Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

145 submissions , 122 unreviewed
3,930 questions , 1,398 unanswered
4,862 answers , 20,637 comments
1,470 users with positive rep
502 active unimported users
More ...

Why do clocks measure arc-length?

+ 4 like - 0 dislike
294 views

Apologies in advance for the long question.

My understanding is that in GR, massive observers move along timelike curves $x^\mu(\lambda)$, and if an observer moves from point $x^\mu(\lambda_a)$ to $x^\mu(\lambda_b)$, then his clock will measure that an amount of time $t_{ba}$ given by the curve's arc length; $$ t_{ba} = \int_{\lambda_a}^{\lambda_b}d\lambda \sqrt{-g_{\mu\nu}(x(\lambda))\dot x^\mu(\lambda)\dot x^\nu(\lambda)} $$ will have elapsed where $g_{\mu\nu}$ is a metric on spacetime with signature $(-,+,+,+)$.

Why is this so?


Here is how I would attempt to justify this fact in special relativity with $g_{\mu\nu} = \eta_{\mu\nu}$. Consider an inertial observer $O$ in $\mathbb R^{3,1}$, and suppose that this observer sees a clock, which I'll call observer $O'$ moving around on a curve $x^\mu(\lambda)$. If $O'$ were also an inertial observer, then given any event with coordinates $x^\mu$ as measured by $O$, observer $O'$ would measure the coordinates of the event to be $x'^\mu = \Lambda^\mu_{\phantom\mu\nu} x^\nu + x_0^\mu$ for some Lorentz transformation $\Lambda$. If $O'$ is not inertial, then this is no longer true, and there is some more complicated family of transformations, say $T_\lambda$ between events as seen by both observers.

I would argue, however, that if we were to partition the interval $[\lambda_a, \lambda_b]$ into a large number $N$ of intervals $I_1=[\lambda_a, \lambda_i], \dots, I_N=[\lambda_{N-1}, \lambda_b]$ with $\lambda_n = \lambda_a+n\epsilon_N$ and $\epsilon_N=(\lambda_b-\lambda_a)/N$, then on each interval $I_n$, $O'$ is approximately an inertial observer in the sense that $$ T_{\lambda_n} = P_n + \mathcal O(\epsilon_N), \qquad (\star) $$ for some Poincare transformation $P_n$. Then we would note that since $O'$ is stationary in his own reference frame, he measures his worldline to have the property $\dot x'^\mu(\lambda) = (\dot t(\lambda), \mathbf 0)$ so that $$ I_{ba}=\int_{\lambda_a}^{\lambda_b}d\lambda \,\sqrt{-\eta_{\mu\nu}\dot x'^\mu\dot x'^\mu} = \int_{\lambda_a}^{\lambda_b} d\lambda \, \sqrt{\dot t^2} = t(\lambda_b) - t(\lambda_a) = t_{ba} $$ On the other hand the integral on the left can be written as a Riemann sum using the partition above, and we can invoke ($\star$) above to get \begin{align} I_{ba} &= \lim_{N\to\infty}\left[\sum_{n=1}^N \epsilon_N\sqrt{-\eta_{\mu\nu}\dot x'^\mu(\lambda_n)\dot x'^\nu(\lambda_n)}\right] \notag\\ &= \lim_{N\to\infty}\left[\sum_{n=1}^N \epsilon_N\sqrt{-\eta_{\mu\nu}\dot x^\mu(\lambda_n)\dot x^\nu(\lambda_n)} + \mathcal{O}(\epsilon_N^2)\right] \notag\\ &= \int_{\lambda_a}^{\lambda_b}d\lambda \,\sqrt{-\eta_{\mu\nu}\dot x^\mu\dot x^\mu} \end{align} Combining these two computations gives the desired result.

How do others feel about this argument?

I'm not completely comfortable with it because of the assumption $(\star)$ I made on $T_\lambda$.

I imagine that in GR a similar argument could be made by invoking local flatness of the metric.

This post imported from StackExchange Physics at 2014-06-14 12:59 (UCT), posted by SE-user joshphysics
asked Feb 7, 2013 in Astronomy by joshphysics (830 points) [ no revision ]
retagged Jun 14, 2014
Most voted comments show all comments
Good point; I'll remove that part since what I'm actually asking is for arguments making that assumption plausible. Let me add that I'm completely comfortable simply taking this fact as an axiom of relativity that is ultimately justified by experiment. What bothers me is that people rarely state the assumption in such terms, and it seems to me that they often even suggest that it's some obvious/trivial consequence of how events transform between inertial frames. For example, the resolution of the twin paradox ultimately rests on this assumption (as far as I can tell).

This post imported from StackExchange Physics at 2014-06-14 12:59 (UCT), posted by SE-user joshphysics

Maybe it is mostly imported because of Eduardo's (he is now here ;-)!) answer, and to me it does not seem that the question is obviously bad or off topic but rather nice and technical. I am not too familiar with the American curriculum to study physics, but when looking at Gerard t'Hoofts guide how to become a Good Theorist, it seems that not dumbed down versions of GR at least and maybe even SR topics in their full technicality do not come that early.

Of course as always, people who think otherwise can vote to close and then we will see...

Also, Joshphysics is a graduate student and rather knowledgeable ...

Sure ...

But even before noting that Josphysics asked this (and hours before your comment) I personally did not see anything in the question that would make it blatantly off topic. With Kimmo's question things are different, that really is very basic...

@physicsnewbie What Dilaton means is that joshphysics would not ask a lower-than-graduate-level question.

It was me who imported this question and I don't feel this is lower than graduate-level, because something is General Relativity doesn't automatically mean that it is undergraduate-level. If you disagree, you can always earn 500 reputation and then vote to close the question.

The argument is fine, it is elementary, and there is no need for an answer, as this is simply the OP showing that he or she is scared to think.

Most recent comments show all comments

@dimension10 Sure Josh asks the occasional undergraduate question: Why are wires in simple circuits approximated as equipotentials?

This is basic electromagnetism, but still a very good question that tests the physical insight of people, rather than their blind mathematical ability to manipulate Maxwell's equations.

Likewise with this question. It's asking if it can be proven that an accelerated clock measures arc length. It's already been correctly answered as no, and we need the "clock postulate".

If I disagree with this question being imported, I can comment my objections here, right?

@physicsnewbie Sure, you can comment your objections. (P.S. I'm not the downvoter of your comment)

4 Answers

+ 7 like - 0 dislike

I think it's obvious from the pic:

enter image description here

(My secondary argument would be that the integral is the parametrization-independed geometrical measure, minimizing the energy functional and thus being the natural generalization of the flat case time translation. You often encounter these exponential maps in dynamics, here given by the Geodesic flow. Locally you can flatten out the metric to obtain $g(x)=\eta+O(x^2)$ and try to find a limit of smaller and smaller patches. But this really is equivalent solving the geodesic equation, which should be viewed as giving the spacetime direction of the moving object as it's pushed through spacetime. It is also equivalent to minimizing the curve between the points, as motivated above.)

This post imported from StackExchange Physics at 2014-06-14 12:59 (UCT), posted by SE-user NikolajK
answered Feb 7, 2013 by NikolajK (195 points) [ no revision ]
Hahhahha. I'm tempted to +1 for the picture alone. As for the secondary argument; it's not clear to me how this answers the question about time measured by an observer moving along a curve given that I'm looking for plausibility of (arc length) = (time measured by observed moving on a curve) even in the flat space case.

This post imported from StackExchange Physics at 2014-06-14 12:59 (UCT), posted by SE-user joshphysics
@joshphysics: The geometric length is parameter independed, so as soon as you've chosen the curve in the manifold (requirement of extremal length) the quantity depends only on the two points $x_a,x_b$ in spacetime. If your Newtonian wolrd is $\mathbb{R}\times\mathbb{R}^3$, then $t_{ab}=f(I_{ab})$ is the obvious choice.

This post imported from StackExchange Physics at 2014-06-14 12:59 (UCT), posted by SE-user NikolajK
That's an interesting argument; I like it for plausibility quite a lot actually.

This post imported from StackExchange Physics at 2014-06-14 12:59 (UCT), posted by SE-user joshphysics
+1: At the very least for your picture answer.

This post imported from StackExchange Physics at 2014-06-14 12:59 (UCT), posted by SE-user joshphysics
What about clocks on non-geodesic timelike paths?

This post imported from StackExchange Physics at 2014-06-14 12:59 (UCT), posted by SE-user Retarded Potential
@RetardedPotential: It's all explained in the movie/documentary.

This post imported from StackExchange Physics at 2014-06-14 12:59 (UCT), posted by SE-user NikolajK
+ 3 like - 0 dislike

[This is now a long answer. In summary, generally you need a physical assumption, the clock postulate, which people tend to omit, but is necessary, and can't be argued for a priori. However sometimes special relativity plus a restricted version of the postulate suffices. Mundane experience is sufficient to verify this restricted version.]

Let $\lambda = t$ the time according to inertial $O$, and let $\vec{x}'$ be the spatial position of $O'$ according to $O$ , while $t'$ is the time measured by $O'$. If $O'$ is piecewise inertial, then along each piece, $$c^2(\Delta t'/\Delta t)^2 = c^2 - (\Delta\vec{x}'/\Delta t)^2\qquad[1]$$ and what you are trying to justify is that, even if $O'$ is not piecewise inertial, $$c^2(dt'/dt)^2 = c^2 - (d\vec{x}'/dt)^2\qquad[2]$$ So, the problem is, special relativity strictly speaking only makes claims about inertial observers. And if you don't make any assumptions whatsoever about the experience of accelerated observers, then I think you're just stuck, mathematically I don't think you can go from $[1]$ to $[2]$. (For example, we can't rule out that proper acceleration itself further contributes to time dilation.) I suggest:

  • The motion of $O'$ is smooth. [A1]

  • For every $\epsilon>0$, there is a $\delta>0$ such that, if, from the point of view of a certain unaccelerated observer $A$, the magnitude of the velocity of another observer $B$ never exceeds $c\delta$ betwen time $t_0$ and $t_1$, then time lapse $\Delta t_B$ on $B$'s clock satisfies $(t_1-t_0)(1-\epsilon) < \Delta t_B < (t_1-t_0)(1+\epsilon)$. [A2]

Pick $\epsilon>0$, use [A2] to get $\delta$; use [A1] to break the motion of $O'$ into intervals small enough such that, from the frame of reference of an interial observer travelling between the endpoints of a piece, the velocity of $O'$ never exceeds $c\delta$; use [A2] to make $[2]$ true within $\epsilon$. Since this works for all $\epsilon>0$, [2] is simply true.

Now [A1] might look suspect, since we've used a piecewise inertial observer, whose motion is obviously not smooth! So we can't even assume anything about what this piecewise inertial observer experiences at the corners! But that's okay, [A2] only refers to the individual pieces and not the whole. Use a family of (truly) inertial observers that meet at the appropriate points.

As for [A2], it is a bit opaque, but what it says that if you're not moving too fast relative to an inertial observer, your experience of time is almost the same. This doesn't follow logically from anything in particular, it's just a physical assumption. But note that special relativity is so hard for many people to accept precisely because [A2] is a fact of life, for reasonably small $\epsilon$. To make it true for even smaller $\epsilon$ requires more than everyday experience, but it is still "common sense", and presumably testable to quite small values.

Now, to believe it literally for arbitrarily small $\epsilon$ requires quite a leap, but don't take differential equations literally.

(Added:) Aha! I found the clock postulate for accelerated observers, and I do believe [A2] is interchangeable with it. And yes, it is often omitted but cannot be derived from other assumptions. It has been tested.

(Second addendum): Even though they're interderivable, mine is better :-) I've given the accuracy of [2] directly in terms of the accuracy of [A2]. For example, we don't need the full clock postulate for the twin paradox (which you mention as a motivating example in a comment):

  • The proper acceleration of $O'$ is continuous and its magnitude is bounded by $a_{max}$. [A1']

(Over any finite interval, [A1] does imply [A1'] for some value of $a_{max}$. And [A1'] is sufficient for the above argument.)

Now, even with mundane accelerations, the twin paradox can produce a sizeable mismatch in ages within a human lifetime. (Besides, if they're not survivable accelerations, the travelling twin's lifetime ends!) So, there's a usable $a_{max}$ for [A1']. And, mundane experience alone proves [A2] up to that $a_{max}$ and down to fairly small $\epsilon$. So [2] holds sufficiently accurately to give the twin paradox. We only need special relativity plus a mundane restricted clock postulate.

(I realise you can bypass the whole acceleration question by altering the paradox so that there are three inertial observers who compare clocks as they pass. But then it's not the twin paradox anymore, duh!)

This post imported from StackExchange Physics at 2014-06-14 12:59 (UCT), posted by SE-user Retarded Potential
answered Feb 7, 2013 by Retarded Potential (30 points) [ no revision ]
Most voted comments show all comments
Are you sure you mean to say "deviation of velocity from inertial"? Recall that even inertial observers can have speeds arbitrarily close to $c$; are you sure you didn't mean to phrase the assumption in terms of acceleration? If so, I think my initial argument is essentially in the same spirit as what you're attempting here, in particular, the assumption I call $(\star)$.

This post imported from StackExchange Physics at 2014-06-14 12:59 (UCT), posted by SE-user joshphysics
Mine is different. For one, it doesn't assume there is any transforms applicable to the accelerated observer, nor any bound on acceleration. From the point of view of the inertial A, their own velocity is 0, and B's velocity is bounded to within $c\delta$ in any direction (in A's frame). Then A2 asserts A and B experience the same time lapse, to within a factor $(1 \pm \epsilon)$. You only need to believe that there is some small $\delta$ for which this is true. Then the only Lorentz transforms you need are for inertial observers, which I took to be given.

This post imported from StackExchange Physics at 2014-06-14 12:59 (UCT), posted by SE-user Retarded Potential
Ok I'll read it more closely and try to get a feel for your assumption [A2]; I still feel that the phrase "deviation of velocity from inertial" is a bit misleading. By the way; thanks for all of your attention and comments; I really do appreciate it: +1.

This post imported from StackExchange Physics at 2014-06-14 12:59 (UCT), posted by SE-user joshphysics
I rephrased. I think you need something like [A2] to connect the experiences of inertial and accelerated observers, or else the latter's experiences can be quite arbitrary.

This post imported from StackExchange Physics at 2014-06-14 12:59 (UCT), posted by SE-user Retarded Potential
Comment on "Aha!": Cool! Thanks for the link!

This post imported from StackExchange Physics at 2014-06-14 12:59 (UCT), posted by SE-user joshphysics
Most recent comments show all comments
@joshphysics The proper time of clocks aren't affected by acceleration is a postulate - isn't that what it boils down to?

This post imported from StackExchange Physics at 2014-06-14 12:59 (UCT), posted by SE-user Larry Harson
@LarryHarson Yeah I think that's the consensus after much discussion, hence the quoted part at the start of this answer. Originally I basically wanted to know if there were arguments one could make to lend plausibility to this postulate. Also Retarded Potential's response shows that there are equivalent, more intuitive ways of formulating the "clock postulate" which is nice.

This post imported from StackExchange Physics at 2014-06-14 12:59 (UCT), posted by SE-user joshphysics
+ 2 like - 0 dislike

I definitely thing this question is much simpler.

It is an a priori assumption upon which GR is built, that this quantity shall be independent of the observer:

$$ds^2=g_{\mu\nu}dx^{\mu}dx^{\nu}$$

that is, two different observers shall measure the same spacetime interval between two infinitesimally close events.

On the other hand, the $x^0$ coordinate of any observer is by definition the lecture in his rest clock, (times $c$ or not, times $+1$ or $-1$, all depending on every modern author's bad taste).

Since the clock of the moving observer is at rest with respect to his proper reference frame, he measures no variation in the spatial coordinates, i.e. $dx^1=dx^2=dx^3=0$. Therefore, as measured from the moving observer, $$ds^2=g_{00}dx^{0}dx^{0}$$

That is his proper time (squared) because $g_{00}=\pm 1$ (the metric of a free falling observer is locally flat). Since $ds^2$ shall be the same for both observers, then the lecture on the free-falling clock equals the spacetime interval measured by the distant observer. That's all.

Now, if you prefer so, you may rewrite the $dx^{\mu}$ as functions of a variation in the parameter $\lambda$ describing the curve, and integrate along that curve, in order to arrive trivially at the expressions in your question.

I don't mean that the other answers are wrong, but merely that this is a straightforward result from the original theory, as exposed very early by Einstein himself in the Princeton lectures of 1921, that doesn't need much mathematical sophistication or additional postulates. Einstein wanted to extend the invariance of $ds^2$ from Special Relativity, to the realm of non-inertial observers. In his effort to find how $g_{\mu\nu}$ had to be for that invariance to hold too in the presence of acceleration/gravity and curved coordinates, he then used parallel transport to arrive at the geodesics equation. That is why the geodesics equation was a postulate in the early formulation of GR (the next step was to relate $g_{\mu\nu}$ to matter/energy, and that is why the Field Equations with the Einstein tensor is the other postulate of the theory, but that is another question)

I think I remember having read somewhere, that the geodesics is no longer a postulate, because it can be derived from the Field equations. I would be thankful to any user that provides the corresponding reference. In 1921, however, it was a postulate.

This Einstein's early exposition of GR of the Princeton lectures is of a raging beauty, full of fresh insights and brilliant heuristic thinking. I don't know why it is almost systematically ignored in every standard GR bibliography.

Now, if I am wrong and this question is not so naive as it seems, I hope that someone points out where I am wrong and what I am missing, so that I learn something new.

This post imported from StackExchange Physics at 2014-06-14 12:59 (UCT), posted by SE-user Eduardo Guerras Valera
answered Feb 8, 2013 by Eduardo Guerras (435 points) [ no revision ]
I agree that it's part of the very construction of GR that the metric is a coordinate-independent quantity in the sense that you can perform arbitrary diffeomorphisms on the spacetime manifold without changing the metric. As far as I can tell, however, it is not a priori clear that such transformations allow one to say something about what arbitrary timelike observers measure. I'd have to think more about your argument to be more clear on where (if at all) you're making the "clock postulate" assumption.

This post imported from StackExchange Physics at 2014-06-14 12:59 (UCT), posted by SE-user joshphysics
Again, there must be something here I don't understand. For me, it is crystal clear that $dx^0$ is the measure on the clock of any observer. That is a definition that goes back to Minkowski. Einstein did a great effort to maintain that definition and postulated that the distance between two infinitesimally close spacetime events $ds^2$ is the same for all observers, even in the presence of gravity and curved coordinates. In his effort to find how $g_{\mu\nu}$ had to be for that assumptions to be true not only in the absence of gravity, he then used parallel transport to arrive at...

This post imported from StackExchange Physics at 2014-06-14 12:59 (UCT), posted by SE-user Eduardo Guerras Valera
(...he then used parallel transport to arrive at) the geodesics equation and eventually postulated the Einstein tensor in the field equations. That is how GR is exposed in the Princeton lectures in 1921. The invariance of $ds^2$ is postulated, and the meaning of $dx^0$ as the clock measure (multiplied with c) is simply an a priori definition.

This post imported from StackExchange Physics at 2014-06-14 12:59 (UCT), posted by SE-user Eduardo Guerras Valera
Yea your point is well taken, and I completely understand what you're saying; let me think about it for a while. Thanks for the detailed comments!

This post imported from StackExchange Physics at 2014-06-14 12:59 (UCT), posted by SE-user joshphysics
@joshphysics, something tells me that you have completely forgotten this question...

This post imported from StackExchange Physics at 2014-06-14 12:59 (UCT), posted by SE-user Eduardo Guerras Valera
Hehe yeah you're right I did; research is killing me man!

This post imported from StackExchange Physics at 2014-06-14 12:59 (UCT), posted by SE-user joshphysics
+ 0 like - 0 dislike

> My understanding is that in GR, massive observers move along timelike curves $x^{\mu}(\lambda)$

 ... where $x^{\mu} : \mathcal E_O \to \mathbb R^n$ appears to be some coordinate assignment to the set of events $\mathcal E_O$ in which the observer under consideration ($O$) took part           
(perhaps not even "just any" coordinate assignment, but rather an assignment which represents the events $\mathcal E_O$ in the order in which observer $O$ took part, as a "path $\mathbb R^n$");

and $\lambda : \text{(subset of) } \mathbb R \to \text{(subset of) } \mathbb R^n $ is a real-valued parametrization of those coordinates
(perhaps not even "just any" parametrization, but again rather a parametrization which represents the order of observer $O$'s indications monotonously) ...

> and if an observer moves from point $x^{\mu}(\lambda_a)$ to $x^{\mu}(\lambda_b)$

 ... or focussing on the physical: if observer $O$ moves from one particular event (which we happened to denote by coordinates $x^{\mu}(\lambda_a)$) to another particular event (denoted by $x^{\mu}(\lambda_b)$) ...

> then his clock will measure that an amount of time $t_{ba}$ given by the curve's arc length;

This seems to indicate a misunderstanding, either of terminology and conventions of notation, or even more profoundly:

in the theory of relativity, "time" means

 - foremost an indication of an observer
(such as the indication $O_A$ of observer $O$ of having been in coincidence with observer $A$, having jointly taken part in coincidence event $\varepsilon_{A O}$),

 - and in a derived sense a (real) number value $t$ assigned as a clock reading to an indication of an observer (or also assigned to the entire event, of which the observer indicated its participance), such as any function $$ t : \mathcal E_O \to \mathbb R$$

(or perhaps not even "just any" such function, but rather only such functions which are monotonous wrt. the order in which observer $O$ took part in the events of set $\mathcal E_O$).

Obviously, there is a tremendously large number of distinct such clock reading functions $t$;
any two of which are not necessarily continuous wrt. each other, much less differentiable, or even smooth (differentiable to any order), or even affine (proportional to each other).

On the other hand, the arc length of the timelike curve of a particular observer (between two particular events, such as between the two events $\varepsilon_{A O}$ and $\varepsilon_{B O}$ in which observer $O$ took part, or between two particular indications of the observer, such as correspondingly between the two indications $O_A$ and $O_B$ of observer $O$),     
a.k.a. the duration of observer $O$ between these two indications,
a.k.a. the proper time of observer $O$ between these two indications
(as far as the term "proper" is permissible at all, since it suggests the possibility of "improper" notions as well),
 is usually denoted by the letter $\tau$;   
for the above case explicitly as $\tau O[ \, \_A, \_B \, ]$.

Ratios of durations are unambiguos real numbers, i.e. quantities to be measured.

Further, the readings $t$ assigned to indications of a given clock provide a measure of its coresponding durations $\tau$ as far as for any three distinct indications $O_H, O_J, O_K$ holds:

$$ (t[ \, O_K \, ] - t[ \, O_H \, ]) \, \tau O[ \, \_H, \_J \, ] = (t[ \, O_J \, ] - t[ \, O_H \, ]) \, \tau O[ \, \_H, \_K \, ].$$

If (and only if) this was satisfied, the clock under consideration (incl. the clock reading assignment $t$) is said to have been "good", or having "run evenly".

> Why is this so?

As a matter of definition;
in particular of the notions "duration $\tau$ (of an observer, between two of its indications)", and of whether a given clock was "good" or to which extent it was not.
 

answered Jun 10 by Frank Wappler (0 points) [ no revision ]

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsOverf$\varnothing$ow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
To avoid this verification in future, please log in or register.




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...