I Relativistic hidden variable quantum mechanics?

[Demystifier]

But we are discussing the present state of affairs.

You are the one who first mentioned future in relation to the value of string theory.

 

[A. Neumaier]

You are the one who first mentioned future in relation to the value of string theory.

Only because you asked a question unrelated to the thread.

 

[martinbn]

He can, which doesn't mean that he will.
You have to be a better motivator. You have to say something like "Wow, you explained that so well, it would be great if you could just fill this little detail, so that my understanding of your great ideas can be complete."

But I am in a good mood these days, so I will give you a sketch of the proof for $n=1$. Suppose that $V^{\mu}V_{\mu}=0$, but $V^{\mu}\neq 0$ and $V^{\mu}\neq \infty$. Then
$$\frac{dX^{\mu}}{ds}=V^{\mu}\neq 0, \infty \;\;\;\; (1)$$
Therefore on a line segment on which $dX^{\mu}\neq 0$ we have $dX^{\mu}dX_{\mu}=0$, but (1) implies that $ds\neq 0$. $\Box$

If you don't accept a proof using infinitesimals, I live it to you to reformulate this in the $\epsilon$-$\delta$ language or in an integral form.

Well, it doesn't imply $ds\neq 0$, it implies $ds=\frac00$.

But you are just assuming something about the world line. The claim is that the dynamics imply that the $(something)$ has the correct sign or order of vanishing. I don't see even a hint of that, let alone a sketch of proof.

 

[Demystifier]

It is still not clear to me. What do you mean by the general solution with infinite number of constants? And how do you pick their values? What information is given?

Here is a simple example. Take the usual wave equation in $1+1$ dimensions. It has a general solution $u(x,t)=f(x-ct)+g(x+ct)$ for arbitrary functions $f$ and $g$. How do you pick your solution? And what happens in higher dimensions, what is the general solution?

Now you ask me questions about standard theory of partial differential equations. I think it's not directly relevant to my relativistic hidden variable theory.

 

[martinbn]

Now you ask me questions about standard theory of partial differential equations. I think it's not directly relevant to my relativistic hidden variable theory.

Ok, answer the same question for equations $(94)$. What is the general solution and how do you pick the constants? What information is given so that you can determine those constants.

 

[Demystifier]

Well, it doesn't imply $ds\neq 0$, it implies $ds=\frac00$.

Sorry, but you are wrong.

 

[Demystifier]

Ok, answer the same question for equations $(94)$. What is the general solution and how do you pick the constants? What information is given so that you can determine those constants.

Let me first make one thing clear. Do you ask because you want to learn something, or do you ask because you want to find my error? I have the feeling that you only want the latter, so you ask nitpicking questions, which are actually easy to me, but I find them pretty much irrelevant.

 

[martinbn]

Sorry, but you are wrong.

Don't you have $V^{\mu}V_{\mu}$ in the denominator?

 

[martinbn]

Let me first make one thing clear. Do you ask because you want to learn something, or do you ask because you want to find my error? I have the feeling that you only want the latter, so you ask nitpicking questions, which are actually easy to me, but I find them pretty much irrelevant.

I don't know if you have an error. I just want to understand the statements, because they make no sense to me. But you don't have to waste time with this, it isn't important.

 

[Demystifier]

Don't you have $V^{\mu}V_{\mu}$ in the denominator?

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

 

[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 $G_{\mu\nu}(x)$.