      # A Generalization Of The Lorentz Ether Interpretation To The Einstein Equations Of Gr

Lorentz ether general relativity

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### #69 Dubbelosix

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Posted 11 November 2019 - 07:46 AM

Yes. It is the positivist idea that what is unobservable does not exist.

Its actually rooted in the philosophy of the scientific method, in which you deduct from nature what you can extrapolate through measurement. If it cannot, then a theory maybe deemed unscientific. The idea is that a theory is able to be falsified.

### #70 Dubbelosix

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Posted 11 November 2019 - 07:50 AM

The Lorentz ether, as an interpretation of SR, gives zero velocity for the Minkowski metric, which is the only metric considered in this theory.
My general formulas also give zero velocity for the Minkowski metric.

I don't see a contradiction. It is you who has to explain where you see a contradiction.

You said on one hand there was motion and on another said the metric velocity is zero - I am usually a clear thinker so I am wondering as to the nature of your aether. You say it is rooted in gravitational physics and maybe that is true, but it has been clear as mud so far. Not to be offensive.

### #71 Schmelzer

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Posted 11 November 2019 - 10:49 AM

You said on one hand there was motion and on another said the metric velocity is zero - I am usually a clear thinker so I am wondering as to the nature of your aether. You say it is rooted in gravitational physics and maybe that is true, but it has been clear as mud so far. Not to be offensive.

I don't understand your problems with the simple formula $v^i(x,t) = g^{0i}(x,t)/g^{00}(x,t)$ which holds in the preferred (harmonic) coordinates.  In general, it defines a nontrivial velocity.  But in particular cases (Minkowski metric, Schwarzschild metric, FLRW ansatz, all in harmonic coordinates) we have $g^{0i}(x,t)=0$ and therefore the ether velocity for these cases will be zero.

SR is the limit of GR where we have no nontrivial gravitational field, so that the metric is the Minkowski metric $\eta^{\mu\nu}$.  This metric is obviously a case where the formula above gives also $v^i(x,t)=0$. So, in the SR limit, where my theory should give the classical Lorentz ether, we have automatically what is required for the classical Lorentz ether, namely zero ether velocity.

Its actually rooted in the philosophy of the scientific method, in which you deduct from nature what you can extrapolate through measurement. If it cannot, then a theory maybe deemed unscientific. The idea is that a theory is able to be falsified.

The first two sentences follow positivism/empiricism, which is outdated and understood to be wrong.  There can be no deduction of theories from nature, and extrapolation always needs some hypotheses how to extrapolate.

Popper's critical rationalism (fallibilism) is much superior in comparison.  It is obviously behind the third sentence.  So, I see a mixture of two theories about the scientific method which contradict each other.

### #72 Dubbelosix

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Posted 11 November 2019 - 11:01 AM

I don't understand your problems with the simple formula $v^i(x,t) = g^{0i}(x,t)/g^{00}(x,t)$ which holds in the preferred (harmonic) coordinates.  In general, it defines a nontrivial velocity.  But in particular cases (Minkowski metric, Schwarzschild metric, FLRW ansatz, all in harmonic coordinates) we have $g^{0i}(x,t)=0$ and therefore the ether velocity for these cases will be zero.

SR is the limit of GR where we have no nontrivial gravitational field, so that the metric is the Minkowski metric $\eta^{\mu\nu}$.  This metric is obviously a case where the formula above gives also $v^i(x,t)=0$. So, in the SR limit, where my theory should give the classical Lorentz ether, we have automatically what is required for the classical Lorentz ether, namely zero ether velocity.

The first two sentences follow positivism/empiricism, which is outdated and understood to be wrong.  There can be no deduction of theories from nature, and extrapolation always needs some hypotheses how to extrapolate.

Popper's critical rationalism (fallibilism) is much superior in comparison.  It is obviously behind the third sentence.  So, I see a mixture of two theories about the scientific method which contradict each other.

Its not so much math, if I had a problem with your math, I'd say so. This is about whether your theory will cohere with modern understanding of aether theories. If it means anything, it's gettinga little clearer, but I need to understand more about the reasons why you chose this.

### #73 Dubbelosix

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Posted 11 November 2019 - 11:02 AM

I'll consider this more and be back.

### #74 Dubbelosix

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Posted 11 November 2019 - 03:29 PM

Just a small question how are the dimensions being constructed on the right hand side here;

v_i(x, t) = g_[oi](x, t) / g_[oo](x, t)

I know the definition 9f the metric terms, so this is just a general question about the dimensions alone. Thanks in advance.

### #75 Schmelzer

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Posted 11 November 2019 - 11:31 PM

Essentially, there is no modern understanding of ether theory.  There is the classical Lorentz ether, where the metric is only the Minkowski metric.  Here the $\eta^{0i}=0$ so that the velocity is zero and the density $\rho=g^{00}\sqrt{-g}$ is constant.  And there is the Leyden lecture, which is nothing than a popular lecture, which has been essentially ignored by the mainstream which made "ether" a bad word.  All this essentially a century old. There are ether freaks who don't even understand SR, unpublishable for good reasons. There is Jacobson's "Einstein ether" which is based on completely different concepts. What else?

Just a small question how are the dimensions being constructed on the right hand side here;

v_i(x, t) = g_[oi](x, t) / g_[oo](x, t)

I know the definition 9f the metric terms, so this is just a general question about the dimensions alone. Thanks in advance.

Usually I follow the c=1 convention, so that a velocity becomes a dimensionless number (simply v/c).  If one would have to handle it with c, then one would have to include the appropriate factors of c to everything.  In order to avoid giving the different components of $g^{\mu\nu}$ different dimensions, the appropriate choice would be to use the $x^0 = ct$  so that all the coordinates $x^\mu$ have the same dimension of length. Then $v_i(x, t)/c = g^{0i}(x, t) / g^{oo}(x, t)$ would be the correct dimensionless  expression.

BTW, the components are the ones with upper indices, the coefficients used in the wave equation $\square = \partial_\mu g^{\mu\nu}\sqrt{-g} \partial_\nu$ and the harmonic condition $\square x^\nu = \partial_\mu g^{\mu\nu}\sqrt{-g} = 0$.  These have to be identified with the continuity and Euler equations of condensed matter theory, applied to the ether:

$\partial_t \rho + \partial_i (\rho v^i) = 0$

$\partial_t (\rho v^j) + \partial_i(\rho v^i v^j - \sigma^{ij}) = 0$

in such a way as to get the appropriate factors of c into it.

Edited by Schmelzer, 12 November 2019 - 01:41 AM.

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### #76 Dubbelosix

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Posted 12 November 2019 - 06:55 AM

Essentially, there is no modern understanding of ether theory.  There is the classical Lorentz ether, where the metric is only the Minkowski metric.  Here the $\eta^{0i}=0$ so that the velocity is zero and the density $\rho=g^{00}\sqrt{-g}$ is constant.  And there is the Leyden lecture, which is nothing than a popular lecture, which has been essentially ignored by the mainstream which made "ether" a bad word.  All this essentially a century old. There are ether freaks who don't even understand SR, unpublishable for good reasons. There is Jacobson's "Einstein ether" which is based on completely different concepts. What else?

Usually I follow the c=1 convention, so that a velocity becomes a dimensionless number (simply v/c).  If one would have to handle it with c, then one would have to include the appropriate factors of c to everything.  In order to avoid giving the different components of $g^{\mu\nu}$ different dimensions, the appropriate choice would be to use the $x^0 = ct$  so that all the coordinates $x^\mu$ have the same dimension of length. Then $v_i(x, t)/c = g^{0i}(x, t) / g^{oo}(x, t)$ would be the correct dimensionless  expression.

OK, this looks good.

### #77 Dubbelosix

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Posted 12 November 2019 - 06:59 AM

Just a bit of advice then for now, if you do come to use natural units, try and explain this in your paper as it can make the understanding a bit clearer. Authors do tend to do this, from time to time to help the reader keep track more efficiently.

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