Relativistic hidden variable quantum mechanics?

[Demystifier]

I just want to understand the statements, because they make no sense to me.

Do statements of ordinary non-relativistic Bohmian mechanics make sense to you? The paper is written with an assumption that the reader is already familiar with non-relativistic Bohmian mechanics, as well as with standard relativistic QM and standard QFT.

 

[Demystifier]

Only because you asked a question unrelated to the thread.

Then let me put it this way. If string theory is unphysical today, then so is my theory.

 

[martinbn]

From equation (1) above one has (no sum over $\mu$)
$$ds=\frac{dX^{\mu}}{V^{\mu}}=\frac{non zero}{non zero}$$

You are using $s$ in a confusing way (for two things). Let's say $X^{\mu}=(\lambda,\lambda,0,0)$. Then $V^{\mu}=(1,1,0,0)$, and $V^\mu V_\mu=0$, and $V^0\neq 0, \infty$. What is $ds$?

 

[Demystifier]

You are using $s$ in a confusing way (for two things). Let's say $X^{\mu}=(\lambda,\lambda,0,0)$. Then $V^{\mu}=(1,1,0,0)$, and $V^\mu V_\mu=0$, and $V^0\neq 0, \infty$. What is $ds$?

$ds=d\lambda$.

 

[martinbn]

You are using $s$ in a confusing way (for two things). Let's say $X^{\mu}=(\lambda,\lambda,0,0)$. Then $V^{\mu}=(1,1,0,0)$, and $V^\mu V_\mu=0$, and $V^0\neq 0, \infty$. What is $ds$?

$ds=d\lambda$.

Why? Equation $(108)$ says

$ds^2=\frac1{V^\mu V_\mu}dX^\mu dX_\mu$

And in the example above both $V^\mu V_\mu$ and $dX^\mu dX_\mu$ are zero.

 

[Demystifier]

Why? Equation $(108)$ says

$ds^2=\frac1{V^\mu V_\mu}dX^\mu dX_\mu$

And in the example above both $V^\mu V_\mu$ and $dX^\mu dX_\mu$ are zero.

Sec. 5.3 contains a detailed answer to your question.

See also the last paragraph in Sec. 5.2, which answers one of your previous questions.

 

[martinbn]

Sec. 5.3 contains a detailed answer to your question.

It doesn't. To go from $(114)$ to $(115)$ you are either assuming that $V^\mu V_\mu$ and $dX^\mu dX_\mu$ are not zero, or you use division by zero.

See also the last paragraph in Sec. 5.2, which answers one of your previous questions.

Not sure which question this answers.

 

[Demystifier]

It doesn't. To go from $(114)$ to $(115)$ you are either assuming that $V^\mu V_\mu$ and $dX^\mu dX_\mu$ are not zero, or you use division by zero.

You can look at it this way. Eq. (116) is derived for any value of $V^\mu V_\mu$, except for $V^\mu V_\mu=0$ in which case there is an ambiguity. Therefore it is natural to resolve the ambiguity by postulating (116) to be valid even in this case. Technically it should need a separate postulate, but to me it seemed too obvious to state it explicitly in the paper. (As you know, physicists and mathematicians have very different standards of "obvious".)

See also Eq. (55). Would you say that Newton time $t$ is not well defined when the 3-velocity of a particle is zero?

Not sure which question this answers.

I think previously you objected that my theory is not really relativistic, or something like that.

 

[A. Neumaier]

Then let me put it this way. If string theory is unphysical today, then so is my theory.

And if string theory is physical (.i.e., describes reality) today then your theory is unphysical since string theory assumes even more symmetry than Lorentz invariance, while you reject it.

Together with what you just conceded, it follows that your theory is always unphysical!

 

[Demystifier]

And if string theory is physical (.i.e., describes reality) today then your theory is unphysical since string theory assumes even more symmetry than Lorentz invariance, while you reject it.

Together with what you just conceded, it follows that your theory is always unphysical!

You made two logical errors in arriving to that conclusion. First, you used two different meanings of "physical". Second, you forgot to consider a logical possibility that my theory describes reality, in which case string theory doesn't.

 

[A. Neumaier]

You made two logical errors in arriving to that conclusion. First, you used two different meanings of "physical". Second, you forgot to consider a logical possibility that my theory describes reality, in which case string theory doesn't.

First - we hadn't discussed the definition of physical; for me, physical only means corresponding to reality.

Second - This is your error, not mine. I concluded logically correctly from what you conceded. Indeed, this already excludes the logical possibility you mentioned since the latter implies that string theory is unphysical, so by your previous mail, your theory is unphysical.

 

[Demystifier]

physical only means corresponding to reality.

Then my theory is ugly, intuitive and possibly physical.

 

[Demystifier]

It doesn't. To go from $(114)$ to $(115)$ you are either assuming that $V^\mu V_\mu$ and $dX^\mu dX_\mu$ are not zero, or you use division by zero.

How about the following proof? We start from
$$ds^2=\frac{dX^{\mu}dX_{\mu}}{V^{\nu}V_{\nu}}$$
Using
$$\frac{dX^{\mu}}{d\lambda}=V^{\mu}$$
this can be written as
$$ds^2=\frac{V^{\mu}d\lambda V_{\mu}d\lambda}{V^{\nu}V_{\nu}}=\frac{V^{\mu}V_{\mu}}{V^{\nu}V_{\nu}}d\lambda^2
=d\lambda^2$$
so
$$ds=d\lambda$$
Do you think this proof is valid in the limit $V^{\mu}V_{\mu}\rightarrow 0$? I think it is, because clearly
$$\lim_{V^{\mu}V_{\mu}\rightarrow 0} \frac{V^{\mu}V_{\mu}}{V^{\nu}V_{\nu}}=1$$

If you think it's invalid, do you think that you can do it better?

 

[DaTario]

I had a teacher who was involved with these ideas. He wrote some papers in this respect that I found interesting. So, I share here for possible reference.

a) L. C. Ryff, Phys. Rev. A 60, 5083 (1999)
<Moderator's note: links to unpublished results removed.>

 

[bhobba]

How about the following proof?

Can you explain this in a bit more detail for a dumb ass like me? As far as I can see its not really saying anything - its true - but more like a tautology with no actual content. What am I missing??

Thanks
Bill

 

[Demystifier]

Can you explain this in a bit more detail for a dumb ass like me? As far as I can see its not really saying anything - its true - but more like a tautology with no actual content. What am I missing??

It shows that quantum proper time $s$ defined by the first equation is non-zero even along null-curves, along which the classical proper time is zero. @martinbn did not believe that it can be so, so I had to show him that explicitly. If this proof looks like a tautology, that probably means that proof is convincing. But the main idea of the proof is to write it in such a form that one can apply it even to the null-curves, i.e. in the limit $V^{\mu}V_{\mu}\rightarrow 0$. Without that limit, the proof is indeed quite trivial. The fact that @martinbn had no further comments also indicates that proof is convincing.

 

[rubi]

$$ds^2=\frac{dX^{\mu}dX_{\mu}}{V^{\nu}V_{\nu}}$$

This expression is nonsensical in the first place. It neither specifies a proper metric, since the coefficients of the $\mathrm d X^\mu$ aren't functions of spacetime, nor a proper line element, because it doesn't define a reparametrization invariant length integral. Only Einstein's expression has this property. If I asked you to calculate the length of a circle with this formula, you couldn't do it, because you'd need me to specify additional unphysical information. The same circle would have have different length, depending on whether I parametrize it as $(\cos(\lambda),\sin(\lambda),0,0)$ or as $(\cos(\lambda^2),\sin(\lambda^2),0,0)$.

 

[Demystifier]

This expression is nonsensical in the first place. It neither specifies a proper metric, since the coefficients of the $\mathrm d X^\mu$ aren't functions of spacetime, nor a proper line element, because it doesn't define a reparametrization invariant length integral. Only Einstein's expression has this property. If I asked you to calculate the length of a circle with this formula, you couldn't do it, because you'd need me to specify additional unphysical information. The same circle would have have different length, depending on whether I parametrize it as $(\cos(\lambda),\sin(\lambda),0,0)$ or as $(\cos(\lambda^2),\sin(\lambda^2),0,0)$.

I disagree. Just as classical proper time depends on the classical dynamical quantity $g_{\mu\nu}$ which itself is found by solving classical (Einstein) equations, this quantum proper time depends on the quantum dynamical quantity $V^{\mu}$ which itself is found by solving quantum (e.g. Klein Gordon or Dirac) equations. As shown in the paper, this quantum proper time does not depend on the parametrization.

 

[rubi]

The standard expression for the length integral of a curve $x : [0,1] \rightarrow \mathbb R^4$ is:
$\int_\gamma \mathrm d s = \int_0^1 \sqrt{g_{\mu\nu}(x(\lambda))\frac{\mathrm d x^\mu(\lambda)}{\mathrm d\lambda}\frac{\mathrm d x^\nu(\lambda)}{\mathrm d\lambda}}\,\mathrm d\lambda$
Now, if we reparametrize $x(\lambda) \rightarrow x(f(\lambda))$ (with $f(0)=0$ and $f(1)=1$), we can evaluate the integral again with the new parametrization:
$\int_0^1 \sqrt{g_{\mu\nu}(x(f(\lambda)))\frac{\mathrm d x^\mu(f(\lambda))}{\mathrm d\lambda}\frac{\mathrm d x^\nu(f(\lambda))}{\mathrm d\lambda}}\,\mathrm d\lambda = \int_0^1 \sqrt{g_{\mu\nu}(x(f(\lambda)))\frac{\mathrm d x^\mu(f(\lambda))}{\mathrm d f(\lambda)}\frac{\mathrm d f(\lambda)}{\mathrm d\lambda}\frac{\mathrm d x^\mu(f(\lambda))}{\mathrm d f(\lambda)}\frac{\mathrm d f(\lambda)}{\mathrm d\lambda}}\,\mathrm d\lambda = \int_0^1 \sqrt{g_{\mu\nu}(x(f(\lambda)))\frac{\mathrm d x^\mu(f(\lambda))}{\mathrm d f(\lambda)}\frac{\mathrm d x^\mu(f(\lambda))}{\mathrm d f(\lambda)}}\frac{\mathrm d f(\lambda)}{\mathrm d\lambda}\,\mathrm d\lambda = \int_{f^{-1}(0)}^{f^{-1}(1)} \sqrt{g_{\mu\nu}(x(f))\frac{\mathrm d x^\mu(f)}{\mathrm d f}\frac{\mathrm d x^\mu(f)}{\mathrm d f}}\,\mathrm d f = \int_0^1 \sqrt{g_{\mu\nu}(x(f))\frac{\mathrm d x^\mu(f)}{\mathrm d f}\frac{\mathrm d x^\mu(f)}{\mathrm d f}}\,\mathrm d f = \int_\gamma\mathrm d s$
So the standard length integral is indeed independent of the choice of reparametrization. We used the substitution rule ($\int_0^1 g(f(\lambda)) \frac{\mathrm d f(\lambda)}{\mathrm d\lambda}\,\mathrm d \lambda = \int_{f^{-1}(0)}^{f^{-1}(1)} g(f) \,\mathrm d f$) in the last step. Now if we use your expression instead, the factor $\frac{\mathrm d f(\lambda)}{\mathrm d \lambda}$ will not come out of the square root, because it would cancel with the factor that you will get from transforming the velocities in the denominator and hence, the substitution rule won't apply, rendering the integral dependent of the choice of paramentrization.

 

[Demystifier]

The standard expression for the length integral is:
$\int_\gamma \mathrm d s = \int_a^b \sqrt{g_{\mu\nu}(x(\lambda))\frac{\mathrm d x^\mu(\lambda)}{\mathrm d\lambda}\frac{\mathrm d x^\nu(\lambda)}{\mathrm d\lambda}}\,\mathrm d\lambda$
Now, if we reparametrize $x(\lambda) \rightarrow x(f(\lambda))$, we can evaluate the integral again with the new parametrization:
$\int_a^b \sqrt{g_{\mu\nu}(x(f(\lambda)))\frac{\mathrm d x^\mu(f(\lambda))}{\mathrm d\lambda}\frac{\mathrm d x^\nu(f(\lambda))}{\mathrm d\lambda}}\,\mathrm d\lambda = \int_a^b \sqrt{g_{\mu\nu}(x(f(\lambda)))\frac{\mathrm d x^\mu(f(\lambda))}{\mathrm d f(\lambda)}\frac{\mathrm d f(\lambda)}{\mathrm d\lambda}\frac{\mathrm d x^\mu(f(\lambda))}{\mathrm d f(\lambda)}\frac{\mathrm d f(\lambda)}{\mathrm d\lambda}}\,\mathrm d\lambda = \int_a^b \sqrt{g_{\mu\nu}(x(f(\lambda)))\frac{\mathrm d x^\mu(f(\lambda))}{\mathrm d f(\lambda)}\frac{\mathrm d x^\mu(f(\lambda))}{\mathrm d f(\lambda)}}\frac{\mathrm d f(\lambda)}{\mathrm d\lambda}\,\mathrm d\lambda = \int_{f^{-1}(a)}^{f^{-1}(b)} \sqrt{g_{\mu\nu}(x(f))\frac{\mathrm d x^\mu(f)}{\mathrm d f}\frac{\mathrm d x^\mu(f)}{\mathrm d f}}\,\mathrm d f = \int_\gamma\mathrm d s$
So the standard length integral is indeed independent of the choice of reparametrization. We used the substitution rule ($\int_a^b g(f(\lambda)) \frac{\mathrm d f(\lambda)}{\mathrm d\lambda}\,\mathrm d \lambda = \int_{f^{-1}(a)}^{f^{-1}(b)} g(f) \,\mathrm d f$) in the last step. Now if we use your expression instead, the factor $\frac{\mathrm d f(\lambda)}{\mathrm d \lambda}$ will not come out of the square root, because it would cancel with the factor that you will get from transforming the velocities in the denominator and hence, the substitution rule won't apply, rendering the integral dependent of the choice of paramentrization.

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
$$G_{\mu\nu}(x)=\frac{g_{\mu\nu}(x)}{g_{\alpha\beta}(x)V^{\alpha}(x)V^{\beta}(x)}$$
You can think of my theory as a geometrical theory with a new effective metric $G_{\mu\nu}(x)$.

 

[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?