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This question already has an answer here:

So I've encountered the site "Alternative Physics" – a website proposing "alternative" theories to theories in modern physics. Of course, the site claims that both SR and GR are wrong. To show "why" SR is wrong, the author presents two apparent contradictions in SR, as following:

enter image description here

Two accurate identical clocks, A and B, move with uniform motion along a shared straight-line path but with different constant velocities such that the distance between them steadily decreases with time.

[...]

We wish to know which clock runs more slowly.

(Keep in mind A will eventually catch up to B and you don't need to change reference frame by accelerating in order to measure the times side-by-side)

I can obviously see the apparent contradiction – one can claim that A is at rest and B is experiencing time dilation or equally claim that B is at rest and A is experiencing time dilation, giving two different results when the two clocks "meets" – but I have no idea to actually solve it.

When solving this, please use real numbers of your choosing.

The second "paradox" involves a circular particle accelerator with equally spaced electrons. Now, if we'd would speed them up to relativistic speeds, say $\gamma = 2$, how would the tube look like?

The initial condition
(source: alternativephysics.org)

Electrons sped up, A
(source: alternativephysics.org)

Electrons sped up, A
(source: alternativephysics.org)

Electrons sped up, A
(source: alternativephysics.org)

My own thought on this is that the electrons have a non-inertial motion and thus alternative A must be right (the electrons shrinking but their spacing staying the same). Is this correct a reasoning?

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marked as duplicate by Dale, John Rennie homework-and-exercises Oct 6 at 8:12

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ "We wish to know which clock runs more slowly." According to which (or what) clock? Do you see the hidden (false) premise in the quoted question? According to SR, there are an infinity of inertial reference frames in which clock A runs slower than B as well as an infinity of inertial reference frames in which clock B runs slower than clock A. But this isn't a paradox in SR which is easy to see using a spacetime diagram. See, for example: physics.stackexchange.com/a/111089/9887 $\endgroup$ – Alfred Centauri Jul 2 '15 at 3:24
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    $\begingroup$ When you see sites such as that, run away. Don't just walk, run. $\endgroup$ – David Hammen Jul 2 '15 at 5:11
  • $\begingroup$ @Madde Anerson: "We wish to know which clock runs more slowly." -- If this request is understood as asking which one (of two "ticking" clocks) was ticking at a lower frequency (itself, properly), then that's a perfectly sensible and legitimate request, and SR is perfectly suited to address it consistently. (While other interpretations may lead to inconsistency.) "[...] is experiencing time dilation -- Time dilation refers to a ratio, comparing frequencies (or, foremost, durations) between participants who are not at rest to each other. It's not for only one individually "to experience". $\endgroup$ – user12262 Jul 2 '15 at 5:11
  • $\begingroup$ @AlfredCentauri Ah, so because we can imagine a reference frame in which the clocks show the same thing, the events are space-like separated and there cannot simply be an universal agreement on the order. Thus, it is equally valid to say that A ticks ahead of B or that B ticks ahead of A. Is this correct? The events in the second paradox is space-like as well; can this solve the paradox? $\endgroup$ – Madde Anerson Jul 2 '15 at 12:52
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    $\begingroup$ @MaddeAnerson - "Let's not fight over semantics." But it was semantics "We wish to know which clock runs more slowly" which was the source of your confusion in the first place. Specifically, the site you refer to talks about "slowly" as if it is an absolute (or can be considered absolute). But SR says there's no such thing. For two clocks with non-zero relative velocity, BOTH clocks run slow as measured by the other, and there is no paradox. "Time ticking" as a phrase makes it easy to fall into the trap of thinking that there is one preferred time (you know, "the" time), and there isn't. $\endgroup$ – WhatRoughBeast Sep 8 '15 at 23:41
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In A's frame of reference, clock B is running more slowly. In B's frame of reference, A is running more slowly. The statement of the problem implies a third frame of reference in which one clock is traveling faster than the other. In that frame of reference, the faster-moving clock will be ticking more slowly than the slower-moving clock.

If the two clocks compare times as they pass each other, and then one of them changes speed to catch back up with the other, then the one who did the accelerating will have experienced fewer ticks.

It takes time seeing examples to get comfortable with the notion that this doesn't lead to any true paradoxes, but there simply is no absolute answer to the question of which clock is running more slowly.

Again, for the second problem, you have to be clear about what your reference frame is. If you use the rest frame of the accelerator, your picture has to be correct. There is a fixed number of particles, and at any moment they are uniformly spaced around a circle. Since you sped them up under your control, you have arranged for this to be true according to what you recognize in your frame of reference as a "specific moment".

In the reference frame of the particles, the ones at the opposite side of the circle are scrunched together, and the ones ahead and behind you are more spread out than your frame would show.

Again, neither of these is more correct than the other, although a non-accelerating frame does have preferred properties.

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  • $\begingroup$ Mark Foskey: "In A's frame of reference, clock B is running more slowly." -- Your formulation is apparently agrammatical, and therefore meaningless. If you're trying to use a comparative then you should continue your sentence with the word than, followed by a description of the comparison reference. $\endgroup$ – user12262 Sep 9 '15 at 5:07
  • $\begingroup$ @user12262 - Technically, you are correct. However, the context is obvious, isn't it? You could have edited the answer instead of ranting. $\endgroup$ – David Hammen Sep 10 '15 at 0:00
  • $\begingroup$ @David Hammen: You've merely replaced Mark Foskey's formulation "clock B is running more slowly." with "clock B is running slow." I doubt that this will stop anybody from asking and considering "Slow -- in comparison to what?" and if so: "By which exact method of comparison?". "You could have edited the answer" -- Hardly, because I don't presume to know how Mark Foskey might (intend to) answer these questions. Perhaps not by "deeming" (as you suggested), but rather by "measuring". $\endgroup$ – user12262 Sep 10 '15 at 18:09
  • $\begingroup$ I really prefer my original wording. The question as stated described two clocks, and asked which one runs more slowly. The comparative was entirely appropriate in that case because the context established two clocks. I am answering in the same context, and so the comparative is still correct (not just acceptable). So I'm going ahead and rolling it back - hope you'll forgive me. $\endgroup$ – Mark Foskey Sep 12 '15 at 18:11
  • $\begingroup$ Mark Foskey: "The question as stated described two clocks, and asked which one runs more slowly. The comparative was entirely appropriate in that case because the context established two clocks." -- The OP context also prescribed "Two accurate identical clocks"; and (as the context seems to suggest) the word "identical" is not supposed to mean "the same (the very same, one-and-the-same, none other than this one)" but two clocks which are equal in some particular sense (some particular measure) which apparently refers to their "running" as well. So: what gives? ... $\endgroup$ – user12262 Sep 12 '15 at 18:39
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This was bumped up by the community user, I think, and answered anew by Mark Foskey. However, I don't really see a satisfactory concise answer to the first. (The second is a canard; electrons don't have size and even if you use protons, a sphere traveling past you always "looks like" a Terrell-rotated sphere, so this asks us to construct something rather complicated due to the acceleration, and then be astonished at how complicated it looks.)

The inconsistency in the first case is resolved by the relativity of simultaneity. The points in the ground frame are $(c t, v_a t)$ and $(ct, L + v_b t),$ colliding at our time $t^* = L / (v_a - v_b)$ which corresponds to $t_a^* = t^*/\gamma_a= t^* \sqrt{1-(v_a/c)^2}$ and, respectively,$t_b^* = t^*/ \gamma_b = t^* \sqrt{1-(v_b/c)^2}.$ So those are the times that they cross as they respectively mark it; it's straightforward time dilation in this frame.

In frame $a$ we can do this with Lorentz transforms. We have points $\gamma_a (ct - v_a^2 t / c, 0) = (ct/\gamma_a, 0) = (c \tau, 0)$ and $r^\mu = \gamma_a (c t- v_a (L+v_b t)/c, L + v_b t - v_a t).$ Equating these yields the same formula for $t^* = L /(v_a - v_b)$ while $c\tau=\gamma_a \,t^*\,(c - v_a (v_a - v_b + v_b)/c).$ After all is cancelled we find $\tau = t^* / \gamma_a$ as expected.

The only really tricky thing is working out the time on the clock at $b$ as seen from the frame at $a$. The modified velocity of $b$ as seen from $a$ is taken by looking at the above with $t \mapsto t + dt,$ where $dr^\mu = dt\, \gamma_a \, (c - v_a v_b / c , v_b - v_a),$ so the speed in these coordinates is $u = (v_b - v_a) / (1 - v_a v_b/c^2).$

Working out all of this becomes easiest if you define $\beta_a = v_a / c$ and $\beta_b = v_b / c,$ so that $\beta_u = u/c = (\beta_b - \beta_a)/(1 - \beta_a \beta_b)$ and hence $$\begin{align}1 - \beta_u^2 ~&= \frac{(1 - \beta_a \beta_b)^2 - (\beta_b - \beta_a)^2 }{(1 - \beta_a \beta_b)^2} = \frac{1 + \beta_a^2 \beta_b^2 - \beta_b^2 - \beta_a^2 }{(1 - \beta_a \beta_b)^2}\\ ~&= \frac{(1 - \beta_a^2) (1 - \beta_b^2)}{(1 - \beta_a \beta_b)^2}.\end{align}$$ Hence $\gamma_u = \gamma_a \, \gamma_b\,(1 - \beta_a \beta_b).$ That's important because the coordinate time in the $a$ frame for the time $dt = t^*$ is $$t^*\, \gamma_a (1 - \beta_a \beta_b) = t^* \,\gamma_u / \gamma_b. $$ Thus our normal process of dividing this by $\gamma_u$ produces the exact same thing as $t^* / \gamma_b,$ what we saw in the ground frame.

Everyone sees clock $a$ dilated more, hence at an earlier time, than clock $b$. More precisely they see it having the time $t^*/\gamma_a$ as opposed to $t^*/\gamma_b$ on the other clock.

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  • $\begingroup$ Chris Drost: "[...] it's straightforward time dilation." -- Yes, of course, the first OP problem involves straightforward time dilation (which, btw., can be derived without appealing to particular coordinates). However, your rather short answer completely failed to address the actual OP problem statement: "We wish to know which clock runs more slowly.". $\endgroup$ – user12262 Sep 9 '15 at 5:15
  • $\begingroup$ @user12262 Yes, it is an intentionally short answer. No, that is not what the OP asked, which was "I can obviously see the apparent contradiction – one can claim that A is at rest and B is experiencing time dilation or equally claim that B is at rest and A is experiencing time dilation, giving two different results when the two clocks "meets" – but I have no idea to actually solve it." This is a short answer about how to actually solve it, which I felt was lacking from this discussion. $\endgroup$ – CR Drost Sep 9 '15 at 7:51
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Two accurate [...] clocks, $\mathcal A$ and $\mathcal B$, [...]

Specificly, we call a given clock $\mathcal A := (A, \theta_{\mathcal A})$ "accurate" (or "good") if for any three of its distinct indications ($A_H$, $A_J$ and $A_K$) clock $\mathcal A$'s durations ($\tau_{\mathcal A}$) between pairs of these indications are proportional to the differences between corresponding pairs of readings ($\theta$):

$$ \frac{\tau_{\mathcal A}[~{}_H, {}_J~]}{\tau_{\mathcal A}[~{}_H, {}_K~]} = \frac{\theta_{\mathcal A}[~A_J~] - \theta_{\mathcal A}[~A_H~]}{\theta_{\mathcal A}[~A_K~] - \theta_{\mathcal A}[~A_H~]},$$

and likewise for clock $\mathcal B := (B, \theta_{\mathcal B})$.

For simplicity, consider in the following the distinct indications of clock $\mathcal A$ and clock $\mathcal B$ as distinctive "ticks"; and the corresponding readings $\theta$ as integer counts of these ticks.

The durations of clock $\mathcal A$ between its successive ticks are consequently equal to each other: for any three of clock $\mathcal A$'s successive tick indications throughout a trial, $A_{(n)}$, $A_{(n + 1)}$ and $A_{(n + 2)}$ for which $$\theta_{\mathcal A}[~A_{(n)}~] = n,$$ $$\theta_{\mathcal A}[~A_{(n + 1)}~] = n + 1$$ and $$\theta_{\mathcal A}[~A_{(n + 2)}~] = n + 2$$ holds $$\tau_{\mathcal A}[~{}_{(n)}, {}_{(n + 1)}~] = \tau_{\mathcal A}[~{}_{(n + 1)}, {}_{(n + 2)}~] = \text{constant} := 1/f_{\mathcal A},$$

where $f_{\mathcal A}$ is the (proper) "tick rate" of clock $\mathcal A$.

Likewise, the durations of clock $\mathcal B$ between its successive ticks are equal to each other; and $f_{\mathcal B}$ is the (proper) "tick rate" of clock $\mathcal B$.

We wish to know which clock runs more slowly.

Considering two accurate, ticking clocks, $\mathcal A$ and $\mathcal B$, the one with the lesser "tick rate $f$" is the one which is said to have "run more slowly". In the following it is therefore addressed how to compare the "tick rates $f$" between clocks; in mutual agreement; especially concerning trials in which such clocks were separated from each other, and not even at rest with respect to each other.

But there's still one particularity to discuss first:

Two [...] identical clocks, A and B,

That seems a very strange, senseless requirement.
Either, by "identical", you mean "the very same". But obviously were going to consider two distinct clocks, $\mathcal A$ and $\mathcal B$; not (only) one-and-the-same identical clock. Or, by "identical", you mean distinct but equal by some (as yet unspecified) measure, or by any measure. But if indeed you meant "(distinct but) of equal "tick rates $f$" then these clocks under consideration were outright required to "run equally"; and therefore you should have known and understood already how to determine whether two given clocks had "run equally" (or otherwise, which one of them had "run more slowly") thus answering your request ("to know which clock runs more slowly") readily yourself.

Consequently I'll ignore your requirement of "identical clocks" (whatever you might have meant by that), and sketch how to compare their rates. Several cases can be distinguished, building up from the simpler to the more complicated:

(1):
$B \equiv A$: the same set of "tick" indications are being considered for both clocks, $\mathcal A$ and $\mathcal B$, throughout the trial under consideration. (The only distinction between clocks $\mathcal A$ and $\mathcal B$ can then be that their corresponding readings differ by some constant integer $k$; $\forall A_s \in A : \theta_{\mathcal A}[~A_s~] - \theta_{\mathcal B}[~A_s~] = k := \text{const.}$.)
Then the rates $f_{\mathcal A}$ and $f_{\mathcal B}$ are (obviously) equal; clocks $\mathcal A$ and $\mathcal B$ "ran" equally.

(2):
$B \subset A$: only some (but not all) of clock $\mathcal A$'s "tick" indications are considered (and counted) as clock $\mathcal B$'s "tick" indications. (Since both clock $\mathcal A$ and clock $\mathcal B$ are supposed to be accurate, therefore set $B$ of indications contains only every second (or only every third, or only every fourth etc.) of set $A$.
Therefore clock $\mathcal B$ is said to have "run more slowly" than clock $\mathcal A$; rate $f_{\mathcal A}$ is an interger multiple of rate $f_{\mathcal B}$, in the trial under consideration.

(3):
The two sets of indications, $A$ and $B$ are disjoint; but together (as union) they constitute the set (or a subset) of indications of one particular participant, $\mathsf P$. For instance, $A$ might be the set of "tick" indications and $B$ the set of the alternating "tock" indications of a pendulum $\mathsf P$. From a trial with $n \ge 2$ "ticks" and $n \pm 1$ "tocks" it can be concluded that $f_{\mathcal A} = \left(1 \pm \frac{1}{n - 1}\right) f_{\mathcal B}$.
If the value $n$ is suitably large, the two clocks may be said to have "practically run equally".

(4):
The two sets of indications, $A$ and $B$ are of distinct participants, $\mathsf A$ and $\mathsf B$, who are separate and at rest to each other throughout the trial under consideration. Therefore the ping durations between $\mathsf A$ and $\mathsf B$, and vice versa, are equal and constant. Considering a trial of $n \ge 1$ successive pings (signal front round trips) during which there were $k_a[~n~] \ge 2$ "tick" indications counted of clock $\mathcal A$, and $k_b[~n~] \ge 2$ "tick" indications counted of clock $\mathcal B$ it can be concluded that $(k_b[~n~] \pm 1) f_{\mathcal A} = (k_a[~n~] \pm 1) f_{\mathcal B}$.
So if it is found that $k_a[~n~] - k_b[~n~] \ge 2$ then clock $\mathcal B$ is said to have "run more slowly" than clock $\mathcal A$.

(5):
The two sets of indications, $A$ and $B$ are of distinct participants, $\mathsf A$ and $\mathsf B$, who are both moving uniformly (straight, without acceleration, in a flat region). For simplicity, require that $\mathsf A$ and $\mathsf B$ had met/passed each other (once), and consider a subsequent trial. Then there can another participant $\mathsf M$ be imagined (in the course of a thought-experiment), or perhaps even be found experimentally, such that

  • $\mathsf M$ also moved uniformly (straight, without acceleration);

  • $\mathsf M$ also took part in the meeting/passing of $\mathsf A$ and $\mathsf B$;

  • for each signal indication stated, $\mathsf M$ observed the corresponding ping echoes of $\mathsf A$ and $\mathsf B$ in coincidence. (In other words: $\mathsf M$ observed the pings to $\mathsf A$ and to $\mathsf B$ being the same.);

  • for each signal indication stated, $\mathsf A$ observed the corresponding ping echo of $\mathsf B$ in coincidence with the corresponding second successive ping echo of $\mathsf M$. (In other words: $\mathsf A$ observed the pings to $\mathsf B$ being the same as two successive pings to $\mathsf M$.); and

  • for each signal indication stated, $\mathsf B$ observed the corresponding ping echo of $\mathsf A$ in coincidence with the corresponding second successive ping echo of $\mathsf M$. (In other words: $\mathsf B$ observed the pings to $\mathsf A$ being the same as two successive pings to $\mathsf M$.)

In consequence, it can be said that

  • $\mathsf A$ and $\mathsf M$ and $\mathsf B$ were (separately) members of distinct inertial systems, and

  • $\mathsf A$ and $\mathsf M$ and $\mathsf B$ were moving with respect to each other on a "common line of sight"; and that the speed of $\mathsf A$ and $\mathsf M$ with respect to each other is equal to the speed of $\mathsf B$ and $\mathsf M$ with respect to each other (let's denote that speed value as $u$).

Let's consider a trial of $\mathsf M$ having stated one particular signal indication $M_s$ and having observed $n + 1$ successive pings to $\mathsf A$ and (with corresponding coincident observations) to $\mathsf B$. In the course of this trial both $\mathsf A$ and $\mathsf B$ observed $\mathsf M$'s indication $M_s$ and $n$ successive pings to $\mathsf M$. Importantly ("due to the symmetry of the setup") $\mathsf A$'s duration from having observed $\mathsf M$'s indication $M_s$ until having observed the corresponding $n$th successive echo from $\mathsf M$ is equal to $\mathsf B$'s duration from having observed $\mathsf M$'s indication $M_s$ until having observed the corresponding $n$th successive echo from $\mathsf M$;

$$\tau \mathsf A[~{}_{\circledR \mathsf M_s}, {}_{(\circledR \mathsf M \circledR \mathsf A)^{\mathbf n} \circledR \mathsf M_s}~] = \tau \mathsf B[~{}_{\circledR \mathsf M_s}, {}_{(\circledR \mathsf M \circledR \mathsf B)^{\mathbf n} \circledR \mathsf M_s}~].$$

Call $k_a[~n~] \ge 2$ the number of "tick" indications counted of clock $\mathcal A$, and $k_b[~n~] \ge 2$ "tick" the number of "tick" indications counted of clock $\mathcal B$. (Require a trial with $n$ large enough to obtain sufficiently large "tick" counts.) Consequently, as in (4) above, it can be concluded that $(k_b[~n~] \pm 1) f_{\mathcal A} = (k_a[~n~] \pm 1) f_{\mathcal B}$; and if it is found that $k_a[~n~] - k_b[~n~] \ge 2$ then clock $\mathcal B$ is said to have "run more slowly" than clock $\mathcal A$.

So far on how to determine "which clock runs more slowly".

But of course there's a closely related consideration still to be addressed:

one can claim that A is at rest and B is experiencing time dilation or equally claim that B is at rest and A is experiencing time dilation,

While the description above (especially the appeal to the "symmetry of the setup" and the evaluation of equal speeds $u$) was, so to say, "centered at $\mathsf M$", it is of course possible and also instructive to describe the setup of (5) specificly "from the perspective" of participant $\mathsf A$ (along with all those participants who were at rest with respect to $\mathsf A$; jointly as members of one common inertial frame); or likewise for $\mathsf B$ (along with participants who were at rest with respect to $\mathsf B$).

This means specificly: expressing the setup relations in terms of distances between $\mathsf A$ and/or between the relevant participants who are at rest wrt. $\mathsf A$ and wrt. each other; and (using simultaneity relations among the indications of these participants, incl. $\mathsf A$) in terms of $\mathsf A$'s durations.

As noted already, $\mathsf A$ and $\mathsf M$ are moving with respect to each other at speed $u$. From the conditions of "motion on a common line of sight" and "$\mathsf M$ always finds same pings to $\mathsf A$ and to $\mathsf B$" follows (by a somewhat tedious but straighforward calculation, the details of which I may add later) the speed value of $\mathsf B$ and $\mathsf A$ with respect to each other: $$v = \frac{2~u}{1 + (u/c)^2}$$.

A related result obtains for ratios of the relevant durations which characterize the trial under consideration:

Let's call $\mathsf P$ the participant who remained at rest wrt. $\mathsf A$ and whom $\mathsf B$ observed meeting and passing in coincidence with observing $\mathsf M$'s signal indication $M_s$,
and call $\mathsf Q$ the participant who remained at rest wrt. $\mathsf A$ and whom $\mathsf B$ observed meeting and passing in coincidence with observing the corresponding $n$th successive echo from $\mathsf M$.

By the simultaneity relation between indications of $\mathsf A$ and indications of $\mathsf P$ and $\mathsf Q$ are identified

  • $\mathsf A$'s indication simultaneous to $\mathsf P$'s indication of having met $\mathsf B$ (i.e. indication $A_{\circledS \mathsf P \circ \mathsf B}$), and

  • $\mathsf A$'s indication simultaneous to $\mathsf Q$'s indication of having met $\mathsf B$ (i.e. indication $A_{\circledS \mathsf Q \circ \mathsf B}$); and consequently

  • $\mathsf A$'s duration between these two indications: $\tau \mathsf A[~{}_{\circledS \mathsf P \circ \mathsf B}, {}_{\circledS \mathsf Q \circ \mathsf B}~]$.

Of particular interest is the comparison with $\mathsf A$'s (own) duration between having observed $\mathsf M$'s signal indication $M_s$ and having observed the corresponding $n$th successive echo from $\mathsf M$; $\tau \mathsf A[~{}_{\circledR \mathsf M_s}, {}_{(\circledR \mathsf M \circledR \mathsf A)^{\mathbf n} \circledR \mathsf M_s}~]$.

It obtains (again by a somewhat tedious but straighforward calculation, the details of which I may add later):

$$\frac{\tau \mathsf A[~{}_{\circledR \mathsf M_s}, {}_{(\circledR \mathsf M \circledR \mathsf A)^{\mathbf n} \circledR \mathsf M_s}~]}{\tau \mathsf A[~{}_{\circledS \mathsf P \circ \mathsf B}, {}_{\circledS \mathsf Q \circ \mathsf B}~]} = \frac{1 - (u/c)^2}{1 + (u/c)^2} := \sqrt{1 - (v/c)^2};$$

i.e. the familiar "dilation factor".
Obviously, a similar derivation obtains "from the perspective" of participant $\mathsf B$ (along with suitable participants who were at rest with respect to $\mathsf B$). Importantly, it would be improper to attribute some duration of $\mathsf A$ to some duration of $\mathsf B$; even if they have a specific relation to each other, as have $\tau \mathsf A[~{}_{\circledS \mathsf P \circ \mathsf B}, {}_{\circledS \mathsf Q \circ \mathsf B}~]$ and $\tau \mathsf B[~{}_{\circ \mathsf P}, {}_{\circ \mathsf Q}~] \equiv \tau \mathsf B[~{}_{\circledR \mathsf M_s}, {}_{(\circledR \mathsf M \circledR \mathsf B)^{\mathbf n} \circledR \mathsf M_s}~]$.

p.s.

The second "paradox" involves a circular particle accelerator

I'm planning to address this second part of your question in a separate answer, perhaps within a few days. Of particular interest should be how to determine whether some given set of participants is "(rigidly) rotating", or (instead) "at rest" with respect to each other.

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I can obviously see the apparent contradiction – one can claim that A is at rest and B is experiencing time dilation or equally claim that B is at rest and A is experiencing time dilation, giving two different results when the two clocks "meets" – but I have no idea to actually solve it.

It is true that A will deem B's clock to be running slow but that B will deem A's clock to be running slow. How to resolve this?

The resolution is simple: That's exactly how things work. There is no true paradox here. This is just an apparent contradiction that results from false expectations.

The second "paradox" involves a circular particle accelerator with equally spaced electrons.

This supposed paradox is so chock full of errors that it's hard to know where to begin. The shape of the electrons is a red herring. Electrons are, as far as we can tell, point particles. They have no shape. Even worse, this supposed paradox ignores relativity of simultaneity. This is the Ehrenfest paradox, writ wrong. Worse yet, it it ignores that things do indeed look different from the perspective of one of those electrons, and that different electrons disagree.

One can naively object that this makes no sense, or one can learn relativity and find that it does make sense. It's just counterintuitive. Our intuition is based on everything we see in our everyday world moving very slowly compared to the speed of light. Our intuition does not apply here.

==============================================================================

The twin paradox has famously been tested by the Hafele and Keating experiment. This experiment flew atomic clocks around the world on a plane heading east and on a plane heading west. On landing at the destination, the clocks disagreed with atomic clocks at the US Naval Observatory that didn't make the flight. The disagreements were inconsistent with the hypothesis that time dilation doesn't exist. They were consistent with the results predicted by relativity. Relativity is tested on a daily basis by physicists running their particle accelerators and by an untold number of people using the GPS receivers in their smart phones. Those particle accelerators and those GPS receivers wouldn't work if relativity was wrong.

All of the paradoxes in relativity are a result of false expectations, misinterpretation of concepts, ignoring relativity of simultaneity, or abusing reference frames.

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  • $\begingroup$ Good answer: did you mean Tolman's Paradox, though? (I don't quite see the link with the Ehrenfest paradox, or maybe there's another Ehrenfest one I don't know of). $\endgroup$ – WetSavannaAnimal Sep 10 '15 at 3:42
  • $\begingroup$ @WetSavannaAnimalakaRodVance - I was definitely thinking of Ehrenfest's paradox. This paradox primarily addresses a rotating rigid body. The shell of rotating electrons looks like a rotating rigid body from the perspective of the lab frame. The crackpot website cited in the question implicitly assumes this must also the case from the perspective of one of the electrons. The resolution is that this is not the case. $\endgroup$ – David Hammen Sep 10 '15 at 3:53
  • $\begingroup$ Ah yes, quite right: I was thinking in relation to the first question. $\endgroup$ – WetSavannaAnimal Sep 10 '15 at 4:08

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