[Jaaanosik (OP)]

I see the problem this way.
The frame (a) sees the rim of the wheel symmetrically. See the figures 13.14 and 13.15.
The frame (b) sees the rim of the wheel asymmetrically. See the figure 2 of the paper.
Both are the inertial frame observers.
The (a) and (b) observers are not on the rim itself though.
If there is an accelerated observer on the rim of the wheel then this local observer will measure either symmetrical centripetal acceleration as predicted by (a) frame or asymmetrical acceleration where the spacing between 'the rim blocks' changes as predicted by (b) or ... completely something else that neither reference frame predicted.
Jano

[Malamute Lover]

I see the problem this way.
The frame (a) sees the rim of the wheel symmetrically. See the figures 13.14 and 13.15.
The frame (b) sees the rim of the wheel asymmetrically. See the figure 2 of the paper.
Both are the inertial frame observers.
The (a) and (b) observers are not on the rim itself though.
If there is an accelerated observer on the rim of the wheel then this local observer will measure either symmetrical centripetal acceleration as predicted by (a) frame or asymmetrical acceleration where the spacing between 'the rim blocks' changes as predicted by (b) or ... completely something else that neither reference frame predicted.
Jano

As already stated, the outside observer who sees (b) is not looking at an inertial reference frame. The spokes are going faster on top and slower on the bottom relative to overall motion.

[Jaaanosik (OP)]

The world lines lose the symmetry in (b), they are asymmetric.
This is the problem, two different frames show different analysis.
They do not agree on the world lines. This is not good.
They do not agree on physics,

Two different reference frames seeing things differently is what relativity is all about.

How can this be reconciled with this: "The laws of physics take the same form in all inertial frames of reference."

The spokes of the wheel are traveling at different speeds as they go around. This is not an inertial frame of reference

.

Nonetheless I fall to see that total angular momentum is not conserved.

Can you, please, elaborate how you see it?
Jano

As I said earlier.
“

This situation is quite similar to the spin Hall effect in various systems, where variations in the intrinsic AM(spin) are compensated at the expense of the centroid shift generating extrinsic AM.
    […]
    However, in addition to the shape deformations, a rotating body also acquires mass deformations.  The y >0 and y <0 sides of the wheel have different velocities in the moving frame and their constituent particles acquire different local γ-factors. Owing to this, the y >0 particles become heavier than the y <0 particles.

    [page 2 of the article, the same page as the spoked wheel picture

The spokes on the top are denser, as seen in the picture and heavier. The energy centroid is higher than the geometric centroid. Looks to me like overall angular momentum is conserved, but redistributed. “

There is a tradeoff between intrinsic and extrinsic AM with the shift in energy centroid generating extrinsic AM compensating for the changes in intrinsic AM, just like the text of the paper says. Where does it say anything about non-conservation of AM? If that had been demonstrated by the paper, it would have been shouted long and loud as something totally new and ultra-important.

The question: Is accelerated observer on the wheel rim going to see/observe/measure the deformation that is not predicted by (a)?
Jano

[Jaaanosik (OP)]

I see the problem this way.
The frame (a) sees the rim of the wheel symmetrically. See the figures 13.14 and 13.15.
The frame (b) sees the rim of the wheel asymmetrically. See the figure 2 of the paper.
Both are the inertial frame observers.
The (a) and (b) observers are not on the rim itself though.
If there is an accelerated observer on the rim of the wheel then this local observer will measure either symmetrical centripetal acceleration as predicted by (a) frame or asymmetrical acceleration where the spacing between 'the rim blocks' changes as predicted by (b) or ... completely something else that neither reference frame predicted.
Jano

As already stated, the outside observer who sees (b) is not looking at an inertial reference frame. The spokes are going faster on top and slower on the bottom relative to overall motion.

My apologies, I introduced a new scenario when the axle is accelerated.
Having said that, my last couple of posts are about the textbook and the paper.
There is no acceleration of the axle here, just to make it clear.
Both, (a) and (b) are inertial observers looking at the rotating wheel.
The question stands, is the accelerated wheel rim observer going to see/observe/measure the deformation or not.
What prediction/observation wins for the accelerated observer on the wheel rim? Is it (a) or (b)?
Jano

[Malamute Lover]

I see the problem this way.
The frame (a) sees the rim of the wheel symmetrically. See the figures 13.14 and 13.15.
The frame (b) sees the rim of the wheel asymmetrically. See the figure 2 of the paper.
Both are the inertial frame observers.
The (a) and (b) observers are not on the rim itself though.
If there is an accelerated observer on the rim of the wheel then this local observer will measure either symmetrical centripetal acceleration as predicted by (a) frame or asymmetrical acceleration where the spacing between 'the rim blocks' changes as predicted by (b) or ... completely something else that neither reference frame predicted.
Jano

As already stated, the outside observer who sees (b) is not looking at an inertial reference frame. The spokes are going faster on top and slower on the bottom relative to overall motion.

My apologies, I introduced a new scenario when the axle is accelerated.
Having said that, my last couple of posts are about the textbook and the paper.
There is no acceleration of the axle here, just to make it clear.
Both, (a) and (b) are inertial observers looking at the rotating wheel.
The question stands, is the accelerated wheel rim observer going to see/observe/measure the deformation or not.
What prediction/observation wins for the accelerated observer on the wheel rim? Is it (a) or (b)?
Jano

I repeat, the observer of (b) is not looking at an inertial reference frame. There are continuous velocity changes as the wheel goes round (velocity being a vector). That is the whole point of the paper is that the relativistic effects seen in (b) are due to acceleration and not even straight line acceleration. This is why the energy centroid moves. And as I have already quoted from the paper, that movement balances the books on AM conservation.

[Jaaanosik (OP)]

...
I repeat, the observer of (b) is not looking at an inertial reference frame. There are continuous velocity changes as the wheel goes round (velocity being a vector). That is the whole point of the paper is that the relativistic effects seen in (b) are due to acceleration and not even straight line acceleration. This is why the energy centroid moves. And as I have already quoted from the paper, that movement balances the books on AM conservation.

The observer (a) is not looking at an inertial reference frame as well.
There is an observed velocity change for (a) observer too.
The (a) sees one centripetal acceleration to the center of the wheel and the rim blocks are equally spaced.
The (b) sees normal and tangential accelerations but their sum should point to where?
Energy centroid, different location?
If an observer is on the rim and a measurement is done, what is the direction of the acceleration?
Is the spacing/contraction change of the rim blocks real for the rim observer or not?
Jano

[Halc]

I see the problem this way.
The frame (a) sees the rim of the wheel symmetrically. See the figures 13.14 and 13.15.
The frame (b) sees the rim of the wheel asymmetrically. See the figure 2 of the paper.
Both are the inertial frame observers.
The (a) and (b) observers are not on the rim itself though.
If there is an accelerated observer on the rim of the wheel then this local observer will measure either symmetrical centripetal acceleration as predicted by (a) frame or asymmetrical acceleration where the spacing between 'the rim blocks' changes as predicted by (b) or ... completely something else that neither reference frame predicted.

The question stands, is the accelerated wheel rim observer going to see/observe/measure the deformation or not.
What prediction/observation wins for the accelerated observer on the wheel rim? Is it (a) or (b)?
Jano

I understand the question. No observer on the rim is going to see a symmentrical wheel. The close parts appear bigger, and the far side appears smaller.  It doesn't need to spin or move for that to be true.

As for the view, it is the same for any observer on the wheel at any point on the wheel.  They're all going to see the exact same thing which is the rest of the wheel going off into the distance to each side and curving up above in the 'up' direction.  There would be no way to tell where on the wheel you are in the picture or if the wheel is moving. This is exactly per principle of relativity which say there is no way, from within a box, to determine inertial motion of the box.

Of course the observer on the rim can tell that the wheel is spinning, which way, and how fast. But those proper facts are the same for any observer on the rim.

[Kryptid]

...
I repeat, the observer of (b) is not looking at an inertial reference frame. There are continuous velocity changes as the wheel goes round (velocity being a vector). That is the whole point of the paper is that the relativistic effects seen in (b) are due to acceleration and not even straight line acceleration. This is why the energy centroid moves. And as I have already quoted from the paper, that movement balances the books on AM conservation.

The observer (a) is not looking at an inertial reference frame as well.

If you are talking about two accelerating frames, those do not necessarily have to agree with each other. Accelerating frames are not inertial.

[Malamute Lover]

...
I repeat, the observer of (b) is not looking at an inertial reference frame. There are continuous velocity changes as the wheel goes round (velocity being a vector). That is the whole point of the paper is that the relativistic effects seen in (b) are due to acceleration and not even straight line acceleration. This is why the energy centroid moves. And as I have already quoted from the paper, that movement balances the books on AM conservation.

The observer (a) is not looking at an inertial reference frame as well.
There is an observed velocity change for (a) observer too.
The (a) sees one centripetal acceleration to the center of the wheel and the rim blocks are equally spaced.
The (b) sees normal and tangential accelerations but their sum should point to where?
Energy centroid, different location?
If an observer is on the rim and a measurement is done, what is the direction of the acceleration?
Is the spacing/contraction change of the rim blocks real for the rim observer or not?
Jano

In (a), that is as seen by an observer who is stationary with respect to the wheel, the AM is constant because the wheel is moving symmetrically. The motion of the wheel does not involve any speed changes, only direction. An observer on the rim will be on a wheel located in an inertial frame of reference and will see no difference in acceleration as the wheel goes around. This is also the case with an observer on the rim in (b). The wheel is still located in an inertial frame of reference to this observer. For an observer on the rim, there is no difference between (a) and (b). The presence or absence of an observer on the rim changes nothing. Neither sees any motion other than the wheel spinning.

In (b), that is as seen by an observer who sees the wheel moving at 0.7 c, the motion of the wheel does involve speed changes, the top goes faster and the bottom goes slower than the axle. The observed relativistic effects are different. The energy centroid that this observer sees has moved upward where the spokes are going faster and have more mass. Yet, as the paper says, the changes in the intrinsic AM are compensated by the changes in the extrinsic AM. Total AM is conserved.

[Jaaanosik (OP)]

The Fig. 13.15 shows 'the rim blocks'. Is the length contraction real? Are the blocks changing the length for real?
... or it is just a visual effect?
We connect the A₀ block and B₀ block with a string.
Is the string going to break when the wheel accelerates and the gap grows between A₁ block and B₁ block?
Jano

[Halc]

The Fig. 13.15 shows 'the rim blocks'. Is the length contraction real?

It's very real. Length contraction is not just a visual effect. It is entirely real. If radius is held constant as depicted, then real gaps must form between the blocks, which will be quite apparent to any observer.
I brought this up in post 3.

We connect the A₀ block and B₀ block with a string.
Is the string going to break when the wheel accelerates and the gap grows between A₁ block and B₁ block?

The string will break, just like in Bell's spaceship 'paradox'.  The only difference here is that they're moving in a circle instead of a straight line, but length contraction is real in both cases.

[Malamute Lover]

The Fig. 13.15 shows 'the rim blocks'. Is the length contraction real?

It's very real. Length contraction is not just a visual effect. It is entirely real. If radius is held constant as depicted, then real gaps must form between the blocks, which will be quite apparent to any observer.
I brought this up in post 3.

Several problems here.

In the Ehrenfest paradox, the circumference contracts, not expands, since the wheel is moving.  From the viewpoint of a non-rotating observer at the center of the circle, the blocks would be seen to shrink along with any gaps between them. This could be seen by comparing the observed length of the blocks compared to their width. An observer riding on the rim would not notice any difference in the size of the blocks or the gaps.

Why are you making the circumference larger? And why are the gaps larger?

The radius is assumed to be constant. This is incorrect. The size of the radius can only be determined with a measuring rod. If we use the rotating stick as the measuring rod, we will see something interesting. It will not appear to be rigid. Light coming the rod from near the rim will take longer to reach the observer in the center and will show the stick at an earlier time in its rotation. The stick will appear to curve and have a length greater than the expected radius.

But if the observer in the center holds a measuring rod out to the circumference until sparks fly when it touches the rim, the measured radius would be as expected for a non-rotating wheel. (Actually, the measurement will be a bit too long as it would take time for the light of the sparks to reach the center. But since the speed of light is known, an adjustment could be made for this.) This result contradicts the radius implied by the observations made of the blocks on the spinning wheel.

Length contraction is not real. It is relative. (Hey, a catchy name for a theory.)

We connect the A₀ block and B₀ block with a string.
Is the string going to break when the wheel accelerates and the gap grows between A₁ block and B₁ block?

The string will break, just like in Bell's spaceship 'paradox'.  The only difference here is that they're moving in a circle instead of a straight line, but length contraction is real in both cases.

Bell’s spaceship ‘paradox’ is not a paradox at all. It is impossible to synchronize separated clocks. The two spaceships cannot start at the same time. There is not such thing as the same time.

[Jaaanosik (OP)]

Halc,
that's the answer I expected.

Mamalute Lover,
if you deny Bell's spaceship 'paradox' then you are going against Einstein and his original 1905 paper.
He starts his paper with a definition of simultaneity:
http://hermes.ffn.ub.es/luisnavarro/nuevo_maletin/Einstein_1905_relativity.pdf

If you deny the simultaneity then you have nothing to back up your statements.
You cannot argue this is like this because the relativity says so. You denied the relativity.
Do you have your own math? Do you have your own hypothesis?
Jano

[Jaaanosik (OP)]

The Fig. 13.15 shows 'the rim blocks'. Is the length contraction real?

It's very real. Length contraction is not just a visual effect. It is entirely real. If radius is held constant as depicted, then real gaps must form between the blocks, which will be quite apparent to any observer.
I brought this up in post 3.

We connect the A₀ block and B₀ block with a string.
Is the string going to break when the wheel accelerates and the gap grows between A₁ block and B₁ block?

The string will break, just like in Bell's spaceship 'paradox'.  The only difference here is that they're moving in a circle instead of a straight line, but length contraction is real in both cases.

Halc,
if you look at the (b) frame analysis:



Is the deformation real then?
The wheel is already accelerated and it has a constant rotation.
If an A₁ rim observer connects a string to B₁ rim observer at the bottom then is it fair to expect the string to be broken at the top?
Is the (a) axle observer going to see the string broken?
Do we have a multiverse here?
The strings not broken for (a) but broken for (b)?
Jano

[Halc]

Several problems here.

In the Ehrenfest paradox, the circumference contracts, not expands, since the wheel is moving.

I didn't state otherwise, in the quote of mine to which you responded or in post 3. Given a spinning radius of 1, it has a contracted circumference of 6.283 and a rest circumference (proper circumference) of 8.796. That's contracting due to spin.  Either the spokes get shorter (reeled in??) as it spins, or the thing perhaps needs to be manufactured already spinning.

  From the viewpoint of a non-rotating observer at the center of the circle, the blocks would be seen to shrink along with any gaps between them.[/quote]Given fixed length radius, the blocks would have no gaps between them when stationary, and as they shrink, gaps would form, allowing more blocks to be inserted in them if you like. You seem to suggest that the gaps shrink as well, which is wrong.  Picture a bunch of roller coaster cars, touching each other while parked in a circular track. As they speed up, the track doesn't change size at all, but gaps must form between the cars as they contract. That's what the recent diagram depicts.  The ring circumference will contract if there's no track and the cars are bolted together. In that case there never are gaps, but the radius goes down as the ring circumference contracts.

This could be seen by comparing the observed length of the blocks compared to their width. An observer riding on the rim would not notice any difference in the size of the blocks or the gaps.

He'd very much notice the gaps forming, which were not there at all before. I agree that he'd not notice the ratio of length/width of his own car changing. We're talking about fixed radius here. Fixed spokes or a track or something.

Why are you making the circumference larger?

I'm not.

And why are the gaps larger?

Different example. Don't mix them. The text of mine you quoted is about gaps forming with blocks moving at a fixed radius.  Post 3 talks about a solid ring contracting as it rotates, reducing the radius. Nothing gets larger with speed.

The radius is assumed to be constant.

Only in the example in post 29, where the radius is held constant with detached objects moving around that fixed path at ever increasing speeds.

The size of the radius can only be determined with a measuring rod.

We have one. You can't measure a stationary track or fixed length spokes moving only perpendicular to their length?  Post 29 shows a sort of track. No spokes. The grey boxes seem to be the track, not spinning with the red stuff.

If we use the rotating stick as the measuring rod, we will see something interesting. It will not appear to be rigid. Light coming the rod from near the rim will take longer to reach the observer in the center and will show the stick at an earlier time in its rotation. The stick will appear to curve and have a length greater than the expected radius.

Appearances or no, the stick (spoke) is in fact straight in an frame where the axis of rotation is stationary. Light 'appearing' to curve is Coriolis effect that you get in a rotating frame. Light does not move in straight lines in a rotating frame. If you photograph the thing from a distant point on the axis, the spoke-stick will appear straight since light takes about equal time to get from any location of the stick to the point of view.

But if the observer in the center holds a measuring rod out to the circumference until sparks fly when it touches the rim, the measured radius would be as expected for a non-rotating wheel. (Actually, the measurement will be a bit too long as it would take time for the light of the sparks to reach the center. But since the speed of light is known, an adjustment could be made for this.)

If the radius is 1, then the light will take time 1 to reach the observer in the center despite it not moving along the spoke. No adjustment is needed.

This result contradicts the radius implied by the observations made of the blocks on the spinning wheel.

The blocks are not making any measurement of the radius. They measure proper circumference.

Length contraction is not real. It is relative.

You're denying that gaps form between the blocks? By your posts above, it seems so. Let me know how that works for you. I stand by my statement that contraction is objectively real and is well illustrated by the Ehrenfest scenario. There is no frame in which those gaps do not form.

Bell’s spaceship ‘paradox’ is not a paradox at all.

With that I'll agree. None of them are paradoxes.

It is impossible to synchronize separated clocks. The two spaceships cannot start at the same time. There is not such thing as the same time.

This is nonsense.
The ships are initially stationary, and begin identical proper acceleration at the same time relative to the frame in which they are stationary. Are you in denial now that clocks can be synced in a given inertial frame? Einstein gives some nice examples of ways to do exactly that.

You appear to be searching for cop-out excuses to avoid explaining a scenario that you apparently don't understand. The ship at the rear could even start out a little before the other, putting initial slack in the string. As the ships and the string gain speed and contract, the string will eventually break, but it takes a bit longer due to that initial slack.

Is the deformation real then?

The deformation into an ellipse is frame dependent, but that doesn't make it not real. M-L seems to think otherwise.

The wheel is already accelerated and it has a constant rotation.

Both wheels have identical proper angular velocity. That's not the same as constant rotation. The angular velocity of an object is frame dependent since it can be used as a clock, and time is dilated in a frame in which the object is moving.

If an A1 rim observer connects a string to B1 rim observer at the bottom then is it fair to expect the string to be broken at the top?

If the two wheels are different wheels in the same frame, then the string breaks same as if I attach a string between a moving car and a parked one.
I don't think you mean that, but I don't know what you mean by 'A1 rim' and 'B1 rim'.  They seem to be references to different objects in relative motion, which probably breaks the string.
If you mean a string from one side of a wheel to the other side of the same wheel, then no, that string will not break for either wheel so long as they keep spinning.  The spokes already serve as such a string.

Is the (a) axle observer going to see the string broken?

I cannot figure out where you are putting your string. The (a) axle observer cannot see anything different than the (b) axle observer since, per principle of relativity, linear motion cannot be locally detected. The two wheels might be the same wheel, just considered in two different inertial frames.

Do we have a multiverse here?

?  What brings this up?

The strings not broken for (a) but broken for (b)?

Your description made it sound like one string between the two wheels, so if one sees it break, the other will see the same string break.

[Jaaanosik (OP)]

...

Is the deformation real then?

The deformation into an ellipse is frame dependent, but that doesn't make it not real. M-L seems to think otherwise.

The wheel is already accelerated and it has a constant rotation.

Both wheels have identical proper angular velocity. That's not the same as constant rotation. The angular velocity of an object is frame dependent since it can be used as a clock, and time is dilated in a frame in which the object is moving.

If an A1 rim observer connects a string to B1 rim observer at the bottom then is it fair to expect the string to be broken at the top?

If the two wheels are different wheels in the same frame, then the string breaks same as if I attach a string between a moving car and a parked one.
I don't think you mean that, but I don't know what you mean by 'A1 rim' and 'B1 rim'.  They seem to be references to different objects in relative motion, which probably breaks the string.
If you mean a string from one side of a wheel to the other side of the same wheel, then no, that string will not break for either wheel so long as they keep spinning.  The spokes already serve as such a string.

Is the (a) axle observer going to see the string broken?

I cannot figure out where you are putting your string. The (a) axle observer cannot see anything different than the (b) axle observer since, per principle of relativity, linear motion cannot be locally detected. The two wheels might be the same wheel, just considered in two different inertial frames.

Do we have a multiverse here?

?  What brings this up?

The strings not broken for (a) but broken for (b)?

Your description made it sound like one string between the two wheels, so if one sees it break, the other will see the same string break.

Halc,
This is the bottom of the cycloid as seen from (b).
Please, ignore A₀ and B₀, it should say A₁ and B₁ because the speed is close to 0 in the (b) frame, the bottom part of the cycloid, therefore there is almost no gap between the A₁ and B₁ rim blocks from (b) point of view.
The string is attached here at the bottom when there is very tiny gap between the A₁ and B₁ rim blocks.



The wheel makes a half a turn.



The gap grows. Is the string going to break?
I agree both observers will see the same result, either it breaks or it does not.
There is no multiverse, two different outcomes for two different observers.
Jano

[Halc]

This is the problem in physics.
Over a hundred years after the SR paper and the relativists do not agree on the Lorentz contraction.

Physicists are in complete agreement over these kinds of trivial implications of SR.
If by 'relativists' you mean the sort of people that hang out on forums like this, then yes, there are always those who either don't understand or simply choose to deny the prevailing view for whatever purpose. Over a thousand years since it's been shown that the world is round, and the Earthlings still do not agree on its shape. So what? The people who matter do agree about it. Neither theory has been in contention for a long time.

This is the bottom of the cycloid as seen from (b).

Ah, thankee for the closeup. I didn't even see the subscripts and realize that those were the points you were talking about. Yes, a string from those two points must break as there is no extra string to bridge the gap that forms between those two blocks when the blocks start moving around that fixed-radius path.

The wheel makes a half a turn.

It probably makes a whole lot of them. There's no need for the angular speed to ramp up almost instantly. The gaps form even if it takes millions of rotations to get up to speed.

The gap grows. Is the string going to break?

Yes.

I agree both observers will see the same result, either it breaks or it does not.

So will any observer not on the ring. The string breaking is an objective fact, meaning length contraction is real.

[Jaaanosik (OP)]

Halc,
I am not sure we understand each other. I'll try again.
There is no more angular acceleration in (a) frame.
The wheel is already up to speed, accelerated and it has the shape on the right from this figure 13.15.
The gaps are already created in the (a) frame.



We also know that the wheel rim blocks in (b) have 0 speed at the bottom of the cycloid.
The question is if it is OK to assume the spacing is like this for the (b) reference frame.




Almost no space at the bottom of the cycloid and a bigger gap at the top.
That's why it is a half a turn, 180 degrees, from the bottom to the top for already rotating wheel as seen from (b).
Jano

[Halc]

Halc,
I am not sure we understand each other. I'll try again.
There is no more angular acceleration in (a) frame.
The wheel is already up to speed, accelerated and it has the shape on the right from this figure 13.15.
The gaps are already created in the (a) frame.

You're mixing two very different examples, so this doesn't make sense.
(a) and (b) are the colorful picture from reply 2.  There are no blocks or gaps in that example.
In that example, (a) and (b) are the same spinning wheel viewed in two different inertial frames. There is no angular acceleration involved. Being the same wheel both pictures spin at the same proper angular velocity (but not at the same angular velocity).

The grey/red picture from post 29 depicts the blocks, stationary on the left (t0) and spinning on the right (t1) at some fixed radius track.  The gaps form as the angular speed increases. Near light speed, the 'rim' is all gaps and an arbitrarily large number of blocks could fit in. Neither of these two wheels has linear motion.

We also know that the wheel rim blocks in (b) have 0 speed at the bottom of the cycloid.

(a) and (b) do not depict blocks.

The question is if it is OK to assume the spacing is like this for the (b) reference frame.

There is no spacing involved in the (a) (b) picture. You're mixing scenarios.  (a) and (b) are the same wheel in different inertial frames. They can't differ in any objective way because they're the same thing.

That's why it is a half a turn, 180 degrees, from the bottom to the top for already rotating wheel as seen from (b).

I think you're talking about A₀ and B₀.  The half turn is irrelevant. The gaps form because the wheel on the right is spinning, and both A₀ and B₀ travel continuously all around the circuit, not just a half turn.

[Jaaanosik (OP)]

Right, if the wheel looked like this:



No need for the outside 'track', spokes keep the rim blocks together.
I hope the same question about the spacing makes sense now,
Jano