It's a well-known fact from special relativity that moving objects are subject to a length contraction: an object with a proper length $L_0$ moving at a high velocity $v$ will appear to a stationary observer as having a length $L=L_0/\gamma$, where $\gamma=(1-v^2/c^2)^{-1/2}$ is the Lorentz factor.

Now consider an electron bunch in a particle accelerator as in the following image.

enter image description here

The particle accelerator is simply a DC voltage ramp. The electron bunch enters with relativistic velocity $v_1$ and a length $L_1$ (for the stationary observer). It will be accelerated to a higher velocity $v_2$. Let's say that we start with $\gamma=5$ (or $v/c=0.98$) and the bunch is accelerated to $\gamma=10$ (or $v/c=0.995$) by a voltage of 2.55 MV. We will ignore space-charge effects (repulsion between the electrons) and effects of field curvature near the openings in the "A" and "B" electrodes.

What is the length $L_2$ of the accelerated electron bunch? Is it further Lorentz-contracted to $L_2=L_1 \gamma_1/\gamma_2$? Does it get stretched because it is not a rigid body?

I had heated discussions with other physicists on this question over a number of coffee breaks. Of course, I'm convinced that my answer is the correct one, but I was not able to convince everyone else. So I'll post my view below and will wait for other viewpoints. If you think you understand special relativity, think a while to make up your own mind before you read my answer.


2 Answers 2


If you're reading this, I assume that you either have no idea how to approach this question, or have already come to an answer of your own. :-)

To answer the question, we first have to agree about the meaning of the term "length". It is not the length that you would see if you made a photograph with a short shutter time, because then you would need to account for the difference in travel time of the light from different parts of the object to the camera lens. That would lead to Terrell rotations, but that's not what we talk about here.

Length is the product $v\Delta t$, where $\Delta t$ is the duration of the bunch, i.e., the time elapsed from the moment that the front of the bunch passes some fixed point to the moment that the trailing edge of the bunch passes that point. We know $v$ because we know the kinetic energy and the mass of the electrons.

Each electron enters the accelerator with the same velocity and subsequently feels exactly the same accelerating field. So, each electron takes exactly the same time to travel from plane A to plane B. If the initial bunch had a duration $\Delta t=L_1/v_1$, then the final bunch will have the same duration $\Delta t$, corresponding to a length $L_2=v_2\Delta t=(v_2/v_1)L_1$. Since $v_1$ and $v_2$ are already very close to the speed of light, the final length differs barely from the initial one: $L_2/L_1=1.015$: a very slight elongation rather than a contraction.

  • $\begingroup$ Yup, this is absolutely correct. I guess the main misconception people would have is that it's tempting to just apply length contraction every time something is moving fast. But this problem involves only one reference frame, so that doesn't make sense. $\endgroup$
    – knzhou
    Commented Jun 11, 2016 at 18:17
  • $\begingroup$ It actually goes to show that an electron bunch is by no means a rigid body: the co-moving/rest lengths of the bunch before and after acceleration are related as $$L^0_2 = \frac{v_2\gamma(v_2)}{v_1\gamma(v_1)}L^0_1$$ which for the figures proposed reads $L^0_2 \approx 2L^0_1$. In fact other frames will also observe a significant elongation. $\endgroup$
    – udrv
    Commented Jun 12, 2016 at 4:18


I do not interpret length contraction but length expansion.

Whether V1 or V2, the lab observer sees the electron bunch, so the bunch is 10 Lo in length.

V = Vo: Lo, Length of electron bunch running at slow speed (approximate)
V = V1: 5Lo when γ = 5
V = V2: 10Lo when γ = 10
Therefore, the length of the bunch is 10 Lo.

The final length(V2), of course, doubles the length when entering.(V1)

$$L={1\over{\sqrt{1-β^2}}}L_o$$ This is an expression of length expansion rather than length contraction.

Do you know the Bell's spacecraft paradox? Your problem is the same as Bell's paradox. If you know this, our story becomes shorter. The main researchers' view of Bell's paradox today is that two laws of length apply together.

The body of the spaceship has a general Lorentz contraction, and the space between the two spaceships will expand. This is not my opinion, but a common view of many scholars. This follows the view of Dewan and Beran.

(1) Rigid body: The distance between two ends of a connected rod
----> Lorentz contraction

(2) Space: The distance between two objects which are not connected but each of which independently and simultaneously move a with same velocity with respect to an inertial frame.
-----> Space expansion

This view, however, presents a serious contradiction. The three objects(two spaceships and string) that move together are subject to two different laws of physics.

enter image description here

This is a direct violation of the Einstein's principle of relativity. The three objects must be governed by a single physical law. I have many methods to prove the length expansion. I will only talk about the simplest. Do you think the time dilation is right? The discussion is very simple if you think it is right.

enter image description here

All problems are solved simply by considering length expansion instead of length contraction. If you have done the experiment, I want to know the result. I also have a lot of experimental evidence of length expansion.

I hope that your research will be honored.

My opinion on Bell's spaceship paradox

  • 1
    $\begingroup$ Hi, welcome to Physics SE! Please don't post formulas as screenshots, but use MathJax instead. MathJax is easy for people on all devices to read, and can show up clearer on different screen sizes and resolutions. $\endgroup$
    – user191954
    Commented Jun 26, 2018 at 10:00
  • $\begingroup$ You explain Bell's spaceship paradox in the framework of the Lorentz Ether theory. It is empirically equivalent to SR. In this theory, if we accelerate two spaceships simultaneously and equally in preferred frame, measuring rods of astronauts will shorten. Though distance between spaceships actually doesn't change, due to contraction of their rulers they will think that spaceships scatter from each other. Speed of light back in forth in their frame will be not the same, but if they are relativists they will say that their clocks no longer synchronous. All the same but very simple and logical. $\endgroup$
    – user139020
    Commented Jun 26, 2018 at 20:27
  • $\begingroup$ researchgate.net/publication/… $\endgroup$
    – user139020
    Commented Jun 26, 2018 at 20:36
  • $\begingroup$ But - yes, even in SR relativistic aberration formula clearly indicates, that "from the point of view" of a moving observer measuring rod “at rest” is $\gamma$ times longer than his own. Look Feynman lectures on relativistic aberration. However, inertial observer, if he thinks, that he is a rest can call the same phenomena as "light time correction" and to avoid the "problem" this way. Good to note, that rotating observer cannot call aberration "light time correction". Thus, rotating observer will always see, that measuring rod in the center of circumference is always $\gamma$ times longer $\endgroup$
    – user139020
    Commented Jun 26, 2018 at 20:50
  • $\begingroup$ The following people agree that the point between the two spaces is expanding. Michael Weiss, Don Koks, Jerrold Franklin, Dewan, Beran, Vesselin Petkov, Brocwell. Jerrold Franklin has also directly derived the L'=γLo equation. Do you deny the opinions of these people? Can you explain Bell's paradoxes perfectly without any inconsistency using relativistic aberration formula? $\endgroup$ Commented Jun 27, 2018 at 1:15

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