I Relativistic hidden variable quantum mechanics?

[rubi]

The velocities are just vector fields, so they don't cancel what you think they do. My length integral is invariant in exactly the same way as the standard length integral in your derivation above, with a replacement $g_{\mu\nu}(x)\rightarrow G_{\mu\nu}(x)$ where
$$G_{\mu\nu}(x)=\frac{g_{\mu\nu}(x)}{V_{\alpha}(x)V^{\alpha}(x)}$$
You can think of my theory as a geometrical theory with a new effective metric $G_{\mu\nu}(x)$.

No I can't. What does $V^\alpha(x)$ even mean? It needs to be $V^\alpha(\lambda)$. Velocities aren't vector fields. They depend on the parametrization of the curve. If you run through the curve at twice the speed, they are twice as long.

 

[Demystifier]

No I can't. What does $V^\alpha(x)$ even mean? It needs to be $V^\alpha(\lambda)$. Velocities aren't vector fields. They depend on the parametrization of the curve. If you run through the curve at twice the speed, they are twice as long.

Obviously you didn't read the paper. My velocity is a vector field defined by Eq. (102) in https://arxiv.org/abs/1309.0400
It is similar to the velocity in the classical Hamilton-Jacobi theory in Eq. (23), which is also a vector field. If you claim the opposite, then you don't understand the Hamilton-Jacobi theory.

 

[rubi]

In your post #83, you used this definition:

$$\frac{dX^{\mu}}{d\lambda}=V^{\mu}$$

So either your post #83 is wrong or your paper is wrong.

 

[Demystifier]

In your post #83, you used this definition:
So either your post #83 is wrong or your paper is wrong.

I never said that it is a definition. It is Eq. (113) in the paper, which is not a definition. Nothing is wrong, you just don't read carefully.

 

[rubi]

I never said that it is a definition. It is Eq. (113) in the paper, which is not a definition.

In response to my post, you argued that $V^\mu(x)$ does not depend on the parametrization of the curve. The formula $V^\mu = \frac{\mathrm d X^\mu}{\mathrm d\lambda}$ does depend on the parametrization of the curve, independent of whether it is a definition or a derived formula. So it's still true that either your post #83 is wrong or your paper is wrong. (Edit: In fact, your paper is wrong, since as you said, the formula is also in your paper.)

 

[Demystifier]

In response to my post, you argued that $V^\mu(x)$ does not depend on the parametrization of the curve. The formula $V^\mu = \frac{\mathrm d X^\mu}{\mathrm d\lambda}$ does depend on the parametrization of the curve, independent of whether it is a definition or a derived formula. So it's still true that either your post #83 is wrong or your paper is wrong.

Or you just refuse to read carefully. Please read my paper if you want to discuss it, otherwise it doesn't make sense. For a start, you need to distinguish $V^{\mu}(x)$ from $V^{\mu}(X(\lambda))$.

 

[rubi]

Or you just refuse to read carefully. Pleas read my paper if you want to discuss it, otherwise it doesn't make sense.

I can already tell that you made a mistake without reading your paper. In post #83 you wrote a formula that definitely depends on the parametrization of the curve and in post #92 you claimed that it doesn't depend on the parametrization of the curve. One of the posts must necessarily be wrong. I don't need to read any paper in order to recognize this contradiction. If you disagree, then surely you are able to explain it here, instead of just claiming that the answer is somewhere to be found in those 53 pages.

 

[Demystifier]

If you disagree, then surely you are able to explain it here, instead of just claiming that the answer is somewhere to be found in those 53 pages.

I will not repeat what I already wrote in the paper. If you don't want to read the whole paper, read only Secs. 5.2 and 5.3. If it is too much for you, then I can't help you.

 

[rubi]

I will not repeat what I already wrote in the paper. If you don't want to read the whole paper, read only Secs. 5.2 and 5.3. If it is too much for you, then I can't help you.

I have read those sections now and I can only guess that you want me to accept that equation (112) and the paragraph above it show that the cuves somehow become independent of the parametrization. However, this is where your paper is wrong. Equation (112) does not prove that the trajectories don't depend on $\lambda$ or $\Omega$. It's simply not true. Different choices of $\Omega$ will in fact lead to different parametrizations of the curve.

 

[Demystifier]

Different choices of $\Omega$ will in fact lead to different parametrizations of the curve.

Of course they will, but different parametrizations of the same curve.

 

[Demystifier]

Equation (112) does not prove that the trajectories don't depend on $\lambda$ or $\Omega$.

I think you should read some standard text on integral curves of a vector field.

 

[rubi]

Of course they will, but different parametrizations of the same curve.

Obviously, but the point is that the parametrizations will be different and hence their tangent vectors will be different. They are scaled by the factor $\Omega$. One can't have a curve with tangent vectors whose length is independent of the parametrization.

I think you should read some standard text on integral curves of a vector field.

Sorry, but it's you who has a serious lack of understanding of differential geometric concepts.

 

[Demystifier]

 

[rubi]

I wonder why you are even in science if you think responses like "read my paper/a book" or "" are appropriate reactions in a scientific discussion. It's a fact that there is a mistake either in your paper or in one of your posts, but I guess I shouldn't waste any more time on it.

 

[Demystifier]

I wonder why you are even in science if you think responses like "read my paper/a book" or "" are appropriate reactions in a scientific discussion. It's a fact that there is a mistake either in your paper or in one of your posts, but I guess I shouldn't waste any more time on it.

Forget about my paper and answer the following question. Given a vector field $V^{\mu}(x)$, do you agree that an integral curve of the vector field does not depend on the parametrization of the curve?

 

[Demystifier]

One can't have a curve with tangent vectors whose length is independent of the parametrization.

I absolutely disagree with this statement. Given a curve and a point on this curve, one can only determine the direction of the tangent vector at this point, not the length of this vector. And all this is totally independent of the parametrization of the curve. In principle, the curve can be defined even without the parametrization.

 

[rubi]

Forget about my paper and answer the following question. Given a vector field $V^{\mu}(x)$, do you agree that an integral curve of the vector field does not depend on the parametrization of the curve?

If you have a vector field $V^\mu(x)$ and calculate an integral curve of it, then it is already given in a specific representation. If you reparametrize the curve, it is no longer an integral curve to the original vector field, but an integral curve to a rescaled vector field.

I absolutely disagree with this statement. Given a curve and a point on this curve, one can only determine the direction of the tangent vector at this point, not the length of this vector. And all this is totally independent of the parametrization of the curve. In principle, the curve can be defined even without the parametrization.

If I give you a curve $\gamma(t) = (\cos(t),\sin(t))$, then I can of course calculate the length of the tangent vectors. for example, the tangent vector at $t=0$ is given by $v = \left.\frac{\mathrm d}{\mathrm d t}\right\lvert_{t=0} \gamma(t) = (0,1)$, which has length $1$. If we reparametrize $\gamma^\prime(t) = (\cos(2t),\sin(2t))$, we get $v^\prime = (0,2)$, which has length $2$.

 

[Demystifier]

@rubi I have found the source of our disagreement. We are both right, but we are using different definitions of "curve". In literature one can find two different definitions:

Definition 1: Let $M$ be a manifold. A curve is a map $\mathbb{R}\rightarrow M$.

Definition 2: Let $M$ be a manifold. A curve is the image of a map $\mathbb{R}\rightarrow M$.

According to Def. 2, a curve is a 1-dimensional submanifold of $M$. The curve in Def. 2 is an equivalence class of all curves in Def. 1 with the same image on $M$. In the book Y. Choquet-Bruhat, C. DeWitt-Morette, Analysis, Manifolds and Physics 1 the first definition of curve is called the parameterized curve, while the second definition of curve is called the geometric curve.

For instance, if $M$ is a 2-dimensional manifold with coordinates $x,y$, consider the parabola defined by
$$y=x^2$$
This parabola is not a curve according to Def. 1, but is a curve according to Def. 2. (To further complicate terminology let me also note that, in algebraic geometry, an object such as that parabola is called a variety.)

Your claims are right if by "curve" one means the first definition. My claims are right if by "curve" one means the second definition. Neither of us presented the explicit definition of the curve because we were not aware that there are two inequivalent definitions. I think that this resolves our disagreement. So instead of my previous , now my face looks more like .

 

[stevendaryl]

@rubi I have found the source of our disagreement. We are both right, but we are using different definitions of "curve". In literature one can find two different definitions:

Definition 1: Let $M$ be a manifold. A curve is a map $\mathbb{R}\rightarrow M$.

Definition 2: Let $M$ be a manifold. A curve is the image of a map $\mathbb{R}\rightarrow M$.

According to Def. 2, a curve is a 1-dimensional submanifold of $M$. The curve in Def. 2 is an equivalence class of all curves in Def. 1 with the same image on $M$. In the book Y. Choquet-Bruhat, C. DeWitt-Morette, Analysis, Manifolds and Physics 1 the first definition of curve is called the parameterized curve, while the second definition of curve is called the geometric curve.

For instance, if $M$ is a 2-dimensional manifold with coordinates $x,y$, consider the parabola defined by
$$y=x^2$$
This parabola is not a curve according to Def. 1, but is a curve according to Def. 2. (To further complicate terminology let me also note that, in algebraic geometry, an object such as that parabola is called a variety.)

Your claims are right if by "curve" one means the first definition. My claims are right if by "curve" one means the second definition. Neither of us presented the explicit definition of the curve because we were not aware that there are two inequivalent definitions. I think that this resolves our disagreement. So instead of my previous , now my face looks more like .

This is an interesting digression. In differential geometry, the notion of a vector as the tangent to a curve only makes sense if a curve is considered a parametrized path, not the image of that map. Just given the image, there is no unique tangent. (Well, I guess you could take an equivalence class of the tangents in the parametrized sense: $V^\mu \approx U^\mu$ if there is a positive real number $\lambda$ such that $V^\mu = \lambda U^\mu$. But after taking equivalence classes, there is no notion of the length of a vector, only a direction.)

 

[Demystifier]

This is an interesting digression. In differential geometry, the notion of a vector as the tangent to a curve only makes sense if a curve is considered a parametrized path, not the image of that map. Just given the image, there is no unique tangent. (Well, I guess you could take an equivalence class of the tangents in the parametrized sense: $V^\mu \approx U^\mu$ if there is a positive real number $\lambda$ such that $V^\mu = \lambda U^\mu$. But after taking equivalence classes, there is no notion of the length of a vector, only a direction.)

Yes, exactly.

 

[A. Neumaier]

Yes, exactly.

But your tangents must be vectors, hence depend on the parameterization.

 

[Demystifier]

But your tangents must be vectors, hence depend on the parameterization.

Not necessarily. I start with a vector field, which are of course vectors, but do not have any parameterization. Then I can found an integral curve in the sense of Definition 2, so now the vectors are tangent to the curve, but I still do not need to take any parameterization.

Here is an example. Consider a 2-dimensional manifold with a constant vector field $V^{\mu}=(V^1,V^2)=(0,1)$. The integral curves in the sense of Definition 2 are vertical lines with constant $x^1$, e.g. $x^1=7$. In this way I have an integral curve without a parametrization.

 

[A. Neumaier]

In this way I have an integral curve without a parametrization.

But without a parameterization you cannot talk about $d/d\lambda$.

 

[Demystifier]

But without a parameterization you cannot talk about $d/d\lambda$.

Of course, but I don't see how it contradicts any of my claims.

 

[bhobba]

I wonder why you are even in science if you think responses like "read my paper/a book" or "" are appropriate reactions in a scientific discussion. It's a fact that there is a mistake either in your paper or in one of your posts, but I guess I shouldn't waste any more time on it.

Please - with my moderators hat on - can we be careful in wording and diplomatic. You both are science advisers and the above is not a good look. I am trying to follow what the exact issue is. It seems to me just re-parameterisation of proper time. I don't really know what's trying to be achieved - but can you do such a re-parameterisation - as far as I know that's OK. Now I will let you guys get back to fleshing out exactly what's going on.

Thanks
Bill

 

[A. Neumaier]

Of course, but I don't see how it contradicts any of my claims.

Well, your argument in post #83 uses $d/d\lambda$ and concludes $ds=d\lambda$. But if one parameterizes martinbn's example in #73 instead as $X^\mu=(2\lambda,2\lambda,0,0)$ then how does your ''argument'' in #74 now imply the necessary $ds=2d\lambda$?

 

[Demystifier]

Well, your argument in post #83 uses $d/d\lambda$ and concludes $ds=d\lambda$. But if one parameterizes martinbn's example in #73 instead as $X^\mu=(2\lambda,2\lambda,0,0)$ then how does your ''argument'' in #74 now imply the necessary $ds=2d\lambda$?

Yes, in this case we would have $ds=2d\lambda$, and I never said that we wouldn't. You are missing two things.
First, you are missing the context of post #83, the purpose of which was to show that quantum proper time is non-zero even for null trajectories.
Second, you (like everybody else in this thread) refuse to read the paper, especially Secs. 5.2 and 5.3. where the meanings of $\lambda$ and $s$ are explained. What I write here is a supplement to the paper, not a substitute for the paper.

 

[martinbn]

@Demystifier along with geometric curve and parametrized curve, the term tangent vector can have different meanings depending on the context. The tangent vector of a curve is by definition $\frac d{d\lambda}\gamma$, here the curve is assumed to be parametrized. On the other hand if you have a submanifold (curve or not) there is the notion of a tangent vector on it, roughly a tangent vector of the manifold that happens to be a tangent vector of the submanifold viewed as a manifold on its own, no parametrizations involved. Not sure if this is relevant to this discussion.

 

[Demystifier]

@Demystifier along with geometric curve and parametrized curve, the term tangent vector can have different meanings depending on the context. The tangent vector of a curve is by definition $\frac d{d\lambda}\gamma$, here the curve is assumed to be parametrized. On the other hand if you have a submanifold (curve or not) there is the notion of a tangent vector on it, roughly a tangent vector of the manifold that happens to be a tangent vector of the submanifold viewed as a manifold on its own, no parametrizations involved. Not sure if this is relevant to this discussion.

Yes, that's relevant and I agree with this.

 

[A. Neumaier]

if you have a submanifold (curve or not) there is the notion of a tangent vector on it, roughly a tangent vector of the manifold that happens to be a tangent vector of the submanifold viewed as a manifold on its own, no parametrizations involved.

Yes, that's relevant and I agree with this.

But (for a curve) this tangent vector is unique only up to a (positive or negative) factor.