# Time in Special Relativity

Discussion in 'Physics & Math' started by Pete, Apr 10, 2010.

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1. ### Jack_BannedBanned

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No problem.

Where is the center of the light sphere in the coords of the stationary frame?

Last edited: May 22, 2010

3. ### PeteIt's not rocket surgeryModerator

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So you agree with my previous post, then. That's good.

Apply the same logic:
Every event (x,y,z,t) on the expanding light sphere is such that x² + y² + z² = ct.
In the stationary frame, light is expanding spherically around the emission point at (x,y,z)=(0,0,0).

Do you agree that any event (x,y,z,t) such that x² + y² + z² = ct transforms to (x',y',z',t') such that x'² + y'² + z'² = ct'² ?
It's not hard to prove.

Last edited: May 22, 2010

5. ### AlphaNumericFully ionizedModerator

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He should know how to do it, seeing as I've walked him through the n dimensional generalisation of it. More than once!

And I can tell you now what he'll say, he'll say that they arent a sphere because the central point (x,y,z)=(0,0,0) doesn't map to the central point (x',y',z')=(0,0,0). The fact the photons are arranged in a sphere is completely lost on him, he seems to think that because some point with no object at it moves to some other point with no object at it then magically the photons, which are all the same distance from one and only one point, are not in a sphere. The fact a sphere is defined as the region which is a specified distance from a single point seems to have no bearing on his brain.

7. ### Jack_BannedBanned

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So what?

What light sphere is this x'² + y'² + z'² = ct'² on? As you transform x² + y² + z² = ct, what are you seeing?

Perhaps you should spend more time on this.

8. ### Jack_BannedBanned

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If you map out LT, you will find two light spheres in the coords of the stationary frame.

The stationary light sphere is s aphere located at (0,0,0).

The moving light sphere is located at ( vt,0,0 ) cept it is an ellipse elongated on the x-axis in the coords of the stationary frame.

This invalidates SR.

I see you have not yet been able to map out space yet.

9. ### PeteIt's not rocket surgeryModerator

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So, we agree that any event (x,y,z,t) such that x² + y² + z² = ct transforms to (x',y',z',t') such that x'² + y'² + z'² = ct'². Excellent.

In the stationary frame, do you agree that the expanding light sphere is completely described by the equation x² + y² + z² = ct ?
I.e. at any time t, all points of the light flash are a distance ct from where the light was emitted at (x,y,z)=(0,0,0) ?

10. ### funkstarratsknufValued Senior Member

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It is obvious that someone didn't learn anything from the previous ban.

I suggest a swiftly enforced and prolonged exile in the land of indifference. This place has enough cranks as it is: I see no reason to tolerate blatant trolling in the Physics & Math subforum.

11. ### PeteIt's not rocket surgeryModerator

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You are an incoherent moron.

At different times. Of course you get a different shape if you consider different sections of an expanding sphere at different times. What would you expect? Are you deliberately being as thick as an overcooked custard?

I see you are still refusing to think about how time transforms under a lorentz transform.

I agree with funkstar. Your willful ignorance is no longer forgivable.

12. ### mikelizziRegistered Senior Member

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I can see that this is a very stress filled thread and at the risk of making it worse I would like to put my two cents in. I had the same problem with Special Relativity as Jack, for 25 years! I could never follow the explanations that other were telling me. They seemed like obfuscation. I think I have a good understanding of SR now. I have resolved all my issues with it. But I only came to that understanding after I took a course in linear algebra, at the age of 58.

If it will help with this thread I would like to explain how SR clicked for me after some practice with linear algebra.

13. ### Jack_BannedBanned

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I will step in for a moment.

The moving light sphere is a ellipsoid elongated on the x-axis in the coords of the stationary frame centered at (vt,0,0).

Yet, it is also centered at (0,0,0) as a sphere.

If you really understand SR, then you uinderstand this.

14. ### Jack_BannedBanned

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sorry pete, this is not up to my standards.

i know you are well respected here and your views are god.

when in the time of the stationary frame does the moving frame see light a distance r.

you have not given an answer and there is one.

if you are so good, you should know it.

then if you know it, what does this do to LT?

15. ### Jack_BannedBanned

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i think you are jealous of me.

why do you follow me around?

I checked my backside and see nothing anyone would want. But, you persist.

16. ### funkstarratsknufValued Senior Member

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Go away, Jack_. You're not fooling anyone.

17. ### AlphaNumericFully ionizedModerator

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No, you don't. I don't know how many times you need this explained to you, we must be past 10 by now.

You are considering the centre of the sphere and where it maps to. This is wrong. There's nothing physical there, your physical predictions will depend only on the relative position and motion of things in the space-time. When someone asks "What does the photon sphere transform into under a Lorentz transformation?" they are asking "If I apply a Lorentz transform to each and every photon in the system then what shape do the results arrange themselves into?". Its a sphere and you've been walked through the proof by at least 3 different people, including myself multiple times in general N dimensions.

The question "Where does the space-time point at the centre of a photon sphere map to under a Lorentz transform" is irrelevant. The position of the photons in the new frame depends only on their individual positions and the transformation being applied. What happens to the centre is immaterial to your physical predictions. Each frame sees one and only one light sphere. Physically this is obvious and mathematically its straightforward. All objects in the space-time transform consistently and in such a way to leave physical results/causality unchanged.

Go back and read my explainations about multiple emitters in one of your threads in pseudo. If all move at a constant velocity and pass through one another at the moment one of them emits a flash of light then its impossible to tell which one did it from the motion of the physical objects. You're arguing that dependent on your frame you get different photon spheres and thus SR, according to you, predicts multiple ones. It doesn't. If it did it would contradict the requirement that light's velocity is independent of the velocity its emitter. If it were dependent then you'd be able to pick out which emitter emitted it. But you shouldn't if the SR postulates are true and SR predicts that you can't by construction.

Apply a Lorentz transformation to all physical objects in the space-time, and nothing else, and you obtain a mathematically consistent description whose physical implications are experimentally tested and verified to enormous levels of accuracy.

Your obsession with "Where does the point (x,y,z) = (0,0,0) map to under (x,y,z) -> (x',y',z')" is utterly without justification and you've been told why many times. The point has no physical properties, it cannot influence any physical predictions since they all arise (in SR at least) from the behaviours of physical objects. If you put in a physical object to sit at the emission point (as viewed by a stipulated frame) then it moves time-like and transforms as such. The space-time point without an object there is defined by the centre of a photon sphere, which transforms light-like (aka null-like). Since an object at rest (in a given frame) is moving time-like it cannot transform in the same manner as photons. As a result it follows that we could have stated apriori that SR must assign no physical attributes to the space-time point and if it did it'd be invalid. But it doesn't and the fact light's velocity is independent of the velocity of its emitter is to be expected.

You are treating space-time as some kind of pseudo-elastic physical material, as if you can 'track' space-time points as you apply Lorentz transformations, as if a photon sphere 'knows' where its centre is mapped to under any given set of transformations and that its relevant. It isn't. If I give you a photon sphere you cannot tell me where the associated emitter should be in space at that moment in time. You can, however, tell me the space-time position of the point of emission. Notice I said 'space-time', as that means you are not saying "The point of emission has moved to there during the time from emission to now!" but instead you're saying "In the four dimensional manifold associated to space-time the event of photon emission happened there" where 'there' now means you're also 'pointing' through time, not just space. All frames must and do agree on that point in space-time as it does define the light cone. The location of the centre of a time slice of the cone to form a sphere is obviously dependent on how you slice the cone (which is equivalent to choice of frame and then solving the relevant geodesic equations). If frames agree on the position of emission in space-time then it doesn't matter they don't agree on the space location that flows to at some later time. And all frames do agree on the space-time location of emission because all frames we're talking about are related via Lorentz transformations defined at that point. If two frames don't agree on the location of the emission event in the four dimensional manifold then they are not related by Lorentz transformations. No doubt you'll whine about this, given you don't really understand how a Lorentz transformation is defined and constructed in terms of vector elements of the fibres of tangent bundles where the base space is the space-time but your ignorance is not my problem. If you don't get it read the multitude of posts I've made talking about bundles, all of which you either ignored or lied about, claiming you understood them and are 'okay' with them.

Wow, the fact you're arguing this is bad, even for you.

The definition of an N-sphere embedded in $\mathbb{R}^{N+1}$ with centre at the origin and radius R is

$\sum_{n=1}^{N+1} x_{n}^{2} = R^{2}$

For a sphere whose radius is time dependent and is increasing at speed c this is

$\sum_{n=1}^{N+1} x_{n}^{2} = (ct)^{2}$

Rpenner, Pete and I (at least) have proven that under a Lorentz transformation of any kind, $x_{n} \to \tilde{x}_{n}$, $t \to \tilde{t}$, the above expression becomes

$\sum_{n=1}^{N+1} \tilde{x}_{n}^{2} = (c\tilde{t})^{2}$

This is obvious from the definition of a Lorentz transformation, which is made clear by the rearrangement of the above to the following :

$\sum_{n=1}^{N+1} x_{n}^{2} - (ct)^{2} = 0$

The left hand side is the very definition of a space-time interval in SR (once you obtain the x from dx's via geodesic equations) and thus by definition is invariant under any and all Lorentz transformations,

$0 = \sum_{n=1}^{N+1} x_{n}^{2} - (ct)^{2} = \sum_{n=1}^{N+1} \tilde{x}_{n}^{2} - (c\tilde{t})^{2} = 0$

So the transformation leaves the defining equation of a sphere unchanged and thus the image of a sphere must be a sphere. If we replaced the photons with some set of massive objects (like a 'flash' of particles which have mass) then this would not be the case as the 'c' dependence is essential. In that case you'd get a squashed sphere but for photons you don't.

You seem to be under the naivy misconception that I, or anyone else who has talked about this with you or in academia doing relativity or differential geometry, am unfamiliar with or don't understand geodesics, light cones, light spheres and Lorentz transformations. Tens of thousands of papers (perhaps hundreds) are written on these things, its taken as assumed knowledge by researchers that anyone doing relativity has done these things a lot, which they have.

Simply repeating debunked failed arguments doesn't make them any truer. I see you still haven't got the balls to submit your work to a journal. Coward.

I shouldn't need to, I've proven my claim by algebra, I just walked you through the various definitions and results which utterly negate your claim.

I really don't get why you try to pretend you've got your head around this and no one else has. All those with any university level knowledge disagree with you, all your arguments negated, you refuse to read any relevant material as you know it won't tell you what you want to here and you continue, after many months, to refuse to submit to a journal. If you think I can't do this stuff then show your work to professors at journals. Until then its hypocritical and intellectually dishonest to try to portray yourself as being in some superior position on a matter you refuse to submit to review and which others are peer reviewed on.

I fail to see in you anything anyone would be jealous of.

Why do people go to freakshows? To point and laugh or think "Thank god I'm not like that".

18. ### Jack_BannedBanned

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do you ever post anything interesting?

19. ### mikelizziRegistered Senior Member

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No, Jack, your conclusion is not correct for SR. Your conclusion is correct for Newtonian physics. It appears you can't concieve of an alternative.

It is a terrible lapse on the part of authors of books on relativity that they do not expressly state that no one should attempt to understand SR unless they understand linear algebra. The scientific community is doing themselves a great disservice by alienating sincere people like yourself.

20. ### Jack_BannedBanned

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If I were you, I would spend more time understanding the facts of SR.

This is the definition of the ellipsoid in the stationary frame that maps to the moving light sphere. When light acquires a distance r in the stationary frame, this object maps to the moving light sphere centered at vr/c in the coords of the stationary frame. As such, it will be centered at the origin of the moving frame and is r/γ in all directions. Use LT to map all these points and verify it is a light sphere at the center of the moving frame.

I trust you can do the math.

$\frac{(x-vr/c)^2}{(r)^2}+\frac{y^2}{(r/\gamma)^2}+\frac{z^2}{(r/\gamma)^2}=1$

$t = \sqrt{ x^2+y^2+z^2}/c$

Last edited: May 26, 2010
21. ### Jack_BannedBanned

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Really, I understand linear algrbra.

What have I missed?

22. ### Jack_BannedBanned

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So the transformation leaves the defining equation of a sphere unchanged and thus the image of a sphere must be a sphere.

You have not shown with your freak sideshow that you map to the complete moving sphere from the stationary light sphere.

Since I know what is going on, you cannot do it.

I want you to show me how to map the stationary light sphere to the moving light sphere in its entirety. I want you to show me for all points on the stationary light sphere, you can map to every point on the moving light sphere.

Do it with a simple proof. If you cannot then SR is refuted.

Well, it requires more.

Anyway, show me the proof. I want a math proof.

23. ### rpennerFully WiredStaff Member

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Dude, you've just been shown again, that the expanding light sphere in one frame $(x,t)$ is also an expanding light sphere in any other frame $(\tilde x, \tilde t)$ and you have failed to refute this. From the perspective of either frame, it is self-consistent to declare itself stationary and the other one is therefore "moving" and you have failed to argue against this.

When you write things like "show me how to map the stationary light sphere to the moving light sphere in its entirety" you are demonstrating that you do not "know what is going on" since everyone has been applying the Lorentz transform to relabel the space-time coordinates of the same expanding light sphere. Further, they helpfully show both observers see the same expanding light sphere as being consistently described as having a center which is a "place" in different coordinate systems. Since the notion of "place" in a foreign coordinate system is a moving target for another coordinate system, this more than adequately demonstrates the "stationary" and "moving" light spheres. All expanding light spheres look to have a center which at rest with the current coordinate system (even if the source was moving in that system), but since what is a stationary place in one coordinate system is a moving target in another coordinate system a foreign coordinate system description of an expanding light sphere is always "moving."

This at the edge of one of great geometrical ideas -- the futures of all massive particles are confined within the bounds of an expanding light sphere. But even moving at 0.99c relative to us, a particle may consider itself at rest and therefore its view of the future has the entire freedom that we think we have -- the whole of the interior of the light cone. Geometrically, the Lorentz transformation between frames is analogous to a rotation which preserves futures within the light cone.

$\left( \begin{array}{c} \tilde x \\ c \tilde t \end{array} \right) = \left( \begin{array}{c} \cosh \rho & \sinh \rho\\ \sinh \rho & \cosh \rho \end{array} \right) \left( \begin{array}{c} x \\ c t \end{array} \right)$ with $\rho = \tanh^{-1} \frac{v}{c}$.

Science isn't about word games and obstinacy. It's about being guided by evidence. From the time of Galileo, the evidence shows that the universe doesn't seem to care about "place" and that all notion of motion is only relative to human concepts of what is "stationary." As AlphaNumeric recently posted, you can't put a thumbtack into the substance of space to leave a mark. Therefore, human experience is misleading if you think it suggests that there is an absolute true stationary coordinate system. Aberration of starlight, Foucault's pendulum, and other lines of detailed observation and reasoning show the Earth's surface, which was once thought to "obviously" be at absolute rest, to be moving and moving non-inertially.

Quoting posts wholesale and failing to tackle these concepts does not indicate that you are making progress. You need to figure out why we think we have (repeatedly) demonstrated that expanding light spheres map to expanding light sphere via the Lorentz transform and you need to figure out why you don't think that's good enough. If you aren't able to communicate these ideas, then you aren't putting enough work into it. And if you aren't putting work into communication, you miss the point of this forum.