# 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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I dumb it down some.

Did you ever prove the light sphere in the moving frame?

3. ### Jack_BannedBanned

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It is quite clear he did not prove a light sphere in the moving frame.

If he has proven a light sphere in the moving frame, I would like to see this light sphere centered at the moving frame's light emission point.

His prove did not supply that.

5. ### Jack_BannedBanned

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You are not demonstrating the light sphere in the moving frame with this logic and this post did not disprove anything of mine.

You failed to understand using my point, the two co-located observers elapsed the same time on their clocks which was r/c.

Given the light postulate, then it must be the case light is a distance r from the light emission point in alll directions in each of the individual frames.

Clearly, this causes light to be outside the light cone of each.

Further, if an observer is in a frame and light is emitted at the origin, light is a distance r in all directions for any given r at some time t.

This is what the moving light sphere needs to look like in the view of the moving frame.

What you are arguing without realizing it is you believe Minkowsky applies to the frame when taken as stationary.

Here is what things look look in the moving frame when taking the view of the moving frame, ie stationary.

Let us take a system of co-ordinates in which the equations of Newtonian mechanics hold good.2 In order to render our presentation more precise and to distinguish this system of co-ordinates verbally from others which will be introduced hereafter, we call it the stationary system.''
http://www.fourmilab.ch/etexts/einstein/specrel/www/

7. ### Jack_BannedBanned

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You deserve a response for this junk.

If we use (-r,0,0) and ( r,0,0) in the stationary frame, any human understands, these are on a sphere centered at origin (0,0,0).

When we map them to the "view" of the moving frame using t=r/c, and we get

(-r,0,0,r/c) maps to (-r √ ((c+v)/(c-v)), 0, 0, r/c √ ((c+v)/(c-v)) )
(r,0,0,r/c) maps to (r √ ((c-v)/(c+v)), 0, 0, r/c √ ((c-v)/(c+v)) )

Clearly, we do not have a sphere centered at the light emission point in the view of the moving frame.

You see, "the view of the moving frame" means taking the moving frame as stationary.

See you all are so very smart, perhaps you can explain how this is a sphere when the moving frame is stationary. Do not forget, a stationary frame is not Minkowsky.

8. ### PeteIt's not rocket surgeryModerator

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Jack, do you notice that the time coordinates of the mapped events are not the same?

At $t' = \frac{r}{c} \sqrt{\frac{c+v}{c-v}}$, you've found where the left-moving light flash is (at x'=-ct'), but not the right (assuming the negative x-axis points left). So you haven't found whether the light flash is spherical at that time in the moving frame.

At $t' = \frac{r}{c} \sqrt{\frac{c-v}{c+v}}$, you've found where the right-moving light flash is (at x'=ct'), but not the left. So you haven't found whether the light flash is spherical at that time either.

For all events on the lightflash worldlines, Jack, x'=ct' for the right-flash, and x'=-ct' for the left-flash.
That's what demonstrates that the light flash is spherical in the moving frame.
This has been explained to you several times. If you continue to ignore people's explanations, you'll probably be permabanned.

Anyway, try this exercise:
What event on the right-moving light flash in the stationary frame maps to $t' = \frac{r}{c} \sqrt{\frac{c+v}{c-v}}$?
If the light flash is spherical in the moving frame at that time, it should be the opposite of what you found above for the left-moving light flash at that time.
Here's a start:
(x,t) = (ct, t)
(x',t') = (???, r/c √ ((c+v)/(c-v)))

Similarly, what event on the left-moving light flash in the stationary frame maps to $t' = \frac{r}{c} \sqrt{\frac{c-v}{c+v}}$?

9. ### Jack_BannedBanned

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This is very good and commendable. You are proving when you take a frame as stationary, it is Minkowsky space-time in the moving frame when taking the view of the moving frame as stationary.

That implies the original points taken from the stationary frame must be Minkowsky.

But, these points were (-r,0,0,r/c) and (r,0,0,r/c) which are Euclidian and not Minkowsky.

When you are able to map the stationary sphere to the moving sphere such that in the view of the moving frame, ie take the moving sphere as stationary, you get a Euclidian light sphere then you have proven the logical consistency of the two postulates.

Hint, don't forget, when "in the view of the moving frame", you are taking the frame as stationary.

Einstein will tell you what that means.

Let us take a system of co-ordinates in which the equations of Newtonian mechanics hold good.2 In order to render our presentation more precise and to distinguish this system of co-ordinates verbally from others which will be introduced hereafter, we call it the stationary system.''

http://www.fourmilab.ch/etexts/einstein/specrel/www/

10. ### Jack_BannedBanned

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OK, I will explain it.

You must pick points from the stationary frame that are not all on the light sphere in the moving frame to create the Euclidian light sphere in the moving frame.

You must know what you are doing and pick well.

11. ### PeteIt's not rocket surgeryModerator

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Jack, it's becoming more and more dificult to untangle your twisted use of the language to decipher what you are actually thinking, and your willful ignorance is intolerable.

You said:
The first line shows that an event on the light sphere in the stationary frame maps to an event on the light sphere in the moving frame at a particular time.
The second line shows that another event on the light sphere in the stationary frame also maps to an event on the light sphere in the moving frame at a particular time.

12. ### AlphaNumericFully ionizedModerator

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No, you're wrong. Stop thinking of the point at the centre of the light sphere as something physical. No object is there, you have no way of 'tagging' it so you can keep track of it in a consistent manner. If you consider only the objects in the space (ie the photons) then they form a sphere in any and all frames.

False. Rpenner and I have proven they form a sphere in any frame. Read the links.

This isn't even coherent.

You ignored links I provided, claim I haven't said what I have said, fail to respond to anything and make claims demonstrated to be false. Do you like demonstrating you're an ignorant liar?

13. ### Jack_BannedBanned

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The light sphere in the view of the moving frame is not equidistant to the light emission point in the moving frame.

Simple basic stuff.

14. ### Jack_BannedBanned

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You and rpenner did not prove a light sphere in the view of the moving frame.

That light sphere must be a distance r in all directions from the light emission point.

That is not what you did as I showed above.

15. ### Jack_BannedBanned

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1,383
This is how you find the light sphere in the view of the moving frame from the context of the stationary fame.

You will note, it will be a constant distance from the origin as required when viewed in the moving frame.

Let r be given, select y such that $-r \leq y\leq r$

$x = (\pm\sqrt{ r^2-y^2}+vr/c)\gamma$

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

Try it out for yourself so you can learn.

16. ### rpennerFully WiredStaff Member

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You left out z, so I will also suppress talk about that dimension, and may correctly use spherical to describe a circle.

Rewriting for arbitrary $\theta$ and $\beta = \frac{v}{c}$ so that $\pm$ and some square roots are superfluous, the content of your setup is:
$\begin{eqnarray} y & = & r \sin \theta \\ x & = & r \gamma ( \beta + \cos \theta ) \\ t & = & \frac{r}{c} \sqrt{\frac{\beta^2 + 2 \beta \cos \theta + \cos^2 \theta + \sin^2 \theta - \beta^2 \sin^2 \theta}{1-\beta^2}} \\ & = & \frac{r}{c} \sqrt{\frac{1 + 2 \beta \cos \theta + \beta^2\cos^2 \theta }{1-\beta^2}} \\ & = &\frac{r \gamma}{c} (1 + \beta \cos \theta) \end{eqnarray}$

Notice that while the observer sees the points parameterized by r and $\theta$ to be on the light cone with origin at (t=0, x=0, y=0) (which is enforced by the line $t = \sqrt{ x^2+y^2}/c$), they don't all happen at the same time for this observer, but they all are points in space in time where only at the speed of light can a signal reach that point from the origin. t is a function of our choice of r and $\theta$.

But if you look at the same light and hold t constant (making r a function of $\theta$), then x and y describe a circle for constant t (which is also enforced by $t = \sqrt{ x^2+y^2}/c$).

So the light pulse originates at (t=0, x=0, y=0) and is spherical for constant timeslices of t.

Transforming to the frame moving at speed v, we have:
$\begin{eqnarray} x' & = & \gamma ( x - vt ) \\ & = & r \gamma^2 ( ( \beta + \cos \theta ) - \beta (1 + \beta \cos \theta) ) \\ & = & r \cos \theta \\ y' & = & y \\ & = & r \sin \theta \\ t' & = & \gamma ( t - v c^{-2} x ) \\ & = & \frac{r \gamma^2}{c} ( (1 + \beta \cos \theta) - \beta ( \beta + \cos \theta ) ) \\ & = & \frac{r \gamma^2}{c} ( 1 - \beta^2 ) \\ & = & \frac{r}{c}\end{eqnarray}$

So t' is a function only of r and is independent of $\theta$. So in a timeslice where t' is constant, r is constrained and $\theta$ is free to take on any value. Further it is trivial to verify that $r^2 = x^2 + y^2$ and so the light pulse originates at (t'=0, x'=0, y'=0) and is spherical for constant timeslices of t'.

Congratulations, even with as convoluted a setup as possible, your muddled thinking gives us an example of self-consistant math that shows the Lorentz transform maps spherically expanding light pulses in one inertial frame to spherically expanding light pulses in the other frame.

17. ### Jack_BannedBanned

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1,383

.

Why did you need all that mess?

I provided a simple and consistent method to reveal the moving light sphere as shown by LT.

Do you have the ability to simulate my equations graphically? I do.

Further, we started with Einstein's consistency proof which I invalidated and replaced with my own. You have agreed.

Now, you are on my bandwagon and agreeing with two light spheres.

You have not figured that part out yet. Further, you have not yet figured out, my equations and muddled mind have refuted Einstein's consistency proof.

I wonder why I did that?

Apply your results above to Einstein's consistency proof. Check it out.

Let me know what you find.

18. ### Jack_BannedBanned

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1,383

So t' is a function only of r and is independent of $\theta$. So in a timeslice where t' is constant, r is constrained and $\theta$ is free to take on any value. Further it is trivial to verify that $r^2 = x^2 + y^2$ and so the light pulse originates at (t'=0, x'=0, y'=0) and is spherical for constant timeslices of t'.

This conclusion is false.

r is free and t' is constrained based on r. The set of t's must be located.

I assumed the the consistency of the stationary light sphere. You are frame mixing.

I proved if the stationary light spere exists, then the moving light sphere is consistent and exists.

This may be beyond you.

19. ### AlphaNumericFully ionizedModerator

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You're making up your own criteria. In the initial frame is there a point which is the same distance, at any given value of t, from all the photons? Yes. Therefore the photons form a sphere. In the moving frame they all map to points which satisfy $-c(dt')^{2}+dx'\cdot dx' =0$ and therefore they all have $|\frac{dx'}{dt'}|=c$, they all are a distance cdt' away from a particular point at time t'. They form a sphere. That's something we've proven to you many times now.

Your argument is that because the Lorentz transformation doesn't map the centre of one sphere to the centre of the other then the photons don't from a sphere. If the photons are, at a given instant in time, all the same distance from some point then they by definition form a sphere. You're making up your own definition of what it means for them to be a sphere. If you could actually do some basic vector calculus you'd have proven it to yourself now. Instead you keep spelling Minkowski with a 'y'. Don't you read any literature on the guy? Of course you don't.

20. ### PeteIt's not rocket surgeryModerator

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Yes it is. The events you transformed confirm it.

Look:
If light in the moving frame is expanding spherically around the emission point at (x',y',z')=(0,0,0), then every event (x',y',z',t') on the light sphere must be such that x'² + y'² + z'² = ct'. Right?

Look at your two transformed events, for example:
Look at the first event:
{x',y',z',t') = (-r √ ((c+v)/(c-v)), 0, 0, r/c √ ((c+v)/(c-v)) )
It's easy to see that x'² + y'² + z'² = ct'²

And the second event:
(x',y',z',t') = (r √ ((c-v)/(c+v)), 0, 0, r/c √ ((c-v)/(c+v)) )
Once again, it's easy to see that x'² + y'² + z'² = ct'²

Both events are ct' from the light emission point - they're on a sphere of radius ct'.

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

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Still beating the dead horse, I see.

22. ### Jack_BannedBanned

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1,383
No, the moving light sphere is centered at rvγ/c in the coords of the stationary frame. That is a different light sphere than the stationary one.

Rpenner sort of got it but not quite. He did not follow through.

If you are good at math, you will find one at the origin of the stationary frame and one at rvγ/c for the light sphere of the moving frame in the coords of the stationary frame.

SR is the theory of multiple light spheres.

23. ### Jack_BannedBanned

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No, you do not have my argument.

I transposed the moving light sphere into the coordinates of the stationary frame to prove there exists two unique light spheres.

This will exceed you.

Given any space-time coord (x,y,z,t) there exists some light sphere in the moving frame that intersects that point. That light sphere is disjoint from the stationary light sphere. In addition, that light sphere may have already gone by that point in the past of the stationary frame. LT is not what you think it is.

Since you have claimed to be so good at math, you will be able to prove this theorem as I have.