Prove no contraction in the orthogonal directions

Question

When talking about relativity, we always mention Lorentz contraction. If a body is moving with velocituy $V$ in the $x$ direction, its length will be contracted in that direction. The length remains the same in the orthogonal directions.

This is usually seen as obvious, as the velocity is zero in those directions so the object might as well be at rest, as long as those directions are concerned.

However, a student today raised an objection. She said "no, this is not obvious; the moving object is not at rest and I cannot be sure what happens; in principle, there could be contraction in all directions".

She wanted a physical reason to convince her that no contraction takes place in the orthogonal directions. She took me off guard and I didn't know what to say. It is sometimes hard to argue the obvious.

Any help?

Answer 1

Thought experiment:

Two rings are flying toward each other at relativistic speed. The rings are perpendicular to the velocity, flat toward each other, and have very thin paper stretched across like a drum head.

Now imagine the high speed makes transverse measures smaller. Ring A sees a tiny ring B coming toward it: B punches a tiny hole in A, in the process being obliterated. But B sees A tiny, and sees A punching just a tiny hole in B.

That’s a contradiction: two different results of the same space-time events doesn’t happen.

Ditto if transverse measures get longer.

For longitudinal measures, simultaneity creates consistency between separate points. But here we see inconsistency at the same space-time point.

Must be that neither happens, because transverse measures don’t change, and the rings exactly hit each other.

Answer 2

If you think about this question in terms of the matrix elements of the Lorentz transformation $\Lambda$, for a boost in the $x$ direction, there are really two separate questions.

(1) First there is the question of why we can't have $\Lambda_{yt}\ne0$. This would violate symmetry under parity or time reversal.

(2) Next there is the question of why we must have $\Lambda_{yy}=1$. There are various arguments that can be constructed:

• Rings or cylinders passing through each other. This is described in Bob Jacobsen's answer.

• The "nails on rulers" argument. Suppose that two observers, in motion relative to one another along the $x$ axis, each carry a stick oriented along the $y$ axis, such that the butts of the sticks coincide at some time. Due to the vanishing of the types of terms in the transformation referred to in (1) above, the sticks are collinear at this time. Therefore these observers must agree whether the sticks are equal in length or, if not, then on whose is longer. But if the sticks are unequal, then this would violate the isotropy of space, since it would distinguish +$x$ from −$x$.

• Arguments from conservation of spacetime volume. Under a Lorentz transformation, one can prove from symmetry arguments that area in the $(t,x)$ plane is preserved. The same proof applies to volume in the spaces $(t, x, y) $ and $(t, x, z)$, hence lengths in the $y$ and $z$ directions are preserved.

Answer 3

To add to Ben Crowell's answer there are several slightly different symmetry arguments that lead to the conclusion you seek and can be taken as fulfilling motivation for believing (aside from experimental evidence) that directions normal to a boost direction are unscaled.

But it basically follows from isotropy of space: the idea that there is no preferred direction.

(1) Homogeneity of space (a distinct concept) assumptions and (2) the assumptions of the Lorentz transformation for any given relative speed is a continuous function of the spacetime co-ordinates and (3) that it is a continuous function of the relative speed can all be combined to show that the Lorentz transformation is linear and moreover must have the form:

$$\Lambda = \exp(\eta\,K)\tag{1}$$

where $\eta$, the rapidity or generalized velocity, is some yet to be determined continuous function of the relative speed and $K$ a constant $4\times 4$ matrix. I outline the steps needed to see this in my answer here and other parts of the arguments are in answers that mine link to.

Let's write $K$ as:

$$K=\left( \begin{array}{cccc} K_{tt} & K_{tx} & K_{ty} & K_{tz} \\ K_{xt} & K_{xx} & K_{xy} & K_{xz} \\ K_{yt} & K_{yx} & K_{yy} & K_{yz} \\ K_{zt} & K_{zx} & K_{zy} & K_{zz} \\ \end{array}\right)\tag{2}$$

and let's call our boost direction the $x$ direction.

So now we assume space is isotropic. In particular this means that if we rotated our co-ordinates any angle about the $x$ axis, our Lorentz transformation would have to be the same. Now a rotation through angle $\theta$ about $x$ is:

$$R(\theta) = \exp(\theta\,H_x)\tag{3}$$

where

$$H_x = \left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 \\ \end{array}\right)\tag{4}$$

and our invariance with respect to rotation is described by:

$$\exp(\theta\,H_x)\,\Lambda\,\exp(-\theta\,H_x) = \Lambda;\quad\forall\theta\in\mathbb{R}\tag{5}$$

a statement which can be shown to be equivalent to:

$$[H_x,\,K] = H_x\,K - K \,H_x = 0\tag{6}$$

(6) can tediously be shown to imply that the most general form of $K$ is now:

$$K=\left( \begin{array}{cccc} K_{tt} & K_{tx} & 0 & 0 \\ K_{xt} & K_{xx} & 0 & 0 \\ 0 & 0 & K_y & -K_{zy} \\ 0 & 0 & K_{zy} & K_y \\ \end{array}\right)\tag{2}$$

Now we apply isotropy of space a second time, and flip our $x$ axis over to the opposite direction by rotating through $180^\circ$ about the $y$ or $z$ axes. Either (or any linear superposition) will do just fine and all give the same result. Given that space is isotropic, the Lorentz transformation must have the same form, but with a subtlety. We need to allow the possibility that the rapidity $\eta$ might be some different value. That is, we have the same kind of motion, but it is in the opposite direction, so we don't yet know what happens to the rapidity in reversing the sense of direction.

Thus we must have:

$$\eta\,\mathrm{diag}(1,\,-1,\,-1,\,1)\,K\,\mathrm{diag}(1,\,-1,\,-1,\,1)^{-1} = \eta^\prime\,K\tag{7}$$

and this, applied to (7) yields several equations that we must consider carefully:

$$K_{tt}\,(\eta -\eta^\prime) = K_{xx}\,(\eta -\eta^\prime) = K_y\,(\eta -\eta^\prime)=0\tag{8}$$ $$K_{tx}\,(\eta +\eta^\prime) = K_{xt}\,(\eta +\eta^\prime) = K_{zy}\,(\eta +\eta^\prime)=0\tag{9}$$

So if any of our $K$ are to be nonzero (i.e. if $\Lambda$ is to be nontrivial) we have two distinct choices: $\eta^\prime = \pm1$.

Both are so far possible, but think what the choice $\eta^\prime = \eta$ would do. We'd have a diagonal Lorentz transformation. That would be a truly weird universe. Boosts would scale space and time co-ordinates, but we'd have, through (9), $K_{xt}=K_{zy}=0$: there'd be no off diagonal terms in the Lorentz transformations and it would not be motion at a velocity as we know it. Instead, everyone would be rooted to the same spot in space, and spatial co-ordinates could be uniformly (isotropically) shrunken and swollen at will (or at least, according to the rapidity) - even without a "Drink Me" bottle. For obvious reasons, Sean Carroll calls this the "Alice in Wonderland" universe. One doesn't need to study this too hard to understand that it simply cannot accord with our everyday experience, at least when we're not under the influence of highly psychotropic substances.

So we are left with the relativistic reciprocity theorem as the only workable alternative $\eta^\prime = -\eta$, i.e. that the rapidity corresponding to relative motion of the same speed but in the opposite direction is the same magnitude but opposite sign. And with this possibility we are forced, from (8), to conclude that:

$$K_{tt}=K_{xx}=K_{yy}=K_{zz}=0\tag{10}$$

and that the only remaining possibility is that the Lorentz transformation is of the form:

$$\Lambda(\eta) = B(\eta)\circ R(\eta) = R(\eta)\circ B(\eta)\tag{11}$$

where $B$ is a boost that leaves the directions orthogonal to the motion untransformed, and $R$ is a rotation about the $x$ direction.

This answers your question, because it shows there is no contraction in directions orthogonal to the motion. To complete the tale and get fully to the Lorentz transformation, though, we tie up the following loose ends.

Since we can, in the physical world impart a rotation without relative motion, we can do so to cancel the rotation part of (11) so that if (11) is a possibility then we can always arrange to have:

$$\Lambda(\eta) = \left( \begin{array}{cc} \cosh \left(\eta \sqrt{K_{tx}\,K_{xt}}\right) & \sqrt{\frac{K_{tx}}{K_{xt}}} \sinh \left(\eta \sqrt{K_{tx}\,K_{xt}}\right) \\ \sqrt{\frac{K_{xt}}{K_{tx}}} \sinh \left(\eta \sqrt{K_{tx}\,K_{xt}}\right) & \cosh \left(\eta \sqrt{K_{tx}\,K_{xt}}\right) \\ \end{array} \right)\tag{12}$$

and most of the constants in (12) can be removed by appropriate choice of units, aside from one crucial point: the sign of $K_{xt}\,K_{tx}$. If this sign is negative, then the Lorentz tranformation is a rotation, and in this case we could therefore always find a boost that would reverse the time direction of the vector joining any two events. It would be very hard to make sense of causality in such a universe, since not only would simultaneity be relative, but also the order of causally linked events. Thus, finally, only the wonted and beloved Lorentz transformation, free of transformation on spatial co-ordinates orthogonal to the boost, remains as the one possibility consistent with our experimental observation.

Answer 4

We can rule out one kind of more complicated relationship than $y' = y$ and $z' = z$, namely a dependency on time analogous to that of $x'$ in $x' = \gamma x - (\gamma v)t$, ie a linear relationship of the form $y' = \alpha y + \beta t$ and $z' = \eta z + \epsilon t$, for certain constants $\alpha, \beta, \eta, \epsilon$ (taking the 'standard setup' of a frame $S'$ moving at $v$ wrt frame $S$).

In other words instead of extending the Lorentz Transform from the 1-dimensional space version $$L = \left[ \begin{array}{cc} \gamma & -\gamma v \\ -\gamma v/c^{2} & \gamma \end{array} \right] $$

to the 3-dimensional space version $$L = \left[ \begin{array}{cccc} \gamma & 0 & 0 & -\gamma v \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ -\gamma v/c^{2} & 0 & 0 & \gamma \end{array} \right] $$

we extend it to $$ L = \left[ \begin{array}{cccc} \gamma & 0 & 0 & -\gamma v \\ 0 & \alpha & 0 & \beta \\ 0 & 0 & \eta & \epsilon \\ -\gamma v/c^{2} & 0 & 0 & \gamma \end{array} \right] $$

We can then show necessarily that $\alpha = 1, \beta = 0, \eta = 1, \epsilon = 0$, ie we are back to the original Lorentz Transformation. This at least rules out length contractions in the $y$ and $z$ directions of a certain kind which are analogous to the case of the $x$-direction, where $x'$ has a linear dependency on $x$ and $t$.

The method of proof is to use the property of preserving 'light-like intervals' which $L$ must possess, ie intervals $\Delta \mathbf{w} = (\Delta x, \Delta y, \Delta z, \Delta t)$ with the property $(\Delta x)^{2} + (\Delta y)^{2} + (\Delta z)^{2} = c^{2}(\Delta t)^{2}$. L is required to map $\Delta \mathbf{w}$ to an interval $\Delta \mathbf{w'} = (\Delta x', \Delta y', \Delta z', \Delta t')$ with the same property, ie $(\Delta x')^{2} + (\Delta y')^{2} + (\Delta z')^{2} = c^{2}(\Delta t')^{2}$, because the speed of light has to be preserved by $L$.

Because $L$ is linear we have $\Delta \mathbf{w'} = L \Delta \mathbf{w}$, and this is the interval between the two events in the frame $S'$ (in the 'standard setup').

Now take $\Delta \mathbf{w} = (c, 0, 0, 1)$, which is a light-like interval, then interval $\Delta \mathbf{w'} = L\Delta \mathbf{w}$ must be light-like also. But $\Delta \mathbf{w'} = (c\gamma - \gamma v, \beta, \epsilon, -\gamma v / c + \gamma)$, so

\begin{eqnarray*} \gamma^{2}(c - v)^{2} + \beta^{2} + \epsilon^{2} & = & c^{2}\gamma^{2}(1 - v/c)^{2} \\ & = & \gamma^{2}(c - v)^{2} \\ \Rightarrow \beta^{2} + \epsilon^{2} = 0 \\ \Rightarrow \beta = \epsilon = 0. \end{eqnarray*}

Taking light-like interval $\Delta \mathbf{w} = (0, c, 0, 1)$, $\Delta \mathbf{w'} = (-\gamma v, c \alpha + \beta, \epsilon, \gamma)$, and so

\begin{eqnarray*} \gamma^{2}v^{2} + (c\alpha + \beta)^{2} + \epsilon^{2} & = & c^{2}\gamma^{2} \\ \Rightarrow \gamma^{2}v^{2} + (c\alpha)^{2} & = & c^{2}\gamma^{2} \\ \Rightarrow \alpha^{2} c^{2} & = & \gamma^{2}(c^{2} - v^{2}) \\ & = & \frac{1}{1 - v^{2}/c^{2}}(c^{2} - v^{2}) \\ & = & c^{2} \\ \Rightarrow \alpha^{2} = 1 \Rightarrow \alpha = \pm 1. \end{eqnarray*}

But we require $L$ to reduce to the Galilean transformation , $$G = \left[ \begin{array}{cccc} 1 & 0 & 0 & -v \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right] $$

in the limit as $v \ll c$, and so $\alpha = -1$ would be impossible, and thus $\alpha = 1$.

Taking light-like interval $\Delta \mathbf{w} = (0, 0, c, 1)$, $\Delta \mathbf{w'} = (-\gamma v, \beta, \eta c + \epsilon, \gamma)$, and so

\begin{eqnarray*} \gamma^{2}v^{2} + \beta^{2} + (\eta c + \epsilon)^{2} & = & c^{2}\gamma^{2} \\ \Rightarrow \gamma^{2}v^{2} + (\eta c)^{2} & = & c^{2}\gamma^{2} \\ \Rightarrow \eta^{2} c^{2} & = & \gamma^{2}(c^{2} - v^{2}) \\ & = & c^{2} \\ \Rightarrow \eta^{2} = 1 \Rightarrow \eta = \pm 1 \end{eqnarray*}

And as with $\alpha$ above, $\eta = -1$ is impossible so $\eta = 1$.

Generally, the condition of preserving the speed of light is quite a strong condition on $L$ from which many equations on the 16 elements of $L$ could be derived, by choosing any light-like $\Delta \mathbf{w} = (kc, lc, mc, 1)$, ie. with $k^{2} + l^{2} + m^{2} = 1$.

Answer 5

TL;DR

Give her evidence that supports the theory (see other answers for examples) AND reinforce to her that this only shows that the phenomenon is obvious if you assume the theory to be true. But it is not at all obvious that the theory is true--this is the case with all theories, as they are models of reality whose performance can be measured, but whose truth can never be 100% confirmed. This discussion of theories and philosophy is what this answer adds to the other answers.

What Physical Reason Can You Provider To Show it's Obvious?

If her question is philosophical (see my reasoning below), then there isn't one per se (but keep reading, I do answer the question!). However, you can give her physical evidence that supports the theory (see the other answers for examples), and while this doesn't show that the theorized phenomenon is obviously the truth, it does show her that the result is obvious in the context of the theory, meaning that if you accept the theory as truth, then the phenomenon is obvious.

Why the distinction? Because there is no way to determine with 100% certainty that any theory is the truth; this is easy to see as new and better theories are constantly developed.

From a pedagogical standpoint this is an important distinction, and as a teacher it is important to reinforce this for your student.

Why am I saying this?

Maybe the Student is Asking a Different Question

Maybe the student is questioning whether the theory itself must be accepted as obviously the truth without question, raising a philosophical question instead of expressing a misunderstanding about the mechanics of the theory.

If So, This is a Good Thing

From a learning perspective, this level of awareness where a student realizes that theories are models with empirical and/or theoretical support instead of the reality itself, is valuable, and something that should be rewarded and fostered. It's the type of thinking that can lead to new out-of-the-box theories in the future.