Shortest Paths

[□h=-16πT]

Is this right?

 

 

Let V denote a vector in SR. We write:

 

[math] \vec V = V^\alpha \hat e_\alpha [/math]

 

?

 

Yes. I'm still a novice in GR' date=' so I don't know how to answer your question on expansion of ST. As to the other, no the Euclidean metric does not give an accurate description of space-time, obviously as it contains no negative time component, and hence fails to predict such phenomenon as time dilation, length contraction etc. which [b']do[/b] occur in "reality". If you didn't know, the space-time of SR is called Minkowskian ST.

 

I also must appologise as I realised an error in my notation last night whilst browsing my posts on this thread. That is that I had inadvertantly written a one-form as a column matrix with a 1 in the top slot; this is not true and it should be in the bottom, as you discovered upon research. This is, of course, because a one-form is a linear mapping whose arguement is one contravarient vector (a vector) and so the notation requires a 1 in the bottom slot (as the bottom slot denotes the number of vectors supplied to the mapping).

[Johnny5]

When you speak of hbar do you actually mean [math]\hbar[/math] (planks constant over [math]2\pi[/math]) or was it [math]\bar{h}[/math'], used to represent the field equations in weak form? If the former I'd like to know what the problem was as I've not seen it used in the field equations (although I am new to GR).

 

 

I meant Planck's constant divided by two pi here.

[Johnny5]

I also must appologise as I realised an error in my notation last night whilst browsing my posts on this thread. That is that I had inadvertantly written a one-form as a column matrix with a 1 in the top slot; this is not true and it should be in the bottom' date=' as you discovered upon research. This is, of course, because a one-form is a linear mapping whose arguement is one contravarient vector (a vector) and so the notation requires a 1 in the bottom slot (as the bottom slot denotes the number of vectors supplied to the mapping).[/quote']

 

Yes I remember that.

 

So what you are saying is that you should have written:

 

[math]

\left(

\begin{array}{c}

0\\

1\\

\end{array} \right)

[/math]

 

to represent a 'one form' or tensor or 'covariant vector'

 

instead of

 

[math]

\left(

\begin{array}{c}

1\\

0\\

\end{array} \right)

[/math]

 

?

 

Also, what do you mean that the bottom slot represents the number of vectors applied to the mapping?

[Johnny5]

Yes. I'm still a novice in GR' date=' so I don't know how to answer your question on expansion of ST. As to the other, no the Euclidean metric does not give an accurate description of space-time, obviously as it contains no negative time component, and hence fails to predict such phenomenon as time dilation, length contraction etc. which [b']do[/b] occur in "reality". If you didn't know, the space-time of SR is called Minkowskian ST.

 

Yes, I know its called "Minkowskian space-time"

 

My question was about 'greek indices' that get used in GR.

 

I remember that you said that I don't have to use the summation symbol, if i stagger the location of the indices.

 

So for an arbitrary vector in an N dimensional space, I would write:

 

[math] v^\alpha \hat e_\alpha [/math]

 

?

 

Do I have that right? Must I use greek letters?

[Meir Achuz]

The standard convention is to use Latin letters for sum from 1 to 3 (just space),

and Greek indices for the sum over 0,1,2,3, where x^0=ct.

[□h=-16πT]

Also' date=' what do you mean that the bottom slot represents the number of vectors applied to the mapping?[/quote']

 

"Bottom slot" was just an informal way for me to say the bottom entry in the column matrix. "The number of vectors applied to the mapping"- how many vectors the tensor has as its arguements.

 

Greek indices are only used to denote the three spacial coordinates and time. Otherwise latin indices are used, usually i, j ,k, and they denote only spacial coordinates. In an N dimensional space one would not use greek indices as they are reserved for the coordinates of our 4-D space-time .

 

It seems to me as if you're paying to much attention to the summation convention, which is a pretty simple case of what is really just notation, and one gets the impression that it's hindering your progress. Or at least that's all you seem to be asking about.

 

To summarize:

 

Greek indices take the value 0, 1, 2, 3 and represent time, x, y, z respectively. Used only in discussions of space-time such as SR and GR. Also in discussions of space-time latin indices are used to denote the spatial coordinates of space-time, namely x, y, z.

 

In general discussions of N-dimensional spaces in topology, linear algebra etc. latin indices are used to take a value from 1 to N. This should not be confused with the index notation used in discussions of space-time.

 

The Einstein summation convention: if the same index is raised and lowered in the same expression then a summation is assumed over that index. This is not explicitly just for relativity and can be used in any mathematical study of geometry in any N-D space (for example I've seen it used in discussions of topology and general tensor algebra).

[Johnny5]

It seems to me as if you're paying to much attention to the summation convention, which is a pretty simple case of what is really just notation, and one gets the impression that it's hindering your progress. Or at least that's all you seem to be asking about.

 

Have to make sure I have the notation right, before I can begin to use it.

 

Kind regards

 

Let me just see what I remember offhand:

 

I remember this equation here:

 

[math] G_\alpha_\beta = \frac{8 \pi G}{c^4} T_\alpha_\beta [/math]

 

Now the LHS is the Einstein tensor, and the RHS is a temporal constant, times the stress-energy tensor.

 

The constant has units of inverse Newtons, when G and c are in SI.

 

Now, the whole equation is called a field equation of the General theory of Relativity. I believe it is Einstein's formulation of GR?

 

Now, I very clearly understand that in any discussions of either SR (special theory of relativity, or GR(general theory of relativity), that the index can only vary over four dimensions. And the numbers 0,1,2,3 are used for this.

 

0 corresponds to time: t

1,2,3 to space: x,y,z (in this order)

 

All of this I get.

 

So thus, when I am specifically having a discussion on GR or SR, i am to use greek indices. And those indices will take on values from 0, to 3.

 

Now, I could write an arbitrary spatiotemporal vector as follows:

 

[math] \vec v \equiv \sum_{\alpha = 0}^{\alpha = 3} v_\alpha \hat e_\alpha [/math]

 

But, if I use the Einstein summation convention, then I could just write an arbitrary spatiotemporal vector as follows:

 

[math] \vec v \equiv v^\alpha \hat e_\alpha [/math]

 

Where I have to stagger where the indices are, upper/lower, like i just did.

 

 

Did I get all this right? See, i was planning to use regular roman letters for indices always, but now I am being told to use greek letters when talking about spacetime in SR or GR. Which is fine with me, I just want to be sure I am understanding this much, before moving on.

 

Also I am concurrently reviewing linear algebra, and studying a few other things.

 

Also, there are two indices used in the Einstein tensor, so why two?

 

Also, I remember this:

 

That the Einstein tensor can be written in terms of the Ricci tensor, and the Ricci scalar, as follows:

 

[math] G_\alpha_\beta = R_\alpha_\beta - g_\alpha_\beta \frac{R}{2} [/math]

 

In the formula above, R sub alpha sub beta is the Ricci tensor, named after Italian mathematician 1853-Gregorio Ricci Curbastro-1925, who invented it. He died just two years before American physicist Arthur Holly Compton recieved the Nobel Prize in Physics for discovery of the Compton effect.

 

Ricci developed the "absolute differential calculus" also called "the Ricci calculus" now called "tensor analysis."

 

Ricci's work in the absolute differential calculus (ten year undertaking 1884-1894) was later used by Einstein, in Einstein's general theory of relativity).

 

Also, R is called the Ricci scalar, and [math] g_\alpha_\beta [/math] is the metric tensor. These things I also remembered.

 

Ok you also talked about "one forms"

 

Here is an example of one:

 

[math] \left ( \begin{array}{c}

0\\

1\\

\end{array} \right )[/math]

 

I believe a one form is an example of a tensor?

[□h=-16πT]

To understand the Ricci tensor you really need to understand the Riemann curvature tensor, geodesics, parrallel transport etc. If I get time in the next week or so I'll write something up on it for you. It's nothing of great difficulty, but it's incredibly elegant. I may add something on Geodesic deviation as well. Oh, read up on covariant differentiation and the Christoffel symbols.

 

The column matrix one-form thing is only notation for the type of tensor. If one were to represent a one-form as a matrix it would be a row matrix. The notation for a general tensor with N one-forms and M vectors as its arguements is

 

[math]

\left ( \begin{array}{c}

N\\

M\\

\end{array} \right )

[/math]

 

I think before you go onto general tensors you need to educate yourself fully in one-forms and their geometrical interpretation and vector spaces (as you're doing now). As I mentioned above, look up the covariant derivative (this is a one-form).

[Johnny5]

The column matrix one-form thing is only notation for the type of tensor. If one were to represent a one-form as a matrix it would be a row matrix. The notation for a general tensor with N one-forms and M vectors as its arguements is

 

[math]

\left ( \begin{array}{c}

N\\

M\\

\end{array} \right )

[/math]

 

Ok' date=' the general notation for a general tensor, comprised of N one-forms, and M vectors as its arguments, is...

 

[math'] \left( \begin{array}{c}

N\\

M\\

\end{array} \right)

[/math]

 

N denotes the number of "one forms"

M denotes the number of "vectors"

 

So, in the case of the following tensor:

 

[math] \left( \begin{array}{c}

0\\

1\\

\end{array} \right)

[/math]

 

There is 0 "one-forms" in the argument, and there is one vector in the argument.

[□h=-16πT]

Could you give me the definition of a tensor? I want to know if you actually know what one is.

[Johnny5]

Could you give me the definition of a tensor? I want to know if you actually know what one is.

 

 

The following is from answer dot com

 

Tensor- A set of quantities that obey certain transformation laws relating the bases in one generalized coordinate system to those of another and involving partial derivative sums. Vectors are simple tensors.

 

And this definition makes sense to me.

 

A tensor, is a set of certain quantities. These quantities obey certain transformation rules, which relate the bases (basis vectors?) of one coordinate system, to the bases of another coordinate system.

 

These quantities involve sums of partial derivatives.

 

A vector is a first rank tensor.

 

Is that definition there ok?

 

I went back and corrected the other thing. N is at the top, M at the bottom, N is the number of one forms in the tensor, and M is the number of vectors which are arguments of the tensor.

[Johnny5]

Oh' date=' read up on covariant differentiation and the Christoffel symbols.

[/quote']

 

 

Christoffel symbols

 

In general relativity, the Christoffel symbols represent "gravitational forces".

 

It is always possible to find a frame of reference on a Riemannian manifold, such that the Christoffel symbol vanishis at a chosen point?

 

The preferred frame of reference mentioned above, would be attached to a body in free-fall.

 

Christoffel symbols of the first kind:

 

[math] \Gamma_i,_j,_k [/math]

 

Christoffel symbols of the second kind:

 

[math] \Gamma^k_i,_j [/math]

[□h=-16πT]

Christoffel symbols

 

You also must be aware that a tensor is linear in its arguements. I will set about writing something on the curvature tensor on thursday or later this evening, the intermediate time being occupied with an art exam I'm doing (God only knows why I chose to do an art GCSE )

[Johnny5]

You also must be aware that a tensor is linear in its arguements.

 

What does that mean?

 

I'll go review more linear algebra.

 

Kind regards

[□h=-16πT]

What does that mean?

 

If T is a (1 0) tensor then (the sum of two one-forms being a one-form)

 

[math]\mathbf{T}(a\tilde{\omega^{\alpha}}+b\tilde{\omega^{\beta}})=a\mathbf{T}(\tilde{\omega^{\alpha}})+b \mathbf{T} (\tilde{\omega^{\beta}})[/math]

 

This is very important. Here I've just used the example of a (1 0) tensor for simplicity, but it's the same for any (N M) tensor. Oh, when one writes the arguements of a general (N M) tensor in a notation similar to the above, the vectors go on the left of the brackets and the one-forms on the right and the two are separated by a semi-colon. For example a (1 1) tensor

 

[math]\mathbf{T}(\vec{V^{\alpha}}; \tilde{\omega^{\beta}})[/math]

[□h=-16πT]

This is what I have so far, Johnny5. I'll work on the rest and you can read through what I have at the moment.

 

Throughout this discussion raised indices denote covariant components; lowered contravariant; and [math]g_{\alpha\beta}[/math] and [math]g^{\alpha\beta}[/math] are the components of the metric and inverse metric respectively. A comma in the index denotes differentiation with respect to the subsequent indices and a semi-colon followed by an index is the something-th component of the covariant derivative.

 

I shall first set down some properties that will prove useful in the construction of the Riemann tensor. The first being that at a point P of any Riemannian manifold

 

[math]g_{\alpha\beta}(P)=\eta_{\alpha\beta}[/math] (1)

 

where [math]\eta[/math] is the metric of SR (obviously we don't have the SR metric in discussions of general manifolds but we still have a flat metric), i.e. that at a point on any Riemannian manifold the space is flat (for a proof see the local flatness theorem). Since at any point the space is flat, we have

 

[math]g_{\alpha\beta}_{,\mu}(P)=0[/math]

and

[math]g_{\alpha\beta}_{,\mu\nu}(P)\neq 0[/math]

 

for any index. The first equality is obvious from (1). The second is such for curved manifolds, as the derivatives change about a point P and so a second derivative exists in general. If you have looked up the Christoffel symbols, a definition you may have found is that they are the coefficients of the basis vectors upon differentiation of the basis, i.e.

 

[math]\frac{\partial\vec{e_{\alpha}}}{\partial x^{\beta}}=\Gamma^{\mu}_{\alpha\beta}\vec{e_{\mu}}[/math]

 

In terms of metric components the symbols are

 

[math] \Gamma^{\gamma}_{\beta\mu}=\tfrac{1}{2}g^{\alpha \gamma}(g_{\alpha\beta,\mu}+ g_{\alpha\mu,\beta}- g_{\beta\mu,\alpha})[/math] (2)

 

You should also have accustomed yourself with the manipulation of indices by now, as proficiency in this will prove useful (such a dummy, free and changing of indices).

[Johnny5]

If T is a (1 0) tensor then (the sum of two one-forms being a one-form)

 

[math]\mathbf{T}(a\tilde{\omega^{\alpha}}+b\tilde{\omega^{\beta}})=a\mathbf{T}(\tilde{\omega^{\alpha}})+b \mathbf{T} (\tilde{\omega^{\beta}})[/math]

 

This is very important.[/math]

 

You say that if T is a (1 0) tensor' date=' then it is a linear operator.

 

So thus:

 

[math'] T \equiv (1 \ \ 0) [/math]

 

In the definition above, T denotes a tensor 1, is the number of vectors/one forms and 0 is the number of oneforms/vectors?

 

1 is the number of ????? 0 is the number of ???? You rotated the notation on me.

 

I recall this from yesterday:

 

[math]

\left (

\begin{array}{c}

N\\

M\\

\end{array} \right ) [/math]

 

Denotes a tensor with N one forms in the argument, and M vectors in the argument. Now you have rotated the notation, so what is what now?

 

Linearity of L means this

 

[math] L[af(x)+b{g(x)] = aL[f(x)] + bL[g(x)] [/math]

 

So in the following equation...

 

 

[math]\mathbf{T}(a\tilde{\omega^{\alpha}}+b\tilde{\omega^{\beta}})=a\mathbf{T}(\tilde{\omega^{\alpha}})+b \mathbf{T} (\tilde{\omega^{\beta}})[/math]

 

 

You have told me:

 

In the case where

 

[math] \mathbf{T} [/math] denotes a (1 0) tensor, [math] \mathbf{T} [/math] is a linear operator.

 

My question now is what does

 

[math] \tilde{\omega^{\alpha}}

[/math]

 

denote?

 

No hurry.

 

Regards

[□h=-16πT]

I turned the comlumn matrix 90 degrees anti-clockwise so that I wouldn't have to bother with letexing the maths for it, so one-forms are on the left and vectors on the right. A letter with a tilde over it denotes a one-form and so omega alpha is the alpha-th one form. However, I have been slightly sloppy with my notation as omega is usually reserved for basis one-forms (just as e is for basis vectors), a notation which I shall use in my discussion of the Riemann tensor. In the example of linearity above just take the omega to be any one-form rather than part of a basis. Yes, tensors are linear operators.

[Johnny5]

I turned the comlumn matrix 90 degrees anti-clockwise so that I wouldn't have to bother with letexing the maths for it, so one-forms are on the left and vectors on the right. A letter with a tilde over it denotes a one-form and so omega alpha is the alpha-th one form. However, I have been slightly sloppy with my notation as omega is usually reserved for basis one-forms (just as e is for basis vectors), a notation which I shall use in my discussion of the Riemann tensor. In the example of linearity above just take the omega to be any one-form rather than part of a basis. Yes, tensors are linear operators.

 

It's ok to be a bit sloppy with notation, as long as you explain it as you go, I can follow. I understand that e is reserved for basis vectors.

 

Ok so... for me... let me see where I am at here:

(this will all be from memory)

Start off with Einstein's field equation of GR, which is this:

 

Thus, we can also write:

 

(N M) to denote the tensor.

 

[math] G_\alpha_\beta = \frac{8 \pi G}{c^4} T_\alpha_\beta [/math]

 

In the equation above, the LHS is called the Einstein tensor, and the RHS is the product of a scalar (with units of inverse Newtons (in SI) ) times the stress-energy tensor. G is the Newtonian gravitational constant, and c is the speed of light relative to an emitter.

 

Tensors are linear operators, that means this...

 

 

For any mathematical operator [math] \text{L} [/math]: [math] \text{L} [/math] is a linear operator if and only if:

 

[math] \text{L}[a f(x) + b g(x)] = a\text{L}[f(x)] + b \text{L}[g(x)] [/math]

 

In the formula above, a,b denote constants, whereas x denotes a variable quantity.

 

In the years from 1884-1894, Gregorio Ricci Curbastro worked out absolute diffferential geometry, later called the Ricci Calculus, and now referred to as Tensor analysis. This was the mathematics used by Albert Einstein, in the formulation of the general theory of relativity.

 

The Einstein tensor can be written in terms of the Ricci scalar, and the Ricci tensor, and the metric tensor as follows:

 

[math] G_\alpha_\beta = R_\alpha_\beta - \frac{1}{2} R g_\alpha_\beta [/math]

 

In the expression above, R sub alpha sub beta is referred to as the Ricci curvature tensor, and R denotes the Ricci curvatur scalar, and g sub alpha sub beta denotes the 'metric tensor'.

 

Now, here is one notation for a tensor with N "one forms" and M vectors for its arguments:

 

[math] \left( \begin{array}{c}

N\\

M\\

\end{array} \right) [/math]

 

In the notation above, N denotes the number of "one-forms" and M denotes the number of vectors that are the arguments of the tensor.

 

And there is also a similiar notation, in which we write out the elements of the tensor horizontally. This will lessen the usage of latex as well:

 

(N M)

 

Ok lets see what else do I remember...

 

Ok, there are Christoffel symbols of the first and second kind, here is what they look like:

 

[math] \Gamma_i,_j,_k [/math]

 

The symbol above is a Christoffel symbol of the first kind.

 

And there is also...

 

[math] \Gamma^k_i,_j [/math]

 

And the symbol above, is a Christoffel symbol of the second kind.

 

 

Lets see, what else is there...

 

Ok, I need to know about Einstein summation convention...

 

Suppose we have some arbitrary vector in three dimensional Euclidean space... the most common notation for an arbitrary vector is:

 

[math] \vec V = V1 \hat i + V2 \hat j + V3 \hat k [/math]

 

Now, the LHS makes sense in the affine plane, that is a plane without coordinates, or an affine space I should say, which would be a frame without coordinates, but the RHS requires that your frame have coordinates. let me explain this:

 

Let us define an affine frame as follows:

 

Definition: An affine frame is three mutually orthogonal infinite straight lines.

 

Definition: A non-affine frame, or simply frame for short, most commonly called a rectangular coordinate system, is a frame in which the lines have coordinates on them. In other words, the axes of a frame are are number lines, instead of just lines.

 

And because we have mapped the set of points on an axis to real numbers, we can pick a unit of distance, like the meter for example, to be our unit of distance measurement.

 

 

 

 

Oh, in post 141, you said...

 

This is what I have so far' date=' Johnny5. I'll work on the rest and you can read through what I have at the moment.

 

Throughout this discussion raised indices denote covariant components; lowered contravariant; and [math']g_{\alpha\beta}[/math] and [math]g^{\alpha\beta}[/math] are the components of the metric and inverse metric respectively. A comma in the index denotes differentiation with respect to the subsequent indices and a semi-colon followed by an index is the something-th component of the covariant derivative.

 

I shall first set down some properties that will prove useful in the construction of the Riemann tensor. The first being that at a point P of any Riemannian manifold

 

[math]g_{\alpha\beta}(P)=\eta_{\alpha\beta}[/math] (1)

 

where [math]\eta[/math] is the metric of SR (obviously we don't have the SR metric in discussions of general manifolds but we still have a flat metric), i.e. that at a point on any Riemannian manifold the space is flat (for a proof see the local flatness theorem). Since at any point the space is flat, we have

 

[math]g_{\alpha\beta}_{,\mu}(P)=0[/math]

and

[math]g_{\alpha\beta}_{,\mu\nu}(P)\neq 0[/math]

 

for any index. The first equality is obvious from (1). The second is such for curved manifolds, as the derivatives change about a point P and so a second derivative exists in general. If you have looked up the Christoffel symbols, a definition you may have found is that they are the coefficients of the basis vectors upon differentiation of the basis, i.e.

 

[math]\frac{\partial\vec{e_{\alpha}}}{\partial x^{\beta}}=\Gamma^{\mu}_{\alpha\beta}\vec{e_{\mu}}[/math]

 

In terms of metric components the symbols are

 

[math] \Gamma^{\gamma}_{\beta\mu}=\tfrac{1}{2}g^{\alpha \gamma}(g_{\alpha\beta,\mu}+ g_{\alpha\mu,\beta}- g_{\beta\mu,\alpha})[/math] (2)

 

You should also have accustomed yourself with the manipulation of indices by now, as proficiency in this will prove useful (such a dummy, free and changing of indices).

 

 

Raised indices denote covariant...

Lowered indicied denote contravariant...

 

"metric inverse metric" respectively...

 

In the general theory of relativity, the following symbol denotes the metric tensor...

 

[math] g_\alpha_\beta [/math]

 

Alpha, beta are called contravariant components.

 

And the following symbol denotes the inverse metric tensor...

 

[math] g^\alpha^\beta [/math]

 

In the previous symbol, alpha,beta are covariant components of the tensor.

 

When a comma is used in the notation, that means that subsequent indices are to be differentiated, rather than previous ones.

[□h=-16πT]

You can leave out your terms with c and G in them if you like. Geometrized units are of no difficulty, and they don't alter the interpretation of an expression at all (other than the dimensions, obviously). I'd also ignore the field equations for now until you know about the Riemann tensor, Christoffel symbols (of the second kind), covariant derivative etc. Also it's the speed of light in a vacuum, theres is no need to define a frame to which the light is travelling relative to. This is because the speed of light is constant in all inertial frames.

[Johnny5]

You can leave out your terms with c and G in them if you like. Geometrized units are of no difficulty, and they don't alter the interpretation of an expression at all (other than the dimensions, obviously).

 

What if that matters? I don't understand this "geometrization" thing which keeps getting referred to. If it doesn't alter the truth value of any statements, then I will learn how to switch to it. Actually, I can learn how to switch to it no matter what. But what do most physicists do?

[Johnny5]

I'd also ignore the field equations for now until you know about the Riemann tensor, Christoffel symbols (of the second kind), covariant derivative etc.

 

Ok, one at a time. What is the Riemann tensor?

[□h=-16πT]

Ok, one at a time. What is the Riemann tensor?

 

The Riemann curvature tensor, it's what I said I'd explain and what I've started to discuss above. It'll be finished by the weekend. Physicists use geometrized units in GR because it saves their wrists a little bit of strain from writing the same two constants out over and over. The physical interpretation of equations is not just in the units, if one becomes accustomed to geometrized form then it's just as simple as SI. The important point about the field equations is that space time is warped by a body described dynamically by the SE tensor, not what units we're using. They're called geometrized units because dealing with the geometry of space-time etc. Just as we may call the units of SR, where just c=1, "natural" because it seems natural to have such a fundamental constant, of which the entire theory is based on, just blend into the background, so to speak.

[Johnny5]

Also it's the speed of light in a vacuum' date=' theres is no need to define a frame to which the light is travelling relative to. This is because the speed of light is constant in all inertial frames.[/quote']

 

The speed of light is only constant in all inertial reference frames IF simultaneity is relative, instead of absolute. However, simultaneity is absolute.

 

However, my statement as it was, could not be false, even if the speed of light is constant in all inertial reference frames, for the following reason:

 

Certainly if the emitter is in an inertial reference frame when it emits a photon, then the speed of light will be equal to c in the frame at which the emitter is at rest(which is all I meant). Whether or not the speed of that photon is c in all other inertial reference frames is another question entirely.

[Johnny5]

The Riemann curvature tensor, it's what I said I'd explain and what I've started to discuss above. It'll be finished by the weekend.

 

Alright, take your time, I am learning this painfully slowly I'm afraid. But I am doing my level headed best.

 

And you are a good teacher, thank you.

 

Kind regards