Quantifying the Physical Property of Direction.

[Mordred]

Why would you need to do the above?

 

Let's take a vector field for example. You don't care which direction the "current" is travelling you only care what direction the stick moves compared to that current.

(Let's assume the stick floats)

 

You don't need to have a specific coordinate system. The above can be accurately described in 2d, 3d or 4d coordinates.

 

This is one of the uses of tensors which describes vector and scalar relations independant on a coordinant system.

 

Here is a good example similar enough to your above post.

 

http://www.feynmanlectures.caltech.edu/II_31.html

 

For example the coordinate system you described works well in Euclidean space or Minkowskii (Special relativity). However it is inadequate in curved polar coordinate metric systems such as The Schwartzchild metric.

 

Or for example some of the dimensions in string theory where the coordinate metric is a rotating vector field. (Which is an application of circular vectors Studiot mentioned above)

PS a vector isn't restricted to 2d. You can have n dimensions in a vector component.

Edited May 16, 2016 by Mordred

[swansont]

 

Even properties that aren't intrinsic, such as momentum and energy, are frame dependent, not coordinate system dependent. Indeed, the behavior of physics (as Mordred mentioned) can't depend on these choices. There is no preferred axis.

[Mordred]

It might help to better understand Studiots post if you look at the four theorems in vector calculus.

 

http://mathinsight.org/fundamental_theorems_vector_calculus_summary

 

The fundamental theorems are:

the gradient theorem for line integrals,

Green's theorem,

Stokes' theorem, and

the divergence theorem.

 

The hyperlinks will give a basic intro into each

Edited May 16, 2016 by Mordred

[steveupson]

Why would you need to do the above?

 

 

 

 

This is my question. I honestly need help on this. I guess anther way to go about it is to ask if you can explain to me how one would go about generating a nine quantity tensor for the model?

 

The way that it looks to me, there is no 'information' about how these two directions relate to one another. Sure, you can pick some arbitrary reference frame, but how would you know that these two directions are 45 degrees to one another, other than by just declaring that it's so?

 

The function is what makes them 45 to one another, not their orientation in any reference frame.

 

In other words, without knowing how either axis is oriented in space, we know they are 45 to one another if they satisfy the equation.

[Mordred]

Well it all boils down to coordinate and metric choice.

Take for example a straight line in two coordinate systems

In Cartesian coordinates its easy to visualize a straight line.

However in polar coordinates this isn't so easy. A good example being the path of a photon following a null geodesic.

 

To answer how a direction is determined you must first understand the coordinate system in use as well as the coordinate position.

 

For example describing direction in 4d with components ct,x,y,z isn't as straightforward as simply describing a direction using just spatial components x,y,z.

 

Either way there is no hard and fast choice of coordinate system, nor axis. The choice is made in the chosen reference point/frame and coordinate system your using.

Direction must have a comparison factor. It is never the same for all observers performing the measurement.

 

No two observers will measure the same direction unless they use the same reference point and coordinate system.

 

For example I can choose to say the vector field is moving in the positive x direction. Then compare object a to that field. However this is a choice. As long as the comparison correctly describes the direction compared to the reference system then it's accurate. I could just as easily state the vector field flows north. Then compare.

I don't know how well your math skills are but here is some of the math involved in converting Cartesian to polar coordinates.

 

http://www.brown.edu/Departments/Engineering/Courses/En221/Notes/Polar_Coords/Polar_Coords.htm

Edited May 16, 2016 by Mordred

[steveupson]

Well it all boils down to coordinate and metric choice.

 

 

What metric choice is chosen in the new model?

 

Also, in my last post I asked how one would show me how to generate the tensor for the new model. Even though I followed your explanation, it didn't seem to respond to my question. I still don't know how one would do it. Can you show me?

 

I sincerely appreciate your patience and you effort on this. It seems like we are talking past each other. If you could actually answer either of these two questions for me it would be extremely helpful in aiding my understanding. Understand, I don't want you to tell me how to do it, I want you to show me.

 

Can you do it for me while I follow along with you?

 

Or, it could be that you've taken me as far as you can using this approach for now. Maybe another approach would make a lot more sense to me. If I were to publish the Mathematica .cdf file again (it was previously published at Wolfram Community but then taken down without any explanation) perhaps someone here with skills that I don't possess can plot the function for us.

 

If it were plotted for 45 degrees, then 44, and so on, and if those results were plotted together, then we could all sit around looking at those plots and wax philosophic about what, exactly, we're looking at. Someone might already know, once they see the graphic representation.

[Mordred]

The last post gives you some of the tools on tensor conversion. Look through the link.

[steveupson]

The last post gives you some of the tools on tensor conversion. Look through the link.

 

Correct.

 

My problem is that even though I have the tools, I don't know which end of the shovel to hold.

 

Can you show me?

 

 

On edit>

 

It's an equality.

 

It's an equality.

 

It's an equality.

 

It's an equality.

 

.....

 

Everyone seems to be saying that there is another way to express this equality using vectors.

 

Can anyone express this equality using vectors?

 

 

and being passing familiar with the number line -- I'm sure if I don't

say passing familiar somebody here will say, "You have to know vector

tensor shmelaculus in 15 triad synergies to really understand the

number line. It's not even called that, it's called the real

torticular space." or something to that effect --

 

 

http://www-m10.ma.tum.de/foswiki/pub/Lehre/ProjektiveGeometrieWS0607/chap5.pdf

 

 

Do we know what an equality is?

Edited May 16, 2016 by steveupson

[Mordred]

Ok well the first question you have to ask is what property is invariant of a vector?

In other words what property will remain the same regardless of coordinate rotation or point of origin?

 

Remember vectors has two components magnitude and direction.

 

No I am not going to simply apply your example above. Instead I'll help guide you to the above.

 

This question alone should provide a clue, In regards to your example above. In particular you've chosen a coordinate system. Yet ignore the arbitrary choices involved.

 

 

Which does not describe the purpose and functionality of a tensor.

Edited May 16, 2016 by Mordred

[steveupson]

 

No I am not going to simply apply your example above. Instead I'll help guide you to the above.

 

 

At some point I tire of the game.

 

If you have derived the function and if you can express the new equality using algebraic terms, then why not just tell us all?

 

If you can express the new equality using vectors, even better.

 

Why the mystery?

 

First I was wrong, now I'm pigheaded? Am I boring you?

 

I don't think this is going to work this way.

 

 

 

Why don't we simply find someone who is willing to graph the function?

 

 

 

I am unable to upload the file containing the model, but perhaps someone has a suggestion about how it can be published here so that anyone who is interested can have a stab at it.

[Mordred]

The problem is your ignoring that tensors work regardless of coordinate system or reference point.

 

Direction requires a reference point and a reference coordinate system.

 

The only property of a vector that is invariant is the magnitude. Not the direction.

There is a difference between Cartesian, polar, spherical vector to tensor conversions. The reason being not due magnitude but due to coordinate choice.

 

Anytime you want to convert to tensors the first question one needs to ask is what doesn't change regardless of coordinate transformation or position.

 

What remains invariant.

 

For example take a fan blade. If you move the fan blade does the shape of the fan blade change? Answer is no, so you can already mathematically describe that fan blade.

 

The other step is to describe its new position and orientation. Regardless you have symmetry. (The fan blade shape)

Edited May 17, 2016 by Mordred

[steveupson]

Direction requires a reference point and a reference coordinate system.

 

 

 

Nothing else is in question. I agree wholeheartedly with all the other claims in your post.

 

 

If we express direction as an equality, as an equation, then we don't need any coordinate system, do we?

 

The reason we need the coordinate system is so that we can express the relationship between our reference and our direction. If this relationship is expressed as an equality instead, haven't we accomplished something completely different?

[Mordred]

Take for example player a throws a ball to player b. You can describe The path the ball takes in two equally accurate descriptions.

 

1) the ball is moving away from player a.

2) the ball is moving toward player b.

 

In this example alone we see direction is not invarient.

How do you express a direction as an equality? Good luck

 

You can express the magnitude of change as an equality not the direction

Edited May 17, 2016 by Mordred

[steveupson]

How do you express a direction as an equality? Good luck

 

 

 

Aha!

 

I apologize for my impatience with you.

 

I thought that you must have read the OP where I link to exactly that!

 

Evidently no.

 

I originally had the .gif displayed in the OP, but the moderators removed it, so as not to offend anyone here, I imagine.

 

Go back the OP and click the link, then I think we'll be on the same page.

 

 

On edit> better yet, go here to see it all on the same page.

 

http://www.thephysicsforum.com/mathematics/9252-defining-new-function.html

Edited May 17, 2016 by steveupson

[Mordred]

Yeah so? The model you posted simply shows rotation symmetry. ( note the model uses the same coordinate system and has the same point of reference, including starting point) not so in defining direction as an equality.

 

 

Observer a measures a ball moving at 20 degrees from his position but another observer can measure a different angle. Both can be correct.

 

Which is precisely what I am trying to get you to understand.

 

Here perhaps this will help

 

http://www.google.ca/url?q=http://www.springer.com/cda/content/document/cda_downloaddocument/9783642329579-c1.pdf%3FSGWID%3D0-0-45-1365411-p174888679&sa=U&ved=0ahUKEwif87rM8N_MAhUC8GMKHUSpBhUQFggcMAM&sig2=dfm5M-If42-scIMIpTsZfw&usg=AFQjCNE5MgmQRPy6mH8FlqwpGhWzeaspHQ

 

It is a basic coverage of vector to tensor conversions.

Edited May 17, 2016 by Mordred

[steveupson]

Yeah so? The model you posted simply shows rotation symmetry.

 

Which is precisely what I am trying to get you to understand.

 

Here perhaps this will help

 

http://www.google.ca/url?q=http://www.springer.com/cda/content/document/cda_downloaddocument/9783642329579-c1.pdf%3FSGWID%3D0-0-45-1365411-p174888679&sa=U&ved=0ahUKEwif87rM8N_MAhUC8GMKHUSpBhUQFggcMAM&sig2=dfm5M-If42-scIMIpTsZfw&usg=AFQjCNE5MgmQRPy6mH8FlqwpGhWzeaspHQ

 

It is a basic coverage of vector to tensor conversions.

 

 

I think that I already understand (sort of) what you are trying to get me to understand.

 

Take a closer look.

 

What is show is not rotation symmetry, it's rotation synchronicity.

 

In two dimensions there exists a condition which causes translation-invariance. In three dimensions, this condition becomes translation-synchronous in much the same way, but with much different effect.

 

I think you should see the equality where alpha is a function of E. This is true for every E. What we've shown is the equivalence of a triangle in plane geometry. Only here, as our relationship between the cardinal and ordinal directions change, the the relationship where alpha is a function of E also changes accordingly.

 

This equality quantifies direction in a completely different way than how it is quantified using plane geometry.

[Mordred]

That part is correct. However if you follow the math in the last link were doing the same thing.

 

On any vector the unchanging portion is the magnitude of the vector. To translate The direction element requires showing the start position and coordinate change according to the chosen coordinate system and reference frame.

 

For example a clock handle has rotation symmetry as it changes angles. However you can remove the clock handle and place it somewhere else in arbitrary space at the same angle and have translational symmetry.

 

Though I see nothing in that model you posted that hints at quantifying direction as a property.

Edited May 17, 2016 by Mordred

[steveupson]

That part is correct. However if you follow the math in the last link were doing the same thing.

 

On any vector the unchanging portion is the magnitude of the vector. To translate The direction element requires showing the start position and coordinate change according to the chosen coordinate system and reference frame.

 

For example a clock handle has rotation symmetry as it changes angles. However you can remove the clock handle and place it somewhere else in arbitrary space at the same angle and have translational symmetry.

 

 

I don't see how it can be the same, mathematically, since vectors and equations respond differently to Lorentz transformations.

[Mordred]

 

 

I don't see how it can be the same, mathematically, since vectors and equations respond differently to Lorentz transformations.

That involves coordinate transformation rules. Ie Cartesian to polar coordinates. Not all transformations are symmetric some are antisymmetric. Edited May 17, 2016 by Mordred

[steveupson]

That involves coordinate transformation rules. Ie Cartesian to polar coordinates. Not all transformations are symmetric some are assymetric.

 

Correct, and equations are what? Unchanged.

[Mordred]

 

Correct, and equations are what? Unchanged.

What do you mean?

[steveupson]

What do you mean?

 

If this is a scalar quantity, then isn't it invariant? I honestly don't know the answer. This might be what I'm not understanding.

[Mordred]

Ah ok certain properties remain invariant under Lorentz boosts.

 

One property is the speed of light.

The second being the laws of physics.

 

Certain transformations will be considered invariant here is a good explanation.

 

According to Einstein's principle of relativity all systems moving with a uniform velocity are equivalent, the laws of physics must obey the same equations in e.g., system S and system [latex]\prime{S}[/latex]. Compare rotations in three

dimensions. The laws of physics are independent from the position of one coordinate system with respect to the other. Physical quantities are described by scalars, vectors

and tensors, and the physical laws are given by combination of those quantities. Since the physics is independent from the position of the coordinate system, the form of the

equation is the same in each coordinate system.

 

For example the length of a vector in four vector or four momentum is invariant.

 

http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/vec4.html

 

In a vector the length is the scalar quantity.

Another good example is the Maxwell equations are invarient to Lorentz boosts.

 

Another key example is the rest mass of a particle is invariant. Hence rest mass has been replaced by invariant mass. Inertial mass however is not invariant.

[steveupson]

In a vector the length is the scalar quantity.

 

 

In the new function, the direction is the scalar quantity.

[Mordred]

Good luck on making direction invariant. That's a handwave if I ever saw one.

Edited May 17, 2016 by Mordred