How specifically does an accelerated uniform rod Lorentz contract?

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Summary:

Does dv/dx=-Fv/mc^2(1-v^2/c^2)^2 describe it adequately?

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I am trying to push the boundaries of special relativity with a self-imposed challenge problem. A common derivation of relativistic kinetic energy involves an object to which a constant force is applied. I want to consider a similar scenario, but instead of a point object we now have a uniform rod accelerating along its axis.

Let's say there is a constant force [itex]F[/itex] applied to the center of the rod. Its momentum is [itex]p=Ft[/itex], from which we get its velocity:
[tex]\frac{v_\mathrm{CM}}{c}=\frac{Ft/mc}{\sqrt{1+(Ft/mc)^2}}.[/tex]
The velocity of the CM of the rod increases, asymptotically approaching [itex]c[/itex]. That means its length contracts more and more, so the rod gets shorter and shorter. But that means the front of the rod must be traveling more slowly than the center, and the back end more quickly, because of the time-dependent Lorentz factor. This, in turn, means that the Lorentz factor at the back of the rod is greater than the Lorentz factor at the front of the rod.
This observation is not unique to the endpoints of the rod; it is true for every distinct pair of points along the rod. We thus reformulate the above statement as follows: The Lorentz factor decreases monotonically as we move along the rod from back to front. In particular, it is nonuniform: [itex]\gamma=\gamma(x)[/itex]. (There is also, obviously, a time-dependence, but we'll get to that later.)
The continuous variation of [itex]\gamma[/itex] with position suggests a differential equations approach. Knowing [itex]v_\mathrm{CM}(t)[/itex], I obtain
[tex]\gamma_\mathrm{CM}(t)=\sqrt{1+\left(\frac{Ft}{mc}\right)^2}.[/tex]
In particular, [itex]\gamma_\mathrm{CM}(t)[/itex] satisfies the differential equation
[tex]\frac{\mathrm{d}}{\mathrm{d}t}\left[\frac{1}{\gamma}\right]=-\frac{(Ft/mc)(F/mc)}{[1+(Ft/mc)^2]^\frac{3}{2}}=-\frac{Fv}{mc^2}\frac{1}{\gamma^2}.[/tex]
A segment of rod of length [itex]\mathrm{d}x[/itex] will Lorentz-contract to a length [itex]\mathrm{d}x'=\mathrm{d}x/\gamma[/itex]. Therefore, the velocity of a point located at position [itex]x+\mathrm{d}x[/itex] relative to a point at position [itex]x[/itex] is
[tex]\frac{\mathrm{d}}{\mathrm{d}t}\left[\frac{\mathrm{d}x}{\gamma}\right]=\frac{\mathrm{d}}{\mathrm{d}t}\left[\frac{1}{\gamma}\right]\,\mathrm{d}x=-\frac{Fv}{mc^2}\frac{\mathrm{d}x}{\gamma^2}=-\frac{Fv}{mc^2}\left(1-\frac{v^2}{c^2}\right)\,\mathrm{d}x.[/tex]
Let us call this velocity [itex]\mathrm{d}u[/itex] to reflect its differential nature. We add velocities relativistically to get the velocity [itex]v(x+\mathrm{d}x)[/itex] of the point at position [itex]x+\mathrm{d}x[/itex] in terms of the velocity [itex]v(x)[itex] of the point at position [itex]x[/itex] and the differential relative velocity [itex]\mathrm{d}u[/itex]:
[tex]v(x+\mathrm{d}x)=\frac{v(x)+\mathrm{d}u}{1+v(x)\,\mathrm{d}u/c^2}.[/tex]
Since [itex]\mathrm{d}u[/itex] is differential and [itex]v(x)<c[/itex], we can Taylor expand and throw away all terms of order [itex]\mathrm{d}u^2=0[/itex]:
[tex]v(x+\mathrm{d}x)=(v(x)+\mathrm{d}u)\left(1-\frac{v(x)\,\mathrm{d}u}{c^2}\right)=v(x)+\mathrm{d}u-\frac{v(x)^2}{c^2}\,\mathrm{d}u[/tex]
[tex]\Rightarrow v(x+\mathrm{d}x)-v(x)=\left(1-\frac{v(x)^2}{c^2}\right)\,\mathrm{d}u
=-\left(1-\frac{v(x)^2}{c^2}\right)\frac{Fv(x)}{mc^2}\left(1-\frac{v(x)^2}{c^2}\right)\mathrm{d}x.[/tex]
Therefore
[tex]\frac{v(x+\mathrm{d}x)-v(x)}{\mathrm{d}x}\equiv\frac{\mathrm{d}v}{\mathrm{d}x}=-\frac{Fv}{mc^2}\left(1-\frac{v^2}{c^2}\right)^2.[/tex]
This is a differential equation, and to determine the unique solution we specify the initial condition that [itex]v(0)=v_\mathrm{CM}[/itex]. Note that the slope field of this expression is as we would expect.

Does this approach look good? Qualitatively, the numerical solutions to this ODE (and the analytic inverse function [itex]x(v)[/itex]) behave exactly as they should; as [itex]v[/itex] increases, [itex]x[/itex] decreases until [itex]\lim_{v\rightarrow c}x(v)=-\infty[/itex]. Thus, although points further back along the rod move more quickly, they never reach [itex]c[/itex]. Similarly, [itex]\lim_{v\rightarrow 0}x(v)=+\infty[/itex], meaning that although points further forward along the rod move more slowly, the velocity never reaches [itex]0[/itex]. If we integrate the Lorentz-contracted length from the point where [itex]v=v_\mathrm{CM}[/itex] to the ``point'' (infinitely far away along the uncontracted rod) where [itex]v=c[/itex], we do indeed obtain an infinite length -- again, this is good, as any finite length would mean the ``end'' of the massive rod would be traveling superluminal velocity.

I want to check this solution, not only mathematically but also physically. Obviously [itex]x(v)[/itex] can be converted to [itex]x(1/\gamma)[/itex] by substituting [itex]\frac{v}{c}=\sqrt{1-(\frac{1}{\gamma})^2}[/itex]. Now, I cannot get the total contracted length of the rod since I don't know how fast its endpoints are moving. However, given that some point has velocity [itex]v[/itex], I can find its [itex]x[/itex]-coordinate (and, thus, whether or not there is a point on my rod with this velocity, using the condition that [itex]|x|<\frac{L}{2}[/itex]). I am confident in my reasoning everywhere except for one step:
[tex]\frac{\mathrm{d}}{\mathrm{d}t}\left[\frac{1}{\gamma}\right]=-\frac{Fv}{mc^2}\frac{1}{\gamma^2}.[/tex]
I know that this is true for [itex]\gamma_\mathrm{CM}[/itex]. But is it logical to assert that it is true for all positions along the rod? Does this assertion, in conjunction with the function [itex]\gamma(x)[/itex], imply that the ratio of [itex]\gamma[/itex] at time [itex]t[/itex] to [itex]\gamma[/itex] at time [itex]0[/itex] must be this same function that was found for [itex]\gamma_\mathrm{CM}[/itex] early on? Because that can't be true -- [itex]\gamma(x)\equiv 1[/itex] for all [itex]x[/itex] at time [itex]t=0[/itex].

But then how should I find the velocity profile for the uniform rod? It looks like the space- and time-dependence of the rod are coupled, so I can't treat them as obeying separate ordinary differential equations. Should I treat [itex]\gamma[/itex] as a function of two variables, [itex]\gamma(x,t)[/itex], and derive a partial differential equation instead? My gut says the derivation would go something like this:
[tex]\Rightarrow v(x+\mathrm{d}x)-v(x)=\left(1-\frac{v(x)^2}{c^2}\right)\,\mathrm{d}u[/tex]
where
[tex]\mathrm{d}u=\frac{\mathrm{d}}{\mathrm{d}t}\left[\frac{1}{\gamma}\right]\,\mathrm{d}x[/tex]
which gives
[tex]\frac{\mathrm{d}v}{\mathrm{d}x}=\left(1-\frac{v^2}{c^2}\right)\frac{\mathrm{d}}{\mathrm{d}t}\left[\frac{1}{\gamma}\right].[/tex]
But
[tex]\frac{\mathrm{d}}{\mathrm{d}t}\left[\frac{1}{\gamma}\right]=\frac{\mathrm{d}}{\mathrm{d}t}\left[\sqrt{1-\frac{v^2}{c^2}}\right]
=-\frac{v}{\sqrt{1-v^2/c^2}}\frac{\mathrm{d}v}{\mathrm{d}t},[/tex]
so
[tex]\frac{\mathrm{d}v}{\mathrm{d}x}=-\left(1-\frac{v^2}{c^2}\right)\frac{v}{\sqrt{1-v^2/c^2}}\frac{\mathrm{d}v}{\mathrm{d}t},[/tex]
i.e.,
[tex]\frac{\partial v}{\partial x}=-v\sqrt{1-\frac{v^2}{c^2}}\frac{\partial v}{\partial t}.[/tex]
This looks like a demented version of the advection equation. I'm not even totally sure what boundary conditions apply here; for sure, [itex]v(x=0,t)=(Ft/m)/\sqrt{1+(Ft/mc)^2}[/itex] and [itex]v(x,t=0)=0[/itex] must be true, but I doubt that will be enough to solve this PDE.

What do you guys think? Was my initial approach right? Did I underestimate this problem, and now have to solve a PDE? Or did I overthink it, and actually the solution is simple?

I acknowledge that special relativity is only really supposed to be about inertial reference frames and so what I'm doing here is a bit of a stretch. Professors I talk to agree that this is a teaser for general relativity. I'm prepared to bring mathematical machinery to bear here. I really just want to know two things: First, can anything be said about the velocity as a function of position (and time) for a rod with constant force applied to its CM? And second, what approach will get me there in a way that is consistent with relativistic kinematics? From the qualitative argument I laid out in the beginning, I simply do not buy that an accelerated uniform rod Lorentz-contracts uniformly.

This is not a homework-style problem, though like my previous questions it is rather phrased like one. Your feedback is appreciated.

Cheers,
QM
 

Answers and Replies

  • #2
Orodruin
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It is unclear what you mean by a "uniform rod". The object that you seem to want to describe cannot exist in the way you seem to believe.
I acknowledge that special relativity is only really supposed to be about inertial reference frames and so what I'm doing here is a bit of a stretch.
Where did you get this idea from? It is simply not true. Special relativity is about flat spacetime, not about inertial frames.

First, can anything be said about the velocity as a function of position (and time) for a rod with constant force applied to its CM?
Not without bringing physics of materials into the game in a fashion that is compatible with SR.

Professors I talk to agree that this is a teaser for general relativity.
I do not know who you have talked to, but they are wrong.
 
  • #3
Ibix
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If you have a rod that is accelerating such that it feels no change (i.e. accelerometers and strain gauges scattered along it give unchanging readings) then fixed points on the rod are Rindler observers with a common origin. If you pick a point on the rod to be undergoing proper acceleration ##\alpha## then it is a distance ##c^2/\alpha## from the origin, as measured in its instantaneous rest frame. Points at other distances, ##x##, from the origin will experience proper acceleration ##c^2/x##. With this, you can deduce the length of the rod at any time in any frame.

Note that I've made assumptions about the acceleration profile (eternal, constant proper acceleration). If that's not a good approximation then you need to introduce a material model of the rod.
 
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  • #4
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I acknowledge that special relativity is only really supposed to be about inertial reference frames and so what I'm doing here is a bit of a stretch
That’s a very common misunderstanding. Special relativity works just fine for accelerated frames (it has to, because the difference is just a coordinate transformation). General relativity is only needed if gravitational effects are not negligible.

If you google for “Born rigid acceleration” you’ll find some good treatments of your accelerated rod problem. You’ll also want to be comfortable with Rindler coordinates.
 
  • #5
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If you google for “Born rigid acceleration” you’ll find some good treatments of your accelerated rod problem
Do you think this rod is Born rigid? Usually I thought that rods accelerated from a single point would be non rigid.
 
  • #6
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Do you think this rod is Born rigid? Usually I thought that rods accelerated from a single point would be non rigid.
of course, but adding that term still improves the search results.
 
  • #7
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of course, but adding that term still improves the search results.
Doesn't it settle into Born rigid motion once any transients have died away? Assuming it doesn't get into some sort of plastic flow and eventual failure, of course.
 
  • #8
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Thanks everybody for your responses. I find Ibix's most helpful:

If you have a rod that is accelerating such that it feels no change (i.e. accelerometers and strain gauges scattered along it give unchanging readings) then fixed points on the rod are Rindler observers with a common origin. If you pick a point on the rod to be undergoing proper acceleration ##\alpha## then it is a distance ##c^2/\alpha## from the origin, as measured in its instantaneous rest frame. Points at other distances, ##x##, from the origin will experience proper acceleration ##c^2/x##. With this, you can deduce the length of the rod at any time in any frame.

Note that I've made assumptions about the acceleration profile (eternal, constant proper acceleration). If that's not a good approximation then you need to introduce a material model of the rod.
You are making the same assumptions I did. I have heard the terms "Rindler coordinates" and "Rindler observers" before, in the context of the Unruh effect; looking at the Wikipedia page for Rindler coordinates seems to elucidate the matter completely! (Specifically, the A "paradoxical" property subsection.)

Thanks again! I'm going to try to square this with my Lorentz-contraction differential equation approach and see where the hiccough was. I'll report back when/if I find a satisfactory answer.

Cheers,
QM
 
  • #9
Ibix
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You are making the same assumptions I did
Are you sure? I haven't followed through your maths in detail, but I think your force is constant in some inertial frame, which will lead to a time varying proper acceleration, which would need a matetial model in general.
 
  • #10
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Summary: Does dv/dx=-Fv/mc^2(1-v^2/c^2)^2 describe it adequately?

Let's say there is a constant force F applied to the center of the rod. Its momentum is p=Ft, from which we get its velocity:
Is that the momentum of a small segment of the rod located at the center?

An identical segment at other position has 1% more momentum, if the reading of an accelerometer is 1% higher there. That's because that segment has been pushed by 1% larger force for the same time as the other segment. p=F*t

Now if we know momentums we can calculate velocities. Right?


(If anyone doubts whether it's correct to infer the accelerating force from the reading of an accelerometer, even if the accelerometer moves, it is correct. Motion does not matter in this case. )



Addition:
Ok, p was obviously the momentum of the whole rod. But it's better to consider small segments of the rod instead.
 
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  • #11
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that segment has been pushed by 1% larger force for the same time as the other segment.
Surely you've been here long enough to recognize that the phrase "the same time" is a big red flag?
 
  • #12
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Thanks again! I'm going to try to square this with my Lorentz-contraction differential equation approach and see where the hiccough was. I'll report back when/if I find a satisfactory answer.
And if you are not already familiar with it, you might also want to work through Bell's spaceship paradox (many threads here, and Google will find some good references).
 
  • #13
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Surely you've been here long enough to recognize that the phrase "the same time" is a big red flag?

Well every point of my rod actually started accelerating at the same time, but at different proper acceleration, in the original rest frame.

Now that I think about it, at the start there was a short time when the rod was not Born-rigid.


Hmm ... well maybe we could say that in this case the acceleration is small enough, and the times we consider are long enough so that we don't need to care about that start of acceleration problem?

Hey now I know: Moving rod decelerates, when it's momentarily still, we start our clocks and the rod continues with the same acceleration as before.
 
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  • #14
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question marks, I think you will probably break SR if you try to find the length contraction values for an accelerating object.

You probably need GR to do these things. And I don't know if there is an equivalent of length contraction ( or expansion) in GR. It would be nice if someone could write an equation as to which kinds of length contraction or expansions would occur in Gr

and I mean an infinite rigid body would prolly be something like the singularity of a black hole....lol
 
  • #15
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You probably need GR to do these things.
Not so - you are the victim of a common misunderstanding.

Most introductory treatments of special relativity do not cover acceleration, but that’s not because SR doesn’t work with acceleration, it’s because the math is more complex and it’s hard enough getting the concepts across with the simpler math used for the unaccelerated case. But as long as the spacetime is flat (meaning no significant tidal gravity effects, and with apologies for tickling @Orodruin’s pet peeve) SR works just fine.

Rindler coordinates, Bell’s spaceship paradox, Ehrenfest’s paradox, and the line integral calculation of proper time are all worth studying here.
 
  • #16
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Alright fair enough.

The rindler metric looks different than the minkowski metric though. (very suspicious)
 
  • #17
PeterDonis
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The rindler metric looks different than the minkowski metric though. (very suspicious)
It's not suspicious at all. It's simply an example of the fact that you can describe the same spacetime geometry using very different coordinates.

If you compute the Riemann curvature tensor for the Rindler metric you will see that it vanishes identically. That is what tells you that the underlying spacetime geometry is flat Minkowski spacetime.

And you should be very careful calling something "very suspicious" when you don't understand it.
 
  • #18
Orodruin
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But as long as the spacetime is flat (meaning no significant tidal gravity effects, and with apologies for tickling @Orodruin’s pet peeve) SR works just fine.
I don’t see why it should tickle my pet peeve. I am fine with the usage here as it in essence refers to the curvature tensor being sufficiently close to zero.

The ”locally flat” I am complaining about is true everywhere despite the curvature tensor being significantly non-zero.
 
  • #19
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well I don't know all the metrics by memory peterdoris. but yeah i guess. it shows in this manual that the ricci tensor goes to zero. jst feels weird.
 
  • #20
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well I don't know all the metrics by memory peterdoris. but yeah i guess. it shows in this manual that the ricci tensor goes to zero. jst feels weird.
If this feels weird, you should start by considering the same ideas in Euclidean space rather than in spacetime. I suggest starting with polar coordinates in ##\mathbb R^2##, where ##ds^2 = dx^2 + dy^2## in Cartesian coordinates. You should find that ##ds^2 = dr^2 + r^2 d\phi^2##. Nothing about ##\mathbb R^2## changed, just the coordinates used to describe it.
 
  • #21
Ibix
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If this feels weird, you should start by considering the same ideas in Euclidean space rather than in spacetime. I suggest starting with polar coordinates in ##\mathbb R^2##, where ##ds^2 = dx^2 + dy^2## in Cartesian coordinates. You should find that ##ds^2 = dr^2 + r^2 d\phi^2##. Nothing about ##\mathbb R^2## changed, just the coordinates used to describe it.
@sqljunkey - this isn't a random suggestion. Rindler coordinates on a Minkowski plane are closely analogous to polar coordinates on a Euclidean plane. They are symmetric under a hyperbolic "rotation" in the same way polar coordinates are symmetric under a Euclidean rotation.
 
  • #22
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I would replace "accelerated uniform rod" with "accelerated rigid rod", myself.

If we accelerate a rod rigidly, it's proper length doesn't change. Though we need a bit of jargon to be precise enough to do a calculation. By "accelerate rigidly" we mean "accelerate in a Born rigid manner", or that the rod undergoes Born rigid motion.

See [wiki link]

It shouldn't be a surprise that if we accelerate a non-rigid object, it's proper length changes. The only reason for the length to stay constant is if the object is rigid.

Born's original notion of rigidity is the one we need here. The wiki link discusses some of the issues with rigidity in special relativity in more detail.

Bell's spaceship paradox is an example of how we might accelerate the ends of a rod in a non-rigid manner. If we have two spaceships at rest in an inertial frame of reference, and start their engines at the same time in said inertial frame so that they each accelerate at a constant proper acceleration, the spaceships won't maintain a constant proper distance. Thus a string connecting the two spaceships will break, and the system of spaceships + string is not undergoing Born rigid motion. It's certainly a possible state of affairs - however, it's not rigid, and the proper length between the two spaceships changes in this scenario.

Born's original papers would probably be a good reference, unfortunately they are written in German and behind a paywall. I don't know if English translations of the original paper are available. Hyperbolic motion is Born rigid motion, though, and is frequently discussed in modern English textbooks in terms of the Rindler metric. However, metrics are typically introduced in the context of tensors, which is an A-level topic.
 
  • #23
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Before this rod accelerates into infinity, I was wondering one more thing. If the observer of this rod was also accelerating and or moving with a constant velocity(in some direction), would we also have a metric tensor that has Riemann curvature tensor that goes to zero?
 
  • #24
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Before this rod accelerates into infinity, I was wondering one more thing. If the observer of this rod was also accelerating and or moving with a constant velocity(in some direction), would we also have a metric tensor that has Riemann curvature tensor that goes to zero?
Yes. The metric tensor is a property of spacetime, not of the point of view of an observer. Suggesting that it might be zero for some observers and not others is like suggesting that the surface of a sheet of paper might be flat when we use Cartesian coordinates to label points on its surface but not flat when we use polar coordinates.
 
  • #25
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Okay, I am within tasting distance of the answer I'm looking for. I found this question posted to PSE:
https://physics.stackexchange.com/questions/175684/lorentz-contraction-in-continuously-accelerating-rodThis is very similar to the question I am asking here.

The last element I need comes from the Wikipedia page for Bell:
https://en.wikipedia.org/wiki/Bell's_spaceship_paradox#Constant_proper_accelerationThis page links arXiv #1003.3022, which contains the formula ##\mathrm{d}x=\mathrm{d}(\frac{c^2}{a})## (I am restoring factors of ##c## for clarity). Can someone please tell me how I might derive this formula? The resulting ODE goes a long way toward answering my initial question.

Thanks again,
QM
 

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