I Relativistic hidden variable quantum mechanics?

[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.

 

[Demystifier]

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

One way to fix this factor is indeed to fix the parametrization. But the point is that it is not the only way.

 

[A. Neumaier]

One way to fix this factor is indeed to fix the parametrization. But the point is that it is not the only way.

So how do you fix it without getting a spurious sign in your ds?

 

[Demystifier]

So how do you fix it without getting a spurious sign in your ds?

If the curve is oriented (which is a much weaker requirement than that it is parameterized), then the sign is chosen such that $s$ increases in the direction of orientation. In my case, the orientation at each point is defined by the direction of the vector $V^{\mu}$, to which the curve is tangent. Note also that in my paper $\Omega$ in Eq. (111) is positive.