely calibrated force to each point on the rod, not just the front.

\In fact, if the force is constant, the proper accelerations $a_n(t)$ and $a_{\rm ball}(t)$ do not depend on $t$.

Again, if this is true at all, it can only be true in the equilibrium state after the rod has stopped oscillating. But you're not asserting it that way. Or are you?

 

[Demystifier]

Again, if this is true at all, it can only be true in the equilibrium state after the rod has stopped oscillating. But you're not asserting it that way. Or are you?

I am. See Sec. 6 in my paper.