What do I see wrong here ? T2 = T1(1 - V/C)

[digi99]

Gwh, ΔB = ΔA .(1 - (V/C) . 2γ . 1/γ.

That what A thinks is a little different what B sees after Lorenz, but the length is going shorter (shorter than factor 1/γ).

 

[digi99]

Gwh, sorry I mean Gwh, ΔB = ΔA . (1 - (V/C) . (γ)2 . 1/γ.

I know than ΔTb = ΔTa . (1 - (V/C) . (γ)2 . 1/γ.

And that is not the same as Δt' = Δt . 1/γ as general in Lorenz, so where I am going wrong ?

You teach maybe the SRT (I still don't), so you tell me

 

[digi99]

Yes the party is over, this subject may be closed.

I made a mistake in my formula, I was just using an X1 and an X2 but not the same as for ΔXa, if you fill in V.T1 and V.T2 than you get ΔXb = 1/γ . ΔXa.

But I have learn now more about light by all this, now I have a better view. I understand now the SRT, it is all relative.

E.g. comparing passing light waves (mass = 0) to a long train (mass > 0). Person A (standing still) sees the train passing but B (has speed) sees the train slower passing. But with light as well A as B (from the meeting point) see the same passing light waves and that problem is solved by time dilation.

That's why I got this problem, you may not see light as another object with a mass > 0.

Funny, light is really magical now for me ... but it represents still time .. the end ..

 

[digi99]

I am not finished with this subject, maybe it is a mathematically problem I don't see.

When I fill in in my equation x1=v.t1 and x2=v.t2, I get the very known formula ΔXb = 1/γ . ΔXa.

If I don't fill in you get my formula and the results are not the same (see time_e4.jpg in #33). E.g. 0.1c than ΔXb = 0.904 . ΔXa and 0.904 < 1/γ (= 0.995).

Which formula is ok ? (see attachment)

 

[digi99]

Solved, this topic is done now.

The first formula is right of course because you must see it for B standing still, where Lorenz is meant for.

In the second I made the mistake to take ΔX twice but they are not the same.

What I did in my equation is only possible for the line c (lightspeed) and not for other speeds (Lorenz is only meant for A standing still and B standing still).

But I learned a lot and my understanding is better now (don't see the lightspeed as an other object with mass > 0, if you have more speed (B) than (A), you see lesser wagons of a passing train (from the meeting point with A) but the same waves of light than A sees, solved by nature by time dilation).

 

[digi99]

Yes, with my topic I found a meaningfull formula, I did not seen sooner. Not different Einstein but it shows better with my pictures.

ΔXb_light = γ . ΔXa_light - γ . V/C . ΔXb_seen_from_a (it's equal ΔXb_light = 1 / γ . ΔXa_light)

or

ΔXb_light = γ . (ΔXa_light - V/C . ΔXb_seen_from_a).

Voila.

So shows there is a relation between the length of the light wave for A, the length of the light wave for B and the movement of B how A it could see. That was my point only the formula is something different. And of course all compared to A, and of course symmetric in SRT terms.

 

[digi99]

Conclusion :

Person A predicted a time dilation for person B without Lorenz and person A was right. Only the values were different of course (Lorenz), but the way person A was thinking is ok (subtract person B's own movement from the total measured light wave by person A, but his device for counting waves has no function here).

The final end (has at least a meaning) ..

 

[digi99]

To make it complete because of the title:

Person A calculated ΔTb_light = ΔTa_light . (1 - V/C).

After Lorenz : ΔTb_light = γ . ΔTa_light . (1 - V2/C2).

 

[digi99]

There are several things wrong with your thinking.

First off, a device that counts the cycles of light can be used as a clock, but only as long as the light source is stationary with respect to the counting device. This would work for observer 1 but since observer 2 is moving away from the light source, he will count fewer cycles than he should for it to be a legitimate clock so the count of the wave cycles won't be correlated with time. This is not the time dilation that Einstein described. It is instead Relativistic Doppler which would result in your "clock" running even slower than it should. But if observer 2 carried an identical light source and a second period counter, it will count at the correct rate predicted by Einstein's time dilation. Then observer 2 could compare the rates of the two devices and that will demonstrate the Relativistic Doppler Factor.

To explain my point of the Doppler effect in this topic. It's an discussion in another topic but completes this one.

If I use this formula ΔXb_light_seen_from_a = ΔXa_light . (1 - V/C) in this formula ΔXb_light = γ . (ΔXa_light - V/C . ΔXb_seen_from_a) I get :

ΔXb_light = SQRT(1 - v2/c2) / (1 - V/C) . ΔXb_light_seen_from_a

This looks indeed the Relativistic Doppler effect (if ΔX can be replaced by frequencies too).

But this was not meant in this topic, it's a coincidence that both formulas are the "same". But in this topic I was expressing time ΔTb_seen_from_a and came to my formula.

If I would talking about counting waves (but I don't), indeed you get the Doppler effect while holding an object (device) in a light wave. So maybe we have to find something in the future that it can be in another way, if it was necessary for me, but it is not.

I have already confirmed that you never will see my effect, a slower going light wave, because by nature distances are immediately corrected, so the light speed will be the same. Finally you will see only smaller light waves (if could be showed to you).

But still can't be proofed that I may not explain that on my website and to other interesting people (compare a light wave to time, the moving one sees the light wave slower passing, and symmetric of course). In the Lorenz formula ΔXb_light = γ . (ΔXa_light - V/C . ΔXb_seen_from_a) you still see that ΔXb_seen_from_a is subtracted (partially) from ΔXa_light, maybe it has no meaning.

 

[digi99]

To be sure that somebody who reads this topic ever thinks, ok relativistic Doppler, next topic, I want to say this is not the case here.

Consider only answer #11 with picture, answer #18 (ignore formula Lorenz, was wrong), and last answers from #39 (rest is nonsense because of mistakes).

Consider an extra dimension and add an y-axes. For (A) (y=0), and for (B) y=p (just a value).

Compare a light wave to (A) and (B) just in the middle, say y=.5p.

Mathematically the same results and B is not moving in a light wave, so no Doppler effect.

 

[digi99]

Eureka, proved !

I was doing difficult with Lorenz and suddenly I saw the simple solution.

Look to the Lorenz transformation : Xb_light = γ . ( x - V . Ta_light).

In all the points from frame A where x = V . Ta_light and time Ta_light for B is the coordinate in frame B zero.

In all other points starts the time in frame B, so in each other point the transformed light wave has missed piece V . Ta_light. That's in frame B pro rato 1/γ . V / C for distance (length wave) and time.

With other words the time dilation calculated by A pro rato in frame A V/C is exactly the time dilation in frame B 1/γ . V / C

So you may see a light wave as the relative time signal (it expresses relative time between two events), if you are standing still you see the time passing at a maximum, if you move (compared to another object/person) you see a little piece lesser passing, and that's exact the time dilation (in reality you see a smaller light wave but you are not aware of).