Editing Special relativity (section)

=== Acceleration ===
{{Further|Acceleration (special relativity)}}

Special relativity does accommodate [[acceleration (special relativity)|accelerations]] as well as [[Rindler coordinates|accelerating frames of reference]].<ref>{{cite book |title=Relativity Made Relatively Easy |edition=illustrated |first1=Andrew M. |last1=Steane |publisher=OUP Oxford |year=2012 |isbn=978-0-19-966286-9 |page=226 |url=https://books.google.com/books?id=75rCErZkh7EC}} [https://books.google.com/books?id=75rCErZkh7EC&pg=PA226 Extract of page 226]</ref>
It is a common misconception that special relativity is applicable only to inertial frames, and that it is unable to handle accelerating objects or accelerating reference frames.<ref>{{cite book |title=Explorations in Mathematical Physics: The Concepts Behind an Elegant Language |edition=illustrated |first1=Don |last1=Koks |publisher=Springer Science & Business Media |year=2006 |isbn=978-0-387-32793-8 |page=234 |url=https://books.google.com/books?id=ObMb7l9-9loC}} [https://books.google.com/books?id=ObMb7l9-9loC&pg=PA234 Extract of page 234]</ref> It is only when gravitation is significant that general relativity is required.<ref name="PhysicsFAQ">{{cite web|last1=Gibbs|first1=Philip|title=Can Special Relativity Handle Acceleration?|url=http://math.ucr.edu/home/baez/physics/Relativity/SR/acceleration.html|website=The Physics and Relativity FAQ|publisher=math.ucr.edu|access-date=28 May 2017|archive-date=7 June 2017|archive-url=https://web.archive.org/web/20170607102331/http://math.ucr.edu/home/baez/physics/Relativity/SR/acceleration.html|url-status=live}}</ref>

Properly handling accelerating frames does require some care, however. The difference between special and general relativity is that (1) In special relativity, all velocities are relative, but acceleration is absolute. (2) In general relativity, all motion is relative, whether inertial, accelerating, or rotating. To accommodate this difference, general relativity uses curved spacetime.<ref name="PhysicsFAQ" />

In this section, we analyze several scenarios involving accelerated reference frames.

{{anchor|Dewan–Beran–Bell spaceship paradox}}

==== Dewan–Beran–Bell spaceship paradox ====
{{Main|Bell's spaceship paradox}}

The Dewan–Beran–Bell spaceship paradox ([[Bell's spaceship paradox]]) is a good example of a problem where intuitive reasoning unassisted by the geometric insight of the spacetime approach can lead to issues.

[[File:Bell's spaceship paradox - two spaceships - initial setup.png|thumb|Figure 7–4. Dewan–Beran–Bell spaceship paradox]]
In Fig.&nbsp;7-4, two identical spaceships float in space and are at rest relative to each other. They are connected by a string that is capable of only a limited amount of stretching before breaking. At a given instant in our frame, the observer frame, both spaceships accelerate in the same direction along the line between them with the same constant proper acceleration.<ref group=note>In relativity theory, proper acceleration is the physical acceleration (i.e., measurable acceleration as by an accelerometer) experienced by an object. It is thus acceleration relative to a free-fall, or inertial, observer who is momentarily at rest relative to the object being measured.</ref> Will the string break?

When the paradox was new and relatively unknown, even professional physicists had difficulty working out the solution. Two lines of reasoning lead to opposite conclusions. Both arguments, which are presented below, are flawed even though one of them yields the correct answer.<ref name="Morin2007" />{{rp|106,120–122}}
# To observers in the rest frame, the spaceships start a distance ''L'' apart and remain the same distance apart during acceleration. During acceleration, ''L'' is a length contracted distance of the distance {{nowrap|1=''L{{'}} = &gamma;L''}} in the frame of the accelerating spaceships. After a sufficiently long time, ''&gamma;'' will increase to a sufficiently large factor that the string must break.
# Let ''A'' and ''B'' be the rear and front spaceships. In the frame of the spaceships, each spaceship sees the other spaceship doing the same thing that it is doing. ''A'' says that ''B'' has the same acceleration that he has, and ''B'' sees that ''A'' matches her every move. So the spaceships stay the same distance apart, and the string does not break.<ref name="Morin2007" />{{rp|106,120–122}}

The problem with the first argument is that there is no "frame of the spaceships." There cannot be, because the two spaceships measure a growing distance between the two. Because there is no common frame of the spaceships, the length of the string is ill-defined. Nevertheless, the conclusion is correct, and the argument is mostly right. The second argument, however, completely ignores the relativity of simultaneity.<ref name="Morin2007" />{{rp|106,120–122}}

[[File:Bell spaceship paradox.svg|thumb|Figure 7–5. The curved lines represent the world lines of two observers A and B who accelerate in the same direction with the same constant magnitude acceleration. At A' and B', the observers stop accelerating. The dashed lines are lines of simultaneity for either observer before acceleration begins and after acceleration stops.]]
A spacetime diagram (Fig.&nbsp;7-5) makes the correct solution to this paradox almost immediately evident. Two observers in Minkowski spacetime accelerate with constant magnitude <math>k</math> acceleration for proper time <math>\sigma</math> (acceleration and elapsed time measured by the observers themselves, not some inertial observer). They are comoving and inertial before and after this phase. In Minkowski geometry, the length along the line of simultaneity <math>A'B''</math> turns out to be greater than the length along the line of simultaneity {{tmath|1= AB }}.

The length increase can be calculated with the help of the Lorentz transformation. If, as illustrated in Fig.&nbsp;7-5, the acceleration is finished, the ships will remain at a constant offset in some frame {{tmath|1= S' }}. If <math>x_{A}</math> and <math>x_{B}=x_{A}+L</math> are the ships' positions in {{tmath|1= S }}, the positions in frame <math>S'</math> are:<ref name="Franklin">{{cite journal |author=Franklin, Jerrold |title=Lorentz contraction, Bell's spaceships, and rigid body motion in special relativity |journal=European Journal of Physics |volume=31 |year=2010 |pages=291–298 |doi=10.1088/0143-0807/31/2/006 |bibcode = 2010EJPh...31..291F |issue=2 |arxiv = 0906.1919|s2cid=18059490 }}</ref>
: <math>\begin{align}
x'_{A}& = \gamma\left(x_{A}-vt\right)\\
x'_{B}& = \gamma\left(x_{A}+L-vt\right)\\
L'& = x'_{B}-x'_{A} =\gamma L
\end{align}</math>

The "paradox", as it were, comes from the way that Bell constructed his example. In the usual discussion of Lorentz contraction, the rest length is fixed and the moving length shortens as measured in frame {{tmath|1= S }}. As shown in Fig.&nbsp;7-5, Bell's example asserts the moving lengths <math>AB</math> and <math>A'B'</math> measured in frame <math>S</math> to be fixed, thereby forcing the rest frame length <math>A'B''</math> in frame <math>S'</math> to increase.

{{anchor|Accelerated observer with horizon}}

==== Accelerated observer with horizon ====
{{Main|Event horizon#Apparent horizon of an accelerated particle|Rindler coordinates}}

Certain special relativity problem setups can lead to insight about phenomena normally associated with general relativity, such as [[event horizons]]. In the text accompanying [[Spacetime#Invariant hyperbola|Section "Invariant hyperbola" of the article Spacetime]], the magenta hyperbolae represented actual paths that are tracked by a constantly accelerating traveler in spacetime. During periods of positive acceleration, the traveler's velocity just ''approaches'' the speed of light, while, measured in our frame, the traveler's acceleration constantly decreases.

[[File:Accelerated relativistic observer with horizon.png|thumb|Figure 7–6. Accelerated relativistic observer with horizon. Another well-drawn illustration of the same topic may be viewed [[:File:ConstantAcceleration02.jpg|'''here''']]. ]]
Fig.&nbsp;7-6 details various features of the traveler's motions with more specificity. At any given moment, her space axis is formed by a line passing through the origin and her current position on the hyperbola, while her time axis is the tangent to the hyperbola at her position. The velocity parameter <math>\beta</math> approaches a limit of one as <math>ct</math> increases. Likewise, <math>\gamma</math> approaches infinity.

The shape of the invariant hyperbola corresponds to a path of constant proper acceleration. This is demonstrable as follows:
# We remember that {{tmath|1= \beta = ct/x }}.
# Since {{tmath|1= c^2 t^2 - x^2 = s^2 }}, we conclude that {{tmath|1= \beta (ct) = ct/ \sqrt{c^2 t^2 - s^2} }}.
# <math>\gamma = 1/\sqrt{1 - \beta ^2} = </math> <math>\sqrt{c^2 t^2 - s^2}/s</math>
# From the relativistic force law, <math>F = dp/dt = </math>{{tmath|1= dpc/d(ct) = d(\beta \gamma m c^2)/d(ct) }}.
# Substituting <math>\beta(ct)</math> from step 2 and the expression for <math>\gamma</math> from step 3 yields {{tmath|1= F = mc^2 / s }}, which is a constant expression.<ref name="Bais">{{cite book|last1=Bais|first1=Sander|title=Very Special Relativity: An Illustrated Guide|url=https://archive.org/details/veryspecialrelat0000bais|url-access=registration|date=2007|publisher=Harvard University Press|location=Cambridge, Massachusetts|isbn=978-0-674-02611-7}}</ref>{{rp|110–113}}

Fig.&nbsp;7-6 illustrates a specific calculated scenario. Terence (A) and Stella (B) initially stand together 100 light&nbsp;hours from the origin. Stella lifts off at time 0, her spacecraft accelerating at 0.01&nbsp;''c'' per hour. Every twenty hours, Terence radios updates to Stella about the situation at home (solid green lines). Stella receives these regular transmissions, but the increasing distance (offset in part by time dilation) causes her to receive Terence's communications later and later as measured on her clock, and she ''never'' receives any communications from Terence after 100 hours on his clock (dashed green lines).<ref name="Bais" />{{rp|110–113}}

After 100 hours according to Terence's clock, Stella enters a dark region. She has traveled outside Terence's timelike future. On the other hand, Terence can continue to '''receive''' Stella's messages to him indefinitely. He just has to wait long enough. Spacetime has been divided into distinct regions separated by an ''apparent'' event horizon. So long as Stella continues to accelerate, she can never know what takes place behind this horizon.<ref name="Bais" />{{rp|110–113}}