How can time only have one direction?

[student34]

Because we'd like to know where you are getting so much misinformation. When you are asked a direct question like that here on PF it is very bad form to ignore it **. How about you just answer the question?

** I've seen threads closed because the OP would not answer a question or questions.

I have been reading a lot of publications related to this subject, and watching Khan videos, but that is not where the misinformation is coming from. My memory is bad, and I sometimes have slopping wording.

 

[vela]

You ask questions that you wouldn't ask if you were working with a textbook. Or at least you would phrase them differently. So why being dishonest?

I think you're being a bit judgmental here. Lots of students struggle with articulating what they're confused about. It's not being dishonest. It's just sometimes hard to put what's already confusing into words.

 

[weirdoguy]

I think you're being a bit judgmental here.

Well, maybe, I am sorry. But the amount of threads started by OP, and his reluctance to say on what textbook is he working on makes me really think this is going nowhere...

 

[student34]

That's like saying the y-axis has more than one dimension because you can choose the orientation of coordinate axes arbitrarily.

But it is not just one arbitrary choice, it seems to be infinite.

 

[vela]

But it is not just one arbitrary choice, it seems to be infinite.

How is that different from how you choose to orient the x- and y- axes when solving a mechanics problem? You also have an infinite number of ways you can choose to do that.

When you solve an inclined plane problem, you typically choose to orient the x-axis along the incline and the y-axis perpendicular to it. Does that mean the y-axis now points in an infinite number of directions because you could have chosen to orient the axes an infinite number of other ways? No. You make your choice, and it points in one direction.

In relativity, the same thing goes. In reference frame S, the t-axis points along one particular direction. In frame S', moving relative to S, the t'-axis points in a different direction but still in only one direction.

 

[student34]

How is that different from how you choose to orient the x- and y- axes when solving a mechanics problem? You also have an infinite number of ways you can choose to do that.

When you solve an inclined plane problem, you typically choose to orient the x-axis along the incline and the y-axis perpendicular to it. Does that mean the y-axis now points in an infinite number of directions because you could have chosen to orient the axes an infinite number of other ways? No. You make your choice, and it points in one direction.

In relativity, the same thing goes. In reference frame S, the t-axis points along one particular direction. In frame S', moving relative to S, the t'-axis points in a different direction but still in only one direction.

The difference is that the x axis is not different than the y axis. The time axis is intrinsically different than the spatial axis.

 

[vela]

You're making a pretty big leap here, and it would help if you could explain your reasoning. In what way is the time axis intrinsically different and why does it make a difference?

 

[phinds]

I have been reading a lot of publications related to this subject, and watching Khan videos, but that is not where the misinformation is coming from.

How do you know that? I suspect some of your misinformation IS coming from such sources. On line videos are usually just entertainment, not education and even decent ones are not a good source of learning.

To learn a subject you need to use a textbook AND work out the problems presented in the textbook. That's what I suggest you do, and, based on this thread, I have every confidence that you are not going to do that.

 

[student34]

You're making a pretty big leap here, and it would help if you could explain your reasoning. In what way is the time axis intrinsically different and why does it make a difference?

I am trying to say that space is different than time. I don't believe that is a controversial statement.

It makes a difference because now we csn have a space of time instead of just a line of time.

 

[PeterDonis]

If I have a line, and it can go in an infinite number of directions, doesn't this have something to do with how many dimensions it has?

No. A line is one dimensional. The space (or spacetime) in which the line exists might have more dimensions, depending on how many parameters it takes to specify the possible directions the line can go; but none of that affects how many dimensions the line itself has.

 

[PeterDonis]

I thought that if time has more than one direction, then it should have more than one dimension.

You thought wrong. In fact you aren't even thinking clearly about what a "dimension" is--despite repeated attempts to get you to do so. Some of the things you say appear to be groping in the right direction, but very slowly. For example:

I am trying to say that space is different than time. I don't believe that is a controversial statement.

A better way of stating this would be that spacelike vectors and curves are fundamentally different from timelike vectors and curves. And there is a third category in spacetime, null (or lightlike) vectors and curves, which are fundamentally different from both.

It makes a difference because now we csn have a space of time instead of just a line of time.

This is where you are sort of groping in the right direction; but you sidetrack yourself by not thinking clearly about what "time" is, what a "dimension" is, and what the different directions in spacetime of the worldlines of different observers in relative motion actually mean.

What you should do is step back from all that and, first of all, ask this question:

(Q1) How many distinct parameters does it take to describe all of the possible directions in spacetime that a timelike worldline (i.e., the worldline of an inertial observer) can have at a particular point?

The answer, of course, is "more than one". (I won't give the exact number right now because I want you to think about the question in those terms.) But why is it more than one?

Consider: suppose spacetime were 2-dimensional. That would mean the spacetime diagrams we draw on 2-dimensional sheets of paper (or the electronic equivalent) would be diagrams of actual spacetime, not just a 2-dimensional "slice" of 4-dimensional spacetime. And in a 2-dimensional spacetime, the answer to question Q1 above would be one. The "directions in spacetime" that a timelike worldline could have could be described by one parameter, which we could think of as the ordinary speed in the x-direction of that worldline in some fixed inertial frame. (Or, if we wanted our parameter to have the range $- \infty < p < \infty$ instead of $-1 < p < 1$, we could use the gamma factor or the rapidity.)

From the above, you should already be able to figure out the answer to question Q1 above for our actual 4-dimensional spacetime.

But now, take another step back, and ask a different question:

(Q2) How many timelike eigenvalues does the metric of spacetime have?

By "timelike eigenvalue" I just mean an eigenvalue of whichever sign we are using for timelike squared intervals in the metric. For example, if we write the metric as $d\tau^2 = dt^2 - dx^2 - dy^2 - dz^2$, timelike squared intervals are positive, so that is the timelike sign. (This "timelike convention" is more common in introductory books on SR and in particle physics. The opposite "spacelike convention" is more common in GR and more advanced relativity literature.)

The answer to Q2 above is one. That is, one always. It doesn't matter what frame you choose. It doesn't matter if spacetime is curved. It doesn't matter if you adopt some weird coordinate chart, even one in which you don't have one timelike and three spacelike coordinates as you do in a standard inertial frame. The metric always has precisely one timelike eigenvalue. And it always has precisely three spacelike eigenvalues.

What do the metric eigenvalues mean? They tell you what kinds of dimensions the space, or spacetime has. And that is why we say that spacetime has one timelike dimension and three spacelike dimensions, for four dimensions total: because of the eigenvalues of the metric. (Similarly, ordinary Euclidean 3-space has a metric with three spacelike eigenvalues--here we always adopt the spacelike convention so the metric is positive definite--and so we say it is a 3-dimensional space.) And note that, since the answers to Q1 and Q2 are different, the question of what kinds of dimensions the space or spacetime has is a different question from the question of how many different directions a particular kind of vector can point. Both are properties of the spacetime and its geometry; but they're different properties.

 

[PeterDonis]

those directions aren't orthogonal, so you aren't counting independent things.

Careful. The proper requirement is not orthogonality, but linear independence. And the kind of "independence" that leads to a count of the dimensions of spacetime is not "independence" of individual pairs of vectors.

The way out of the confusion is to require that the basis directions be orthogonal if you are trying to classify your space/spacetime by looking at them.

I would phrase this as: the fact that a standard inertial frame has four mutually orthogonal axes is an easy shorthand way to remember how many dimensions spacetime has: just count the axes. And the fact that one axis is timelike and three are spacelike gives a handy way to remember how many of what kinds of dimensions there are. But the ultimate count of dimensions doesn't come from that: it comes from counting the metric eigenvalues and their signs. The fact that it is possible to define a standard inertial frame with four mutually orthogonal axes, one timelike and three spacelike, is a consequence of the eigenvalues of the metric.

But counting dimensions that way is not the same as counting possible directions of curves in spacetime, as I made clear in post #81 just now. There are distinct, linearly independent timelike directions (vectors at a point) in spacetime even though there is only one timelike dimension (one timelike eigenvalue of the metric). But the criterion for "independence" of directions in spacetime is not and cannot be orthogonality: no pair of timelike vectors can possibly be orthogonal, but they can be linearly independent.

 

[vela]

I am trying to say that space is different than time. I don't believe that is a controversial statement.

Sure. A separation in time is measured using clocks, and a separation in space is measured using meter sticks. But I still don't see how you're getting from there to "therefore, a time axis points in more than one direction" and why any such argument wouldn't also apply to a spatial axis.

It makes a difference because now we csn have a space of time instead of just a line of time.

What do you mean by "a space of time"? I thought I knew what you meant, but I realized that perhaps you meant something different and this is where your confusion is stemming from.

 

[student34]

You thought wrong. In fact you aren't even thinking clearly about what a "dimension" is--despite repeated attempts to get you to do so. Some of the things you say appear to be groping in the right direction, but very slowly. For example:A better way of stating this would be that spacelike vectors and curves are fundamentally different from timelike vectors and curves. And there is a third category in spacetime, null (or lightlike) vectors and curves, which are fundamentally different from both.This is where you are sort of groping in the right direction; but you sidetrack yourself by not thinking clearly about what "time" is, what a "dimension" is, and what the different directions in spacetime of the worldlines of different observers in relative motion actually mean.

What you should do is step back from all that and, first of all, ask this question:

(Q1) How many distinct parameters does it take to describe all of the possible directions in spacetime that a timelike worldline (i.e., the worldline of an inertial observer) can have at a particular point?

The answer, of course, is "more than one". (I won't give the exact number right now because I want you to think about the question in those terms.) But why is it more than one?

Consider: suppose spacetime were 2-dimensional. That would mean the spacetime diagrams we draw on 2-dimensional sheets of paper (or the electronic equivalent) would be diagrams of actual spacetime, not just a 2-dimensional "slice" of 4-dimensional spacetime. And in a 2-dimensional spacetime, the answer to question Q1 above would be one. The "directions in spacetime" that a timelike worldline could have could be described by one parameter, which we could think of as the ordinary speed in the x-direction of that worldline in some fixed inertial frame. (Or, if we wanted our parameter to have the range $- \infty < p < \infty$ instead of $-1 < p < 1$, we could use the gamma factor or the rapidity.)

From the above, you should already be able to figure out the answer to question Q1 above for our actual 4-dimensional spacetime.

But now, take another step back, and ask a different question:

(Q2) How many timelike eigenvalues does the metric of spacetime have?

By "timelike eigenvalue" I just mean an eigenvalue of whichever sign we are using for timelike squared intervals in the metric. For example, if we write the metric as $d\tau^2 = dt^2 - dx^2 - dy^2 - dz^2$, timelike squared intervals are positive, so that is the timelike sign. (This "timelike convention" is more common in introductory books on SR and in particle physics. The opposite "spacelike convention" is more common in GR and more advanced relativity literature.)

The answer to Q2 above is one. That is, one always. It doesn't matter what frame you choose. It doesn't matter if spacetime is curved. It doesn't matter if you adopt some weird coordinate chart, even one in which you don't have one timelike and three spacelike coordinates as you do in a standard inertial frame. The metric always has precisely one timelike eigenvalue. And it always has precisely three spacelike eigenvalues.

What do the metric eigenvalues mean? They tell you what kinds of dimensions the space, or spacetime has. And that is why we say that spacetime has one timelike dimension and three spacelike dimensions, for four dimensions total: because of the eigenvalues of the metric. (Similarly, ordinary Euclidean 3-space has a metric with three spacelike eigenvalues--here we always adopt the spacelike convention so the metric is positive definite--and so we say it is a 3-dimensional space.) And note that, since the answers to Q1 and Q2 are different, the question of what kinds of dimensions the space or spacetime has is a different question from the question of how many different directions a particular kind of vector can point. Both are properties of the spacetime and its geometry; but they're different properties.

Thanks for all of this, and I will try to keep it in mind moving forward as best I can.

 

[student34]

Sure. A separation in time is measured using clocks, and a separation in space is measured using meter sticks. But I still don't see how you're getting from there to "therefore, a time axis points in more than one direction" and why any such argument wouldn't also apply to a spatial axis.

(I couldn't respond because this website wasn't working well with my cellphone)

Using the example in the OP, the object in the middle would calculate the time axis, of the objects moving away from it, pointing in different directions, no?

And to answer the second part, I would think that the time and space axis would have to be fixed if time is not going to point in a different direction for other observers in motion. Again, the example I gave would seem to have 3 different directions of time.

 

[student34]

What do you mean by "a space of time"? I thought I knew what you meant, but I realized that perhaps you meant something different and this is where your confusion is stemming from.

The worldlines of other objects seem to be able to fill an object's light cone, using multiple time axis. So any point in an object's light cone can be reached by lightlike worldlines. This seems to give time 2 dimensions, or space instead of just a line.

 

[jbriggs444]

The worldlines of other objects seem to be able to fill an object's light cone, using multiple time axis. So any point in an object's light cone can be reached by lightlike worldlines. This seems to give time 2 dimensions, or space instead of just a line.

Not exactly. I would not parameterize the space with [pairs of] lightlike worldlines. Instead, I would parameterize the space in terms of timelike directions.

The future light cone from a particular event includes a multitude of time-like directions. I count three dimensions. Any standard of rest corresponds to one of those directions. You have velocity in the x, y and z directions to define a standard of rest. So it takes three dimensions to identify a particular timelike direction.

If we are sticking to special relativity, "inertial reference frame" is nearly synonymous with both "standard of rest" and "timelike direction".

It is tempting to say "4 dimensions" since every event in the 4 dimensional future light cone is associated with a direction. But any given direction has infinitely many events all lined up on a straight world-line in that direction. So it is only three dimensions to identify a time-like direction/standard of rest/inertial reference frame.

It might be worth asking again -- do you even know what "dimension" means

 

[robphy]

In the plane, given a set of xy-axes, you have to specify two numbers to locate a point… a displacement along the given x-axis and a displacement along the given y-axis. (…Not two numbers along x-axes since you don’t have two x-axes.) This number of coordinates is the dimensionality of the plane…. but not the dimensionality of the x-axis.
One can certainly choose other orientations of the xy-axes. But the dimensionality of the plane is still two and the dimensionality of the given x-axis is still one.In a position vs time diagram, given a ty-plane associated with an inertial frame, you have to specify two numbers to locate an event… a displacement along the given t-axis (the reading of a clock) and a displacement along the given y-axis (the reading along a ruler). (…Not two numbers along t-axes since you don’t have two t-axes (you don’t have two clocks).) This number of coordinates is the dimensionality of the position-vs-time diagram…. but not the dimensionality of the t-axis.
One can certainly choose other spacetime-orientations of the ty-axes for different inertial frames. But the dimensionality of the position-vs-time diagram is still two and the dimensionality of the given t-axis is still one.

One can choose other pairs of axes… but that is a mathematical complication that will likely distract if the above is not first understood.

(crawl before walking and running…
note there is no need for special relativity to make the above points)

 

[student34]

Not exactly. I would not parameterize the space with [pairs of] lightlike worldlines. Instead, I would parameterize the space in terms of timelike directions.

The future light cone from a particular event includes a multitude of time-like directions. I count three dimensions. Any standard of rest corresponds to one of those directions. You have velocity in the x, y and z directions to define a standard of rest. So it takes three dimensions to identify a particular timelike direction.

If we are sticking to special relativity, "inertial reference frame" is nearly synonymous with both "standard of rest" and "timelike direction".

It is tempting to say "4 dimensions" since every event in the 4 dimensional future light cone is associated with a direction. But any given direction has infinitely many events all lined up on a straight world-line in that direction. So it is only three dimensions to identify a time-like direction/standard of rest/inertial reference frame.

It might be worth asking again -- do you even know what "dimension" means

Damn, I meant to put "timelike", not "lightlike". But anyways, I agree with what you are saying, but I still don't see it resolving my issue.

I do not know the rigorous definition of a dimension. I believe that I have a good idea though.

 

[phinds]

I do not know the rigorous definition of a dimension. I believe that I have a good idea though.

And have you made any attempt to find out the rigorous definition or do you prefer continuing to not know?

 

[PeterDonis]

Using the example in the OP, the object in the middle would calculate the time axis, of the objects moving away from it, pointing in different directions, no?

"Time axis" is a property of frames, not objects. So this question is unanswerable as you ask it.

I could try to guess what you actually mean, but I won't. Instead, you need to rephrase your question so it makes sense. It is very frustrating that, even after multiple threads with you on topics like this, you are still unable to use standard terminology in a standard way.

I would think that the time and space axis would have to be fixed if time is not going to point in a different direction for other observers in motion. Again, the example I gave would seem to have 3 different directions of time.

This is all nonsense, and the reason why has already been explained to you. There is no point in continuing to go around in circles.

The worldlines of other objects seem to be able to fill an object's light cone

Objects don't have light cones. Events do. So this is nonsense as you state it.

See my remarks above about rephrasing things so they make sense and using standard terminology in a standard way.

using multiple time axis.

Nonsense. You've already been told why.

This seems to give time 2 dimensions, or space instead of just a line.

Nonsense. You've already been told why.

I agree with what you are saying, but I still don't see it resolving my issue.

Then what's the point of further discussion? How many times do we have to see you making the same errors and giving you the same corrections?

I do not know the rigorous definition of a dimension. I believe that I have a good idea though.

Then why can't you answer the question you've already been asked, multiple times, by giving whatever "good idea" you have?

 

[PeterDonis]

it takes three dimensions to identify a particular timelike direction.

To put it another way: the future-pointing unit timelike vectors at a point in spacetime form a three-dimensional vector space (more precisely, they form a three-dimensional subspace of the space of all possible vectors at a point in spacetime).

It is tempting to say "4 dimensions" since every event in the 4 dimensional future light cone is associated with a direction. But any given direction has infinitely many events all lined up on a straight world-line in that direction.

That is why we need to put the qualifier "unit vectors" in the statement I made above. That is what "divides out", so to speak, the variation along each individual direction, and leaves only the variation in directions.

 

[vela]

I do not know the rigorous definition of a dimension. I believe that I have a good idea though.

I can't say I agree that you have a good idea since you keep suggesting time has two dimensions. Reread @robphy's post above. What he wrote directly conflicts with your claim. If you truly understand his post, you should be able to understand the errors in your reasoning.

 

[student34]

In the plane, given a set of xy-axes, you have to specify two numbers to locate a point… a displacement along the given x-axis and a displacement along the given y-axis. (…Not two numbers along x-axes since you don’t have two x-axes.) This number of coordinates is the dimensionality of the plane…. but not the dimensionality of the x-axis.
One can certainly choose other orientations of the xy-axes. But the dimensionality of the plane is still two and the dimensionality of the given x-axis is still one.In a position vs time diagram, given a ty-plane associated with an inertial frame, you have to specify two numbers to locate an event… a displacement along the given t-axis (the reading of a clock) and a displacement along the given y-axis (the reading along a ruler). (…Not two numbers along t-axes since you don’t have two t-axes (you don’t have two clocks).) This number of coordinates is the dimensionality of the position-vs-time diagram…. but not the dimensionality of the t-axis.
One can certainly choose other spacetime-orientations of the ty-axes for different inertial frames. But the dimensionality of the position-vs-time diagram is still two and the dimensionality of the given t-axis is still one.

One can choose other pairs of axes… but that is a mathematical complication that will likely distract if the above is not first understood.

(crawl before walking and running…
note there is no need for special relativity to make the above points)

Ok, I believe that I understand what you are saying. I will strip my issue down to make it as clear as possible.

If we can, assume only a universe of one space dimension and one time dimension. Two rocks suddenly separate left and right from event e. After 1 second, we want to draw a Minkowski diagram, and it of course is a V shape.

Now how can we make a single graph to satisfy the V? If this universe actually has only one time and one space dimension, how can we now "superimpose" (sorry but I do not know the correct terminology. I hope it is obvious what I am trying to say) a graph onto the V?

As is the procedure, we give both rocks' worldlines the time axis. Which Worldline gets the time axis? And because of time dilation, it seems that which ever rock gets the time axis is the side of the V that has a longer worldline than the other.

 

[robphy]

As is the procedure, we give both rocks' worldlines the time axis. Which Worldline gets the time axis? And because of time dilation, it seems that which ever rock gets the time axis is the side of the V that has a longer worldline than the other.

Each rock has its own time axis in a spacetime diagram, just like two different surveyors on a plane have their own x-axes (their own forward directions).

On a plane, the projection of one rock’s unit x vector is shorter than one on the other rock’s x-axis.

 

[PeterDonis]

how can we make a single graph to satisfy the V?

You can't. There are an infinite number of possible "graphs" (spacetime diagrams) that will "satisfy" the V, corresponding to the infinite number of possible inertial frames in which to draw the diagrams.

As is the procedure, we give both rocks' worldlines the time axis.

No, that is not "the procedure". A given inertial frame can only have one time axis. So you have to pick one. You can't draw a single diagram with both rocks' worldlines as the time axis.

because of time dilation, it seems that which ever rock gets the time axis is the side of the V that has a longer worldline than the other.

How a length looks in a spacetime diagram does not always correspond to the actual length.

The lengths along worldlines are invariant. No spacetime diagram can make all lengths "look" like their actual lengths, because the piece of paper you are drawing the diagram on has Euclidean geometry, but spacetime itself has Minkowski geometry. So any diagram will distort some lengths.

And with all that said, your actual description of what the V will look like (leaving aside that how it looks is distorted compared to actual lengths) is wrong. If you just go by what each arm of the V looks like, and we assume that the actual lengths of the arms are equal (i.e., we mark the ends of each arm by the same elapsed proper time of the rock whose worldline is along that arm), then the arm corresponding to the time axis will look shorter than the other, not longer.

 

[PeterDonis]

Each rock has its own time axis in a its own spacetime diagram drawn in its own rest frame.

Please see the corrections in the above quote. We don't want to confuse the OP any more than he already is.

 

[robphy]

In response to @PeterDonis,
let me also suggest an edit of the analogy:
two different surveyors on a plane have their own x-axes (their own forward directions) in their own xy-planes.

 

[PeterDonis]

In response to @PeterDonis,
let me also suggest an edit the analogy:
two different surveyors on a plane have their own x-axes (their own forward directions) in their own xy-planes.

If the "xy-planes" are the diagrams the surveyors are drawing, yes. But they're both surveying the same actual space.

 

[robphy]

If the "xy-planes" are the diagrams the surveyors are drawing, yes. But they're both surveying the same actual space.

Yup.

 

[Nugatory]

As is the procedure, we give both rocks' worldlines the time axis. Which Worldline gets the time axis?

It doesn't matter as far as counting dimensions is concerned. Up/down on the sheet of graph paper is forwards/backwards in the time dimension, left/right is the space dimension.

 

[student34]

"Time axis" is a property of frames, not objects. So this question is unanswerable as you ask it.

This was from your other post to the ultimate question of this thread, "Using the example in the OP, the object in the middle would calculate the time axis, of the objects moving away from it, pointing in different directions, no?"

I think I have narrowed down closer to my confusion. Here is my conclusion from my premise.

P1: Either rock can claim to be travelling through only time and no space.
Conclusion: Then it would seem to be true to say that both are travelling through only time and no space.

Of course my conclusion must be wrong, but I don't know why. I don't know how it can be either, but not both.

No, that is not "the procedure". A given inertial frame can only have one time axis. So you have to pick one. You can't draw a single diagram with both rocks' worldlines as the time axis.

I meant that each gets its own time axis in its own diagram.

How a length looks in a spacetime diagram does not always correspond to the actual length.

The lengths along worldlines are invariant. No spacetime diagram can make all lengths "look" like their actual lengths, because the piece of paper you are drawing the diagram on has Euclidean geometry, but spacetime itself has Minkowski geometry. So any diagram will distort some lengths.

And with all that said, your actual description of what the V will look like (leaving aside that how it looks is distorted compared to actual lengths) is wrong. If you just go by what each arm of the V looks like, and we assume that the actual lengths of the arms are equal (i.e., we mark the ends of each arm by the same elapsed proper time of the rock whose worldline is along that arm), then the arm corresponding to the time axis will look shorter than the other, not longer.

Oh right, I made a mistake. The point that I am trying to make is that the structure changes depending on which rock gets the time axis.

 

[PeterDonis]

I don't know how it can be either, but not both.

Because "traveling through time and not space" is frame-dependent. Each one, to make his claim that he is only traveling through time and not space, has to choose his own rest frame. But those frames are different. There is no frame in which they are both traveling through time and not space.

The point that I am trying to make is that the structure changes depending on which rock gets the time axis.

How the V looks changes. But the actual, physical lengths of its arms do not.

 

[jbriggs444]

The point that I am trying to make is that the structure changes depending on which rock gets the time axis.

The structure does not change. The description changes. One can describe all three rocks from the point of view of the inertial rest frame of any of the three rocks. Or from any other inertial frame not associated with any of the rocks.

There is a three dimensional space from which one can choose inertial frames that all correctly describe the three rock system.

All of the descriptions work. None of these descriptions are any more correct than any other.

 

[PeterDonis]

Thread closed for moderation.