# Proof Minkowski Spacetime is Poorly Conceived

Discussion in 'Alternative Theories' started by danshawen, Apr 21, 2016.

1. ### The GodValued Senior Member

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......It can but you don't know how...

Like mass warps spacetime but don't know how....well thats the century old trend in science !!

If you know the time dilation at rim and at center, then of course it will have some value at other radial points as well...maths can be worked out by maths guys, but the problem is are you not making time dilation dependent on density instead of mass ? How do you extend this rim/center stuff to non spherical objects ?

And mind you equivalence principle is flawless. In fact it is the most profound aspect of early 20th century Physics....I will tell you what all it can offer. Physics need to be rewritten around this only.

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3. ### danshawenValued Senior Member

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Yes it will, but it won't be anything like either relativity or even Euclid predicted.

It is just possible, you may be correct about the principle of equivalence. It was the defiance of that law, under the guise of explaining inertia without it right after the discovery of the Higgs boson, that brought me to these forums.

Finally, there are at least a few people here who have stopped thinking like Mikowski. G-d be praised. Do you realize where science would be if he had not gifted the Greeks with solid geometry to ponder? I for one, am in awe, and not of Einstein, and certainly not Minkowski.

G-d still loves geometry, just not Minkowski's brand.

Some will no doubt eventually observe, perhaps Minkowsky rotation was trying to point us in this general direction, or that the little hourglass shapes of his light cones suggested the factors of time and inertia that his part of relativity theory was lacking.

I'm fine with that. Einstein certainly bore him no grudge, and neither do I. After a lifetime of pondering his useless physics, I finally found his principle mistake. It was not as bad as what happened with Ptolemy or Aristotle; only harder to find.

Last edited: May 5, 2016
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5. ### The GodValued Senior Member

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Yes, EM field is real. Where is the ambiguity and where is the EM field similarity with spacetime ?

This one is good

This one is good too. [are you the author of this brief ?]

This one is adulterated BS....You can ask me why adulterated why not pure ?

Last edited: May 5, 2016
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7. ### rpennerFully WiredValued Senior Member

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⭕️
What is speed? It is the magnitude of velocity. So this is saying if $| \vec{u} |$ is the speed of a ray of light in one frame, that $| \vec{u}' |$ is the speed of the same ray of light in another frame and the same result is obtained. How do you show that from the Lorentz transformation?

Obviously we need two space-time events to measure a speed, a start, A, and a stop, B. They must be related by
$\begin{pmatrix} t_B \\ \vec{x}_B \end{pmatrix} - \begin{pmatrix} t_A \\ \vec{x}_A \end{pmatrix} = \begin{pmatrix} (t_B - t_A) \\ (t_B - t_A) \vec{u} \end{pmatrix}$
This we operate on by the Lorentz transformation, $\Lambda_{\vec{v}}$, to describe events A and B in a new frame where objects which the first reference frame considered at rest are now considered moving with velocity $\vec{v}$.

$\Lambda_{\vec{v}} \begin{pmatrix} t_B \\ \vec{x}_B \end{pmatrix} - \Lambda_{\vec{v}} \begin{pmatrix} t_A \\ \vec{x}_A \end{pmatrix} = \Lambda_{\vec{v}} \left( \begin{pmatrix} t_B \\ \vec{x}_B \end{pmatrix} - \begin{pmatrix} t_A \\ \vec{x}_A \end{pmatrix} \right) = \Lambda_{\vec{v}} \begin{pmatrix} (t_B - t_A) \\ (t_B - t_A) \vec{u} \end{pmatrix} = \begin{pmatrix} (t'_B - t'_A) \\ (t'_B - t'_A) \vec{u}' \end{pmatrix}$

So we have
$(t'_B - t'_A) = \frac{c}{\sqrt{c^2 - \vec{v}^2}} \left( (t_B - t_A) + (t_B - t_A) c^{-2} \vec{v} \cdot \vec{u} \right) \\ (t'_B - t'_A) \vec{u}' = (t_B - t_A) \vec{u} + \frac{c - \sqrt{c^2 - \vec{v}^2}}{\vec{v}^2 \sqrt{c^2 - \vec{v}^2}} \left( (t_B - t_A) \vec{v} \cdot \vec{u} \right) \vec{v} + \frac{c}{\sqrt{c^2 - \vec{v}^2}} (t_B - t_A) \vec{v}$
or
$\vec{u}' = \frac{\sqrt{c^2 - \vec{v}^2} c^2 \vec{v}^2 \vec{u} + \left( \left( c - \sqrt{c^2 - \vec{v}^2} \right) \left( \vec{v} \cdot \vec{u} \right) + c \vec{v}^2 \right) c^2 \vec{v} }{ \left( c \vec{v}^2 \right) \left( c^2 + \vec{v} \cdot \vec{u} \right) }$

If $\vec{v} = v \hat{x}, \vec{u} = u \hat{x}$ this reduces to
$\vec{u}' = \frac{\sqrt{c^2 - v^2} c^2 v^2 u + c^3 v^2 u - \sqrt{c^2 - v^2} c^2 v^2 u + c^3 v^3 }{ \left( c v^2 \right) \left( c^2 + v u \right) } \hat{x} \\= \frac{ u + v }{ 1 + c^{-2} v u } \hat{x}$
the celebrated velocity composition law in one dimension of space.

But for the full formula, we compute $\vec{u}'^2$
$\vec{u}'^2 = \frac{ c^4 \vec{v}^2 + c^4 \vec{u}^2 - c^2 \vec{v}^2 \vec{u}^2 + 2 c^4 ( \vec{v} \cdot \vec{u} ) + c^2 ( \vec{v} \cdot \vec{u} )^2 }{ c^4 + 2 c^2 ( \vec{v} \cdot \vec{u} ) + ( \vec{v} \cdot \vec{u} )^2 }$

Now, finally, substitute in the assumption that $\vec{u}^2 = c^2$ and we get:
$\vec{u}'^2 = \frac{ c^6 + 2 c^4 ( \vec{v} \cdot \vec{u} ) + c^2 ( \vec{v} \cdot \vec{u} )^2 }{ c^4 + 2 c^2 ( \vec{v} \cdot \vec{u} ) + ( \vec{v} \cdot \vec{u} )^2 } = c^2$

That's how you prove the Lorentz transformation preserves the speed of light -- you do math to support a math claim.

❌ There is not "a" rest frame for any given piece of matter. Inertial reference frames are nothing more or less than imaginary coordinate systems for labelling events in space-time in a systematic manner that happens to be compatible with Newton's law of inertia in that bodies not subject to external forces are described as having straight trajectories.

There is no invariance absent a transformation, so saying the Lorentz transformation preserves the invariance of the Lorentz transformation indicates unclear thinking at best.

⚠️ Your addition in parenthesis makes people doubt you know your subject matter. Since for a free (aka inertial) ponderable particle,
$E = \gamma mc^2, \vec{p} = \gamma m \vec{v}, \vec{v} = \frac{\Delta \vec{x}}{\Delta t}, \gamma = \left( 1 - \frac{\vec{v}^2}{c^2} \right)^{-\frac{1}{2}}$
it follows that
$\Delta t \, \times \, (E/c^2, \vec{p} / c) = ( \gamma m \Delta t , \gamma m c^{-1} \vec{v} \Delta t ) = m \gamma ( \Delta t, c^{-1} \Delta \vec{x} )$ .
The invariant measure of this 4-vector can be computed in this present system of coordinates:
$m^2 \gamma^2 \left( ( \Delta t )^2 - ( c^{-1} \Delta \vec{x} )^2 \right) = m^2 \left( 1 - \frac{\vec{v}^2}{c^2} \right)^{-1} \left( 1 - \frac{\vec{v}^2}{c^2} \right) ( \Delta t )^2 = m^2 ( \Delta t )^2$ .

And so since this is invariant under the Lorentz transform, the quantity $m^2 ( \Delta t )^2$ is more special than either $\Delta t E/c^2$ or $\Delta t \vec{p} / c$ which can change depending on one's choice of reference frame. But since $\Delta t$ was just any made up duration, that means $m^2$ is more basic than $E$ or $\vec{p}$. So even though the Lorentz transform says nothing overtly about mass, energy or momentum, it follows that $(E/c^2, \vec{p} / c)$ transforms like the four-vector $(c t, \vec{x})$ under the Lorentz transform.

But this is great because instead of the clumsy expressions: $E = \frac{mc^2}{\sqrt{1 - \frac{\vec{v}^2}{c^2}}}, \vec{p} = \frac{m \vec{v}}{\sqrt{1 - \frac{\vec{v}^2}{c^2}}}$ we are free to write: $(m c^2) = E^2 - (c \vec{p})^2, E \vec{v} = c^2 \vec{p}$ where we know that regardless which standard of rest $E, \vec{p}, \vec{v}$ came from, $m$ has the same value, and these equations have the additional benefit of working for massless phenomena which convey energy and momentum as well.

⚠️ This is at best clumsily expressed.

It is better to say the relativistic laws conserving energy and momentum are valid in any inertial frame because one does not want to give the impression that the same energy or momentum is conserved in different frames.

Last edited: May 5, 2016
8. ### danshawenValued Senior Member

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I don't see any links in your last post I can access, but it is likely you are responding to someone I am currently ignoring.

No attempt has been made, either by me or Minkowski's version of spacetime, to explain anything about the EM field. I have no model(s) of electric charge or fields. I understand that Special Relativity is already compatible with those, and the necessary math is well developed and worn and in no manner objectionable.

The idea that nothing actually propagates other than an excitation of a field when an unbound photon propagates in linear mode is explicit in our model. "A photon traveling at c" is therefore not an actual reference frame.

9. ### danshawenValued Senior Member

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There is a frame in which invariant rest mass is measured. These are very close to your own words, rpenner. This would be the rest frame, but it is not a single or preferred one.
I agree. The Lorentz transformation preserves the invariance of rest mass and the speed of light. Not intervals, if one also takes into account time dilation, which is not invariant. It is covariance (combining space and time into an interval) which got Minkowski into trouble.

The Lorentz transformations have no issues like Minkowski spacetime, because space and time are considered separately, and because it actually does make sense to do geometry and coordinate systems the way Lorentz did in cases where the situation could be analogous to the geometry of a straight, infinite paved road and a straight, infinite length vehicle to measure the transformations against, and to nail any origins of any coordinate system reference frames to.

When Minkowski uses the same delta t term for time dilation at or near the simultaneous events, the math breaks down because there is a possibility the time dilation near the events may be different or the events may be widely separated, or both, and time dilations are certainly not invariant, under the Lorentz transformations (which deal with INSTANTS, NOT INTERVALS of relative time), or anything else.

I don't need to do very much math of my own to point out where a mistake in math is. A problem in addition is simply a problem in addition. A problem in counting a quantity as invariant when it is not, is simply a problem in counting something invariant when it is not.

The math needed to completely correct the mistakes that were made as a result of this miscalculation will take many years to set straight, no doubt. I'm under no obligation to do so. I don't own the mistakes.

Last edited: May 5, 2016
10. ### rpennerFully WiredValued Senior Member

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So there are multiple coordinate systems where a given ponderable particle is at rest. But the topic of the post was not rest frames, but the Lorentz transform. And the Lorentz transformation is not constrained to use a rest frame as either starting or ending. So then, what language best describes what IS preserved by the Lorentz transform? The invariant mass: $c^{-2} \sqrt{E^2 - ( c \vec{p} )^2} = c^{-2} \sqrt{E'^2 - ( c \vec{p}' )^2} = m$ which can be measured in ANY frame, which is useful, because muons and the like rarely stay still long enough to get them on a scale.

Then you have misunderstood the material. The invariant measure of a time-like interval can be expressed as the proper time between events. $\Delta \tau = \sqrt{ \left( \Delta t \right)^2 - c^{-2} \left( \Delta \vec{x} \right)^2 } = \sqrt{ \left( \Delta t' \right)^2 - c^{-2} \left( \Delta \vec{x}' \right)^2 }$ which pretty much proves the Lorentz transformation preserves the invariance of this measure of intervals, following the same line of reasoning as above. You have to use the Lorentz transformation to learn what it preserves and what it does not.

You need math to support a claim in the field of math. Please give any example of the Lorentz transformation not preserving the measure of a space-time interval. James R has basically asked the same thing of you.

Stop pretending your bad experience in a physics class left you better able to explain physics. It's arrogant and it blinds you to the complete lack of persuasiveness of your baseless claims.

Invariance is only the simplest form of covariance.

11. ### danshawenValued Senior Member

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Turns out, it is a very good thing that it didn't.

Let's do some more math then.

if v = omega x r, (tangential velocity formula), and
L = (moment of inertia, I) x omega (angular momentum formula for unform circular motion), it follows that
L = I x (v^2/r^2)

This looks like what I might have been fishing for in the first post. The moment of inertia I is the rotational equivalent of mass. Looks like mc^2 is defining angular momentum here. What a surprise.

12. ### danshawenValued Senior Member

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You know, for quite a while, as I got older, the price of gasoline kept going up. Someone actually made a formula that related the passage of time to the price of gasoline, and for a while that formula even worked.

Then Arab spring came and the bottom fell out of the price of a barrel of oil.

You could relate time to the price of gasoline. You could relate time to space, too. It's really the same specious argument.

The Lorentz transformation derives length contraction and time dilation without positing a relationship between space and time. That is exactly why it works.

13. ### rpennerFully WiredValued Senior Member

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Try again.

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14. ### danshawenValued Senior Member

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Right.

L = I x omega
v = omega x r
L = I x (v/r)

Never said I was perfect. Where did that error come from? Still requires two photons to do what it has to. Angular momentum must be conserved for both, obviously. For some kinds of particles (Fermions) the angular momentum cannot sum to zero.

Another wrinkle.

Too classical, whatever that means.

15. ### przyksquishyValued Senior Member

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Non sequitur. It is perfectly possible for the spacetime interval to be invariant while time intervals on their own are not.

Completely wrong. For starters the spacetime interval has nothing to do with simultaneity. It can be defined for any two events which may or may not be simultaneous in any given reference frame.

That is an opinion, not a rational criticism.

You've given no reason that the spacetime interval between two events can't be defined. As to its invariance, that follows from the Lorentz transformation, like I already told you.

This is nonsense. The Lorentz transformation is linear, so it applies just as well to coordinate intervals as it does to the coordinates themselves. If the coordinates of event 1 in two different reference frames are related by a Lorentz transformation by

$\begin{eqnarray} t'_{1} &=& \gamma \bigl( t_{1} - \tfrac{v}{c^{2}} x_{1} \bigr) \,, \\ x'_{1} &=& \gamma (x_{1} - v t_{1}) \,, \end{eqnarray}$​

and the coordinates of another event 2 are related by

$\begin{eqnarray} t'_{2} &=& \gamma \bigl( t_{2} - \tfrac{v}{c^{2}} x_{2} \bigr) \,, \\ x'_{2} &=& \gamma (x_{2} - v t_{2}) \,, \end{eqnarray}$​

then you can take the difference of these two sets of equations to find that the intervals are related by

$\begin{eqnarray} t'_{1} - t'_{2} &=& \gamma \bigl( (t_{1} - t_{2}) - \tfrac{v}{c^{2}} (x_{1} - x_{2}) \bigr) \,, \\ x'_{1} - x'_{2} &=& \gamma \bigl( (x_{1} - x_{2}) - v (t_{1} - t_{2}) \bigr) \,. \end{eqnarray}$​

So it is perfectly justified to apply a Lorentz transformation to the coordinate intervals $\Delta t = t_{1} - t_{2}$ and $\Delta x = x_{1} - x_{2}$ between two events.

This is nonsense. Since time dilation is a special case and prediction of the Lorentz transformation, it is already accounted for when the Lorentz transformation is used to prove the invariance of the spacetime interval. Same with length contraction, relativity of simultaneity, and the fact that one reference frame is in motion relative to the other.

Last edited: May 5, 2016
16. ### paddoboyValued Senior Member

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An EMF is dynamic just as spacetime is dynamic.
Neither is really physical but both certainly interact with mass.

I know, just as all of it is good, thank you for the compliment.

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I know...and if I wasn't the author as you put it, I would have indicated that.

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I don't really need to ask you anything that is not obvious from the majority of your posts.
On the subject at hand, no it is not bullshit, but you are entitled to call it that, as obviously you are in the alternative section: It is a fact that spacetime is dynamic...It warps, it curves, it twists it lenses, it makes BH's, it expands, and it even waves......and your spacetime is different to mine also.
Got it?

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17. ### rpennerFully WiredValued Senior Member

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In addition to przyk demonstrating this for you just now in post #192, I showed how this was used to derive the velocity composition law in post #184.
The linearity of the transform is explicit and is ordained by its property of preserving the straightness of trajectories where Newton's first law of motion applies.

Because of this property, it has to be a special linear transform. Which leaves it in the form:

$\begin{pmatrix} t' \\ \vec{x}' \end{pmatrix} = \Lambda \begin{pmatrix} t \\ \vec{x} \end{pmatrix} = e^{ \begin{pmatrix} 0 & K \vec{\rho}^{\textrm{T}} \\ \vec{\rho} & 0 \end{pmatrix} } \begin{pmatrix} t \\ \vec{x} \end{pmatrix}$
or
$t' = \frac{1}{\sqrt{1 - K \vec{v}^2}} \left( t + K \vec{v} \cdot \vec{x} \right) \\ \vec{x}' = \vec{x} + \left( \frac{1}{\sqrt{1 - K \vec{v}^2}} - 1 \right) \frac{1}{ \vec{v}^2} \left( \vec{v} \cdot \vec{x} \right) \vec{v} + \frac{1}{\sqrt{1 - K \vec{v}^2}} t \vec{v}$

So if $\eta = k \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -K & 0 & 0 \\ 0 & 0 & -K & 0 \\ 0 & 0 & 0 & -K \end{pmatrix}$ we have
$\left( (A' - B') \right)^{\textrm{T}} \eta (C'-D') = \left( (\Lambda A - \Lambda B) \right)^{\textrm{T}} \eta (\Lambda C- \Lambda D) = \left( \Lambda (A - B) \right)^{\textrm{T}} \eta \Lambda (C-D) = (A - B)^{\textrm{T}} \Lambda^{\textrm{T}} \eta \Lambda (C-D) = (A - B)^{\textrm{T}} \eta' (C-D)$

With $\eta' = e^{ \begin{pmatrix} 0 & \vec{\rho}^{\textrm{T}} \\ K \vec{\rho} & 0 \end{pmatrix} } \eta e^{ \begin{pmatrix} 0 & K \vec{\rho}^{\textrm{T}} \\ \vec{\rho} & 0 \end{pmatrix} } \\ = \sum_{k=0}^{\infty} \sum_{\ell=0}^{\infty} \frac{1}{k! \ell !} \begin{pmatrix} 0 & \vec{\rho}^{\textrm{T}} \\ K \vec{\rho} & 0 \end{pmatrix}^k \eta \begin{pmatrix} 0 & K \vec{\rho}^{\textrm{T}} \\ \vec{\rho} & 0 \end{pmatrix}^{\ell} \\ = \sum_{m=0}^{\infty} \sum_{n=0}^{m} \frac{1}{ n! (m-n) !} \begin{pmatrix} 0 & \vec{\rho}^{\textrm{T}} \\ K \vec{\rho} & 0 \end{pmatrix}^n \eta \begin{pmatrix} 0 & K \vec{\rho}^{\textrm{T}} \\ \vec{\rho} & 0 \end{pmatrix}^{m-n} \\ = \eta + \sum_{m=1}^{\infty} \sum_{n=0}^{m} \frac{1}{ n! (m-n) !} \begin{pmatrix} 0 & \vec{\rho}^{\textrm{T}} \\ K \vec{\rho} & 0 \end{pmatrix}^n \eta \begin{pmatrix} 0 & K \vec{\rho}^{\textrm{T}} \\ \vec{\rho} & 0 \end{pmatrix}^{m-n} \\ = \eta + \sum_{m=1}^{\infty} \begin{pmatrix} 0 & \vec{\rho}^{\textrm{T}} \\ K \vec{\rho} & 0 \end{pmatrix}^m \eta \sum_{n=0}^{m} \frac{(-1)^n}{ n! (m-n) !} \\ = \eta + \sum_{m=1}^{\infty} \begin{pmatrix} 0 & \vec{\rho}^{\textrm{T}} \\ K \vec{\rho} & 0 \end{pmatrix}^m \eta \sum_{n=0}^{m} \frac{1}{m!} {m \choose n } (-1)^n \\ = \eta + \sum_{m=1}^{\infty} \frac{1}{m!} ( 1 + (-1) )^m \begin{pmatrix} 0 & \vec{\rho}^{\textrm{T}} \\ K \vec{\rho} & 0 \end{pmatrix}^m \eta \\ = \eta + 0 \\ = \eta$

Because $\begin{pmatrix} 0 & \vec{\rho}^{\textrm{T}} \\ K \vec{\rho} & 0 \end{pmatrix} \eta + \eta \begin{pmatrix} 0 & K \vec{\rho}^{\textrm{T}} \\ \vec{\rho} & 0 \end{pmatrix} = 0$ as you may verify.

So K=0 means relativity of standard of rest gives Galilean Relativity with Galilean transforms, Newtonian velocity composition and absolute time, while $K=c^{-2}$ gives Special Relativity with Lorentz transforms, Einstein velocity composition, and an invariant measure of space-time intervals.

Last edited: May 5, 2016
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18. ### danshawenValued Senior Member

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Velocity composition is not in question here, but if you plan to throw that into the mix, there are even deeper problems, because velocities are not invariant either, and the speed of light is much more than just a velocity.

Lorentz transforms provide a means to calculate length contractions and time dilations, and this is not in doubt. But they do not provide a means for handling an implicit covariance, particularly one that has deliberately combined something invariant (the speed of light) with something not invariant (time dilation) in a manner crafted only for reconciling issues of simultanaeity

What does simultanaeity even mean, and what does it mean for time intervals (NOT INSTANTS) between reference frames in relative motion with different rates of time dilation? That is not what Lorentz transforms were crafted to accomplish. It is, all by itself, contrary even to Minkowski's conception of simultanaeity. The speed of light may be invariant, but when combined with time intervals, you have neither invariance nor simultanaeity.

Simultanaeity is about time instants, not intervals.

Last edited: May 6, 2016
19. ### danshawenValued Senior Member

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A pair of quantum entangled electrons in an atom spontaneously flip spins. It requires a time interval delta t = t2-t1 = 0, as simultaneity demands.

A pair of quantum entangled photons created by a beamsplitter and separated by .005 light seconds spontaneously flip polarization states. Again, delta t = 0, as simultaneity demands.

The events in both cases are simultaneous. Your spacetime interval just got divided by delta t = zero in both cases. Handle it. Is the interval still invariant? Is the speed of light still invariant? There is more to simultaneity than photonics, beam splitters, and interferometers.

We know there can be simultaneous events at points that are separated and also synchronized faster than a photon can travel between them. How do you propose we fathom this quandry, using Minkowski spacetime intervals?

Last edited: May 6, 2016
20. ### przyksquishyValued Senior Member

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Simultaneity is ultimately a matter of definition and is defined in a frame-dependent way in relativity. Einstein gave a frame-dependent definition of synchronicity, and therefore also simultaneity, in his 1905 paper in which he introduced special relativity. We still use the same definition, or definitions equivalent to it, today.

I have no idea why you are obsessed with this since simultaneity is not defined any differently in Minkowski geometry than it is in relativity in general.

No. The speed of light is, like any other speed, the distance $\Delta r$ travelled by a light ray divided by the time $\Delta t$ it took to travel that distance. $\Delta r$ and $\Delta t$ are different in different reference frames (because of the different standards of length and time used in different reference frames), but this does not prevent the speed $\displaystyle c = \frac{\Delta r}{\Delta t}$ from being invariant.

That is a mischaracterisation of entanglement. Flipping the spin of an entangled particle does not instantaneously flip the spin of other particles it is entangled with.

You are probably confusing with measurement and quantum state collapse: according to at least a naive reading of the postulates of quantum mechanics, if you measure the spin of a particle then its entangled partner "instantaneously" collapses to a definite spin state, even if it is a great distance away. It is still unsettled whether quantum state collapse even occurs at all as a physical process, let alone questions like which reference frame the collapse is supposed to be happening instantaneously in, so making grand claims about what entanglement and collapse mean for relativity is dubious at best. What is generally agreed upon at the moment is that the act of measuring entangled particles cannot be used for faster-than-light communication, since the correlations associated with measurements on spatially separated entangled particles satisfy the no-signalling constraints.

Either way, whatever the implications of entanglement are for relativity -- and this is far from clear -- they would concern relativity in general and not only Minkowski geometry.

Last edited: May 6, 2016
21. ### danshawenValued Senior Member

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Matter may TRAVEL or matter may flip spin states. Energy that is unbound propagates (but does not actually 'travel' in bulk the way that matter does), and may flip polarization states. In this manner, propagation or travel is what excitations of a quantum field do.

But an EVENT is an energy exchange. Flipping spin states only changes orientation, not location, and this is why it may occur FTL can propagate. Quantum spin and quantum rotation comprised of more than one photon in the aggregate, are capable of both FTL rotations, and in the case of the bound energy that is matter, FTL spin flips or changes of polarization states. If a zero spin boson did not exist, we would know nothing at all about quantum spin states. G-d needed us to find his particle. Minkowski was not involved.

And Minkowski's ignorant 19th century math is not going to help you fathom this even in small measure.

The universe comprised of energy exchange events is relativistic at its very core. The fastest energy exchange events in it are changes of entanglement states, not events synchronized by means of the limit of photonic excitation of a quantum field. The static and continuous vector geometry of Minkowski has no place and no inertia here. Apply it to bulk solids if you like.

The only absolute space is INERTIALESS LIGHT PROPAGATION TIME connecting energy exchange events, OR the time dilation at the exact centers of discrete packets of the bound energy we refer to as matter. The only absolute time is the discrete instant of 'NOW' of quantum spin state changes, an area of knowledge that 21st century science is poised to explore. But it will require a divorce from 19th century concepts of absolute space and absolute time, and also Minkowski's 4D intervals which tried to make them permanent geometric fixtures in an inertialess relativistic quantum field where there is only energy, light propagation time, bound energy travel time, and quantum spin flips which we refer to as events.

You will be needing an understanding of what time is, and you have a choice between a hare and a tortoise to help you figure it out. If you choose the tortoise, you probably will be a slower learner. Good luck with your 19th century tortoises. At least, they age well.

Last edited: May 6, 2016
22. ### James RJust this guy, you know?Staff Member

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danshawen,

Why did you completely ignore my previous post? Where is your mathematics?

While I wait for that (and it's looking like it will be a very long wait), I thought I'd respond to one of your later posts.

You say the Lorentz transformations are 100% correct.
Using the Lorentz transformations, rpenner proved that the interval is invariant when you change reference frames. He did it right here in this thread.

You can't have it both ways. If you accept Lorentz, then you cannot deny rpenner's proof the invariance of the interval (unless you can find a flaw in that proof). If, on the other hand, you reject the invariance of the interval, then you also reject the Lorentz transformations. One follows from the other; you have been given a mathematical proof that this is the case, which is the gold standard.

So, please let us know. Do you (a) reject both the interval and the Lorentz transformations, or (b) accept both the interval and the Lorentz transformations? You can't accept one and reject the other without being logically inconsistent.

But you've been given proof that what Minkowski said is true.

Do you have a disproof? Or do you reject the Lorentz transformations?

Who cares what you think about what is "clean" and what isn't? Minkowski's interval has been proven to be invariant, right here in this thread! That's all that matters.

Consider the Lorentz transformation of the time coordinates for two events:
$t_1' = \gamma (t_1 - vx_1/c^2)$
$t_2' = \gamma (t_2 - vx_2/c^2)$

Now look:

$t_2' - t_1' = \gamma (t_2 - vx_2/c^2 - (t_1 - vx_1/c^2))$
$= \gamma ((t_2 -t_1) - v(x_2-x_1)/c^2)$

Notice that the time interval $\Delta t=t_2 - t_1$ transforms in exactly the same way as either of the time coordinates taken separately. (This follows directly from the fact that the Lorentz transformation is a linear transformation in the coordinates.)

In other words, the Lorentz transformations work just as well for time intervals as they do for "instants" of time.

Who cares? I've just proven for you that we can transform time intervals in the same way that we can transform time coordinates. And that's all that matters.

Wrong! It's right there, built into the Lorentz transformation of the time interval that I just showed you.

No. The spacetime interval is the same in different frames. In general, $\Delta t'$ will be different from $\Delta t$.

rpenner explained this point to you three times. Now I've added a fourth time. When will you learn?

You have been given a proof that the invariance of the interval follows directly from the Lorentz transformations. If one stands, the other stands. If one falls, both fall.

It is logically impossible to have one and not the other. Can you see that?

You have been presented with a proof that those 4D intervals are invariant in exactly the same way that the speed of light is. And that's all that matters. Your opinion counts for nothing when up against the mathematical proof.

23. ### James RJust this guy, you know?Staff Member

Messages:
37,359
After posting the above, I notice that przyk already covered the same ground, in this previous post.

Still, it's worth emphasising how silly it is to argue against a mathematical proof on the grounds that you don't find some aspect aesthetically pleasing.