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I am studying a book about relativistic equations and special relativity, and I keep seeing $\sqrt{1-{v^2/c^2}}$ everywhere. It is not, as with most of the concepts in special relativity, simply a mathematical construct; it is a logical consequence of accepting the experimental fact that the speed of light is the same in every inertial reference frame. Why, then, is this expression so significant?

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    $\begingroup$ It's significant because it's a logical consequence of accepting the experimental fact that the speed of light is the same in every inertial reference frame. Why would you expect it not to be significant? $\endgroup$ – probably_someone Jan 30 '18 at 21:06
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    $\begingroup$ Check out Lorentz factor $\endgroup$ – Stéphane Rollandin Jan 30 '18 at 21:07
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    $\begingroup$ Can anyone please explain why is this getting so many downvotes? Sure, it's not the best question I've ever seen, but it is not terrible either. In fact, it is to some extent a fairly natural question for a beginner. $\endgroup$ – AccidentalFourierTransform Jan 30 '18 at 21:18
  • $\begingroup$ It's the scale on your Lorentz-transformed axes. It's like how $\cos\theta$ is important for rotations (with angle $\theta$) -- $\gamma=\cosh\xi$ is important for skews (with rapidity $\xi$). $\endgroup$ – Abhimanyu Pallavi Sudhir May 10 '18 at 14:56
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1) At least at low speeds, you expect $x'=x-vt$, just from elementary considerations. ($vt$ is, after all, the distance traveled in time $t$, so a person traveling at speed $v$ will have his origin displaced by the amount $vt$.

2) If you believe space and time should be treated symmetrically, then you are led to expect something like $t'=t-vx$.

3) So in matrix terms, our first guess is $$\pmatrix{x'\cr t'}=\pmatrix{1&-v\cr -v&1\cr}\pmatrix{x\cr t}$$

4) But if the transformation matrix is to preserve geometric structure (or, pretty much equivalently, if you want the matrix associated to $-v$ to be the inverse of the matrix associated to $v$) you want its determinant to be $1$, whereas it currently has determinant $\Delta=1-v^2$.

5) So to fix the determinant (while making changes that are negligible when $v$ is small, you multiply the transformation matrix by the appropriate constant, which is $1/\sqrt{\Delta}$. That explains where the constant comes from.

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    $\begingroup$ The usual composition of infinitesimal transformations: Let $M(v)$ be the matrix in 3). Then $M(\eta/n)^n$ tends to $\begin{pmatrix}\cosh\eta&-\sinh\eta\\-\sinh\eta&\cosh\eta\end{pmatrix}$ as $n\to\infty$, nicely connecting to Rod's answer. $\endgroup$ – AccidentalFourierTransform Jan 31 '18 at 3:53
  • $\begingroup$ I like this answer better than my own. So simple and elegant $\endgroup$ – WetSavannaAnimal Jan 31 '18 at 10:00
  • $\begingroup$ I do not see why putting matrix in an answer to a question posed by a very beginner. This is almost as homework, as surely his her book shows some easy explaino $\endgroup$ – Alchimista Jan 31 '18 at 10:46
  • $\begingroup$ @Alchimista maybe, but the "determinant needed to conserve geometric structure" is the deepest, pithiest and most insight-yielding characterization I've ever heard. I'm not sure it is a homework question: I'm pretty sure I recall groping for an elegant explanation for the factor's ubiquity as a late teenager too. Insights like these, especially conveyed to a beginner, are never forgotten. $\endgroup$ – WetSavannaAnimal Jan 31 '18 at 10:54
  • $\begingroup$ Ok. I have nothing against this answer in itself. As from the Q is formulated, I guess(ed) the answer could be something like in the comments and link by @Stephane Rollandin. Perhaps OP would benefit from everything in this thread and your and other answers will do a great job. :) $\endgroup$ – Alchimista Jan 31 '18 at 12:05
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The reciprocal Lorentz factor $$\gamma^{-1}~=~\sqrt{1-{v^2/c^2}}~=~\frac{d\tau}{dt}~<~1$$ is e.g. the ratio between proper time $d\tau$ and coordinate time $dt$.

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Let's confine ourselves to relativity in one spatial dimension: along the $x$-axis say. Velocities are then simply signed real numbers.

Owing to the observer dependence of time and distance measurements that arise between relatively moving observers, "velocity" becomes an awkward and unnatural quantifier of relative motion between frames. A more natural quantification of relative motion is the rapidity $\eta = \mathrm{artanh} \frac{v}{c}$: this takes account of the observer dependence of time and distance measurements such that relative rapidities are linear additive at relativistic velocities in the same way that everyday velocities are at everyday speed.

The co-ordinate transformation factors in the Lorentz transformation are all multiples of $\cosh\eta$ and $\sinh\eta$, which can be expanded:

$$\cosh\eta = \frac{1}{\sqrt{1-\tanh^2 \eta}} = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$

$$\sinh\eta = \frac{\tanh\eta}{\sqrt{1-\tanh^2 \eta}} = \frac{\frac{v}{c}}{\sqrt{1-\frac{v^2}{c^2}}}$$

and so the factor that mystifies you simply arises by dint of the identity $\cosh^2\eta - \sinh^2 \eta = 1$, given that the most natural measure of relative motion is related to velocity by $v = c \tanh \eta$.

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Here's a diagram to elaborate on what some of the other answers are saying.

In relativity, where there is a lot of emphasis on "what observers measure", the computation of components of vectors is important. If a vector-of-interest makes an angle with a reference-vector, then the component of that vector-of-interest along the reference-direction is related to the cosine of that angle [in that geometry].
If, instead of the angle, you wish to use a slope,
then expressions involving square-roots show up.

In relativity, the geometry is Minkowski geometry.

Let's draw a spacetime diagram.
I'm using rotated graph paper so that it is easier to visualize the ticks along the legs of the triangle.

RRGP-Triangle-1

Here the velocity (slope) $v=6/10$. The hypotenuse (of the vector-of-interest) is $T=8$. In terms of the slope, the components are as shown. $\frac{1}{\sqrt{1-v^2}}T$ is time component of that vector.

It might be more geometrically-intuitive to express this in terms of the Minkowski-angle (called the rapidity) between the vector-of-interest and the observer's time-axis. If $v=\tanh\theta$, then it turns out that $(\cosh\theta)T$ is the time-component of that vector.

That is, $$\cosh\theta=\frac{1}{\sqrt{1-v^2}}.$$ In addition, $\sinh\theta=(\tanh\theta)\cosh\theta= v\cosh\theta=\frac{v}{\sqrt{1-v^2}}$

RRPG-Triangle-2

Exercise: In ordinary Euclidean geometry, where $v=\tan\theta$, write $\cos\theta$ and $\sin\theta$ in terms of $v$.

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You know the pythagorean theorem? Take a right triangle of hypotenuse 1. Let $v/c$ be the length of one side. Then that formula is the length of the other side.

Another way to look at it: suppose the hypotenuse is $c$, the speed of light, and $v$ is one side of the triangle, the speed something is moving. The closer $v$ comes to $c$, the smaller the third side is, and $v$ can never be greater than $c$ without breaking the right triangle.

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    $\begingroup$ As it stands this is a pretty meaningless answer. This does not supply the OP with any information relevant to understanding relativity any better. $\endgroup$ – StephenG Jan 30 '18 at 21:12
  • $\begingroup$ @StephenG: I fleshed it out a bit. Remember, the OP is just a newbie/kid. $\endgroup$ – Mike Dunlavey Jan 30 '18 at 21:18
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    $\begingroup$ This page explains how one can get the Lorentz factor via Pythagorea's theorem. It could maybe complement this answer. $\endgroup$ – Stéphane Rollandin Jan 30 '18 at 21:31
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    $\begingroup$ Thank Mike. It's because the OP is a newbie that I figure it needed more info rather than less. YMMV of course. $\endgroup$ – StephenG Jan 30 '18 at 21:48
  • $\begingroup$ @StephenG, I would argue that this is actually key to the most straightforward and intuitive explanation of relativity, though it perhaps needs more elaboration for that to be seen. $\endgroup$ – Steve Jan 31 '18 at 6:49

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