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

1. Oct 5, 2011

### digi99

SRT = the theory of special relativity

I can't find for myself what is wrong with this ?

A <---------------------- C x T1 ----------------------------------------> B : system S1 observer 1

A <--- V x T1 ---><----------------------------- C x T2 -----------------> B : system S2 observer 2

At observer 1 a light wave starts in A and arrives 1 seond later over a distance of 300.000 km in B.

Observer 2 travels with speed V from A with the same light wave in direction B.

Observer 2 uses the light wave to meassure the distance between A and B (when the light wave arrives in B, the meassurement ends), so counts the periods, the counted periods

will be lesser (than the measurement for observer 1 in A) because of his higher speed, so the expectation is like in the SRT, the distance and time will be shorter for observer

2, his time goes slower.

So C . T1 = V . T1 + C . T2, so T2 = T1 . (1 - V/C). It fits with the expectation of the SRT, when V=C than T2 = 0, the time stands still. When V= 0, T2 = T1.
In both situations S1 and S2 is the speed of light C, the same value.

But this formula is not the same as in the SRT.

What do I see here wrong (maybe my view of light waves), but if fits in the SRT, seems it (it is clearly visible that distances and times are shorter as the SRT says) ?

I think that a light wave is always the same (independed for speeds of the observers, because of mass = 0) anywhere in the universe ...

2. Oct 5, 2011

### ghwellsjr

When you use a light wave to measure a distance, you have to wait for the light to reflect off your target and come back to you. But if you are moving with respect to the target, what distance are you actually measuring, the one where you were when the light left you, or the one where you were when it returned to you or something in between?

Last edited: Oct 5, 2011
3. Oct 5, 2011

### digi99

Thanks for the answer, purely measurement, this is not correct.

I mean the distance where observer 2 is when the light wave arrives in B (one way only).

Then it looks shorter for him than for the observer 1 in A (observer 2 moves).

B is exactly the same end point for both observers.

I was looking for an example to get/present a view for shorter times/distances in the SRT.

Maybe this is not the right one, do you know a better one ?

Strange but in some way my example gives a feeling ... purely using the characteristics of a light wave ...

4. Oct 5, 2011

### ghwellsjr

You are thinking that some measurement of time (T2) correlates to some distance (from Observer 2 at then end to B), correct? I would agree, if Observer 2 were stationary at his end location and he used light to measure his distance from there to B, that he would get a meaningful result. But when the light starts at A and goes to B, why should that provide information about a different distance?

But in addition to that, how do the observers know when the light arrives at B unless they wait for a reflection or some other signal to get back to them to let them know when to stop their timers?

The normal example that people use to demonstrate the correct timing relationship between two observers in relative motion is the "light clock". Why don't you look that up in wikipedia?

5. Oct 5, 2011

### digi99

I know now better what I want to demonstrate with this example in fact, not the distance measure.

I want to show that somebody with a higher speed (compared to the other) goes slower in time.

I see than just a light wave from 1 second long between A and B so the distance is 300.000km. This one second illustrates a time frame in the universe, so illustrates time (light speed is always the same everywhere, so illustrates time on a light clock). A and B are for both observers exact the same locations.

When observer 2 has a higher speed passing A the light wave goes slower when he/she "looks" to it than for observer 1 in A.

When the light wave arrives in B, for oberserver 1 is it 1 second later, but for observer 2 is it I thought 1 - V/C seconds passed.

Imagine there is a light clock since the beginning of the universe and this 1 second is a part of it, but observer 2 was traveling in the direction of the light wave, so his time is shorter (and the distance traveled in that period shorter compared with the light speed.

Do you agree, but the time difference is not the same as calculated in the SRT, what is wrong in my thinking ?

But in my thinkings I see maybe a wrong view of light, so I think that the crossing of two light waves (send at the same location on the same time) will always be that location, if somebody is in that location and still there after a while he/she stands completely still, if he/she is not there anymore he/she moves with a speed away. So A and B could be such a locations. Maybe I go wrong here, and that is maybe the reason I go wrong in my formula (but it illustrates maybe something like slower going through time ...).

6. Oct 6, 2011

### digi99

I try to describe it more clearly now. This is just theoretically (experiment in thought).

E.g. one mile before location A we let start a long light wave passing A and B and further.

Observer 1 stands in A, starts a special clock (which can count periods of the lightwave) and counts the periods which are passing and stops the clock after 1 second. The starting point of the light wave on the moment of starting the clock has arrived in B, 300.000km further. If observer 1 has a speed too is not important, this is always possible in his system S1.

Observer 2 has a speed V (> observer 1 and goes direction B) and starts his ingenious clock too at the same time with observer 1 when passing A, his clock counts lesser periods until the starting point of the lightwave from observer 1 has arrived in B. Exactly what Einstein tells us, his time (observer 2) goes slower. Near the shorter time, observer 2 could measure the one-way distance to B too with the counted periods and of course is shorter too (Einstein tells us too).

We are using here just the lightspeed with his fixed periods for our clock (and one long light wave), and the clock for observer 2 goes slower.

Once again (read other anwers too), what do I wrong in my thinkings ?

Last edited: Oct 6, 2011
7. Oct 6, 2011

### ghwellsjr

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.

A second problem is what I asked you about in post #4:

"how do the observers know when the light arrives at B unless they wait for a reflection or some other signal to get back to them to let them know when to stop their timers?"

Of course, observe 1 could stop his counter after one second but this presupposes that he has another clock. If instead, he is simply waiting until his counter reaches the frequency of the light source (the number of cycle in one second), then he is simply stating that the front of the light wave has reached one-light second away and he is not measuring anything.

Even worse for observer 2, what is he supposed to count to "measure" his length contracted distance to point B?

8. Oct 7, 2011

### digi99

But this is an experiment in thought, not a question how do I measure.

So I have changed quickly my ingenious clock (was not easy), my clock can measure the length of the passing light wave, for observer 1 is 300.000 km wave passing.

For observer 2 (1 - V/C).C light wave. So my clock eliminates the doppler effect (the light speed is always the same).

Everything has to do with the constant lightspeed in Einsteins theory (and so with light), I want to visualize this (if possible) in details ... and may use 300.000km as 1 second (like they always do with a light clock, but I want to avoid this because very confusing, so I use the same light wave for both observers to be sure and to learn of course where I am wrong in my thinkings).

9. Oct 7, 2011

### ghwellsjr

If you want to use a rest frame for observer 1, then in your thought experiment you can talk about the time it takes for the light to propagate from A to B and you can call that one second and you can say the distance is 1 light second or 300.000 km. No problem there. You can also say that observer 2 travels at speed V and you can identify where he will be after 1 second according to your specified rest frame. Let's call that position X. No problem there. Then you can also talk about how long it will take for light to go from X to B according to this same rest frame. No problem there. Now you can create your formula as you did in your first post and it will be correct.

But it has nothing to do with what observer 2 experiences. At the time that observer 2 reaches point X, the light has already reached point B. So why do you want to make an equation involving a fictitious time interval for how long it took for your light wave to go from point X, long before observer 2 arrived there, to point B?

10. Oct 7, 2011

### ghwellsjr

After thinking about your scenario some more, I think I understand what you are talking about. You are treating the light wave as a sort of tape measure with your ingenious device keeping track of how many waves, and therefore how far the light has traveled past each observer, correct?

Let's say the light source has one wave per meter (way too few for real light, but this is a thought experiment so we can pretend). Then for observer 1, at point A, after he counts out 300.000.000 waves, he knows that the beginning of the light wave will be one light second away from him, at point B.

But since observer 2 is traveling from A toward B at speed V, he will count the waves at a lesser rate than observer 1 and when the front of the light wave reaches point B, he will be at point X and will have counted out the correct number of waves corresponding to the distance he is from point B.

So we could demonstrate that your equation from post #1 would fit the data as just described.

But here's the problem: we can talk about where observer 2 is when the front of the light wave reaches point B, because we can see everything that is going on as defined according to a frame of reference in which observer 1 is at rest but observer 2 cannot know any of this. In fact, he will not consider the waves to be one meter apart. Since his rulers are length contracted by one over gamma, he will think the waves are too long by a factor of gamma. So when he determines how far away the front of the light wave is when he reaches point X, he will think it is past point B. Furthermore, he will then think point X is closer to point A by a factor of one over gamma. So from the point of view of observer 2, your equation is no longer valid.

11. Oct 8, 2011

### digi99

Forget this situation and consider my space-time-diagram (attached).

So my originally thought was, if two persons A and B are "looking" to the same light wave and person B with a higher speed than A (and in the direction of the light wave), person B goes slower in time because the light wave is passing him slower. I see this light wave as time (it represents time, if your speed is equal to the lightspeed, times stand still for you.

So I try to make Einstein visible to show to other people.

My diagram:

- one/same system S
- light wave starts in Xoc
- B has speed V compared with A
- when B meets A, we start our enginous clocks for measure the passing light wave length
- on any moment time ΔTb = ΔTa . (1 - Vb/C) so slower (see attachment)

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12. Oct 9, 2011

### digi99

Yes you are right, Ghwellsjr, thank you very much for your answer.

I know my error now I think.

I calculate the situation for B, being A (seen from A). how it would be for B probably.

And you see than already that time goes slower for B.

But I have to calculate it too for being B (seen from B) and that will be like Einstein calculates it (a refinement).

But my idea to explain it to others fits I think (compare the lightspeed with time, when you go faster the light waves are going slower passing you, so time will go slower).

If you agree, how do you think about the crossing of two light waves resolving in a point standing still, other objects around are moving ?

Last edited: Oct 9, 2011
13. Oct 10, 2011

### ghwellsjr

Digi99, instead of using your diagrams and analyses to describe what happens when two observers count the waves of light as they go by them, consider a similar scenario on the surface of water where two observers are counting the waves of water going by them from a distant wave maker. Won't your equation apply in this situation as well? And yet I don't think you will conclude that time is going slower for an observer moving on the water, just because he counts the waves passing him at a slower rate than for the stationary observer, would you?

14. Oct 10, 2011

### digi99

Attached the complete diagram, so you can see ΔTb as well.

I find this very teachable, what you can do wrong in mathematics compared with physics.

Ghw, the lightspeed has always the same value in the universe, so it represents time.

My clock is not counting waves but the length of the passing lightwave (divided by time difference always C). You can say the lightspeed represents time.

This all is not true for other moving objects like water waves.

I go to work this out with 2 rest systems and will come back with other diagrams (but takes a few weeks) ...

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15. Oct 10, 2011

### ghwellsjr

Digi99, I'm trying to get you to see that your explanation is no different than normal Doppler which is true for water waves but not true for light waves. It does not teach time dilation. Light waves are like water waves when you view them from a single reference frame and only consider what an observer at rest in that reference frame analyzes using that reference frame and only pay attention to what that one observer sees and measures.

But if you take into account what the moving observer sees and measures, then your explanation, like water waves, does not fit. You have mistakenly interpreted Doppler to be the same as time dilation. You can easily see that this is wrong if you were to do your same thought experiment with the moving observer approaching the stationary light source. Then his count of the waves would be higher than the stationary observer and you would wrongly conclude that time was going faster for the moving observer than for the stationary observer.

You need to learn Special Relativity before you try to teach it to others. And this is the place to learn it if you are willing.

Last edited: Oct 10, 2011
16. Oct 11, 2011

### digi99

Thanks Gwh, also for the other answer (crossing lightwaves), that answer helped me a lot in the way as confirmation how it see it.

Still reading the SRT and I will come back later (after weeks) with this subject (my wrong diagram). For now it is difficult (for me) to agree with you .. I have to find out more ... but "time" in daily life is a problem for me generally ..

17. Oct 12, 2011

### digi99

Gwh, thinking about you say here, is it not true that every object B with a higher speed as object A is always going slower in time (is that a misunderstanding from me) ? (Δt' = γ . Δt in Einstein, A and B in their own rest frames)

So this is also true for water waves except that the speed of a water wave seen from A and B is not constant as it is for lightspeed.

18. Oct 13, 2011

### digi99

In my attachments I have proofed my statement.

With and without Lorenz transformations.

In timediagram 1 seen by oberserver 1 + seen by observer 1 for observer 2, in timediagram 2 seen by observer 2.
Both in/from their own rest frames.

So if an observer 2 (B) has more speed than an observer 1 (A), observer 2 sees the light wave slower passing and his/her time goes slower.

So you can explain to somebody, sees time as an passing light wave in thought, if you go faster, the light wave in thought will be slower passing you, so time will go slower. If
the light wave in thougt is passing you with maximum time, you are standing still in the universe, if the light wave stands still for you, time stands still for you. I think
that lightspeed represents our understanding of time.

In my second attachment I have something strange in a Lorenz equation, I get V = -V, anybody an explanation ?

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19. Oct 15, 2011

### digi99

In time_e2.jpg I had changed t1 and t2 by mistake, that was the reason.

In time_e1.jpg :

it seems that space time diagram 1 was correct, for diagram 2 I was thinking for B (in his own rest frame) to be sure but gives the same result (dividing by γ, the length contraction).
So without Lorenz is the time dialation already visible in diagram 1.

I hope that I am thinking completely wrong, than I learn at most.

Conclusion:

If you are standing still somewhere in the universe, time flies at maximum speed, if you move times flies too but slower.

20. Oct 16, 2011

### ghwellsjr

I've been trying to tell you that your ideas only work for one observer. It is true that if B is traveling with respect to A, then A will determine that B is experiencing time at a lower rate than A. But the same thing is true in the opposite direction: as far as B is concerned, A is the one that is traveling (in the opposite direction) and B will determine that A is experiencing time at a lower rate than B. How does your diagram illustrate this point?

21. Oct 16, 2011

### digi99

Ghw, this is not easy but I have an answer (maybe wrong).

In the diagram I consider A in an absolute rest point (standing still) otherwise the diagram is wrong (while the light wave is moving, A must be moving and it is not). So compared to A, B is really moving.

My story fits when you measure the total length of the passing light wave in the real direction of the movement, that means the moving object moves off the light source.

If you suppose A moves off B (B is standing still), this is not the real situation (A is not moving off the light source). B is moving off the light source so you have to compare it with a light wave in direction B.

If A is moving off B and B is also moving off A (both moving in the opposite direction), than you take a rest point C (standing still) and calculate the time differences compared to C with a separate light wave in direction B and one in direction C. Than in B goes time slower than in A (speed B greater A).

So I created an additional feature in our clocks for A and B, it must measure the total length of a light wave in the direction of the movement (for A or B). If B has an higher speed than A compared with a rest point (standing still, it is an experiment in thought), the clock for B is going slower than for A.

You can also forget the rest point (makes it more difficult than it is) and measure a light wave length in the same direction of the movement (I suppose all light waves are the same for measure the length because the speed is equal). Our clocks must be started on the exact moment when B meets A.

Last edited: Oct 16, 2011
22. Oct 17, 2011

### digi99

It is difficult to see but true I think (tired now, difficult to see), the direction of the light wave in the clock is also not important. If your clock goes in the opposite direction of the movement, the clock still measure the right total length of a light wave (maybe presented as negative because of direction).

So maybe for A and B in the diagram is the direction of the light wave important, but for the measurement not in the clock .. but tired I have to see again ..

23. Oct 17, 2011

### ghwellsjr

Like I said before, you are demonstrating normal Doppler, the same as is true for waves in other media like water or sound. In both normal Doppler and Relativistic Doppler, when an observer is moving away from the wave source, he will count fewer waves than if he were stationary with respect to the source, but this does not demonstrate time dilation. I asked you before to think about an observer moving toward the wave source. In both normal Doppler and Relativistic Doppler, the observer will count more waves than if he were stationary with respect to the source but would you then say he was experiencing time contraction?

24. Oct 17, 2011

### digi99

Gwh, attached the situation.

Yes I compare the speeds V1 (A) and V2 (B) to the point of the light source (C). The point of the light source C is a rest point (standing still, thought experiment).
So you can calculate the time diffference to ΔC, with Lorenz I can proof that this calculation is correct for as well A as B in their own rest frames.

So B has an higher speed than A, B goes slower in time. When you only consider speed compared to A for B you get something same.

In practice you know only the speed B compared to A (or the other way). But compared with any light wave, direction not important (I thought about), B will go slower in time than A because B has an higher speed compared to the light wave.

My diagram sees light speed as a normal but unchangable speed and is not aware from the Doppler effect. But maybe my clocks are ... but that is a problem in measurement .. in this thought experiment I measure the total length of the passing lightwave (if that would be possible in my thought experiment) ..

Thanks Gwh, maybe you are right, I hope anybody else can this confirm too because my understanding of the SRT is now far far away again ... I see light as an oirdinairy light wave with a fixed form and a constant speed and you can compare other (real) speeds from objects with it ...

One more question for you (this is still my problem how I must see it, the mathematics I understand):

If in a diagram eg. A and B are representing as standing still "in relation to each other" and a light wave is reflecting to show points for the same moments (length light wave go and return the same), is it true that when the light wave comes back at A, A and B are moved in the meantime because of the system where they are part of (and have a mass > 0). If yes, this does not matter for the whole concept (they never talk in this way in documentation or books and that is confusing for me how I must see light, and in this case are the length of the light wave go and return not equal) ?

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25. Oct 18, 2011

### digi99

If this would not be the Doppler effect (I get the same formula from Lorenz?):

So V1 and V2 are the real speeds compared to the point where the light wave was started (a kind of absolute point) so speeds are compared to the light speed.

We see in reality only speed V2 - V1 for B (seen from A) or V1 - V2 for A (seen from B).

So Δ B = ΔA . (1 - (V2-V1)/C) or ΔA = ΔB . (1 - (V1-V2)/C).

And here you see clearly that ΔB in both cases is lower ΔA, so B goes slower in time even if you see it from B.

I think that this effect gives time dilation, just a result from the constant lightspeed under all circumstances (no influence from speeds other objects with a mass > 0).

Otherwise how do you explain that I get this same formula via Lorenz (divided by the length contraction) ?