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What does the following mean with respect to special relativity?

$$\frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$$

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It is the Lorentz factor: en.wikipedia.org/wiki/Lorentz_factor and it means that c is an invariant speed en.wikipedia.org/wiki/Invariant_speed –  Alfred Centauri May 26 at 1:54
    
I did not even know it had a name. –  Michael Lee May 26 at 21:02
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It just occurred to me. If the Lorentz factor is true, as I'm sure it is, then the reason why a body cannot travel at the speed of light is because the denominator here would be zero. –  Michael Lee May 26 at 22:18
    
@MichaelLee, don't forget that you can accept answers if they answer your questions. –  Colin McFaul May 31 at 18:26

4 Answers 4

up vote 8 down vote accepted

Colin McFaul's answer is accurate. If you're wondering why it's in that weird form, it has to do with a fundamental postulate of special relativity.

In 2D, say you're given the endpoint of a line that you know starts at the origin, and you want to find its length. According to the diagram below, if you're given a vector in $xy$ coordinates, $(a, b)$ (represented by the red line), you can look at it in an arbitrary coordinate system by rotating your head, but physically, the length of the arrow must remain unchanged. If the coordinates in this other coordinate system are $(c,d)$, we can phrase this with the pythagorean theorem: $a^2+b^2=c^2+d^2=\mbox{distance}$

If we have a special coordinate system drawn, $x'y'$, then $d=0$ and we just have $a^2+b^2=c^2$. In this new frame, its length is just the position on one axis.

diagram of euclidean coordinate system

In relativity, we have one axis $ct$ (where $c$ is the speed of light), and another axis $x$. $ct$ and $x$ are both measured in meters. Instead of a Euclidean distance relation, the factor that remains unchanged from "rotations" is $(ct)^2-x^2$. This is the crux of relativity. It means you can "rotate" space into time and vice versa.

In one frame, the particle may have a constant velocity, so $x=v t$ and the law is conservation of $(ct)^2-(v t)^2=(c^2-v^2)t^2$. If we choose a primed reference frame so that the speed is zero, then we must have $(c^2-v^2)t^2=(ct')^2$. With some algebra we find that this equation implies that $$t=t' \frac{1}{\sqrt{1-\left( \frac{v}{c}\right)^2 }}$$

This is written as $t=\gamma t'$. It all stems from the invariance of $(ct)^2-x^2$. You can do analogous things when talking about $x$ and $x'$.

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Nice geometrical interpretation! –  Davidmh May 26 at 7:29

This is called the Lorentz factor, and is usually denoted by the symbol $\gamma$. The Lorentz factor shows up in the formulas for the Lorentz transformation and the expressions for relativistic mass, momentum, and energy. It also appears in the equations that describe how the electromagnetic fields transform due to motion of an observer.

It means that all of the unusual behavior due to special relativity increases as speed increases. As $v\rightarrow c, v/c \rightarrow 1$, so $\gamma \rightarrow \infty$. So all of these effects increase without bound as the speed approaches $c$. That is, as an observer's speed approaches $c$ relative to their surroundings, they will observe the length of other objects to shrink towards zero, the time of other objects to stand still, other objects with any mass at all to have infinite mass, energy, and momentum. Two consequences of all that is that $c$ is a universal speed limit, and that all observers agree on the value of $c$.

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I get to re-use an old blog post!

Rindler demonstrates[2] the best way ever to derive the relativistic Lorentz contraction/time dilation factor, gamma, (of twin paradox fame), from four velocity! I've been working through Rindler's paper on hyperbolic motion as a result of constant acceleration[1] lately.

Starting from the constancy of the speed of light, Einstein's theory of special relativity posited, and Minkowski refined the idea that the universe is actually four dimensional with space and time sitting on an equal footing. Starting from here, we can write an expression for the distance squared along an infinitesimal line element in four dimensions, (think Pythagorean theorem)...

$ds^2 = c^2dt^2-dx^2-dy^2-dz^2$

$=c^2dt^2-(dx^2+dy^2+dz^2)$

Brian Greene popularized the idea of four velocity[3] in one of his books and although it isn't mentioned as much as some of the other aspects of special rel, it's a simple idea. Everything is moving at the speed of light. Something might have more of it's velocity pointed into either the space dimensions, or the time dimension, but at the end of the day, its four dimensional speed is equal to the speed of light, c. So, let's write down the infinitesimal distance along a four dimensional line divided by an infinitesimal amount of time as a velocity

$\frac{ds^2}{d\tau^2} = c^2$

which can be rearranged to give

$d\tau^2 = \frac{ds^2}{c^2} = dt^2-\frac{1}{c^2}(dx^2+dy^2+dz^2)$

From here on out, it's all Rindler. Factor the increment of time out of the above expression

$dt^2-\frac{1}{c^2}\frac{(dx^2+dy^2+dz^2)}{dt^2} = dt^2\left(1-\frac{1}{c^2}\frac{(dx^2+dy^2+dz^2)}{dt^2}\right) $

At this point, the velocity, (distance over time), squared is sitting in the second term, so we can write

$dt^2\left(1-\frac{1}{c^2}\frac{(dx^2+dy^2+dz^2)}{dt^2}\right) = dt^2\left(1-\frac{1}{c^2}v^2\right) $

Leaving us with

$dt^2\left(1-\frac{1}{c^2}v^2\right) $

$d\tau = dt\sqrt{1-\frac{c^2}{v^2}}$

Which can be rearranged once again to give

$dt = \frac{d\tau}{\sqrt{1-\frac{c^2}{v^2}}} = \gamma d\tau$

Which is the formula for time dilation, (the time in the moving frame is equal to gamma times the time in the rest frame), so

$\gamma = \frac{1}{\sqrt{1-\frac{c^2}{v^2}}}$

and we're done!... or are we?

If we keep staring at $\gamma$ we might be tempted to do more with it than just quote Rinler's derivation

We can write: $\gamma = \frac{1}{\frac{1}{c}\sqrt{c^2-v^2}} = \frac{c}{\sqrt{c^2-v^2}}$,

which anyone in high school trig would think looked an awful lot like a sine or cosine except for the minus sign in the denominator. For hyperbolic trigonometry, they'd be exactly right. The length of the hypotenuse in hyperbolic trig is the adjacent side squared minus the opposite side squared. Since the expression for the hyptoenuse is assymetric, we know that we're dealing with the adjacent side over the hypotenuse, als known as the hyperbolic cosine, so

$\gamma = cosh(\phi)$,

We might also notice that we can write

$\frac{v}{\sqrt{c^2 - v^2}} = sinh(\phi)$

and we would have arrived at the two components of four velocity. We've shown that the rapidity angle can be used to project the invariant velocity c of a projectile into its proper time and space velocities, $\frac{dt}{d\tau}$ and $\frac{dx}{d\tau}$.

Historical Notes Hyperbolic velocity angles were dubbed rapidities by Alfred Robb in a pamphlet he printed in 1911. What the above notation points out is something that Minkowski emphasized only in his first lecture on spacetime. The velocities of all particles in special relativity are constant at the value of c. This viewpoint was developed further by Sommerfeld who explicitly pointed out that adding rapidities in more than one dimension was analogous to adding angles on a sphere and amounted to performing spherical hyperbolic trigonometry. Karapetoff built the same set of hyperbolic formulas, but did not mention the spherical nature of the four space geometry. To the best of my knowledge the concept of the hyperbolic velocity sphere would go largely unknown and unmentioned until Brehme and/or Adler revived it in the 1960s.

References

  1. http://dx.doi.org/10.1103%2FPhysRev.119.2082 Rindler W. (1960). Hyperbolic Motion in Curved Space Time, Physical Review, 119 (6) 2082-2089. DOI: 10.1103/PhysRev.119.2082

  2. Special Relativity by Wolfgang Rindler http://books.google.com/books?id=8kqBAAAAIAAJ

  3. On four velocity http://copaseticflow.blogspot.com/2013/02/spheres-special-relativity-and-rotations.html

  4. Robb on Rapidity https://archive.org/details/opticalgeometryo00robbrich

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+1 really nice exposition. And a nice embellishment using hyperbolic trig. (As a self-studier) I ran into that problem to show $x'/t' = 1$ and had to fake it using a minus sign rather than a plus sign in calculating the hypotenuse. Now I know what's legit. Thanks very much. Regards, –  Andrew May 27 at 15:27
    
@Andrew I'm glad you liked it! Thanks! I ran into the hyperbolic interpretation doing self study as well. –  dolphus333 May 27 at 17:37

The (real number) expression $\sqrt{1 - \left(\frac{v}{c}\right)^2}$, or its inverse, arises in SRT when trying to draw comparisons between two (distinct) inertial systems; i.e. if the members of one are moving with respect to the members of the other, straight and uniformly, mutually with speed $v := \beta ~ c$.

My favorite derivation:
Let $A$, $B$ and $F$ be members of one inertial system, i.e. pairwise at rest to each other; and (in that order) straight to each other, i.e. such that their distance ratios satisfy

$ \frac{AB}{AF} + \frac{BF}{AF} = 1 $.

Let $J$, $K$ and $Q$ be members of one inertial system, i.e. pairwise at rest to each other; also (in that order) straight to each other, i.e. such that their distance ratios satisfy

$ \frac{JK}{JQ} + \frac{KQ}{JQ} = 1 $;

such that each of $J$, $K$ and $Q$ are passing by $A$, $B$ and $F$ (in that order) throughout the course of án experiment, and vice versa each of $A$, $B$ and $F$ are passing by $Q$, $K$ and $J$ (in that order); further

  • $F$ and $Q$ were passing each other in coincidence with observing that $A$ and $K$ had passed each other,

  • $B$'s indication of having been passed by $K$ was simultaneous to $F$'s indication of having been passed by $Q$ (and observing that $A$ and $K$ had passed each other),

  • $J$'s indication of having been passed by $A$ was simultaneous to $Q$'s indication of having been passed by $F$ (and observing that $A$ and $K$ had passed each other),

and finally (consistent with all other conditions above)

$ \frac{AB}{AF} = \frac{JK}{KQ} := \beta $.

Therefore $\frac{BF}{AF} = 1 - \frac{AB}{AF} = 1 - \beta$ and $\frac{JQ}{KQ} = 1 + \frac{JK}{KQ} = 1 + \beta$.

Considering the two explicit requirements of simultaneity above, the corresponding ratios of distances should be equal:

$\frac{BF}{KQ} = \frac{JQ}{AF}$.

Inserting expressions from above:

$\frac{BF}{KQ} = \frac{JQ}{AF} = \frac{JQ}{KQ} \frac{KQ}{BF} \frac{BF}{AF} = (1 + \beta) \frac{KQ}{BF} (1 - \beta) = \frac{KQ}{BF} (1 - \beta^2) = \sqrt{ 1 - \beta^2 }$.

(This relation between distances, of participants in the described configuration, is also called "length contraction".)

Consequently also: $\frac{KQ}{AF} = \frac{KQ}{BF} \frac{BF}{AF} = \frac{1 - \beta}{ \sqrt{ 1 - \beta^2 }} = \sqrt{ \frac{1 - \beta}{1 + \beta} }$.

Following the experiment still further, until $A$ in turn observed that $F$ and $Q$ had been passing each other, in coincidence with passing $H$ therefore

$\frac{KQ}{AF} + \frac{HQ}{AF} = \sqrt{ \frac{1 - \beta}{1 + \beta} } + \sqrt{ \frac{1 + \beta}{1 - \beta} } = \frac{2}{\sqrt{ 1 - \beta^2 }}$.

Substituting the definition of "distance" as "$c/2$ Ping duration" therefore

$\frac{KQ}{AF} + \frac{HQ}{AF} := \frac{c/2 ~ \tau K_{\text{ping} Q}}{c/2 ~ \tau A_{\text{ping} F}} + \frac{c/2 ~ \tau H_{\text{ping} Q}}{c/2 ~ \tau A_{\text{ping} F}} = \frac{\tau K_{\text{ping} Q} + \tau H_{\text{ping} Q}}{\tau A_{\text{ping} F}} = 2 \frac{\tau K[ \circ_A, \circledS H_A ]}{\tau A[\circ_K, \circ_H ]}$

where $\tau K[ \circ_A, \circledS H_A ]$ denotes the duration of $K$ from the indication having been passed by $A$ until the indication simultaneous to $H$'s indication having been passed by A,
and $\tau A[\circ_K, \circ_H ]$ denotes the duration of $A$ from the indication having been passed by $K$ until the indication having been passed by $H$. Comparison of the last two equations yields
$\frac{\tau K[ \circ_A, \circledS H_A ]}{\tau A[\circ_K, \circ_H ]} = \frac{1}{\sqrt{ 1 - \beta^2 }}$.

(This relation between durations, of participants in the described configuration, is also called a special, basic case of "time dilation".)

Two final points to emphasize (also regarding other answers submitted already):

  • the derivation as shown is independent of any (subsequent) assignment of coordinates to the participants involved, or to their indications; and

  • with $\frac{KQ}{HK} + \frac{HQ}{HK} = \frac{1}{\beta}$ follows

$\left(\frac{\tau A[\circ_K, \circ_H ]}{\tau K[ \circ_A, \circledS H_A ]}\right)^2 = 1 - \left(1 / (\frac{KQ}{HK} + \frac{HQ}{HK}) \right)^2$,

or (formally)

$\left(c ~ \tau A[\circ_K, \circ_H ]\right)^2 = \left(c ~ \tau K[ \circ_A, \circledS H_A ]\right)^2 - HK^2$

as a result (rather than being used or even required as an input to the derivation).

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Very interesting. This has a decidely 19th century flavour to it (not meaning to be critical at all): did you get this from an early paper? Also, your questions on your user page: try submitting them to Physics Overflow or Maths SE: you might get some joy from either of these: I suggest the latter because you seem to be asking fairly deep questions about geometrical assumptions. I haven't had the time to think about them in detail, but they certainly seem fair questions to me. The only possible reason I can see for their lack of interest here ... –  WetSavannaAnimal aka Rod Vance May 27 at 0:19
    
... is your slightly unwonted notation: I find, for example, your expressions containing $\circ$ a bit hard to read because $\circ$ is conventionally he function composition operator. Likewise, square brackets generally stand for either closed intervals or Lie brackets: so their use here tends to bend my mind a bit. –  WetSavannaAnimal aka Rod Vance May 27 at 0:21
    
WetSavannaAnimal aka Rod Vance: "did you get this from an early paper?" -- No, I thought of these derivations myself (based on Einstein's thought-experimental descriptions, of course); please let me/us know if you should find others having expressed similar thoughts earlier already. "your expressions containing $\circ$" -- The $\circ$ symbol in the expressions above is best read as "coincides with". (I prefer that to using the $\copyright$ symbol there. [p.s.: which seems to suffer even more "technical difficulties".] The $\circledS$ symbol is read as "simultaneous to", btw) [tbc'd] –  user12262 May 27 at 5:49
    
WetSavannaAnimal aka Rod Vance: "square brackets" -- Using square brackets in math expressions (rather than, say, parentheses) to enclose arguments to functions or formulas corresponds to Mathematica syntax, for instance. (Commutators may be read this way as well. In any case, parentheses are thereby reserved for grouping purposes only; which is my preference.) "questions on your user page: try submitting them to Physics Overflow or Maths SE" -- It's nice to still have options ... (having apparently already lost my account with Math Overflow). –  user12262 May 27 at 5:49
    
p.s. The Mathjax code which is rendered as the copyright symbol is \unicode{xa9}; viz. $\unicode{xa9}$. –  user12262 Jun 1 at 19:39

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