A little math quiz

[QuarkHead]

I confess I remain confused about $$ \mathbb C \otimes \mathbb C $$.
In $$ \mathbb C^2 $$ inner products and tensor products describe entirely different objects.

Now I am confused. Why do you jump from $$\mathbb{C} \otimes \mathbb{C}$$ to $$\mathbb{C}^2$$? Surely you are not under the impression they are interchangeable?

$$\mathbb{C}^2$$ is defined by $$\mathbb{C} \times \mathbb{C}$$.

Just to remind you, the inner product is a scalar - a number - whereas the tensor product is, well, umm, a tensor.

In the first case we take the sum of products of vector components in an ordered form - for $$v=\sum\nolimits_{j}^n\alpha^j e_j$$ and $$w= \sum\nolimits_{k}^n\beta^k e_k$$, the inner product is $$\sum\nolimits_i^n\alpha^i\beta^i$$.

Whereas the tensor product is the sum of unordered products of components, viz. $$\sum\nolimits_{j,k}\gamma^{jk}e_j \otimes e_k$$. If you are familiar with matrix multiplication, you will see this can only be achieved when the first factor in the product is transposed. rpenner showed you this

 

[someguy1]

If you are familiar with matrix multiplication, you will see this can only be achieved when the first factor in the product is transposed. rpenner showed you this

Hi there QH. Isn't this the other way 'round? That is, if we have column vectors $$u, v \in \mathbb{R}^n$$, then the inner (or scalar) product is given by the matrix multiplication $$u^Tv$$. And if $$u \in \mathbb{R}^n$$ and $$v \in \mathbb{R}^m$$, then the outer (or tensor) product is given by $$uv^T$$. In the latter case, the outer product is an $$n \times 1$$ column vector (on the left) times a $$1 \times m$$ row vector, which results in an $$n \times m$$ matrix. Such a matrix represents a linear transformation from $$\mathbb{R}^m$$ to $$\mathbb{R}^n$$, in other words it goes back the other way. In math I believe this is called contravariant, although my understanding is that physicists call that covariant. I may be confused about this.

That's my understanding, can you please confirm or supply corrections?

I'm currently taking a Coursera MOOC on Galois theory so tensor products are heavily on my mind this week. I'm following this thread with great interest and limited understanding. It seems like a long way from the engineering/physics perspective of tensors as representing forces to the abstract algebra perspective of a tensor product as a universal bilinear gadget. I'm trying to get a handle on at least the mathematical end of this.

 

[QuarkHead]

Hi there QH. Isn't this the other way 'round? That is, if we have column vectors $$u, v \in \mathbb{R}^n$$, then the inner (or scalar) product is given by the matrix multiplication $$u^Tv$$. And if $$u \in \mathbb{R}^n$$ and $$v \in \mathbb{R}^m$$, then the outer (or tensor) product is given by $$uv^T$$. In the latter case, the outer product is an $$n \times 1$$ column vector (on the left) times a $$1 \times m$$ row vector, which results in an $$n \times m$$ matrix.

Eeek!! You are right, I was wrong. If you are feeling charitable you could say I mispoke myself. Otherwise throw me to the pit with the abuse mongers here.....

In math I believe this is called contravariant, although my understanding is that physicists call that covariant. I may be confused about this.

As well you might be - my advice, for the reason you gave, avoid these two "c-words" altogether!

 

[arfa brane]

I think one of my difficulties with the math is that I started in $$ \mathbb C^2 $$.

QuarkHead has told us that $$ \otimes $$ is an operation (operator) that gives the outer product. If either of the two elements it operates on is a scalar, the result should not change the number of dimensions so that $$ \mathbb R \otimes \mathbb R \cong \mathbb R $$. My experience with this operator is with vectors and matrices as the Kronecker product, and the result is always a matrix with more dimensions.

For instance, just using it with a standard basis for $$ \mathbb R^2 $$, we have: $$ \begin{pmatrix} 1 \\ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix} $$.

Moreover (I have to stop using that word), $$ \mathbb C^2 $$ has an inner product, an outer product, and tensor products.
So, yes, so sorry, I am A Bit Confused.

 

[arfa brane]

. . . we can make VC into a complex vector space by defining complex multiplication as follows:

. . . every vector v in VC can be written uniquely in the form

Suppose we let $$ V = \mathbb R^2 $$, so we have say, $$ \alpha (\begin{pmatrix} a \\ b \end{pmatrix} \otimes \beta) = \begin{pmatrix} a \\ b \end{pmatrix} \otimes (\alpha \beta) $$, $$ \begin{pmatrix} a \\ b \end{pmatrix} \in \mathbb R^2$$, and $$ \alpha, \beta \in \mathbb C $$.

 

[someguy1]

If you are feeling charitable you could say I mispoke myself. Otherwise throw me to the pit with the abuse mongers here..... As well you might be - my advice, for the reason you gave, avoid these two "c-words" altogether!

Oh I'm sure you misspoke yourself. I'm just trying to get my mind around tensors and tensor products. They're very much like the blind men and the elephant. So many different perspectives.

For instance, just using it with a standard basis for $$ \mathbb R^2 $$, we have: $$ \begin{pmatrix} 1 \\ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix} $$.

That makes sense. If $$\{e_i\}_{i=1}^2$$ and $$\{f_i\}_{j=1}^2$$ are each bases for two copies of $$\mathbb R^2$$, respectively, then the tensor product $$\mathbb R^2 \otimes \mathbb R^2$$ is the set of all finite linear combinations with scalars in $$\mathbb R$$ of the formal gadgets $$e_i \otimes f_j$$, where the scalar multiplication is subject to the rules of bilinearity: $$\alpha e_i \otimes f_j = (\alpha e_i) \otimes f_j,\ \alpha \in \mathbb R$$, etc.

Viewed this way it makes perfect sense that the tensor product of two real vector spaces each of dim 2 is a real vector space of dim 4; and that the tensor product of any two basis vectors is one of the four basis vectors in the tensor product.

 

[QuarkHead]

That makes sense. If $$\{e_i\}_{i=1}^2$$ and $$\{f_i\}_{j=1}^2$$ are each bases.

Look, I know I am a notation Nazi, but this really makes no sense.

$$\{e_i\}$$, say, is a set. Are you applying bounds to some operation on this set? If so, what is this operation?

 

[someguy1]

Look, I know I am a notation Nazi, but this really makes no sense.

$$\{e_i\}$$, say, is a set. Are you applying bounds to some operation on this set? If so, what is this operation?

Sorry if this wasn't clear. $$\{e_1, e_2\}$$ is a basis of $$\mathbb R^2$$ and $$\{f_1, f_2\}$$ is another. They're just bases of a finite-dimensional vector space.

In general if $$\{e_i\}_{i=1}^n$$ is a basis of $$\mathbb R^m$$ and $$\{f_i\}_{j=1}^m$$ is a basis of $$\mathbb R^n$$ then $$\mathbb R^n \otimes \mathbb R^m$$ is an $$nm$$-dimensional vector space with basis $$\{e_i \otimes f_j\}_{1 \leq i \leq n,\ 0 \leq j \leq m}$$. Is that notation ok?

(Now that you mention it, putting indexes outside of set brackets is not proper set notation, but people do this all the time).

The individual gadgets $$e_i \otimes f_j$$ are formal symbols with no meaning at all except that they're required to behave properly with respect to scalar multiplication by elements of $$\mathbb{R}$$.

Didn't you say something similar a few posts back? I seem to remember your writing a sum of scalars times the formal tensor gadgets.

Here's some wikipedia stuff that I can make some sense of, which might be germane to QuarkHead's posts

Can you link the specific Wiki page? This is exactly what I'm studying in class this week and I'm trying to hit this from every direction. Base change aka extension of scalars is a very simple-looking idea that magically leads to tensor products.

So, a complex number is an extension of scalars . . .? Or what mathematicians call a field extension?
And

No, two totally different things, only vaguely related.

A field extension is just a smaller field inside another. For example
$$\mathbb C$$ is a field extension of $$\mathbb R$$. Field extensions are used in understanding the structure of the roots of polynomials. For example in the rationals $$\mathbb Q$$ there is no root of $$x^2 - 2$$. However we can extend $$\mathbb Q$$ by defining $$\mathbb Q[\sqrt{2}] = \{a + b\sqrt{2} : a, b \in \mathbb R\}$$. Then we can show that $$\mathbb Q[\sqrt{2}]$$ is an extension field of $$\mathbb Q$$ in which $$x^2 - 2$$ does have a root.

That's field extensions.

Now, extension of scalars (which I'm just learning, so don't quote me here), also starts with one field inside another, like $$\mathbb R$$ inside $$\mathbb C$$.

Then say you have a vector space over the larger field, in this case $$\mathbb C^n$$. By "forgetting" about the scalars in $$\mathbb C$$, we can view $$\mathbb C^n$$ as a vector space over $$\mathbb R$$. That's reasonably straightforward, it's called restriction of scalars.

Extension of scalars goes the other way with a little more complication. Same setup, $$\mathbb R$$ inside $$\mathbb C$$, $$\mathbb C^n$$ a vector space over the larger field. Now, off to the side, we have some $$\mathbb R$$ vector space, say $$\mathbb R^n$$. It allows scalar multipication by elements of $$\mathbb R$$. Since $$\mathbb C$$ extends $$\mathbb R$$, can we somehow "extend the scalars" so that $$\mathbb R^n$$ becomes a vector space over $$\mathbb C$$?

In general, no.

But if we take some appropriate tensor product, we get a vector space in which we can extend the scalars. And it turns out that if we never heard of a tensor product and we tried to construct a solution to this problem, we would invent the tensor product.

This is what I know about it. I found a really good paper, actually it's a copy of a chapter of Dummitt and Foote here http://math.rice.edu/~evanmb/math465spring11/dummit_foote-tensorproducts.pdf. It describes very clearly and in nice detail this construction.

If I can tear myself away from the Internet for a few hours I'm going to work through this and hopefully come back with some more clarity.

 

[arfa brane]

someguy: --https://en.wikipedia.org/wiki/Complexification

 

[arfa brane]

Before I forget about it: since quantum interference implies that particles don't interact, because you get the same pattern with a beam of single particles as you do with lots of particles (where that means "in a cross section of the beam at any time"), hence coherence isn't relevant before the beam reaches the slits. This implies each particle interacts with itself as a wave between the slits and the screen.

No particles interact, but individual wavefunctions do interact with themselves; in a unitary space--the particle goes through one or the other slit, or the particle as a wavefunction goes through both slits, and becomes two interacting wavefunctions (entangled somehow with something or other).

 

[arfa brane]

I think it's important to note that Young's experiment and in fact any quantum interference experiment has a classical explanation. You can predict, for instance, where the centres of the maxima and minima will appear if you know the wavelength, and the geometry of your apparatus--you triangulate from the centre of the screen to the centre of each slit.
Now, classically the slits are sources of wavefronts which interfere according to a phase relation.

But where is this classical structure, when one electron will make/makes/has made a dot on a screen?
It must look the same for each electron, but the buildup of dots over time seems to be not just random, but maximally random. We might, as I tried to, imagine that each electron is a single wave front, or part of one; it has degrees of freedom, in three spatial dimensions, that means it can randomly choose a maximum to be moving towards (its direction of motion has been chosen already), and it can randomize where it sits relative to that centre in two more dimensions.

 

[arfa brane]

Back the the lab for a bit, where those spring constant measurements are waiting.
I wanted to demonstrate how, for any experiment, it's about what you know. It seems that, because you have a set of different weights with no indication what they weigh (in standard kilograms) there is information missing.

But you realise that since this means you will have to use your own choice of a unit of mass, you can choose your own unit of distance too. After all, metres are an arbitrary local choice of distance.
So, you next realise that you can use your unit of mass (let's make this the lightest weight), to define a unit of distance using one of the springs, and this unit distance can be subdivided using the marked ruler. Anyway, despite the lack of precision inherent, you should be able to determine the relative spring constants after defining a unit spring (one that extends a unit distance with a unit mass attached). And we're done. Sort of.

The spring constant of a spring is what gives it tension--pull one end of a spring with the other end fixed, and it pulls back with an equal (and opposite force). Springs are linear if they have the same dimensions along their whole length. Adding springs together is linear, as is connecting them in parallel, a lot like resistors in electronics.
So, if you shorten a spring (a linear operation, say) its spring constant will get bigger. Cut a spring in two and each part has twice the k of the original; join the two parts back together in series and the original k is restored.

So, you realise that if you can fix a spring anywhere along its length you can change the k, and then use the same spring to find a range of linear solutions for each spring over the set of weights (possibly overdoing things a bit).

 

[arfa brane]

To the results. What you find out is that, no matter how you cut the experiment, you get a plot of a set of distinct lines with mass vs spring length. All these lines have an x intercept, because the length of a spring isn't zero. And they should all intercept the same point on the 'y' axis, where y = mg. This is Newton's constant of gravitation showing up.

But now you need to know the radius and mass of the earth in terms of local constants of mass and distance (the ones you just used). Another day, perhaps. Knowing something must exist (and must be really bloody small), is also a solution. Sometimes.

 

[arfa brane]

The above experiment is about the difference between measuring the change in length (the distance the end of the spring moves), and measuring the total length of the spring. If you want to calculate the k for each spring you only need the changes in length, each plot should be a line through the origin whose slope is = k.

But you might have two springs the same length with different k values, in which case including the total length of each spring will give you two lines with the same x intercept but different y intercepts. To correct this you need to 'normalize' each spring length, which is a matter of dividing the length by k. So each line is of the form: y = kx + c, except for the lines representing the rest lengths x' which are x'/k. The constant c is negative and represents the negative force (normalized to your unit spring) required to compress the spring to zero length.


So you can consider the experimental apparatus as a system which is linear. You 'input' a mass value which is of course scalar, and the system 'outputs' a distance value, also a scalar, because of the physical relation mg = kx at equilibrium. To reach equilibrium, the system has to settle, or you have to wait for oscillations to die out, or ensure oscillations in the system don't start by damping accelerations, i.e. lower the weight by hand at constant velocity.

Likewise, in a double-slit experiment with particles, the system has inputs (identical particles with identical wavelengths), and outputs (dots on a screen), and constant velocities.

The first experiment is about linear response to an applied force, in a frame where accelerations are excluded. The second appears to be about non-linear responses or "randomness", the constant velocities of particles become part of an unseen (and undetectable) region of interference which is determined by the constant wavelengths--a region of "constant interference".

 

[arfa brane]

Back to the math quiz.
Why, if you fix one end of a reasonably loose spring, does the distance between successive coils vary over the length when you apply a force to the free end?? What do you get if you plot the sequence vs units of tension (assuming tension is equal everywhere in an otherwise homogenous spring)?

And here we go with the multi-choice:

The spring is more extended at the end the force is applied (if the force makes the spring longer), because:

1) the end closest to the applied force stores more potential energy than the fixed end, because forces are local.
2) if the spring is hanging vertically with the weight attached at the lower end, it's because there is more of a gradient in gravitational potential at the lower end.
3) it's because of tidal forces, but don't ask me to explain it.

 

[arfa brane]

M'kay, there's a differential separation between windings, increasing towards the weighted end of the spring.

Potential energy is given by $$ \frac {1} {2} kx^2 $$. This looks a lot like the integral $$ k \int{x}\, dx $$. But $$ dx $$ is a constant, though small distance, and there are different distances between windings, albeit ordered distances.

So, that kind of proves there is more potential energy in a winding near the weighted end of a spring, than in a winding near the fixed end. So 1) is correct more or less, the others are wrong because the spring extension can easily be shown to have nothing to do with gravity, only with an applied force. It means the whole spring is involved, which means you should (?) include the rest length (unless you only want to measure an unknown k).

It means the force is felt the most at the weighted end, and the least at the fixed end. A spring extended by a weight is a kind of frozen picture of acceleration "due" to gravity (depending on the spring having constant tension along its length). It's frozen because a weighted spring at equilibrium isn't a clock of any kind, and time isn't measured, you measure mass and distance and find a relation.

 

[BrianHarwarespecialist]

The above experiment is about the difference between measuring the change in length (the distance the end of the spring moves), and measuring the total length of the spring. If you want to calculate the k for each spring you only need the changes in length, each plot should be a line through the origin whose slope is = k.

But you might have two springs the same length with different k values, in which case including the total length of each spring will give you two lines with the same x intercept but different y intercepts. To correct this you need to 'normalize' each spring length, which is a matter of dividing the length by k. So each line is of the form: y = kx + c, except for the lines representing the rest lengths x' which are x'/k. The constant c is negative and represents the negative force (normalized to your unit spring) required to compress the spring to zero length.


So you can consider the experimental apparatus as a system which is linear. You 'input' a mass value which is of course scalar, and the system 'outputs' a distance value, also a scalar, because of the physical relation mg = kx at equilibrium. To reach equilibrium, the system has to settle, or you have to wait for oscillations to die out, or ensure oscillations in the system don't start by damping accelerations, i.e. lower the weight by hand at constant velocity.

Likewise, in a double-slit experiment with particles, the system has inputs (identical particles with identical wavelengths), and outputs (dots on a screen), and constant velocities.

The first experiment is about linear response to an applied force, in a frame where accelerations are excluded. The second appears to be about non-linear responses or "randomness", the constant velocities of particles become part of an unseen (and undetectable) region of interference which is determined by the constant wavelengths--a region of "constant interference".

I know I might be taking you of topic, but am giving you a hypothetical scenario, I want your mathematical opinion.

Suppose we are launching an object into space most likely a satellite. But instead of using rocket fuel to launch it we will use a giant spring. First what materials would you use to make the spring? We don't want the spring to calapse on its own weight.

Second is there a material available that the spring can be made of that will produce enough vertical or horizontal velocity to make it into orbit? At this point wheater or not the object can be stabalised into orbit is not relevant.

Are you able to make relevant calculations using the appropriate materials of your choice ? Just curious I just find this part interesting. As I was thinking about literal springs.

 

[QuarkHead]

Returning to mathematics......

someguy1 and I were talking about tensor products and field extensions. Lemme try and connect these 2 notions.

For the general case, suppose the field $$\mathbb{F}$$ is an extension for the field $$\mathbb{K}$$. Further suppose a Real vector space of dimension $$n$$ - I write $$V_n$$ to emphasize this fact.

I assert that then $$\mathbb{F} \otimes V_n \simeq \mathbb{F}^n$$

To see this, note first for a Real vector space, and the isomorphism $$V_n \simeq \mathbb{R}^n$$.

Set the Real space $$V_1 \equiv \mathbb{R}$$ and define the field extension $$\imath : \mathbb{R} \hookrightarrow \mathbb{C} \simeq \mathbb{R}^2$$.

Then clearly $$\mathbb{R}^{2n} \simeq \mathbb{C}^n$$. So write $$\mathbb{C} \otimes \mathbb{R}^n \simeq \mathbb{C}^n \simeq \mathbb{R}^{2n}$$

 

[rpenner]

M'kay, there's a differential separation between windings, increasing towards the weighted end of the spring.

Potential energy is given by $$ \frac {1} {2} kx^2 $$. This looks a lot like the integral $$ k \int{x}\, dx $$. But $$ dx $$ is a constant, though small distance, and there are different distances between windings, albeit ordered distances.

That's not right. A spring has a natural length, $$L_0$$ and a natural lowest point for potential energy at this length, which will equate with zero.

For a material of constant properties, a spring of half length is twice as stiff, so $$k L_0 = K$$ must be a microscopic material property.

Microscopically, because there is a certain amount of strain per element of the spring, $$\frac{L - L_0}{L_0}$$, then as this strain changes from zero, then each microscopic change in the length of the spring increases the potential energy per element by the force per element at a given strain, $$ K \epsilon$$, times the change in dimension of the element, $$d \epsilon$$, giving the total potential energy per element as $$\int \limits_{0}^{\frac{L - L_0}{L_0}} K \epsilon \; d \epsilon = \frac{1}{2} k L_0 \left( \frac{L - L_0}{L_0} \right)^2$$ and so integrating over each element of the spring, we get $$ \int \limits_{0}^{L_0} \int \limits_{0}^{\frac{L - L_0}{L_0}} K \epsilon \; d \epsilon \; d x_0 = \frac{1}{2} k \left( L - L_0 \right)^2$$.

$$\int \limits_{0}^{L_0} \int \limits_{0}^{\frac{L - L_0}{L_0}} K \epsilon \; d \epsilon \; d x_0 = L_0^2 k \int \limits_{0}^{\frac{L - L_0}{L_0}} \epsilon \; d \epsilon = k \int \limits_{L_0 - L_0}^{L - L_0} x \; d x = \frac{ k \left( L - L_0 \right)^2}{2} $$

So while macroscopically, typically the position free end of the spring is the most important part in describing the dynamics of a spring system, the forces and displacements that make up that potential energy are distributed throughout.

 

[arfa brane]

That's not right.

Yeah, I know. I had another think about it and decided it was bollocks. Dividing a spring by its constant won't do what I said either.
I actually did an experiment like this a long time ago, I remember the goal was to show that there was a constant involved. But I've forgotten the details, and I thought I could recreate them by considering the linear properties of springs.

The important part was that you get a y intercept, this must be a constant of integration.


About the distribution of potential energy in a weighted (simple) spring: the windings near the weighted end are farther apart than the windings near the fixed end.

If you pull down on the weight, the separation between windings increases. Now let the weight go, which part of the spring accelerates the most? The lowest part of the spring gets the most kinetic energy which suggests that part has the most potential energy.