# B Time dilation and Einstein's theorem

#### worlov

Hello!

Einstein's theorem is in the last sentence of the following quote (bold) :

"If at the points A and B of K there are stationary clocks which, viewed in the stationary system, are synchronous; and if the clock at A is moved with the velocity v along the line AB to B, then on its arrival at B the two clocks no longer synchronize, but the clock moved from A to B lags behind the other which has remained at B by 1/2 tv2/c2 (up to magnitudes of fourth and higher order), t being the time occupied in the journey from A to B. It is at once apparent that this result still holds good if the clock moves from A to B in any polygonal line, and also when the points A and B coincide."

This theorem is better known as the twin paradox. Einstein did not stop at a preliminary remark and wrote
in more detail:

"If we assume that the result proved for a polygonal line is also valid for a continuously curved line, we arrive at this result: If one of two synchronous clocks at A is moved in a closed curve with constant velocity until it returns to A, the journey lasting t seconds, then by the clock which has remained at rest the travelled clock on its arrival at A will be 1/2tv2/c2 second slow. Thence we conclude that a balance-clock at the equator must go more slowly, by a very small amount, than a precisely similar clock situated at one of the poles under otherwise identical conditions."

Each observer determines that all clocks in motion relative to that observer run slower than that observer’s own clock. If the moving observer returns to point A, he will notice the following: the clock at point A is slower than his own clock. but this clock shows that more time has passed since his departure than his own clock. For moving observers this conjucture makes no sense. If a clock is slower than its own clock, it should also show less time. Therefore, the clock in point A should be synchronized according to the rules of relativity ... but how?

 ON THE ELECTRODYNAMICS OF MOVING BODIES By A. EINSTEIN June 30, 1905.

Related Special and General Relativity News on Phys.org

#### Ibix

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For moving observers this conjucture makes no sense. If a clock is slower than its own clock, it should also show less time. Therefore, the clock in point A should be synchronized according to the rules of relativity ... but how?
Time dilation and length contraction are well known. A third phenomenon, much more important but less cool-sounding, is the relativity of simultaneity. This resolves the problem. When the moving clock turns round it changes synchronisation convention, and the change in synchronisation exactly accounts for the "extra" time that the stationary clock shows compared to a naive time-dilation analysis done by the moving clock.

The simplest way to see this is to use the Lorentz transforms to work out what time the outbound frame says clock A will show at turnaround and what time the inbound frame will say it shows. The difference is the "extra" time you need.

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#### worlov

So, the moving observer always watches the clock at point A. If he moves away from point A, in his view the clock at point A goes slower. If he moves to point A, the clock at point A goes faster. If he reaches point A, the clock will go renewed slower than his own clock.

Here I miss the actual function of a clock - even measurement of time.

#### PeroK

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Each observer determines that all clocks in motion relative to that observer run slower than that observer’s own clock. If the moving observer returns to point A, he will notice the following: the clock at point A is slower than his own clock. but this clock shows that more time has passed since his departure than his own clock. For moving observers this conjucture makes no sense. If a clock is slower than its own clock, it should also show less time. Therefore, the clock in point A should be synchronized according to the rules of relativity ... but how?
In the case of a circular orbit, the "moving" clock is not moving inertially but continuously accelerating. This means that this clock is not an inertial observer and there is a critical and fundamental difference between that clock and a clock that is moving inertially.

Certainly, if the laws of SR could be applied directly and equally to both clocks, then a true contradiction must result. But, the moving observer (if he is accelerating in circular motion) must apply SR taking this constant acceleration into account. When he does that he agrees that his own clock runs continuously slower than clock A. There is an analysis of that scenario here:

https://www.physicsforums.com/threads/two-twins-moving-towards-each-other-in-a-circular-orbit.896607/#post-5660815

PS In the opening of part I of the Einstein paper, he says:

"Let us take a system of co-ordinates in which the equations of Newtonian mechanics hold good."

This excludes the case where an observer is moving in a circle, where the system of coordinates do not meet this criterion.

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#### Ibix

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So, the moving observer always watches the clock at point A. If he moves away from point A, in his view the clock at point A goes slower. If he moves to point A, the clock at point A goes faster. If he reaches point A, the clock will go renewed slower than his own clock.
That's an accurate description of what the observer will see. But the speed of light is finite, so what she sees is not what the clock is currently showing. She needs to correct for the travel time of light in order to work out what the clock is showing now. Once she does that, she will find that the clock always runs slow - but that the time it is showing "now" changes at turnaround. The amount of that change accounts for the "extra" time the clock at A shows compared to a naive time-dilation-only analysis.

#### worlov

When he does that he agrees that his own clock runs continuously slower than clock A.
Exactly! One of the twins has to admit that his system is so to say subordinate. Then, why should not he believe his eyes, if he sees that the clock in point A is slower than his own clock?

#### PeroK

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Exactly! One of the twins has to admit that his system is so to say subordinate. Then, why should not he believe his eyes, if he sees that the clock in point A is slower than his own clock?
He doesn't measure clock A as running slower. He measures clock A as running faster than his own.

#### Ibix

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One of the twins has to admit that his system is so to say subordinate.
No - one of the twins has a non-inertial coordinate system. It's not "subordinate", whatever you might mean by that. It's different, so naive application of time dilation formulae (which only apply between inertial coordinate systems) won't work.

#### David Byrden

This "paradox" is not paradoxical at all - the predicted results are consistent from all viewpoints. But what it is, is confusing.

Some of you ask, why should an observer not believe what he sees with his own eyes? Because what he sees is carried to him by photons, travelling at "c", and his own velocity is of a similar magnitude. There's a Doppler Effect.
For example, suppose that you are travelling towards a clock which, objectively, according to Lorentz rules, is running slower than yours. You may see it running faster because you are racing into its stream of photons.

Let's make the Doppler effect vanishingly small. Place a "stationary" clock on a "ground" surface and place another "moving" clock one light-year above it. We will move this clock horizontally for only an hour. It will remain almost exactly above the "stationary" clock, not moving toward or away from it by any significant amount.

At a certain time, the "moving" clock sees the "stationary" clock displaying Noon. (of course, this display happened a year before). At this moment we give the "moving" clock an almighty whack with a cricket bat, accelerating it to relativistic speed. After the hour has passed, a friend whacks it back toward us, and an hour later we stop it where it originally was, with a third whack. Thanks to the cricket bats, the clock has spent very little time accelerating.

In our "stationary" reference frame, this "moving" clock was running at half speed. It now reads 1 o'clock but we can see the "stationary" clock, far below us, displaying 2 o'clock.

Because the photons rising up from the "stationary" clock are effectively parallel at both ends of the journey, and orthogonal to the path of the journey, we can be assured that when the "moving" clock read 12:30 o'clock, it was being hit by photons from the "stationary" clock that carried an image of 1 o'clock.

Surely an observer on the "moving" clock would think that the "stationary" clock was the faster one? And isn't that contrary to Relativity? How could an observer on the moving clock possibly conclude that the "stationary" clock was running slower than his own?

Because when he looks out his window while moving, he does not see the "stationary" clock below him. He sees it close to being ahead of him. This is a consequence of light always travelling at "c" relative to inertial observers; the angle of the light incoming to the "moving" clock depends on what reference frame you use.

We tried to elminate the Doppler effect, but we eliminated it only from the "stationary" frame. It's present in the "moving" frame.

Now, suppose that he measures the angle at which he sees this clock ahead of him, and calculates the Doppler effect, and compensates for it, what will he discover? He will discover that the "stationary" clock is actually running slow. It simply looks like it's running fast.

When he gets whacked back in the opposite direction, the clock (and the whole universe) appears to swing around so that it's ahead of him again. He continues to see a clock running at double speed, while his calculations tell him it's really at half speed.

That is how the paradox resolves itself.

David

#### worlov

so naive application of time dilation formulae (which only apply between inertial coordinate systems) won't work.
What does "naive" mean? It is a basis of the theory of relativity: the principle of relativity! Every observer can consider himself at rest. Therefore he uses the Lorentz transformations. In this way every observer gets the same result. This also affects time dilation.

By the way, the movement in the circle is accelerated, but this has no effect on the formula for time dilation:
https://www.researchgate.net/publication/30398795_Measurements_of_relativistic_time_dilatation_for_positive_and_negative_muons_in_a_circular_orbit

#### PeroK

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What does "naive" mean? It is a basis of the theory of relativity: the principle of relativity! Every observer can consider himself at rest. Therefore he uses the Lorentz transformations. In this way every observer gets the same result. This also affects time dilation.

By the way, the movement in the circle is accelerated, but this has no effect on the formula for time dilation:
https://www.researchgate.net/publication/30398795_Measurements_of_relativistic_time_dilatation_for_positive_and_negative_muons_in_a_circular_orbit
What Einstein wrote, essentially:

"Consider a system of coordinates in which the laws of Newtonian mechanics hold good ... we assume (as a postulate) that the speed of light is independent of the source ... we show that time dilation applies to a clock moving in this system of coordinates".

How @worlov interpreted this:

"Consider any system of coordinates (whether Newton's laws apply or not) ... assume time dilation and the Lorentz Transformation apply with this coordinate system ... state that any observer can consider himself at rest and consider himself an inertial observer - reach a contradiction."

You are ignoring the critical criterion of Einstein's paper that his result apply to observations made in an inertial reference frame. Not in any reference frame.

Movement in a circle has no effect on the formula for time dilation for an inertial obsever, observing the object in circular motion. But, the object in circular motion is not an inertial observer and cannot apply the time dilation formula to objects moving in his system of coordiates.

That is, incidentally, the same in Newtonian mechanics. For example, if you are an inertial observer and an object is moving in a circle, you can conclude and calculate that the object is subject to a centripetal force. However, for the object moving in a circle, it sees you moving in a circle, but if it concludes that you must be subject to a force, then it is wrong. There is no force on you. Your circular motion is due to its acceleration, not your own.

Or, think of someone jumping up and down on a trampoline. You can study the trampolinist's motion using Netwon's laws (or SR if your prefer). But, the person on the trampoline cannot directly use Newton's laws (or SR) to explain why the rest of the world is bouncing up and down!

If you do not accept this, then you can go ahead and find contradictions all over physics. No physics will hold together if you take results derived in inertial reference frames and apply them in non-inertial reference frames. Time dilation is one example. Newton's laws of motion are another example.

You must understand the difference between observing accelerated motion (from an inertial reference frame) and an observer in a state of acceleration trying to use the laws of physics that do not apply in an accelerated coordinate system.

• Dale, vela and m4r35n357

#### Nugatory

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What does "naive" mean? It is a basis of the theory of relativity: the principle of relativity! Every observer can consider himself at rest.
Yes, every observer can consider themselves at rest, but that is not what the principle of relativity says. The principle of relativity says that all inertial frames are equivalent. For some observers, the frame in which they are at rest is not inertial. And....
Therefore he uses the Lorentz transformations. In this way every observer gets the same result. This also affects time dilation.
The derivation of the Lorentz transformations assumes that the frames in question are inertial; thus they cannot be correctly applied to transform between non-inertial frames. The researchgate paper you've cited doesn't suggest otherwise.

• PeroK

#### Ibix

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What does "naive" mean? It is a basis of the theory of relativity: the principle of relativity! Every observer can consider himself at rest.
Yes, everybody can regard themself as at rest. Assuming that this means that all observers can regard themself as inertial is naive.
Therefore he uses the Lorentz transformations.
No. Not unless he remains in an inertial frame.
In this way every observer gets the same result. This also affects time dilation.
Clearly not - or the twin paradox would be genuinely paradoxical.
By the way, the movement in the circle is accelerated, but this has no effect on the formula for time dilation:
Yes it does. Between inertial frames, both frames regard the other's clock as running slow by a factor of $1/\gamma$. In the case of circular motion the orbiting observer sees the inertial clock running fast by a factor of $\gamma$ while the inertial observer sees the orbiting clock running slow by a factor of $1/\gamma$. The situation is not symmetric, unlike two inertial observers.

Edit: I can't seem to get the ResearchGate paper you cite to download (on the move - dodgy internet connection). I rather suspect from the abstract that it does not consider the perspective of the muons, since there are no physicists co-moving with them. Thus I presume it won't do the bit of maths that would show the asymmetry. Itxs easy enough to do if you want to do it.

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• PeroK

#### Janus

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What does "naive" mean? It is a basis of the theory of relativity: the principle of relativity! Every observer can consider himself at rest. Therefore he uses the Lorentz transformations. In this way every observer gets the same result. This also affects time dilation.
But How he uses the Lorentz transformations are different when he is accelerating than if he is in an inertial frame. To know what another clock is doing, an observer in inertial motion only has to know the relative velocity of the other clock with respect to himself. If he is accelerating there is an additional factor he has to account for. This factor depends on the magnitude of the acceleration, the direction of the other clock with respect to the acceleration, and the distance between them along the line of acceleration. If you are accelerating towards a the clock, then this factor has the clock running fast, the further away, the faster it runs. Accelerating away slows the clock according to the observer*
By the way, the movement in the circle is accelerated, but this has no effect on the formula for time dilation:
https://www.researchgate.net/publication/30398795_Measurements_of_relativistic_time_dilatation_for_positive_and_negative_muons_in_a_circular_orbit
But it does. Imagine three clocks: one at the center of the circle, one circling it at some speed, and one passing it in a straight line at the same speed relative to the center clock. Assume that the third clock travels at a tangent to the second clock's path, and at the moment of interest they are right next to each other. Both clocks will be getting the same information in terms of light information from the center clock at that moment. However, the circling clock will determine that the center clock is running fast compared to itself, while the inertially moving clock will determine that the center clock is running slow compared to itself.
However both the circling clock and the inertial moving clock will agree that they are running at the same rate at the moment they pass each other.

* this is true even if the clocks are not moving relative to each other. Take two clocks, put them both in the same accelerating frame, with the clocks separated along the line of the acceleration, and the clock in the direction of the acceleration will run fast compared to the other, even though they are at rest with respect to each other and under the same acceleration.

This does no

#### worlov

How @worlov interpreted this
It should not be forgotten that Einstein formulated his theorem in the context of the special theory of relativity. He begins with a polygonal route:

"...It is at once apparent that this result still holds good if the clock moves from A to B in any polygonal line..."

and only after passes to the circle:

"If we assume that the result proved for a polygonal line is also valid for a continuously curved line... Thence we conclude that a balance-clock at the equator must go more slowly, by a very small amount, than a precisely similar clock situated at one of the poles under otherwise identical conditions."

Well, Einstein ignored the acceleration phases.

#### jbriggs444

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But How he uses the Lorentz transformations are different when he is accelerating than if he is in an inertial frame.
Let me try to expand on that a bit.

The Lorentz transforms let you take the coordinates (x,y,z,t) of an event in one frame and obtain the coordinates (x', y', z', t') of the same event in some other frame. There is no particular problem in using the Lorentz transforms to do this for an accelerating frame. It is a simple process.

Suppose that we start with the coordinates (x,y,z,t) in the accelerating frame. We translate to coordinates in the inertial frame that is momentarily co-moving with the accelerating frame at time t. This is dead easy since it is the identity translation. We still have coordinates (x,y,z,t). Then we apply the Lorentz transform to shift to (x',y',z',t') coordinates in the target inertial frame. For the "v" term in the transform, we, of course, use the relative velocity between the two inertial frames.

Easy, peazy. But there is a catch. This works for translating coordinates of events. If we want to do something fancier -- like computing a time dilation factor, a velocity or an acceleration things are not so simple. Let's us do a time dilation.

In principle, computing time dilation is fairly simple. We pick out two events that are separated by $\Delta t$ in our accelerated coordinate system. Then we find the coordinates for the same two events in the inertial coordinate system and look for the $\Delta t'$.

So let us do that. We have two events $(x_1, y_1, z_1, t_1)$ and $(x_2, y_2, z_2, t_1 + \Delta t)$. We do the Lorentz transform on the first one and get $(x'_1, y'_1, z'_1, t'_1)$. We do the Lorentz transform on the second one and get $(x'_2, y'_2, z'_2, t'_1 + something)$.

We might naively expect to find a time dilation factor of $\gamma$. But there is a gotcha.

The gotcha is that in the interval between the first event and the second, our starting frame accelerated. The relative velocity between the frames changed. The second Lorentz transform does not use the same v as the first. That means that the time dilation factor is not as simple as $\gamma$. Nor is it something simple-minded like $\frac{\gamma_1 + \gamma_2}{2}$. The Lorentz transform for time looks like:
$$t' = \gamma (t-\frac{vx}{c^2})$$
To a first order approximation, if you change v between the two transforms, that gives rise to a change of $\frac{\gamma \ x \ \Delta v}{c^2}$. The presence of the $\Delta v$ is a clue that acceleration factors in. The presence of the x means that separation in the x direction also factors in.

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#### PeroK

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Well, Einstein ignored the acceleration phases.
Yes, and by doing this an observer moving with the clock, in a polygon or a circle, is not an inertial observer. So, that observer cannot directly and simply employ the theory of SR as set out in the previous sections of the paper. In the polygon approximation, which is how I analysed the problem in the link I posted earlier, the "moving" clock is instantaneously jumping from one inertial frame to another. You must, therefore, apply the Lorentz Transformation separately for each side of the polygon and add up all those phases. When you do that, you see that the clock at A is advancing more quickly than the "moving" clock.

Check out the link I posted, where as others have mentioned, the relativity of simultaneity is critical to the calculations.

#### Ibix

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Well, Einstein ignored the acceleration phases
But not the relativity of simultaneity. Which is the source of the "extra time" I've been banging on about since #2. You can either treat the polygonal lines using inertial frames and correct for the relativity of simultaneity at the corners, or regard the whole thing as a non-inertial frame. Either way, simply applying the inertial frame time dilation formula isn't correct for describing the inertial clock from the traveller's point of view.

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#### worlov

Check out the link I posted, where as others have mentioned, the relativity of simultaneity is critical to the calculations.
We can mark all clocks. The moving observer will then enter in a table which clock which time has indicated (compared with his own time). When the moving observer sees the first clock again, he can check his table. He will notice that this clock shows too much time, although it is just as slow as before.

#### Janus

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We can mark all clocks. The moving observer will then enter in a table which clock which time has indicated (compared with his own time). When the moving observer sees the first clock again, he can check his table. He will notice that this clock shows too much time, although it is just as slow as before.
Let's try to clarify just what is meant by how the Relativity of Simultaneity is critical.
You have three clocks, Red, Green, and Blue.
Red and Green remain at rest with respect to each other, and it their rest frame are synchronized. Blue travels from Red to Green and then back to Red.
Thus according to the Rest frame of Red and Green the sequence of events unfolds like this assuming that blue has a relative velocity of 0.8c, the distance between Red and Green is 1 light hour, as measured in the rest frame of Red and Green. Blue leaves Red while All three clocks read zero.
Blue arrives at green when both Red and Green reads 1.25 hr, and it reads 0.75 hr due to time dilation.
Blue reverses direction. All the clock readings remain the same.
Blue arrives back at Red, reading 1.5 hrs, while Red and Green read 2.5 hrs.

Now the same events according to Blue. Blue is at rest while Red and Green are moving. Since Red and Green are moving, they and the distance between them is length contracted to 0.6 light hr. Red and Blue still both read zero when they start off next to each other. However, due to the Relativity of Simultaneity, Green already reads 0.8 hr
It takes 0.75 hr for Green to reach Blue (0.6 light hr/ 0.8 c) During which time both Red and Green run 0.6 as fast as Blue and advance by 0.45 hr. Green now reads 1.25 and Blue reads 0.75, just like when we looked at things from Red and Green's frame. Red however only reads 0.45 hr. (0.8 hr behind Green)
Red and Green reverse direction. The reading on Red and Green do not change, But now that the relative velocity of Red and Green with respect to Blue have changed, Relativity of Simultaneity requires that Red is the clock that is ahead of Green by 0.8 hr. Thus now Red read 2.05 hr.
Again, it take 0.75 for Red to reach Blue, during which time both Red and Green advance 0.45 hr. Upon rejoining, Blue reads 1.5 hr and Red reads 2.5 hr. In perfect agreement with the end result when we assumed that Red and Green were at rest the whole time.

Red and Blue are in perfect agreement as to the start and end of the scenario in terms of what each of them reads, but they disagree as to what happened between start and finish.

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#### Dale

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Here I miss the actual function of a clock - even measurement of time.
The type of time measured by a clock is called proper time. It is different from coordinate time. It is a relativistic invariant that can be calculated by the integral of $\sqrt{dt^2-dx^2-dy^2-dz^2}$. All inertial frames agree on this formula and even non inertial frames agree on its value but not the formula.

One of the twins has to admit that his system is so to say subordinate.
Not subordinate. The term is “non inertial”. Everybody agrees that the traveling twin’s frame is non inertial, including the traveling twin himself.

Well, Einstein ignored the acceleration phases.
True. He also performed all of his calculations in the inertial frame only.

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• vanhees71

#### worlov

In perfect agreement with the end result when we assumed that Red and Green were at rest the whole time.
You presented one of the possible calculation for the twin paradox. But that is not the point. Einstein assumes the existence of such a clock, which indicates more time than would have passed according to its speed. In nature there is no such clock. Therefore his assumptions should be reconsidered anew.

#### Ibix

Science Advisor
@worlov - can you explain to me what is the relativity of simultaneity? Because it seems to me that you are totally failing to grasp this and its implications for twin paradox-like scenarios.

#### worlov

To determine the simultaneity at different locations, the running time of the light must be considered... I do not have against the twin paradox - the calculation is correct. But ... it's about natural processes. Physically, no clock can exist that displays more time than it counts with "ticks".

Suppose the moving observer before departure has hidden the clock at point A in a safe. No one can see or even manipulate this clock during his travel. If the moving observer returns and opens the safe, which displayed time he will expect? He sees that the clock is just as slow as when he left. Of course, he assumes that this clock was always uniformly in his absence. And he knows how long he was gone.

#### Ibix

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Physically, no clock can exist that displays more time than it counts with "ticks".
And no clock is doing so anywhere in this setup. The point about the relativity of simultaneity is that if you naively apply the time dilation formula to a polygonal path (edit: or, indeed, a curved path) from the perspective of the person following that path, then there are ticks of the inertial clock that you forgot to account for.
Suppose the moving observer before departure has hidden the clock at point A in a safe. No one can see or even manipulate this clock during his travel. If the moving observer returns and opens the safe, which displayed time he will expect? He sees that the clock is just as slow as when he left. Of course, he assumes that this clock was always uniformly in his absence. And he knows how long he was gone.
You are asking what time will the travelling twin expect to see on the inertial clock? If he sees $\Delta t$ on his clock then he'll expect to see $\gamma\Delta t$ on the inertial clock. Edit 2: Or, at least, if he accounts correctly for ALL relativistic effects he will.

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