# Space-time is a reality

Discussion in 'Physics & Math' started by Fork, Sep 28, 2013.

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1. ### ForkBannedBanned

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319
Loop quantum gravity

Even if it is correct, the thermodynamic approach says nothing about what the fundamental constituents of space and time might be. If space-time is a fabric, so to speak, then what are its threads?

One possible answer is quite literal. The theory of loop quantum gravity, which has been under development since the mid-1980s by Ashtekar and others, describes the fabric of space-time as an evolving spider's web of strands that carry information about the quantized areas and volumes of the regions they pass through. The individual strands of the web must eventually join their ends to form loops — hence the theory's name — but have nothing to do with the much better-known strings of string theory. The latter move around in space-time, whereas strands actually are space-time: the information they carry defines the shape of the space-time fabric in their vicinity.

http://www.nature.com/news/theoretical-physics-the-origins-of-space-and-time-1.13613

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3. ### ForkBannedBanned

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We live in a 3+1 space-time that is intended as a description of the universe with three space dimensions and one time dimension. Space-time dimensionality seems so natural that it is rarely criticized. Experiments and the highly successful relativistic theories teach us that there are four fundamental dimensions, among them is time that is treated as a special dimension. The specialty of time can be removed, leading to the concept that time is simply a function of four new fundamental dimensions, which have now identical properties, in combination with Lorentz invariance. A model is deduced in which a 4-space, characterized by four space-like coordinates, may host four "equivalent but orthogonal" space-times, each with three spatial coordinates and one temporal coordinate. Coordinates are shared; therefore the 4-space and the four space-times are all in one. Electromagnetic interaction is confined in each space-time and the role of the speed of light appears to be that of a barrier for the electromagnetic interaction. The motion of objects can be described by four-dimensional optics in the 4-space. Each of the four space-times may host a universe and, in agreement with recent observations, the proposed model can be directly applied to problems like the cosmological matter-antimatter asymmetry and dark-matter issues. Space travel may also benefit from the concepts presented.

http://arxiv.org/abs/physics/0410054

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5. ### chingluValued Senior Member

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Hi,

Can you explain something to me?

Assume a light pulse is emitted.

Then, for any (x,y,z), the time coordinate is absolutely dependent on x, y, z.

Normally, when we think of dimensions, we think of them as absolutely independent.

Yet, the relativity time dimension is absolutely dependent on the space coordinate.

Thanks for explaining this.

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7. ### AlphaNumericFully ionizedRegistered Senior Member

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Chinglu, if you want to endlessly talk about the same damn ignorance you have about light spheres then do so in the pseudo section. You had your chance, you've shown you neither understand relativity nor wish to understand relativity, all of these questions of yours are completely disingenuous. As such if you cannot stop yourself from revisiting the same tired talking points do not post in this sub-forum. If you cannot stop yourself I'll stop you.

8. ### eramSciengineerValued Senior Member

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What the hell are you saying?

9. ### rpennerFully WiredValued Senior Member

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Chinglu is saying that in any particular inertial coordinate frame the 3-dimensional space-time hypersurface of the light-cone from a pulse of light starting at a particular event $(t_0,x_0,y_0,z_0)$ is given by the locus of all events $(t,x,y,z)$ that satisfy the relationship: $t = t_0 + \frac{1}{c} \sqrt{(x - x_0)^2+(y - y_0)^2+(z - z_0)^2}$ ( or more symmetrically, $t \geq t_0 \quad \textrm{and} \quad c^2 (t - t_0)^2 - (x - x_0)^2 - (y - y_0)^2 -(z - z_0)^2 = 0$ ), which is correct.

What is incorrect is that Chinglu thinks this is not relative. This is probably because Chinglu typically only works with light cones that start at the origin of the coordinate system $(t_0,x_0,y_0,z_0) = (0,0,0,0)$ and so when he Lorentz-transforms this to $(t'_0,x'_0,y'_0,z'_0) = (0,0,0,0)$ he thinks that he has found an absolute relationship for all the events in the light cone, there is a corresponding event in the Lorentz-transformed light cone with the same coordinates. What Chinglu hasn't figured out is that they are different events (except for origin). The Lorentz transform preserves coordinates the light cone at the origin and it preserves the origin and it preserves certain space-time hyperbolas which are part of a light cone from the origin. But this is as unremarkable as saying a Euclidean rotation preserves the coordinates of the center and circles centers on that center.

What Chinglu is objecting to is literally the consequences of the postulate of special relativity, that the speed of light is always measured to be c in any inertial frame. But he's not showing that it leads to inconsistencies, so he isn't making a cogent argument. Indeed, as Eram points out, Chinglu isn't actually communicating an understanding of the theory he objects to.

10. ### chingluValued Senior Member

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I simply asked a question and AN comes pounding around.

I would assume if he had the answer, he would just give it.

And, you have no idea what I am talking about as usual.

If we consider pure Euclidean space, then given any x or y then z can be any real number. It is absolutely independent. Any child knows this.

Now, let us consider space-time where only light flashes and light spheres are present. To be a real geometry, all dimensions must be independent of one another.

However, the equation below under SR must hold true in the world of only light flashes.

$c^2 (t - t_0)^2 - (x - x_0)^2 - (y - y_0)^2 -(z - z_0)^2 = 0$

So, if I pick x, y, z, then t is determined and is therefore not independent as required by a true geometry.

Now is this true of false?

Remember, in a pure Euclidean space, if I pick x, y, then z can be anything.

Anyway, can you and AN answer this with all of your "deep" understanding? I get confused by all this confusion.

11. ### rpennerFully WiredValued Senior Member

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4,833
You did not ask a simple question but rather a leading question that already assumed physics contrary to special relativity therefore you were trolling pseudo-physics instead of precise and communicable predictive models of observable phenomena. You also misstate the burden of proof. It is your burden to describe and argue for your thesis, while AlphaNumeric has no burden to respond to you no matter how wrong you are. You simply don't matter enough for your wrong ideas to demand a reply. Likewise you don't matter enough to get an education in logical reasoning, math and physics when by rights you should pay for such an education so you will properly value the time taken to explain topics and assign homework to you.

Actually, I know exactly what you are saying. I also know it's nonsense.
Seriously? Special relativity has the postulate that the speed of light is fixed at c and you are complaining that if you fix $t_0, x_0, y_0, z_0, x, y, \quad \textrm{and} \quad z$ that there is only one solution for the light moving from $(x_0, y_0, z_0)$ to $(x,y,z)$? If there was more than one solution that would violate the postulate that the speed of light is fixed at c and would prove Special Relativity inconsistent.

So now you are reduced to complaining that Special Relativity is mathematically self-consistent. You might as well complain that 2 + 3 = 3 + 2. The alternative is nonsense and therefore you advocate nonsense.

That's completely the wrong way to think about it. If you consider a "pure" 3-dimensional Euclidean space then you have distinct points. If you have any two distinct points, you can classify all the points in the space as either co-linear with the two points or not-co-linear with the two points. The set of all points which are colinear with the two points is a 1-dimensional sub-space of the full 3-dimensional space. In this 1-dimensional sub-space you can describe all the points in relation how far along the line they are in terms of the first two points. $(\xi -O ) = k_{\xi}(A -O)$ describes the point $\xi$, which must be on the line, in terms of the orgin O and the directed distance to the first point A and a scalar multiplier, k. This can also be written as $\xi = O + k_{\xi}(A-O)$.

If you have two points and a third point which is not co-linear with the first two, you can similarly form a 2-dimension sub-space of all points co-planar with the three points. If you have four points, not all mutually co-planar, you can form a 3-dimensional sub-space which is identical to the original space. Now if you have 4 points, not all mutually co-planar, you can assign one as an origin and the others as foundation for basis vectors. You learn that any point in the whole 3-dimensional space can be written in terms of the points O, A, B and C as: $\xi = O + k_{\xi}(A-O)+ \ell_{\xi} (B-O)+ m_{\xi}(C-O)$. This description of the location of $\xi$ in terms of three numbers (k,l,m) is the Euclidean base concept behind coordinate systems.

The Cartesian coordinate system has an origin and three basis vectors, $\hat{x}, \hat{y}, \hat{z}$ so that any point in the 3-dimensional space can be written in terms of the orgin O as $\xi = O + x_{\xi} \hat{x} + y_{\xi} \hat{y} + z_{\xi} \hat{z}$ where the basis vectors are all of the same unit length: $\hat{x}^2 = \hat{y}^2 = \hat{z}^2 = 1$ and are all mutually orthogonal so that $\hat{x} \cdot \hat{y} = \hat{x} \cdot \hat{z} = \hat{y} \cdot \hat{z} = 0$. This in turn gives the Cartesian coordinate system its greatest strength over the klm arbitrary system -- you can easily figure the distance between points based on just the differences of coordinates.
$\left| \xi - \eta \right| = \left| \left( O + x_{\xi} \hat{x} + y_{\xi} \hat{y} + z_{\xi} \hat{z} \right) - \left( O + x_{\eta} \hat{x} + y_{\eta} \hat{y} + z_{\eta} \hat{z} \right) \right| = \left| \left( x_{\xi} - x_{\eta} \right) \hat{x} + \left( y_{\xi} - y_{\eta} \right) \hat{y} + \left( z_{\xi} - z_{\eta} \right) \hat{z} \right| \quad = \sqrt{ \left| \left( x_{\xi} - x_{\eta} \right) \hat{x} + \left( y_{\xi} - y_{\eta} \right) \hat{y} + \left( z_{\xi} - z_{\eta} \right) \hat{z} \right|^2 } = \sqrt{\left( x_{\xi} - x_{\eta} \right)^2 + \left( y_{\xi} - y_{\eta} \right)^2 + \left( z_{\xi} - z_{\eta} \right)^2 }$

So in summary, a N-dimensional space (or sub-space) requires N (possibly Cartesian) coordinates to describe each point. And an unbounded N-dimensional space (or sub-space) places no limits on what those coordinates might be. But it is incorrect to claim coordinates are "absolutely independent" if we put constrains on them. If we consider the 1-dimensional subspace which is colinear with points O and A, then in klm coordinates, the l and m coordinates are always zero. If we consider the 1-dimensional subspace which is colinear with points (1,2,3) and (2,4,8) in a Cartesian coordinate system, if we know any one coordinate we know the other two. Any two of the follow equations completely constrain two of the coordinates in terms of the first known.
$10x - 5y = 0 \\ 10x - 2z = 4 \\ 5y - 2z = 4$
Alternately one may describe the constraint is that every point on the 1-dimensional subspace can be written in terms of a single invented coordinate k.
$\xi = O + ( k (2-1) + 1 ) \hat{x} + ( k (4-2) + 2 ) \hat{y} + ( k (8-3) + 3 ) \hat{z} = ( k + 1 ) \hat{x} + ( 2 k + 2 ) \hat{y} + ( 5 k + 3 ) \hat{z}$.

This is basic geometry that has been part of physics since Newton wrote the first book on the topic.

You mean light cones. A sphere is defined by its fixed radius, while a cone has a radius which is not fixed.
Here you confuse "dimension" which is a geometric description of a space with "coordinates of a point" which are just convenient labels determined in relationships to other points.

Correct. The light cone in 4-dimensional space-time is a 3-dimensional subspace. By picking 3 coordinates you have eliminated all the degrees of freedom.
We can write the Cartesian coordinates of every point in the 3-dimensional subspace by using 3 spherical coordinates θΦr thus proving it has 3 degrees of freedom.
$t = t_0 + \frac{r}{c} \\ x = x_0 + r \cos \phi \cos \theta \\ y = y_0 + r \cos \phi \sin \theta \\ z = z_0 + r \sin \phi$
Or alternately in xyz:
$t = t_0 + \frac{1}{c}\sqrt{(x - x_0)^2 + (y - y_0)^2 +(z - z_0)^2} \\ x = x\\ y = y \\ z = z$

You don't know geometry or how to support your claims in a discussion.
Part true, part false.
In an unbounded 3-dimensional space or subspace, you can pick 3 coordinates from their full range which may or may not be the whole set of real numbers. Typically, adding an additional equality as constraint reduces the dimensionally of the allowable points by 1. So a 3 dimensional space with 2 equations constraining coordinates results in a 1 dimensional space (but for some exceptional circumstances). The constraint on the propagation speed results in creating a 3-dimensional subspace of a 4-dimensional space-time.

I think this understanding of algebraic geometric is a prerequisite for taking calculus instruction in high school or first year university and that calculus is a prerequisite for even Newtonian physics. So it's not so very "deep".

12. ### chingluValued Senior Member

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1,637

I can simplify for readers.

You have failed to prove space-time is an actual geometry since even you agree if only light flashes are considered, then the time dimension is bound by the space coordinate (x,y,z) and is therefore not independent as required by a true geometry.

So, you have not demonstrated space-time is a real geometry.

Can you explain why you believe in space-time and yet do not agree that space-time is a geometry?

13. ### eramSciengineerValued Senior Member

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1,877
What do you mean by an "actual", "true" or "real" geometry? Why does it require the time dimension to be independent?

He does, only you don't.

If you wanna argue, at least argue over a point that isn't nonsensical. That's the least one can do.
I don't mean to be insulting, but you need to brush up your English as well.

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15. ### arfa branecall me arfValued Senior Member

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That makes as much sense as saying that a cube with a set of coordinates only has a "true" geometry if it has an independent time coordinate. Which could mean that nobody can align their time coordinate with the cube's "independent" time, so therefore the cube doesn't exist.

Something like that . . .

16. ### eramSciengineerValued Senior Member

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Exactly. He makes NO sense whatsoever.

17. ### chingluValued Senior Member

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Well, that's not true. Normally, if we allow non-light motion in a space we have a Minkowski space under SR.
http://en.wikipedia.org/wiki/Minkowski_space

Yet, for a space containing only light flashes, the Minkowski space does not hold true.

In this space, given x,y,z, there is only one t. Hence, it is not a 4-D space. It is simply a solution to an equation based on 3-D space.

18. ### arfa branecall me arfValued Senior Member

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Minkowski spacetime has a geometry that depends on the notion of separation between events.
Rubbish. If the universe contained only light, Minkowski spacetime would not be "false". Your claim makes no sense.

19. ### LaymanTotally Internally ReflectedValued Senior Member

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I would think that if the universe only contained light, the universe wouldn't exist. Minkowski spacetime may only be as real as our inability to not travel the speed of light. If everything there only traveled the speed of light, everything would be contracted to zero. There wouldn't be any spacetime to measure.

They love to talk about Minkowski spacetime all the time in those pop physics books. I was surprised that it seemed so basic from the information on the internet. I think there is a lot more too it than what you can find on the net or they just like to talk big in those books. But, it seems like they are always writing about how someone use Minkowski spacetime to do this and that or whatever...

I think he means that it just wouldn't be necessary. NASA still uses Newtons theory of gravity to put things up in space. They don't use Einsteins theories because they don't really need too. They are not sending anything up there close to the speed of light. If you didn't have any objects moving close to the speed of light, why would you need to worry about Minkowski spacetime? You would just be doing a lot more work than what was actually necessary.

20. ### rpennerFully WiredValued Senior Member

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This sentence is so painfully untrue that it could well form the basis of a site-wide banning. All that is demonstrated above is that the space-time locus of one particular light flash is one dimension less than the dimensionality of the space-time as a whole. This is true of any hypersurface of propagation of any phenomenon at any speed, including non-isotropic speeds or a finite number of coherent reflections. You did not address the properties of 2-or-more light flashes. Indeed, as the light flash in no way changes the Lorentzian properties of the space, your argument cannot succeed. In fact, you didn't even address the Lorentzian properties of the space.

Kind of off the point since we are talking about the model of phenomena called Minkowski spacetime and not the actual observed universe. Being able to understand the concept of Minkowski spacetime greatly simplifies the intellectual exercise of understanding the consequences of Einstein's physical theory of special relativity. Since Chinglu hasn't yet mastered even Euclidean descriptions of space-time (algebraic geometry) he isn't capable of expressing sentiments more advanced than "I don't like SR" and "I think nonsense is to be preferred to SR."

And SR textbooks and GR textbooks and advanced papers in GR and cosmology where they express findings in deviations from Minkowski spacetime.
Actually it is quite basic.
• You have four dimensions,
• one of them is not like the others,
• this is reflected in the generalization of "distance between points" used which has different signs for squared temporal differences and squared spatial differences,
• trigonometric functions are used for spatial rotations and hyperbolic functions are used for space-time analogues of rotation,
• both trig- and hyperbolic-rotations preserve
• this generalization of "distance between points" and
• an analogous generalization of the dot product.
Where a rotation moves parts of certain circles to other parts of the same circle, a hyperbolic rotation moves parts of certain hyperbolas to other parts of the same hyperbola. No part of that geometry was discussed by chinglu so how is he arguing against it?

But this site is biased towards algebra so it is easier to write algebraic proofs in a particular coordinate system rather than geometric proofs.
Example:
$e^A \equiv I + \sum_{k=1}^{\infty} \frac{1}{k!} A^k ; \quad \quad \quad \textrm{for any square matrix} \, A \\ e^{ \tiny \begin{pmatrix} 0 & -\theta \\ \theta & 0 \end{pmatrix} } = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \\ e^{ \tiny \begin{pmatrix} 0 & \kappa \\ \kappa & 0 \end{pmatrix} } = \begin{pmatrix} \cosh \kappa & \sinh \kappa \\ \sinh \kappa & \cosh \kappa \end{pmatrix} \begin{pmatrix} \cos \theta_1 & -\sin \theta_1 \\ \sin \theta_1 & \cos \theta_1 \end{pmatrix} \begin{pmatrix} \cos \theta_2 & -\sin \theta_2 \\ \sin \theta_2 & \cos \theta_2 \end{pmatrix} = \begin{pmatrix} \cos ( \theta_1 + \theta_2 ) & -\sin ( \theta_1 + \theta_2 ) \\ \sin ( \theta_1 + \theta_2 ) & \cos ( \theta_1 + \theta_2 ) \end{pmatrix} \begin{pmatrix} \cosh \kappa_1 & \sinh \kappa_1 \\ \sinh \kappa_1 & \cosh \kappa_1 \end{pmatrix} \begin{pmatrix} \cosh \kappa_2 & \sinh \kappa_2 \\ \sinh \kappa_2 & \cosh \kappa_2 \end{pmatrix} = \begin{pmatrix} \cosh ( \kappa_1 + \kappa_2 ) & \sinh ( \kappa_1 + \kappa_2 ) \\ \sinh ( \kappa_1 + \kappa_2 ) & \cosh ( \kappa_1 + \kappa_2 ) \end{pmatrix}$
Or [post=3101532]this post where I algebraically proved that Lorentz transforms (and spatial rotations) preserve the generalization of the dot product[/post].

But when I [post=3102488]make the effort to provide geometric demonstrations[/post], chinglu ignores that as well.

Special Relativity's most visible usage is in the non-NASA, space-based GPS system which also uses GR. NASA and Stanford University collaborated on Gravity Probe B, which directly tested the relevance of GR to stuff in orbit. Finally, in 1969 NASA put the first of 3 retroreflectors on the moon allowing for precise laser ranging which rules out a number of theories close to, but distinct from, GR. But if chinglu was arguing that SR is not necessary for some practical purpose then he would be arguing from physical evidence and not just against the mathematical model that he doesn't yet understand.
SR is needed in synchronizing satellite feeds of TV broadcasts. I once shot down a startup plan because the fundamental assumption ignored the minimum latency of a globe-spanning Internet connection. I'm not a fiber optic or copper or waveguide engineer, but simple SR says not everyone in the world can get the same info in 30 milliseconds from creation as the Earth has a radius of about 20 light-milliseconds.

21. ### chingluValued Senior Member

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1,637
Let's see, you have yet to express a proof that the Minkowski space with the time dimension of only light flashes is anything more than a discrete finite space in the time dimension.

Since geometric spaces that contain real components must contain only dimensions of real cardinality, then your argument fails.

You attempted to extend your failed argument by considering additional flashes. Yet, this logic does not add a real cardinality of choices in the time dimension unless you can prove there is an uncountable real cardinality of light flashes in the universe. Let me know when you think you can prove this so that I can correct you.

In any event you have not yet shown that the time dimension in the Minkowski space with only light flashes contains a real cardinality of possibilities for the time dimension given light flashes.

How long will you take for this?

If you cannot prove this, then the Minkowski space with only light flashes is not a geometric space.

22. ### chingluValued Senior Member

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Also, as far as GPS sans GR proving SR is valid, we have a current thread on this subject.

Please feel free to join so that you can show readers that your thoughts are valid while mine are invalid. I welcome your input in that thread. Let's see how it goes.

However, I am surprised you changed the subject from your failed argument that a light flash only Minkowski space is valid. I would have stuck to that if I felt my argument was valid.

Why did you change the subject?

23. ### rpennerFully WiredValued Senior Member

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4,833
My argument is that the light flash represents an algebraic constraint and doesn't tell you about the geometry of space-time.

If you want an example, if we replace the light flash with a light pulse of finite duration, the finite-duration light cone is bounded past-and-future by the light cones of the first and last light. This illuminated subspace has dimensionality 4.