Room for J. Murray and Shashaank

J. Murray

16:15

Hi. If you want to ping me, you can navigate to my chat profile by going to chat.stackexchange.com and searching for my username. Once you're at my profile, you can choose Start a New Room With This User to create a chatroom and invite me to it.

J. Murray

17:47

@Shashaank What about that question would you like to discuss?

Shashaank

@J.Murray . Sorry I saw your message just now.

Sure I would like to discuss the solution

Have you already seen the question or would you like me copy the link here

@J.Murray I was trying that...couldn’t work that through

@J.Murray can we proceed then

J. Murray

I've seen the question

Shashaank

Cool then I am struggling a bit on how the coordinates in FLRW taken ( as you would have already seen, particularly my comments on the answer there by Umaxo)

@J.Murray Maybe you can tell me the correct thing. Being quite confused there, I think it would be better if I raise questions after hearing you out.

J. Murray

Okay.

To construct any solution to the Einstein equations, we first need to impose some symmetries on the spacetime.

In the case of the FLRW spacetime, those symmetry constraints are extremely strong.

Shashaank

18:03

Okay yeah I am reading it alongside

J. Murray

The constraints on our spacetime are that it is spatially homogeneous and isotropic. Loosely speaking, homogeneous means that on every time slice, every point in space looks like every other point. Isotropic means that on every time slice, every direction looks like every other direction.

Now, if we think carefully, those loose definitions don't make any sense.

The reason is that different observers would have different time slices, and so defining our symmetries by what should be observed on any particular time slice is doomed to fail.

Does this make sense?

Shashaank

Yes I agree with the last sentence you said.

J. Murray

Okay. So the precise definitions go like this:

A spacetime is spatially homogeneous if there exists a one-parameter family of spacelike hypersurfaces $\Sigma_t$ which foliate the spacetime such that for each $t$ and for any two points $p,q\in\Sigma_t$, there exists an isometry of the metric which maps $p$ to $q$.

(reminder that ChatJax can be found here: math.ucla.edu/~robjohn/math/mathjax.html)

Shashaank

What troubles me is that why is this slice “GLOBAL”. An inertial frame is local. Key question - So a space like hypersurface should be LOCAL because a Lorentz frame has just a local existence. The space like surface surface of simultaneity exists locally ( and on top that what you say - every other observer will draw his own time slice/ space like hypersurface ) which should have a local existence and t should be time on that local surface ....

J. Murray

Nobody has said anything about simultaneity.

Shashaank

18:12

Cont ( after reading)....Is the $t$ you are defining the physical time. If so again it should be local

J. Murray

It is a parameter which indexes the hypersurfaces. It has no meaning at the moment.

Shashaank

@J.Murray okay but then what is $t$ which is constant over the space like hypersurface ....is it not the physical time...

J. Murray

It's a number which tells you which hypersurface you're on.

Shashaank

@J.Murray So it is not the physical time (proper time)

J. Murray

t=3 is one hypersurface. t=17 is another.

t=2.3457820 is yet another.

Shashaank

18:14

Okayama I see now...yes now there is no question of simultaneity because t is not time ....okay just an arbitrary coordinate.....I will come back to this later

J. Murray

Let me be explicit here. Consider the manifold $\mathbb R^2$.

Shashaank

Sure I am reading through

J. Murray

Pick any number you'd like. Go ahead, just choose a number.

Shashaank

Umm say 11

J. Murray

Okay. $\mathbb R^2$ is foliated by a one-parameter family of lines defined as follows:

$\Sigma_t = \{(x,y)\in \mathbb R^2 \ | \ y = 11 x + t \}$

Shashaank

18:17

What is foliated mathematically ( just to follow your argument)

J. Murray

Foliated means that every point in $\mathbb R^2$ lies on exactly one of those lines

Shashaank

Okay

J. Murray

My choice of possible foliations is endless. I could have chosen lines with a different slope, or I could have chosen curves like $y=x^2 + t$, etc

Do you agree?

Shashaank

Are arbitrary functions y= f(x) allowed

J. Murray

Yes

Genuinely any function you'd like.

Shashaank

18:21

And does $R^2$ need to be foliated by just one of these lines or curves or more that one are allowed ( just to follow the discussion)

Other than that yes I agree

J. Murray

Well, not quite any function, but the category is broad

I don't know what you mean by that. You just need a one-parameter family of curves which completely covers $\mathbb R^2$, and where no two curves ever intersect.

Shashaank

And can it be foliated by more than one “curve” or “line” simultaneously. It looks pretty hard that all point in $R^2$ will need to lie on just a single line

J. Murray

It isn't foliated by a line, it's foliated by a family of lines.

Shashaank

Ohhh I see and the parameter is “t” (of that family)

J. Murray

In the example we chose, $\Sigma_0$ is the set of all points which lie on the line $y=11x + 0$.

Yes.

Shashaank

18:25

yes yes I get your point.please continue, I agree

J. Murray

Alright. Now, for curve we can look at the tangent vectors at each point. In our example, all of the tangent vectors are the same everywhere; they are simply multiples of the vector $\mathbf v = (1, 11)$

any constant multiple would do

Shashaank

and $\Sigma_11$ is the the set of all points in the manifold which lie on the line y=11x+ 11...in that way all point on the manifold can be covered by changing “t” the arbitrary parameter

J. Murray

Yes that's right

For a more interesting example, consider the foliation defined by the curves $y = x^2 + t$

Shashaank

@J.Murray yes I agree, alright

J. Murray

The tangent vectors to these curves change across the manifold. At some arbitrary point $(a,b)\in \mathbb R^2$, the tangent space consists of all vectors parallel to $(1,2x)$

sorry

parallel to $(1, 2b)$

The point of all of this is that I can now say something like "Consider a family of curves $\Sigma_t$ which foliates the space in such a way that the tangent vectors all have a certain property"

Shashaank

18:32

Okay yeah

J. Murray

If the space is equipped with a Lorentzian metric, I could say "Consider a family of curves $\Sigma_t$ which foliates the space in such a way that the tangent vectors to $\Sigma_t$ are all spacelike, i.e. $g(v,v) >0$"

where I'm assuming metric signature $(-+++)$

Each $\Sigma_t$ is then called a spacelike hypersurface

Shashaank

and all the way till now, “t” has just been a parameter and arbitrarily one without any physical significance, right

J. Murray

Correct.

Okay. Now consider again the definition of homogeneity.

Shashaank

Okay yeah I get your point. So the whole of this space like hypersurface (because yo7 have introduced a Lorentz metric) has a constant “t”

J. Murray

Yes.

So we're now looking for such a foliation which additionally has the following property:

Given each hypersurface and any two points $p,q\in\Sigma_t$, there must exist some transformation which maps $p$ to $q$ and leaves the distances between any two points unchanged.

Shashaank

18:40

Okay yeah

J. Murray

In loose terms, the metric can't depend on position

Well.... that's probably not the right way to think about it

But no particular points can be special.

Shashaank

Why that’s right. There would be 3 obvious killing vectors

J. Murray

Which ones?

Shashaank

You said the metric won’t depend on position. There would be 3 position coordinates. So if the metric is independent of all 3, then there would be 3 killing vectors corresponding to these 3 “space” coordinates

J. Murray

The metric doesn't have to be independent of all three coordinates, which is why I took a step backward there

Shashaank

18:44

Ok yeah the FLRW depends on theta

J. Murray

If we use polar coordinates on $\mathbb R^2$, the metric depends on $r$

Shashaank

Yeah I agree

J. Murray

I've already stated the rigorous definition - that there must exist some isometry which maps $p$ to $q$ for any $p,q\in\mathbb R^2$.

As an analogy, in Euclidean space, such isometries would be rotations or translations

Shashaank

Yes I agree with that.

J. Murray

On a sphere, such as $S^2$, translations don't make any sense but we can still apply rotations

which are isometries of the normal metric on $S^2$

Okay, so let's recap. We foliated the space by a one-parameter family of spacelike hypersurfaces $\Sigma_t$. We also demanded that for any two points $p,q\in\mathbb R^2$, there must be some isometry of the metric which maps $p$ to $q$.

Shashaank

18:50

Perfect, I agree

J. Murray

Now, $t$ does not yet have a physical meaning. However, we can clearly choose it as one of the 4 coordinates for our spacetime, while choosing the other three to coordinates $\Sigma_t$.

That is, every point in spacetime can be labeled by $t$ (which tells you which hypersurface it's on) and three other coordinates which tell us where the point is on that particular hypersurface.

sorry, to coordinatize $\Sigma_t$ *

Shashaank

Yes that looks all good. We can of course do that

J. Murray

Okay. Now the last thing to do is to think about isometry, and that's what will give us the most powerful constraints on our metric.

Shashaank

Okay

J. Murray

The definition for this is somewhat technical so I'll just give you the idea. What we'll do is consider a family of timelike curves which fills the whole spacetime.

So this is a bit like a foliation, except a foliation a bunch of surfaces indexed by one parameter

and now we're talking about a bunch of curves indexed by 3 parameters

Shashaank

18:59

But we generally in GR take just one parameter to define a worldline...is this completely different from that

J. Murray

Again, this is a family of curves

think of them like blades of grass

every point in the spacetime lies on one of them

Shashaank

Okay but why indexing by 3 parameters. Up till now the only parameter we used to index the space like hypersurfaces was just “t”

Do we wan

*sorry I sent that by mistake

J. Murray

Let's be more explicit. Let's consider a very simply case - $\mathbb R^3$. We can define a very simple foliation:

$\Sigma_t := \{(x,y,z)\in \mathbb R^3 \ | \ z=t$

Just the surfaces of constant $t$

sorry

surfaces of constant $z$

That's a foliation. But now consider a congruence of curves $\gamma_{ab}$, where $\gamma_{ab}:= \{(x,y,z)\in \mathbb R^3 \ | \ x=a\text{ and }y=b\}$

These curves are simply straight vertical lines (in the $z$ direction)

But to specify these curves, I need two parameters $a$ and $b$

So in a 1+2 dimensional spacetime, I need a 1 parameter family of 2D surfaces to define a foliation, and a 2 parameter family of 1D curves to define a congruence.

Shashaank

Ohh okay so in full generalisation where z is not a constant we will need three paramters

J. Murray

No

What I just wrote was for 1+2 dimensional spacetime

We're talking about a 1+3 dimensional spacetime

The hypersurfaces are like like pages in a book, while the curves are like trees in a forest

Shashaank

19:08

Just for full clarification could you please write the set gamma for 1+3 dimension just like you wrote above

J. Murray

For a similar example we could define the following foliation for $\mathbb R^4$:

$\Sigma_t := \{(x,y,z,w)\in \mathbb R^4 \ | \ w=t\}$

$\gamma_{abc}:=\{(x,y,z,w)\in \mathbb R^4 \ | \ x=a,y=b,z=c\}$

^ which is the congruence

Shashaank

Perfect

Please continue sorry for the digression

J. Murray

So the point about isotropy is that there can be no special direction in space. What this means for us is that the curves must all be orthogonal to the hypersurfaces that they poke through.

To see this, imagine a curve poking through a hypersurface in a way that was not orthogonal, and simply observe that you could project its tangent vector down onto the hypersurface.

Shashaank

And that projection would be having a direction on the hypersurface.....so we would like the curves to poke orthogonally

*particular direction

J. Murray

Yes, that's precisely correct

So what this means is that if we choose $t$ to parameterize our spacetime, then we must have that the metric takes the following form:

$ds^2 = b(t) dt^2 + a(t)d\Sigma^2$

where $d\Sigma^2$ is the metric on $\Sigma_t$

Shashaank

19:19

Perfect because the dtdx or dtdy terms won’t come because our curves are orthogonal

J. Murray

Yes

Shashaank

Looks good, please continue

J. Murray

Sorry, wanted to draw one more diagram

The last thing that we will do is note that these curves are all timelike (because we chose them to be so), which means we can parameterize them by the proper time $\tau$ along them

Looking at the form of the metric, that simply equates to absorbing the $b(t)$ factor into the $t$ coordinate: $b(t) dt^2 \mapsto d\tau^2$

and so finally our metric takes the form $ds^2 = d\tau^2 + a(\tau) d\Sigma^2$ where $d\Sigma^2$ is the spatial metric of a space of constant curvature (it must be constant - that is, the same at every point - because of the homogeneity requirement we imposed earlier)

Shashaank

Okay that all looks good. I have one doubt and one follow up. Please let me state it

J. Murray

So to put this in physical terms, we have a family of spacelike hypersurfaces and a family of timelike curves (i.e. observers). Select any hypersurface you like as your initial surface. Now every point in spacetime can be labeled by 4 coordinates: 3 spatial coordinates to tell you which curve you're on, and one time coordinate which is the proper time along that curve.

Okay, go ahead.

Shashaank

19:33

1) follow up - way back up you said we could index hypersurfaces by say t=2 0r t= 1000. And in your 1st comment you said “ different observers will have different time slices”...at this point too did “t” have no physical significance and was just a parameter. Like observer A chooses t=2 hypersurface and t= 1000 hypersurface is chosen by observer B....is that consistent

I will ask the doubt after your comment to this , coupling it with your ,last comment

J. Murray

These $\Sigma_t$'s are not surfaces of constant time in the rest frame of any particular observers.

The observers we have chosen are very special - they are chosen based on our isotropy requirement.

Shashaank

Yeah but the point is different observer can index different hypersurface or not

J. Murray

I'm afraid I don't know what that means.

You say "Observer A chooses t=2 hypersurface " which doesn't make sense

The hypersurfaces are just slices through spacetime, observers have no role in this

Shashaank

Yeah I see your point,I am asking the wrong thing. I see your point. Forget that

I will come to the doubt

J. Murray

Okay

Shashaank

19:39

The follow up is clear

In the last diagram you drew. Pick up the t1 hypersurface.

till now t was just a parameter indexing a hypersurface. So no issue or problem of simultaneity. Perfect.

in the end you state “ spacetime can be labeled by 4 coordinates: 3 spatial coordinates to tell you which curve you're on, and one time coordinate which is the proper time along that curve.“

Consider two curves. Imagine one to be drawn on one end of your drawn space like hypersurface and one to be on the other end

Will $\ tau$ be the same at both points on the hyoersurface

Will $\tau$ take on the same value at both points on the hypersurface

J. Murray

Yes

this follows from the isotropy and homogeneity requirements

Shashaank

if it will take the same value, aren’t we back on the problem of simultaneity. Two very far away points have the same tau. But this shouldn’t have been because simultaneity should be a local concept here. So where am going wrong here...

J. Murray

I'm not sure I understand. Given any set of coordinates, we can always talk about the set of spacetime points with the same time coordinate

In general that isn't meaningful, because two different observers will disagree on those sets of points

However, in this discussion we have distinguished a very special class of observers, namely those for whom the universe appears to be isotropic

and a very special class of hypersurfaces, namely those along which the universe appears to be homogeneous

Shashaank

This “time” you state in just the last comment, you mean proper time right

J. Murray

Proper time is something you measure along a curve. Coordinate time is part of a label for a point in spacetime.

Shashaank

19:53

Yes I see the coordinate time will be the same on the whole hysoersurface.

J. Murray

The point I was making was that we argued that if we impose the requirements of homogeneity and isotropy, then we can fill all of spacetime with these special curves.

For observers traveling along those curves, the universe will appear to be homogeneous and isotropic.

For other observers this will not be the case.

Shashaank

Is this statement meaningless- “ the proper time at 2 very far away points (on the drawn) hypersurface will be the same”

J. Murray

Without any other context, yes.

Shashaank

Yes I think then I get it

J. Murray

Proper time is the "distance" along a spacetime curve

Here's the sense in which it does make sense.

Pick two curves on opposite sides of the drawn surfaces

now look at the bottom surface. We'll call that $\tau = 0$ for every curve poking through it.

Shashaank

19:56

So the thing we have to look for is how the distance ( read proper time ) elapses along those special curves and NOT along a curve drawn on the hypersurface ( or just not along the hysoersurface)

J. Murray

Now trace one of the curves up to the next surface

we can ask, "what is the proper time until this curve intersects the next surface?"

And we can do the same for the other curve

and we can ask if those two numbers will be the same

Shashaank

Yes

J. Murray

and the answer to that question is yes.

Shashaank

Perfect

This last example puts everything clean and clear

J. Murray

The coordinates which we have defined are called comoving coordinates

The observers we have defined - for whom the universe appears homogeneous and isotropic - are called comoving observers.

Shashaank

19:58

@J.Murray yes I can see that why now

J. Murray

Lastly

Shashaank

This is right -
So the thing we have to look for is how the distance ( read proper time ) elapses along those special curves and NOT along a curve drawn on the hypersurface ( or just not along the hysoersurface

J. Murray

Yes, that's right.

The comoving spatial coordinates simply tell you which curve you're on

So for example:

Shashaank

The proper time elapsed for observers moving along both these curves would be same

J. Murray

the comoving distance between $a_1$ and $b_1$ is the same as the comoving distance between $a_2$ and $b_2$

because again, comoving coordinates just label the curve

(yes, you're right)

Shashaank

20:03

Yes I get your last comment too

J. Murray

The proper distance, on the other hand, is the minimum distance between the points along the hypersurface

Shashaank

That will change of ourse

J. Murray

Yes, that's right.

Shashaank

*of course

So the physics of time ( what ever that means; I guess you can understand what I mean) like

* lies along those special curves ( proper distances along those special curves) ....except perhaps distance measures ( like maybe luminosity distance etc)

I get it finally. Many thanks .

J. Murray

I'm not totally sure I understand that last point, but I think you have the right idea, yes.

So in summary

I wanted to give you an idea of how symmetry requirements allow us to assert the existence of special coordinates

usually by constructing them explicitly in terms of the worldlines of special observers for whom the universe appears to obey the requirements in question

Shashaank

20:08

Having just given the exam of GR, and taking advance GR this sem, I feel University courses should be much more active in the side of discussions

@J.Murray yes I get that now perfectly

J. Murray

Glad to hear it. Yes, I did not learn much from my GR course I'm afraid

Shashaank

Just like coordinates need not have any physical significance, so do the components of a vector in those coordinates, right

J. Murray

Correct.

However, things like the magnitude of the vector are given by the metric, and these things do have physical significance

Shashaank

Like in Cartesian coordinates, the P^0 is the energy measured locally by an observer. In a totally arbitrarily coordinate ( may using all time like coordinates ), the P^0 won’t be the energy, right

@J.Murray that’s why we look for scalars

J. Murray

Well, the energy of a particle is defined to be the 0th component of its momentum 4-vector, at least in the absence of any potentials

Different observers may disagree on the value of $P^0$, which reflects the fact that energy is not a Lorentz scalar

Shashaank

20:14

Well then would $m_0(1-2GM/r)^{-1/2}$ be the energy measured by “ some” observer which would be the same as the one using a Lorentz local coordinate

*sorry I meant not the same as the one measured by a Lorentz local coordinate

* oops ....not the same as measured by the observer who uses a local Lorentz coordinate system...... lots of typos

J. Murray

So we're talking about Swarzschild coordinates now, right?

Shashaank

Yeah

J. Murray

So the metric components $ds^2 = -(1- r_s/r)^{-1/2}dt^2 +(1-r_s/r)dr^2 + r^2(d\theta^2+\sin^2(\theta)d\phi^2)$ are given in terms of coordinates $(t,r,\theta,\phi)$ called Swarzschild coordinates

Shashaank

Yeah

J. Murray

whoops

exponents aren't right, but whatever

Shashaank

20:22

Yeah lol ...just -1’s

J. Murray

so those coordinates refer to clocks and rulers carried by observers at spatial infinity

Shashaank

Yeah

So what would be the quantity - $m_0(1-2GM/r)^{-1/2}$ ...will it the energy observed by some observer

J. Murray

So the Swarzschild solution is stationary, which means that it has a timelike killing vector field given by $\partial_t$ in swarzschild coordinates.

Shashaank

Yes

J. Murray

And this implies that along a geodesic, the quantity $g_{00}u^0 = (1-r_s/r)\frac{dt}{d\tau}$ is constant, so this is what we call the energy

Shashaank

20:29

But the conserved quantity won’t be the “energy”....it will be $p_0$....the actual energy being E=-p.U

J. Murray

Well, now we're getting into the question of what you should call energy.

Shashaank

Yeah true maybe you can tell me what will be the quantity “ $m_0(1-2GM/r)^{-1/2}$ “. Will it be energy measured by some observer in the sense of E= -p.U

J. Murray

I don't think so, no

Well

Shashaank

So this quantity ain’t have any physical significance - but you said that **by definition P^0 is (defined) as the energy

What I wrote was just $P^0$

so according to you shouldn’t it by definition be energy measured by some observer

J. Murray

Oh I see what you're saying

Firstly, I meant $P_0$ rather than $P^0$

This is the quantity which is conserved under time translation symmetry

Shashaank

20:39

yeah I agree with that...

J. Murray

So now $P^0 = m U^0 = m \frac{dt}{d\tau}$

and $\big(\frac{d\tau}{dt}\big)^2=(1-r_s/r) - \frac{v^2}{1-r_s/r} = \big(1-\frac{r_s}{r}\big)\big(1- v^2\big)$

which implies that

$P^0 = m\gamma \left(1-\frac{r_s}{r}\right)^{-1/2}$, where $\gamma = (1-v^2)^{-1/2}$.

And then that $P_0 = g_{00} P^0 = m\gamma \big(1-\frac{r_s}{r}\big)^{1/2}$

Oh wait, whoops

$\left(\frac{d\tau}{dt}\right)^2 = \big(1-\frac{r_s}{r}\big)\left(1-\frac{v^2}{(1-r_s/r)^2}\right)$

Again, these things are measured in Swarzschild coordinates, not local ones.

Shashaank

Yeah sure...I was just referring to the simpler case where gamma is 1. I mean $U^\mu=(U^0,0,0) ... but that’s alright

J. Murray

The $v$ written in my expression above is the Shapiro-delayed velocity, see here:

Schwarzschild geodesics

In general relativity, Schwarzschild geodesics describe the motion of particles of infinitesimal mass in the gravitational field of a central fixed mass M {\textstyle M} . Schwarzschild geodesics have been pivotal in the validation of Einstein's theory of general relativity. For example, they provide accurate predictions of the anomalous precession of the planets in the Solar System, and of the deflection of light by gravity. Schwarzschild geodesics pertain only to the motion of particles of infinitesimal mass m...

Well if the particle isn't moving, then $U= (1,0,0,0)$, so $P\cdot U$ is just $P_0$.

Shashaank

20:55

If the particle isn’t moving then $U= (U^0, 0,0) where U^0 is not equal to 1 but is determined by the metric’s 0th component.....which is (1-2GM/r) here

if I am not in an LIF

J. Murray

Ah okay I understand what you're saying.

Shashaank

Well the point I was trying to say was that P^0 doesn’t have any physical significance...right...it’s as simple as that....when the coordinates don’t have any physical significance there is no point that their derivative (wrt a scalar) will be having any physical significance...right..to get a physical sense go to an LIF.

J. Murray

So then yes - the quantity $m_0 \sqrt{1-\frac{r_s}{r}}$ is the total energy of a stationary particle at $r$ as observed by an observer at spatial infinity

Well, yeah. If you just tell me $P^0 = 17$ but you don't tell me what coordinates you're using to calculate it, then that doesn't tell me much

The quantity $m_0 \sqrt{1-r_s/r}$ has meaning because I know I'm working in Swarzschild coordinates.

But I need to know the coordinates in order to interpret what physical significance the expression has

Shashaank

@J.Murray But you initially said that $m_0(1-2GM/r)^{1/2} isn’t the energy measured by any observer.....precisely here....was that wrong

could you please state what’s the final correct thing....just to avoid confusion that might have crept in

i think what you have just now about it being the energy as measured by the observer at spatial infinity is the best thing

i was thinking just that

J. Murray

That's the energy of a stationary object as measured by an observer using Swarzschild coordinates

it's not a general expression for the energy

because obviously most things aren't stationary.

Shashaank

21:06

And this energy measured by observer at infinity ( of a particle at rest) won’t be the same as that measured by an observer next to the particle using a LIF....right

there is no reason for it to be

J. Murray

I don't want to get into a super long discussion of this, but again this is running into what it means to measure something

How do you measure the energy of a particle at rest?

I suppose you could watch it decay into two photons traveling in opposite directions and ask what their energies are, and in that context yes, the answer would be different.

Shashaank

@J.Murray yes I was exactly thinking something like that

the rest energy :)

but never mind. Many thanks for the wonderful discussion. I wish my university professors had time and patience to discuss a lot...so many thanks again

J. Murray

Well I have to go for now, I hope this discussion has been helpful. I would suggest re-reading the answer to your question given by Umaxo in light of this conversation.

As it is a very good one, I think.

Shashaank

Yes indeed you are right.

Next time I will try and ping you not chat.stackexchange.com like you said

Have a good day. Bye

J. Murray

Bye

Transcript for

Room for J. Murray and Shashaank