Why are dimensions always at right angles?

[Sangam Swadik]

I asked my teacher , i didnt get any satisfactory answers , can u tell me why dimensions are always at right angles .

 

[Giant]

Not true at all. Dimensions aren't always at right angles. They can be whatever they want.
Not true at all. Dimensions aren't always at right angles. They can be whatever they want.
We *choose* 90 degrees because a lot of problems are avoided by doing so. For example, assume that 'a' is the angle which the axes make, then the distance formula changed to d = square_root((x2 - x1)^2 + (y2 - y1)^2 + 2*(x2 - x1)*(y2 - y1) * cos(a)); Now by choosing a=90 degrees we eliminate the cos term and get a beautiful formula, instead of an overcomplicated one. Similarly a LOT of trigonometric terms are avoided, sin(90) becomes 1 as well. If complications don't arise and instead stuff gets simplified, mathematicians do indeed take axes which are inclined at an angle.

 

[Dr. Courtney]

Your assumption is in error. There are skew coordinate systems. See:

https://en.wikipedia.org/wiki/Skew_coordinates

https://en.wikipedia.org/wiki/Orthogonal_coordinates

https://en.wikipedia.org/wiki/Curvilinear_coordinates

Orthogonal coordinate systems are in wider use, because they are easier to use for most problems.

 

[HallsofIvy]

I will also point out that "dimension" is the wrong word here. "Dimension" is a number; what you are calling "dimensions" are the coordinate axes. As others say here, they are not necessarily at right angles. They are often chosen to be at right angles because that simplifies the calculations.

 

[Hornbein]

I asked my teacher , i didnt get any satisfactory answers , can u tell me why dimensions are always at right angles .

There all all sorts of spaces. In some of those spaces, dimensions are not at right angles. But the great majority of people are not interested in these, because in our Universe the physical dimensions ARE at right angles. This is true of any space with the Euclidian metric of a^2 + b^2 + c^2 = d^2, where d is the distance between two points and a, b, and c are at right angles to one another.

Some responders are trying to tell you that the right angles are just a convention and convenience. It is possible to use coordinate axes that are not at right angles, but as far as I know no one actually does this in the real world. The real world really does have this right angle character.

 

[FactChecker]

The real world really does have this right angle character.

I'm afraid I have to disagree. The real world does not measure things in coordinate systems. People do. The advantage of measuring coordinates in right angles is so great that there are few exceptions. One example is magnetic North/South versus polar North/South. Magnetic North does not line up with Polar North. So if you go in the magnetic North direction, you are (usually) also moving in the East/West direction. Because of this, Latitude / Longitude measurements are not based on magnetic North even though that is the easiest thing to determine in the "real world". But humans made that decision, not the "real world".

 

[Hornbein]

I'm afraid I have to disagree. The real world does not measure things in coordinate systems. People do. The advantage of measuring coordinates in right angles is so great that there are few exceptions. One example is magnetic North/South versus polar North/South. Magnetic North does not line up with Polar North. So if you go in the magnetic North direction, you are (usually) also moving in the East/West direction. Because of this, Latitude / Longitude measurements are not based on magnetic North even though that is the easiest thing to determine in the "real world". But humans made that decision, not the "real world".

The OP asked about dimensions, not coordinate systems.

 

[micromass]

The OP asked about dimensions, not coordinate systems.

Well, dimensions are just numbers. So dimensions being at right angles makes no sense really.

 

[Hornbein]

Well, dimensions are just numbers. So dimensions being at right angles makes no sense really.

A mathematician can say that dimensions are just numbers related in arbitrary ways. That's valid, but physicists deal with real objects and measurements inside of a real universe that is overwhelmingly preferential to 3 dimensions of space and one of time, all at right angles to one another. In the context of physicsforums such may be assumed.

 

[pwsnafu]

A mathematician can say that dimensions are just numbers related in arbitrary ways. That's valid, but physicists deal with real objects and measurements inside of a real universe that is overwhelmingly preferential to 3 dimensions of space and one of time, all at right angles to one another. In the context of physicsforums such may be assumed.

This is the mathematics section of the website, hence people are answering in the context of mathematics. By that logic do you assume that the Earth Sciences forum or the Computing forum are also physics discussions just because the website is named Physics Forums?

 

[pbuk]

When dealing with "real objects", in what sense does time form an angle of 90 degrees with any of the dimensions of space?

 

[Giant]

When dealing with "real objects", in what sense does time form an angle of 90 degrees with any of the dimensions of space?

He's talking about relativity. It assumes a 4 dimensional space where time is another dimension. Time dilation tell us that time doesn't flow at equal rate depending on your velocity and gravitational potential. So time rate has to be measures. This is simplified by choosing a 4 dimensional space. Time does indeed flow at different rates. Experiments have been done and it's verified.

 

[micromass]

He's talking about relativity. It assumes a 4 dimensional space where time is another dimension. Time dilation tell us that time doesn't flow at equal rate depending on your velocity and gravitational potential. So time rate has to be measures. This is simplified by choosing a 4 dimensional space. Time does indeed flow at different rates. Experiments have been done and it's verified.

OK, how do you measure 90° angles in GR?

 

[Hornbein]

This is the mathematics section of the website, hence people are answering in the context of mathematics. By that logic do you assume that the Earth Sciences forum or the Computing forum are also physics discussions just because the website is named Physics Forums?

You've got a point there.

The question makes sense from the point of physics but not from the mathematics side. So I'd charitably assume that he was asking from the physics point of view.

 

[Hornbein]

When dealing with "real objects", in what sense does time form an angle of 90 degrees with any of the dimensions of space?

It isn't 90 degrees, but it is orthogonal. The Minkowski metric is x^2+y^2+z^2+it^2, which is pretty similar to the Euclidean metric. So one may loosely think of the angle as begin 90 degrees.

But I'm sure the OP wasn't asking about that.

 

[micromass]

It isn't 90 degrees, but it is orthogonal. The Minkowski metric is x^2+y^2+z^2+it^2, which is pretty similar to the Euclidean metric. So one may loosely think of the angle as begin 90 degrees.

But I'm sure the OP wasn't asking about that.

Are you sure the imaginary number $i$ should be in there?

And what about vectors for which this metric is zero. Should you interpret it as being orthogonal on itself?

 

[Hornbein]

Are you sure the imaginary number $i$ should be in there?

And what about vectors for which this metric is zero. Should you interpret it as being orthogonal on itself?

Yep. Albert himself sometimes used that notation. It can also be written as x^2+y^2+z^2-t^2. So on second thought, it should have been x^2+y^2+z^2+(it)^2 with standard operator precedence.

The lines for which the metric is zero are null lines. These are the lines traveled by light in a vacuum. As far as light is concerned, it takes zero proper time to travel anywhere in a vacuum. So it is a pseudometric.

If you are interested, you are invited to look up Minkowski spacetime. I'm sure that there are many others who can explain this better than can I. If that isn't enough, there is the Wick rotation.

 

[micromass]

Yep. Albert himself sometimes used that notation. It can also be written as x^2+y^2+z^2-t^2. So on second thought, it should have been x^2+y^2+z^2+(it)^2 with standard operator precedence.

The lines for which the metric is zero are null lines. These are the lines traveled by light in a vacuum. As far as light is concerned, it takes zero proper time to travel anywhere in a vacuum. So it is a pseudometric.

If you are interested, look up Minkowski spacetime. I'm sure that there are many others who can explain this better than can I.

I know Minkowski spacetime. I just disagree you can meaningfully talk about angles and orthogonality there. And when you move to GR it becomes even more problematic.

 

[pbuk]

It isn't 90 degrees, but it is orthogonal.

Only if you choose a particular definition for the inner product. Coordinate systems do not need to be orthogonal, but they do need to be linearly independent.

Which, getting back to the OP, is the point - it is meaningless to talk about dimensions being at right angles, however the coordinate systems we choose to describe position in any geometry are usually more convenient if their bases are orthogonal which, in the case of spatial coordinates, does imply they are at right angles.

 

[Cruz Martinez]

A mathematician can say that dimensions are just numbers related in arbitrary ways. That's valid, but physicists deal with real objects and measurements inside of a real universe that is overwhelmingly preferential to 3 dimensions of space and one of time, all at right angles to one another. In the context of physicsforums such may be assumed.

How does this make any sense if we are to accept mathematics as the language of physics? The dimension of spacetime (Minkowski a special case) is defined as the number of parameters needed to specify an event biunivocally in a sufficiently small neighborhood. So the dimension is 4, nothing more, how is it meaningfull to mention angles here? Also, a mathematician would not say dimensions are ''numbers related in abitrary ways''.

 

[Giant]

OK, how do you measure 90° angles in GR?

Ok I thought orthogonal meant 90 degrees. Sorry about that. But orthogonality test would be dot product? I'm not well read about relativity so I won't make further arguments.

 

[weirdoguy]

Albert himself sometimes used that notation.

But physicists don't use it for several decades now, it's problematic - it's been discussed in Relativity forum multiple times.

 

[Mark Harder]

Ok I thought orthogonal meant 90 degrees. Sorry about that. But orthogonality test would be dot product? I'm not well read about relativity so I won't make further arguments.

Yes, two vectors are orthogonal if their dot product is zero, by definition. But orthogonal and 90° need not be the same. For one thing, the angle refers to a geometric situation, where the unit vectors are directions in space. Angles have no meaning in other abstract notions of vector spaces. Secondly, one could define different types of "dot" products, for which orthogonal directions are not 90°. A physical and geometric example of this would be a crystal. As you might guess from looking at different minerals that have different shapes, quartz and salt (halite) for example, not all crystals have natural coordinate systems in which the axes meet at 90°. The periodic structures within the crystal, known as unit cells, consist of just enough atoms or molecules so that when the unit cell is translated in 3 directions by the lengths of the unit cell sides in those directions, the entire crystal is generated; and these 3 directions need not be at 90° to each other. It's analogous to periodic functions, like trig functions, that repeat themselves over every period along the axis of the independent variable.

 

[Giant]

Yes, two vectors are orthogonal if their dot product is zero, by definition. But orthogonal and 90° need not be the same. For one thing, the angle refers to a geometric situation, where the unit vectors are directions in space. Angles have no meaning in other abstract notions of vector spaces. Secondly, one could define different types of "dot" products, for which orthogonal directions are not 90°. A physical and geometric example of this would be a crystal. As you might guess from looking at different minerals that have different shapes, quartz and salt (halite) for example, not all crystals have natural coordinate systems in which the axes meet at 90°. The periodic structures within the crystal, known as unit cells, consist of just enough atoms or molecules so that when the unit cell is translated in 3 directions by the lengths of the unit cell sides in those directions, the entire crystal is generated; and these 3 directions need not be at 90° to each other. It's analogous to periodic functions, like trig functions, that repeat themselves over every period along the axis of the independent variable.

Wow. crystal example makes it a lot clear. Thanks thanks!

 

[suremarc]

I suppose that spacetime is "orthogonal" in the sense that the metric tensor has a smoothly varying orthogonal eigenbasis. But that's true for any pseudo-Riemannian manifold, so it's more a statement about a class of topological structures than the physical world.

 

[jerromyjon]

Don't the Pythagorean Theorem and calculus depend upon a right angle? Or should I say rely on...

 

[WWGD]

Don't the Pythagorean Theorem and calculus depend upon a right angle? Or should I say rely on...

Riemannian metrics of the form $adx^2+bdy^2+cdz^2$ where $a,b,c$ not all $1$ refer to, or depict non-Euclidean geometries, or geometries where

$a^2+b^2=c^2$ does not quite hold. But, yes, in the Calculus done in Euclidean space, the generalized Pythagorean theorem does hold. But the Pythagorean theorem does not hold in general when your Calculus is done on a general manifold.

 

[Sangam Swadik]

OK, how do you measure 90° angles in GR?

https://en.wikipedia.org/wiki/Relativistic_Doppler_effect

i had asked the question from both , maths and physics perspective , do u know a good book where i can learn , analytical geometry , perfectly . ??

 

[Mark Harder]

Riemannian metrics of the form $adx^2+bdy^2+cdz^2$ where $a,b,c$ not all $1$ refer to, or depict non-Euclidean geometries, or geometries where

$a^2+b^2=c^2$ does not quite hold. But, yes, in the Calculus done in Euclidean space, the generalized Pythagorean theorem does hold. But the Pythagorean theorem does not hold in general when your Calculus is done on a general manifold.

That looks like a generalized inner product using a quadratic form:
d×(x,y,z)⋅((a,0,0),(0,b,0),(0,0,c))⋅(x,y,z)^T, where (...) indicates a row matrix and ((...),(...)) is a matrix, in this case a diagonal matrix. Is this the right way to think of the metrics you refer to?

 

[WWGD]

That looks like a generalized inner product using a quadratic form:
d×(x,y,z)⋅((a,0,0),(0,b,0),(0,0,c))⋅(x,y,z)^T, where (...) indicates a row matrix and ((...),(...)) is a matrix, in this case a diagonal matrix. Is this the right way to think of the metrics you refer to?

Yes, you can see it that way; this is the Riemannian metric tensor, a quadratic form , often represented as a matrix $g(X_i, X_j)$ where the {$X_i$} is a basis for the tangent space. The Riemannian metric tensor associated with the standard Euclidean metric is, like you said, the identity. The Riemannian metric is an inner product, defined on tangent vectors at each point in a manifold.

 

[Mark Harder]

Yes, you can see it that way; this is the Riemannian metric tensor, a quadratic form , often represented as a matrix $g(X_i, X_j)$ where the {$X_i$} is a basis for the tangent space. The Riemannian metric tensor associated with the standard Euclidean metric is, like you said, the identity. The Riemannian metric is an inner product, defined on tangent vectors at each point in a manifold.

I occurred to me after I wrote that post that by dx², you meant the the square of the differential of x, not multiplication by the scalar, d. Is that correct? And so, as you imply, the vector of differentials, (dx,dy,dz) is a basis for the tangent space at (x,y,z)? Thanks for your help. I only know enough differential geometry to be dangerous, as they say...

 

[WWGD]

I occurred to me after I wrote that post that by dx², you meant the the square of the differential of x, not multiplication by the scalar, d. Is that correct? And so, as you imply, the vector of differentials, (dx,dy,dz) is a basis for the tangent space at (x,y,z)? Thanks for your help. I only know enough differential geometry to be dangerous, as they say...

No problem, feel free to ask, I will do my best, and ask for clarification if needed.
Formally, $dx^2$ is $dx\otimes dx$, the tensor product of dx with itself. And dx,dy,dz are usually define as covectors, as a basis dual to the basis $\partial/\partial x, \partial / \partial y, \partial /\partial z$ of basis tangent vector fields meaning $dx( \partial /\partial x) =1 , dx (\partial /\partial y)=0 , i.e., dx_i (\partial / \partial_j):= \delta^i_j$ , etc. This means that linear operators on tangent vectors are written as $a(x,y,z)dx+b(x,y,z)dy+c(x,y,z)dz$ (notice that in the expression for your metric , you can have "mixed terms" $dx \otimes dy$ , etc., but this is a somewhat-simplified version) , i.e., as linear combinations of these covector basis elements. dx, dy, dz are differential forms, i.e., linear maps defined on tangent vectors. This allows you to compute the length of curves.
Maybe more precisely, this metric or second fundamental form, and it allows you to find the length
of a parametrized curve under a given choice of metric. Here $dx \otimes dx (v_1,v_2):=dx(v_1)dx(v_2)$ , meaning scalar multiplication, for vectors $v_1, v_2$. This allows you to define forms on n-ples of vectors ( in this case, on pairs of vectors ), i.e., given linear maps $dx, dy$ defined on vectors $v_1, v_2$ , you can form a bilinear map $dx \otimes dy$ defined on the ordered pair of vectors $(v_1, v_2)$, and so on, i.e., you can define k-linear maps on ordered k-ples of vectors. The collection of linear, bilinear,..k-linear maps defined on the exterior product of a vector space is called the exterior algebra of the exterior product. This can be defined on (the exterior product of) any vector space, not just the tangent space , exterior powers of the tangent space.

EDIT: the idea is that the fundamental form gives you a "local" version of length, so that, when integrated gives you the length of a curve, in the same sense that you define , when you do Riemann integration, a length element which you integrate to find the overall length of a curve.

EDIT 2: As you said, the metric is a quadratic form, defined on pairs of vector fields. In the differential
geometry version, a quadratic form is usually called a 2-form, which is a bilinear form ( meaning linear
separately in each variable ) defined on a pair of vector fields.

EDIT3: The post may have sprawled out of control, feel free to ask for clarifications.