Length contraction in special relativity

Question

Hello I was just wondering whether the center (origin) is conserved along the axes of an object moving at relativistic speeds. For example if you have a long train moving at relativistic speeds to the right, with its center defined to be at position (0,0), will its center shift or change from (0,0) due to the speed, for example it may shift towards the right and become (0,3) for example. In General is the contraction uniform, if I had another point that is 1/3 away from the front of the train in its rest frame, will it remain as 1/3 away from the front of the train after contraction.

Answer 1

Suppose the train is 2 meters long (in its own frame). You can just as well think of this as two 1-meter long trains that happen to be traveling nose-to-tail.

For someone standing in the station, each of these 1-meter long trains is traveling at the same speed, and therefore each is equally length-contracted --- say to 3/4 of a meter each.

In other words, if an observer on the train mentally divides the train into a front half and a back half, then an observer on the platform must say that those two halves are equally long, and hence must agree about where the center lies.

(More precisely, the worldline of the center-as-perceived-from-the-train and the worldline of thee center-as-perceived-from-the-station must be the same line.)

Answer 2

Your attempt to pose the question in terms of an origin is misconstrued. However, the basic answer is yes.

If the observer is on the train, such that it makes sense to define the origin to move with the train, the observer measures no length contraction of the train; rather, the rest of the universe is length-contracted along the axis of the train's motion.

If the observer is on the station, it does not make sense to define the origin to move with the train. However, if the observer measures the displacement from her to the front of the train to be $x+a$ and the displacement from her to the back of the train to be $x-a$ when the train is stationary, then when the train is in motion, she observes the displacement from her to the front of the train to be $x+a/\gamma$ and the displacement from her to the rear of the train to be $x-a/\gamma$. Where $a$ is some constant with units of length.

In short: if Alice, on the station, measures Bob at the center of the train at any relative velocity, or if Bob measures himself to be at the center of the train at any relative velocity to Alice, Alice and Bob will always agree that Bob is at the center of the train regardless of any subsequent changes to the relative velocity, provided the measurement is taken while the train is not accelerating.

This answer assumes that the problem is in one dimension plus time, that is, the distance from the Alice to Bob when Bob's train drives right past her is negligible and the train moves in a straight line.

Answer 3

So, you have confused some people by using a strange phrase, “conservation of center of mass.”

Yes, Lorentz transforms are linear.

By using various different pairs of masses $m_{1,2}$ at different points $\mathbf r_{1,2}$ you seek to know if the Lorentz transforms of special relativity $\mathbf v'=T(\mathbf v)$ always ppreserve the simplest form for center of mass, $$T\left( {m_1\over m_1+m_2}\mathbf v_1 + {m_2\over m_1+m_2}\mathbf v_2 \right) = {m_1\over m_1+m_2}T(\mathbf v_1) + {m_2\over m_1+m_2}T(\mathbf v_2). $$ This property is called linearity and the answer is yes, in special relativity the Lorentz transforms are linear transforms.

No, they do not in general preserve center of mass.

However, this property also requires all of the masses to be moving uniformly (like a train does!) for the resultant to be recognizable as the center of mass of the moving system. If they are not then you run into a problem with the relativity of simultaneity... The things you have are technically worldlines and you are skewing them diagonally to the side, but you are also evaluating them with a new plane specifying present... I think it's fine if they're all parallel but when they're not parallel it becomes weird.

Let's not be abstract but concrete with an example: Consider two equal masses moving away from each other on some line, in some reference frame they are both moving at the same speed, we can put the point halfway between them as the origin. (For the time, choose the time to be zero by backtracking to a collision that they're coming away from... could be fictitious if they accelerated in the past, doesn't matter much.) So one is moving at $+v$, one at $-v$, on some $x$-axis. The center of mass is at rest on the origin.

Apply a standard Lorentz boost with velocity $+v$ to this situation: now,

• the mass that was moving at $-v$ is now at rest at the origin,
• the center of mass that was not moving is now moving at speed $+v$,
• but, the other mass is not now moving at speed $+2v$. It's slower, at $2v/(1 + v^2/c^2)$, and,
• this doesn't even have a good interpretation by using the so-called relativistic mass rather than the rest mass.

So the broader question that you have asked is easily seen to be false, but for just the purposes that you were looking at it, all you cared about was linearity and yeah, Lorentz transforms are linear.

Answer 4

Suppose in the rest frame of the train, the world-line of the center-of-mass is parameterized by coordinates

$$x_0^{\mu}(t)=(t,0,0,0)$$

so for all times $t$, the train is at the origin.

In the rest frame of an observer, this train is moving with velocity $v$ to the right. Let's suppose that in this frame, at $t'=0$ the train's center-of-mass aligns with the observer's origin $x'=y'=z'=0$. The new world-line of the train's center-of-mass, as seen by the observer, will be $x_0'^\mu=\left(t',x',y',z'\right)$. In terms of the old coordinates, it will be given by the standard Lorentz transformation formulas:

$$\begin{align} t'&=\gamma t\\ x'&=\gamma v t = v t'\\ y'&=0 \\ z'&=0 \end{align}$$

where $\gamma=1/\sqrt{1-\left(\frac{v}{c}\right)^2 }$. So you see, the center-of-mass doesn't "shift right". In the observer's frame, it still crosses the origin at $t'=0$.

To see length contraction, let's see how the distance between the front of the train and the center changes between the two frames. In the train's rest frame, the front is defined by:

$$x_1^\mu = \left( t , d, 0, 0 \right)$$

so at for all times, the separation between the center and front is $\delta x = d$. In the observer's rest frame, it's given by:

$$\begin{align} t'&=\gamma \left( t+\frac{v d}{c^2}\right) \\ x'&=\gamma\left(d+v t \right)\\ y'&=0 \\ z'&=0 \end{align}$$

If the observer measures the distance between the center and front of the train at $t'=0$, they will get

$$ \delta x' = \frac{d}{\gamma} = \frac{\delta x}{\gamma}$$

So they will see a shortened train, i.e. the front is squished back by a factor $1/\gamma$.

Answer 5

Yes, the observed length contraction is uniform. If the moving object were a meter rule, say, and its length in the moving frame was measured to be 0.9m, then every millimetre gap marked upon the rule would be measured to be 0.9mm in the moving frame.

The effect applies to distance itself, so you don't need to introduce physical objects in order to talk about it. If you have two points that are a metre apart in the stationary frame, the distance between them will in general be less in any moving frame. The difference in the distance between the two frames depends on the component of their relative speed parallel to the line joining the two points.