The Schwarzschild Metric - A Simple Case

[starthaus]

What you call my "hack on the SR length contraction" was an equation that applied to an infinitesimal particle where "spagettification" does not occur.

A hack is always a hack no matter what extenuating circumstances you are trying to invoke.

Infinitesimal is very small, even smaller than an atom and is in fact as small is required to make the local spacetime flat (so negligable tidal forces). For a non infinitesimal rod some form of integrated length is required and I am aware as you are, as to how non trivial that is.

This simply means that your attempt at hacking length contraction in GR is unusable.

Now all the equations you give are for free falling particles. For a rigid rod parts of the rod are not free falling. By definition the rigidity of the rod prevents that. You said "front of the rod passes has a higher speed than the tail at any given moment", but Passionf has specified in the OP that we can use rockets to accelerate parts of the rod

The "rocket" bit is over the top, especially since the OP was about a free-falling rod.

to ensure that its proper length remains constant by whatever means necessary.

This only means that the problem cannot be treated kinematically since you are now dealing not only with the tidal and electromagnetic (internal) forces but also with the forces exerted by the two rockets.

So while natural free fall might mean that the rear clock is moving slower than the front clock at any given moment (and here you should define "moment" according to what reference frame) we can attach a rocket to the rear clock to make sure it goes at the same speed (insert according to what observer here) as the front clock if that is convenient to us.

Why entitles you to apply different rules to the spatial coordinate than to the time coordinate?

It is not unlike Born rigid acceleration, where all parts of a system are individually accelerated by the exact required amount to ensure constant proper length is obtained in the accelerating system. We are just trying to make the GR analogue of that.

I see, you plan to put a little rocket at infinitesimal distances from each other.
The point is that the OP problem is much more complicated so it cannot be reduced to simple kinematics.

 

[starthaus]

DrGreg derives that equation in this thread. https://www.physicsforums.com/showthread.php?t=248015 There is no need to spend any more time on this diversion.

All I can see is the post where he tells you that you got things wrong.

 

[yuiop]

You can't do that since it is contradicted by the physics of the problem.
Indeed, the coordinate speed of a particle is :

v=\frac{dr}{dt}=(1-r_s/r)\sqrt{r_s/r}

It is easy to show that the above speed increases as r decreases up to the point r=3r_s

Indeed:

\frac{dv}{dr}=1/2(3r_s/r-1)\sqrt{r_s/r^3}

This means that the front of the rod passes has a higher speed than the tail at any given moment.
It also means that the distance between the front and the tail increases as r decreases, i.e. the rod is getting "spagettified" (extruded).

As you say, the after r = 3Rs the coordinate speed of the falling particle decreases, and the following particles start to catch up and in terms of coordinate distance de-spagettification happens as the event horizon approaches. This however is coordinate distance and no doubt in the proper distance between particles is still increasing even below r = 3Rs and all the way to the event horizon, but we do not have an equation to prove that. that is one of thing Passionflower and myself are trying to establish. What is the proper distance in the rest frame of the falling particles? Does a ball of coffee granules maintain constant volume (as Weyl curvature suggests it should) as measured in the rest frame of the falling granules? it is easy to work out from what we have already established that the volume does not remain constant from a Schwarzschild observers point of view.

 

[yuiop]

All I can see is the post where he tells you that you got things wrong.

That is because you see only what you want to see and stopped reading. In the very next line DrGreg says

"However the ruler distance to the event horizon is the integral ..."

and gives a formula. This formula for the integral is established and confirmed by other posters later in the thread. That thread is a model of cooperation and constructive contributions by members of this forum in the good old days. You could learn a lot from it. Back then I foolishly believed that it was not possible to have a ruler distance to the event horizon. As soon as DrGreg demonstrated how it was calculated I realized he was right. Do you also now realize that his formula is right?

 

[yuiop]

All I can see is the post where he tells you that you got things wrong.

There are 3 pages and 39 posts in that thread and all you can see is one line of one single post where someone said I got something wrong. Obviously you did not see post 33 where I went to the trouble to post the full correct equation for everyone's future reference. Now why is it that all you see is that one line where DrGreg says I got something wrong and why do you feel the need to post that in this thread? Ah yes, it is called an "ad hominem" attack. It is the fallacious logic that if you establish that someone got something wrong in the past then EVERYTHING that person says must be wrong. I consider it a personal attack. Consider yourself reported.

 

[pervect]

This has been quoted from MTW, which I don't have. The mathpages 'Reflections on Relativity' has a very similar treatment and also gives the worldline parameterized by t. I thought this would be a way to compare the two frames but I'm not so sure now, having tried some calculations. It is difficult to compare separated events in GR.

But if an ideal rod falls radially, aligned along a radius, then the velocity differential must cause stresses and there must be a way to put a number on it. It is just the rr component of the tidal tensor, but what does that component ( +2m/r^3 ) signify, quantitatively ? Is it an acceleration, i.e. a velocity gradient of dr/dtau in the r direction ?

What the tidal tensor (or the appropriate component of the Riemann ) measures is the relative acceleration of geodesics. So, if you have two particles, initially at relative rest in their "local lorentz frame" (defined because they are both close together and moving at the same velocity), both following geodesics, the tidal tensor is basically the second derivative of the distance between geodesics with respect to the proper time of a particle following the geodesic.

Take a look at most any textbook derivation of the geodesic deviation equation to see the actual mathematical definition.

One generally considers that one has a one-parameter group of geodesics, i.e. one has some selector parameter n that defines a unique geodesic curve for every value of n. You select which geodesic by some selector parameter n, and you select "how far" along the geodesic by some affine parameter s.

The tangent vector of the geodesic, the partial derivative with respect to the affine parameter s, \partial / \partial s at any point can be intuitively thought of as the "local time vector" of the particle following the geodesic. One needs to exploit some "gauge degree of freedom" to make the separation vector between geodesics independent of the specific affine parameterization of s for each geodesic. When this is done, the separation vector between geodesics, the \partial / \partial n, becomes perpendicular to the tangent vector \partial / \partial s. (The issue with the affine paramterization is sometimes called stretch-out).

In less formal, more "feel good" but less precise language (some might find the less precise language puzzling, for which I apologize, I hope most consider it useful) one makes the local distance between geodesics perpendicular to the "local time" along the geodesic.

The stress measured on a born-rigid rod that's falling is the flip size of this equation. The amount of proper acceleration needed to keep the two ends of the rod the same distance apart is the same as the relative acceleration of the two geodesics passing through their endpoints at zero velocity (the zero velocity relative to the instantaneous local Lorentz frame of the falling rod).

 

[starthaus]

As you say, the after r = 3Rs the coordinate speed of the falling particle decreases, and the following particles start to catch up and in terms of coordinate distance de-spagettification happens as the event horizon approaches. This however is coordinate distance and no doubt in the proper distance between particles is still increasing even below r = 3Rs and all the way to the event horizon, but we do not have an equation to prove that. that is one of thing Passionflower and myself are trying to establish.

Given the fact that r_s is much smaller than the radius of the celestial bodies (for example, for Earth, r_s=9mm. this is not relevant to our discussion.

What is the proper distance in the rest frame of the falling particles?

I don't think that this problem is tractable for reasons explained above.

 

[starthaus]

That is because you see only what you want to see and stopped reading. In the very next line DrGreg says

"However the ruler distance to the event horizon is the integral ..."

He never derived anything close to what you are claiming. In fact, in post 35, he's still telling you that you still have things wrong in here, after your post 33 where you claim that you obtained the correct formula. The reason is simple, the correct formula is:

ds=\frac{dr}{\sqrt{r_s/r-r_s/r_0}}, r<r_0

I showed it earlier in this thread (post 54). For r_0=\infty the formula reduces to:


ds=\frac{dr}{\sqrt{r_s/r}}

This formula for the integral is established and confirmed by other posters later in the thread.

I don't see anyone else "establishing" and/or "confirming" your formula. This is irrelevant anyway since the formula you are attempting to use is incorrect to begin with.

That thread is a model of cooperation and constructive contributions by members of this forum in the good old days. You could learn a lot from it. Back then I foolishly believed that it was not possible to have a ruler distance to the event horizon.

You believe a lot of things. Unfortunately, "belief" is not a good approach to physics, proper derivation from base principles is.

 

[Passionflower]

It looks like yuiop was absolutely correct when he last week mentioned that things can magically can cancel out (at least for a free falling particle from infinity):

For our example (r_s = 1):

Since the proper distance from r₁ to r_s for a stationary observer (v_local=0) is:

<br /> <br /> \int _{1}^{r_1}\!\sqrt {1-{r}^{-1}}{dr} = \sqrt {r \left( r-1 \right) }+\ln \left( \sqrt {r}+\sqrt {r-1}<br /> \right) <br /> <br />

we can calculate the proper distance for an observer who has a non zero local velocity.

We need to length contract this distance by the following factor:

<br /> <br /> {1 \over \gamma} = \sqrt{1-v_{local}^2}<br /> <br />

Thus for instance for a free falling particle from infinity (E=1) we have a local velocity of:

<br /> <br /> \sqrt{1 \over r}<br /> <br />

Now the 'magic' when we multiply the two before we integrate we get:

<br /> <br /> \int _{1}^{r_1}\!{\sqrt {1-{r}^{-1}}\over \sqrt {1-{r}^{-1}}}{dr}<br /> <br />

Thus the proper distance for a free falling particle from infinity (E=1) is simply dr!

One thing I did not realize, like many things, is that one could calculate the proper time till ultimate doom based on the tidal acceleration assuming the distance is small irrespective of the mass of the Schwarzschild metric. So when we are in a rocket radially falling into a black hole (from infinity) and we measure the tidal acceleration between the top and the bottom and we know the length of the rocket we can calculate how much longer we are going to be alive.

So this is the first step the proper distance for a particle falling from infinity is simply r. So how do we go from here?

 

[yuiop]

He never derived anything close to what you are claiming. In fact, in post 35, he's still telling you that you still have things wrong in here, after your post 33 where you claim that you obtained the correct formula. The reason is simple, the correct formula is:

ds=\frac{dr}{\sqrt{r_s/r-r_s/r_0}}, r&lt;r_0

The formula I gave in post 33 is:

<br /> \sqrt{r2*(r2-r_s)} - \sqrt{r1*(r1-r_s)}<br /> + r_s*\left(LN\left(\sqrt{r2} + \sqrt{(r2-r_s)}\right)- LN\left(\sqrt{r1} + \sqrt{(r1-r_s)}\right)\right)<br />

This is the definite integral. The indefinite integral is:

<br /> \sqrt{r*(r-r_s)} + r_s*LN\left(\sqrt{r} + \sqrt{(r-r_s)}\right)<br />

Now some simple algebraic manipulation of the above expression:

\Rightarrow \sqrt{(1-r_s/r)}*r + r_s*LN\left(\sqrt{r} + \sqrt{(r-r_s)}\right)<br />

\Rightarrow \sqrt{(1-r_s/r)}*r + \frac{r_s}{2}*LN\left(\left(\sqrt{r} + \sqrt{(r-r_s)}\right)^2}\right)<br />

\Rightarrow \sqrt{(1-r_s/r)}*r + \frac{r_s}{2}*LN\left(r + 2\sqrt{r(r-r_s)}+(r-r_s)\right)<br />

\Rightarrow \sqrt{(1-r_s/r)}*r + \frac{r_s}{2}*LN\left(2r + 2\sqrt{r(r-r_s)}-r_s\right)<br />

\Rightarrow \sqrt{(1-r_s/r)}*r + \frac{r_s}{2}*LN\left(2r + 2r\sqrt{(1-r_s/r)}-r_s\right)<br />

\Rightarrow \sqrt{(1-r_s/r)}*r + \left(r_s*LN\left(\, -r_s + 2*\left(1 + \sqrt{(1-r_s/r)}\right)*r \right)\right)/2<br />

This is the same as the equation posted by DrGreg in post 18 https://www.physicsforums.com/showpost.php?p=1827990&postcount=18 of that thread where he works out the integral as:

If you replace r by x and r_s by a, the site gives formula that works perfectly well at x=a.

Integrate[1/Sqrt[1 - a/x], x] == Sqrt[1 - a/x]*x + (a*Log[-a + 2*(1 + Sqrt[1 - a/x])*x])/2

This is only one of many times that you have claimed that two equations that look different are not equivalent, when they are. If you are having trouble with the algebraic manipulations then here is a tip for you. Try sample variables in both equations and if you get the same numerical result every time, then the two equations are probably equivalent. This is not proof that they are equivalent, but it is a strong clue that you should look closer before declaring the equations are not equivalent. You claim to be a mathematician. I should not have to keep showing how to do these algebraic manipulations. In your haste to try and prove me wrong at every opportunity, you have shot yourself in the foot again.

 

[Passionflower]

Right and in addition to what yuiop writes if we take r_s=1 and want the proper distance to from r to r_s we get an even simpler formula:

<br /> \sqrt {r \left( r-1 \right) }+\ln \left( \sqrt {r}+\sqrt {r-1} \right) <br />

In continuation we should probably address the issue of translation the acceleration differential (e.g. the tidal acceleration) in terms of some stretching factor in terms of the r coordinate, the local velocity at r and the height of the object in question, e.g. the spaghetti factor.

The tidal acceleration at r is (as always r_s=1):

<br /> {1 \over r^3} dr<br />

 

[starthaus]

The formula I gave in post 33 is:

<br /> \sqrt{r2*(r2-r_s)} - \sqrt{r1*(r1-r_s)}<br /> + r_s*\left(LN\left(\sqrt{r2} + \sqrt{(r2-r_s)}\right)- LN\left(\sqrt{r1} + \sqrt{(r1-r_s)}\right)\right)<br />

This is the definite integral. The indefinite integral is:

<br /> \sqrt{r*(r-r_s)} + r_s*LN\left(\sqrt{r} + \sqrt{(r-r_s)}\right)<br />

...which are both equally incorrect since they have a common incorrect starting point, the integrand:

ds=\frac{dr}{\sqrt{1-r_s/r}}

Try solving the problem starting from the correct integrand:

ds=\frac{dr}{\sqrt{r_s/r-r_s/r_0}}

This is only one of many times that you have claimed that two equations that look different are not equivalent, when they are. If you are having trouble with the algebraic manipulations then here is a tip for you. Try sample variables in both equations and if you get the same numerical result every time, then the two equations are probably equivalent. This is not proof that they are equivalent, but it is a strong clue that you should look closer before declaring the equations are not equivalent. You claim to be a mathematician. I should not have to keep showing how to do these algebraic manipulations. In your haste to try and prove me wrong at every opportunity, you have shot yourself in the foot again.

Physics is a lot more than applying software packages in blindly integrating meaningless expressions. If you start with the wrong integrand, don't be surprised when you get a useless expression, no matter how much you turn the crank on your integrating software package.

 

[starthaus]

It looks like yuiop was absolutely correct when he last week mentioned that things can magically can cancel out (at least for a free falling particle from infinity):

For our example (r_s = 1):

Since the proper distance from r₁ to r_s for a stationary observer (v_local=0) is:

<br /> <br /> \int _{1}^{r_1}\!\sqrt {1-{r}^{-1}}{dr} = \sqrt {r \left( r-1 \right) }+\ln \left( \sqrt {r}+\sqrt {r-1}<br /> \right) <br /> <br />

we can calculate the proper distance for an observer who has a non zero local velocity.

We need to length contract this distance by the following factor:

<br /> <br /> {1 \over \gamma} = \sqrt{1-v_{local}^2}<br /> <br />

Thus for instance for a free falling particle from infinity (E=1) we have a local velocity of:

<br /> <br /> \sqrt{1 \over r}<br /> <br />

Now the 'magic' when we multiply the two before we integrate we get:

<br /> <br /> \int _{1}^{r_1}\!{\sqrt {1-{r}^{-1}}\over \sqrt {1-{r}^{-1}}}{dr}<br /> <br />

Thus the proper distance for a free falling particle from infinity (E=1) is simply dr!

...which is totally incorrect.

One thing I did not realize, like many things, is that one could calculate the proper time till ultimate doom based on the tidal acceleration assuming the distance is small irrespective of the mass of the Schwarzschild metric. So when we are in a rocket radially falling into a black hole (from infinity) and we measure the tidal acceleration between the top and the bottom and we know the length of the rocket we can calculate how much longer we are going to be alive.

So this is the first step the proper distance for a particle falling from infinity is simply r. So how do we go from here?

You don't go anywhere since the formula you keep using is not correct.

 

[Passionflower]

You don't go anywhere since the formula you keep using is not correct.

Which formula do you think is not correct?

 

[starthaus]

Which formula do you think is not correct?

I asked you where the integrand

\frac{dr}{\sqrt{1-r_s/r}}

was coming from. I told you that it was most likely incorrect (turns out that it is, thus making the integral and all your and yuoip's calculations useless). You never answered. I gave you the correct integrand several times already. Try using it.

 

[Passionflower]

I asked you where the integrand

\frac{dr}{\sqrt{1-r_s/r}}

was coming from. I told you that it was most likely incorrect (turns out that it is, thus making the integral and all your and yuoip's calculations useless). You never answered. I gave you the correct integrand several times already. Try using it.

This is very disruptive Starthaus you need to read better before you claim something is wrong. I am talking about a particle not an extended object, the integral is perfectly correct.

 

[starthaus]

This is very disruptive Starthaus you need to read better before you claim something is wrong. I am talking about a particle not an extended object, the integral is perfectly correct.

No, it isn't correct for a particle either. I asked you several times where you got it from and you never answered. Where did you get it from?

 

[Passionflower]

No, it isn't correct for a particle either. I asked you several times where you got it from and you never answered. Where did you get it from?

Looking at the formula that you think is right it appears you simply do not understand the difference between taking the integral from r2 to r1 for a particle free falling from infinity and from a particle free falling from an apogee at r2. Both yuiop and I are talking about a situation free falling from infinity.

Please in the future be more careful in calling other people wrong, first read carefully before you jump into conclusions.

 

[starthaus]

Looking at the formula that you think is right it appears you simply do not understand the difference between taking the integral from r2 to r1 for a particle free falling from infinity and from a particle free falling from an apogee at r2. Both yuiop and I are talking about a situation free falling from infinity.

The formula that you are trying to use is also wrong if you consider a particle free-falling from infinity, I also gave you the correct formula for a particle free-falling from infinity (posts 54, 98). I am asking you again, where did you get your formula from? You claimed it is from a textbook, what textbook?

Please in the future be more careful in calling other people wrong, first read carefully before you jump into conclusions.

I am being very careful.

 

[Passionflower]

The formula that you are trying to use is also wrong if you consider a particle free-falling from infinity, I also gave you the correct formula for a particle free-falling from infinity (posts 54, 98). I am asking you again, where did you get your formula from? You claimed it is from a textbook, what textbook?

Since I remember you have Rindler (2nd edition) take a look in chapter 11.5 page 236 at the top you see the integral for ruler distance. If you make 2m=1 you get exactly the same integrand as I use.

I attached an image of the page in question.

 

[starthaus]

Since I remember you have Rindler (2nd edition) take a look in chapter 11.5 page 236 at the top you see the integral for ruler distance. If you make 2m=1 you get exactly the same integrand as I use.

Thank you, I had a look. You are misunderstanding chapter 11.5, it has nothing to do with free-falling particles. If you want to learn about free-falling particles (falling from a fixed distance or from infinity), this comes up later, in chapter 11.8.

 

[Passionflower]

You are misunderstanding chapter 11.5, it has nothing to do with free-falling particles. If you want to learn about free-falling particles (falling from a fixed distance or from infinity), this comes up later, in chapter 11.8.

As I wrote before you should really read carefully:

Check my posting https://www.physicsforums.com/showpost.php?p=2915076&postcount=99 again.

I quote myself from that posting:

Since the proper distance from r₁ to r_s for a stationary observer (v_local=0) is:

<br /> <br /> \int _{1}^{r_1}\!\sqrt {1-{r}^{-1}}{dr} = \sqrt {r \left( r-1 \right) }+\ln \left( \sqrt {r}+\sqrt {r-1}<br /> \right) <br /> <br />

As I said read carefully before you call wrong.

 

[starthaus]

It looks like yuiop was absolutely correct when he last week mentioned that things can magically can cancel out (at least for a free falling particle from infinity):

For our example (r_s = 1):

Since the proper distance from r₁ to r_s for a stationary observer (v_local=0) is:

<br /> <br /> \int _{1}^{r_1}\!\sqrt {1-{r}^{-1}}{dr} = \sqrt {r \left( r-1 \right) }+\ln \left( \sqrt {r}+\sqrt {r-1}<br /> \right) <br /> <br />

...which is just as wrong since, in this case, the correct integrand is:

ds=\frac{dr}{\sqrt{1-r_s/r}}=(1+r_s/(2r))dr
(see Rindler 11.5)

Since you used the wrong integrand, you also obtained the incorrect integral. If you did it right, you would have obtained a much nicer formula:

\Delta S=r_2-r_1+r_s/2(ln(r_2)-ln(r_1))

The above is a transcendental equation in r_2-r_1, so, it cannot be solved symbolically. The only thing one can say with certitude is that

r_2-r_1&lt;\Delta S

i.e. the coordinate length is less than the proper length.

With r_s=1 you should get:


r_2-r_1+1/2*(ln(r_2)-ln(r_1))=r_2-r_1+ln(\sqrt{r_2/r_1})

Anyway, if you want a comparison between coordinate, radar and proper distances, then Rindler 11.5 gives the answer to your question.
You need to pay attention, the above derivation applies only for a static (unmoving) rod, so all your attempts at using this kind of math, do NOT answer the case of a falling rod, as described in your OP. If you want to find out the length of a ruler falling radially in a gravitational field, then 11.5 is not your answer, you need to proceed to 11.8. A much more difficult problem as I tried to point out to you several times. If you want to know how to answer the problem of a moving rod, then you need to start with the integrand I gave you a few times already.

 

[Passionflower]

...which is just as wrong since, in this case, the correct integrand is:

ds=\frac{dr}{\sqrt{1-r_s/r}}=(1+r_s/2r)dr
(see Rindler)

I admit I made a typing mistake in entering the integrand, but that does not invalidate the rest.

Since you used the wrong integrand, you also obtained the incorrect integral. If you did it right, you would have obtained a much nicer formula:

r_2-r_1+r_s/2(ln(r_2)-ln(r_1))

No my formula is ok for r_s=1

With r_1=1 you should get:

r_2-1+r_s/2*ln(r_2)

No just take a numerical example: if we take rs=1, r1=1 and r2=2 I get: 2.295587149 while your formula gives: 1.346573590 which is wrong.

So yes, I admit there was one mistake typing in the integrand but that has no further influence on the rest.

 

[starthaus]

I admit I made a typing mistake in entering the integrand, but that does not invalidate the rest.

You got the wrong integrand, meaning that you got the wrong integral.

No my formula is ok for r_s=1

I don't think so, I adjusted the answer in order to deal with making r_s rather than r_1 equal to 1 and your formula is just as wrong.

 

[Passionflower]

You got the wrong integrand, meaning that you got the wrong integral.

No, I told you I typed it in wrong the rest is fine.

I don't think so, I adjusted the answer in order to deal with making r_s rather than r_1 equal to 1 and your formula is just as wrong.

So are you saying that 2.295587149, the answer I get is wrong?

This is my formula:

<br /> \sqrt {r \left( r-1 \right) }+\ln \left( \sqrt {r}+\sqrt {r-1}<br /> \right) <br />

for r=2 I get 2.29558714, which is correct.

 

[starthaus]

No, I told you I typed it in wrong the rest is fine.So are you saying that 2.295587149, the answer I get is wrong?

Read the results in post 113. Carefully.

 

[starthaus]

I admit I made a typing mistake in entering the integrand, but that does not invalidate the rest.No my formula is ok for r_s=1No just take a numerical example: if we take rs=1, r1=1 and r2=2 I get: 2.295587149 while your formula gives: 1.346573590 which is wrong.

Umm, r_s&lt;&lt;r_1, remember? So you can't make r_1=r_s.
Rather than pasting numbers at random, I suggest that you read chapter 11.5 from end to end.

 

[Passionflower]

Read the results in post 113. Carefully.

Well yes or no? If we have rs=1, r1=1 and r2 is 2 is the answer 2.295587149 or is it 1.346573590?

Umm, r_s&lt;&lt;r_1, remember? So you can't make r_1=r_s.
Rather than pasting numbers at random, I suggest that you read chapter 11.5 from end to end.

What are you talking about the distance all the way up to rs is finite.

Please answer the question, what is the right answer 2.295587149 or is it 1.346573590?

 

[starthaus]

Well yes or no? If we have rs=1, r1=1 and r2 is 2 is the answer 2.295587149 or is it 1.346573590?

The formula assumes r_s&lt;&lt;r so your attempt to make r_1=r_s makes no sense. Like I said, instead of wasting your time pasting numbers into formulas, why don't you read the one page chapter 11.5? I promise you, you would be learning a lot more.