Time and Light

Quantum Quack

Quantum Quack

Life's a tease...
Valued Senior Member
Wanting to clarify the time and light issue I drew this diagram that shows a light source [star] and a speckled light field shown as a snapshot.
Every speckle shown can be analoguous with a photon particles position relative to the star.

I have placed in this light field 6 relative Observers all traveling at relative velocity with differring vectors to each other and the light source.


diag01.jpg


The question is do they all share the same Hyper Surface Of The Present ?[t=hsp]

I would think that they do but to do so I think would contradict the issue of non-simultaneity for SRT.

Can it be unambiguously stated that :
"If light speed is invariant no matter where a photon event occurs it must be always at the hyper surface of the present ot t=hsp" [ or something to that effect?]

Care to discuss?
Is there a better way to phrase the question?
 
What is "hyper surface of the present"?

Isn't "present" defined locally for each observer?
 
BenTheMan said:
What is "hyper surface of the present"?

Isn't "present" defined locally for each observer?
Click to expand...
According to SRT Yes I guess it is defined locally however as you can see in the image I posted all observers seem to share the same hsp or present moment.

I found to be true then this implies a contra to SRT's non- simultaneity as every light event must occur at the same universal presnt moment and thus the "NOW" is an absolute constant.

diag01.jpg


Can you tell me if the above "speckle" diagram image makes sense to you Ben?

Can you see how every speckle of light must share the same present moment [hsp]

the diagram referred to with regards to the HSP is:
lightc.png

courtecy wiki
 
rpenner said:
The content of the SRT includes that observers in relative motion can disagree that events not happening in the same place happen at the same time.
Click to expand...
and this is exactly what is in contention.
As shown in the "speckle" diagram 01 all observers regardless of relative Velocity must experience their light events simultaneously. Even though their tick rates are different.
And as the entire universe can be considered as a light field then all light events must happen at the same universal moment making the HSP an absolute constant reagrdless of relative velocity of RF.

SRT forbids the use of an absolute reference frame or aether. However light itself is commony used as an absolute reference frame and although not an aether it certainly could be mistaken for one.

What I would like to determine is how my contention is invalid as I fail to see a flaw in the logic. [ which puts me in a difficult position and at odds with SRT]
 
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That is addressed in the Demo Reel above, betrween marks 7:00 and 8:03.
 
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rpenner said:
That is addressed in the Demo Reel above, betrween marks 7:00 and 8:03.
Click to expand...
no it doesn't address it unfortunately.
It only shows SRT's position on light and time.
I am asking that we review this position.

Using light as the absolute reference frame rather than an object of matter.

I do realise that SRT disallows the use of light as a reference frame and that I guess is what is in contention.
Why is it that the most important part about a light effect model is excluded from being a reference frame?
It is afterall the whole reason for SRT existance, that being the invariance of light.

I shall refer to the image I posted again:

diag01.jpg


Again I ask how the logic or reasoning of a universal t= t[sub]hsp[/sub] constant is an unavoidable conclusion.
 
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I have a sort of off topic question, but didnt want to start a whole new thread.

Does light have mass? Doesnt the Einstein's relativistic mass equation require light to be massless?
 
camilus said:
I have a sort of off topic question, but didnt want to start a whole new thread.

Does light have mass? Doesnt the Einstein's relativistic mass equation require light to be massless?
Click to expand...

Not really Camilus. By theory, light acts as thought it is massless, but their are mathematical equations (designed my i add), that can describe the momentum of a photon even if it is massless for $$E^2-pc^2=(Mc^2)^2$$ so $$M=0$$ for photons, or any massless particles.
 
Reiku said:
Not really Camilus. By theory, light acts as thought it is massless, but their are mathematical equations (designed my i add), that can describe the momentum of a photon even if it is massless for $$E^2-pc^2=(Mc^2)^2$$ so $$M=0$$ for photons, or any massless particles.
Click to expand...

I was talking about this equation:

$$m = {m_0 \over sqrt{1 - (\frac{v}{c})^2}$$

obviously, if v=c, then the bottom is zero. as is the case with light.
 
The biggest problem with that equation is that it doesn't work in equations like $$\mathbf{F} = m \mathbf{a}$$.

$$\mathbf{F} = \left(
\mathbf{1} +
{
{\mathbf{v}\mathbf{v}^t}
\over
{
1 - ( \frac{v}{c} )^2
}
}
\right) {{m_0 \mathbf{a}}\over{\sqrt{1 - ( \frac{v}{c} )^2 }}}$$ where $$\mathbf{v}\mathbf{v}^t = {{1}\over{c^2}}\left( \begin{array}{ccc} v_x^2 & v_x v_y & v_x v_z \\ v_x v_y & v_y^2 & v_y v_z \\ v_x v_z & v_y v_z & v_z^2 \end{array} \right)$$.

http://math.ucr.edu/home/baez/physics/Relativity/SR/mass.html

// Edit, or:
$$\mathbf{F} = {{m_0 \mathbf{a}}\over{\sqrt{1 - ( \frac{v}{c} )^2 }}} + {{v}\over{c^2}} {{m_0 \mathbf{v} \cdot \mathbf{a}}\over{ \left(1 - ( \frac{v}{c} )^2 \right) ^{\frac{3}{2}}} }$$
 
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But that's incorrect.

$$\begin{eqnarray}
F = {{dp}\over{dt}} = {{d(\gamma(v) m_0 v)}\over{dt}} = m_0 ( \gamma(v) {{dv}\over{dt}} + {{d\gamma(v)}\over{dt}} v ) & = & m_0 ( \gamma(v) {{dv}\over{dt}} + {{d\gamma(v)}\over{dv}}{{dv}\over{dt}} v ) \\ & = & ( 1 + {{1}\over{\gamma(v)}} {{d\gamma(v)}\over{dv}} v ) m_0 \gamma(v) {{dv}\over{dt}} \\ & = & ( 1 + {{1}\over{\gamma(v)}} \gamma^3(v) {{v}\over{c^2}} v ) m_0 \gamma(v) {{dv}\over{dt}} \\ & = & ( 1 + \gamma^2(v) {{v^2}\over{c^2}} ) m_0 \gamma(v) {{dv}\over{dt}} \end{eqnarray}$$
 
Yes. Everyone agrees on $$F = {{dp}\over{dt}}$$ but since $$ {{dp}\over{dt}} \neq m_0 \gamma a$$ it is generally incorrect to say $$F = m a$$ when relativity is important.

If you really want to memorize $$F = m a + m{{\gamma^2 v \cdot a}\over{c^2}}v $$, that is your privilege. Myself, I find it more useful to memorize $$p = m_0 \gamma v$$, since I never attach a relativistic mass to a set of spring scales.
 
Given that my post [#7] has yet to be addressed the workings you are showing may actually prove to be irrelevant
 
No. The workings I am showing are textbook. The textbook is based on the historical and experimental tests of SRT and theories which are indistinguishable from SRT. Specifically, it is an engineering fact that clocks in orbit cannot be synchronized with clocks on the ground because there can be no agreement on simultanity.

But if you would like to dig below the theory to the experimental backing which the theory currently enjoys, you are in luck:
http://relativity.livingreviews.org/Articles/lrr-2005-5/index.html
http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html
 
rpenner said:
No. The workings I am showing are textbook. The textbook is based on the historical and experimental tests of SRT and theories which are indistinguishable from SRT. Specifically, it is an engineering fact that clocks in orbit cannot be synchronized with clocks on the ground because there can be no agreement on simultanity.

But if you would like to dig below the theory to the experimental backing which the theory currently enjoys, you are in luck:
http://relativity.livingreviews.org/Articles/lrr-2005-5/index.html
http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html
Click to expand...
so I guess that means you are not prepared to address the topic?
 
Quantum Quack said:
Given that my post [#7] has yet to be addressed the workings you are showing may actually prove to be irrelevant
Click to expand...

Using light as the absolute reference frame rather than an object of matter.
Click to expand...

You can't use light as an absolute reference frame, because it moves with the same speed in all reference frames. Doing so would imply that light has a rest frame, which is not true by postulate.
 
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