# C maximized in its own direction, so v > c possible ?

1. Nov 20, 2011

### digi99

It's just an idea if V could be greater than C (proofed in the future maybe).

Look to this derivation of Lorenz (university Australia) : http://www.phys.unsw.edu.au/einsteinlight/jw/module4_time_dilation.htm" [Broken]

Here you can see how Jasper looks to light, is it ever proofed that the speed of light is also C on this way ? But the derivation leads to the same results as in Lorenz ..

I can understand that C is maximized in its own direction, but if you look from an other view to photons maybe it's that not the case anymore, Japser's light speed could be SQRT(V2 + C2).

I should think a light beam is only a light beam in its own direction, with other words if the front of the photons goes in the same direction of the light beam.

So in the case it would not be valid to look to light in this way, V is also not maximized mathematically.

And if this would be correct, and this derivation is not valid but leads also to the same results, somewhere goes something not valid in the Lorenz derivation ...

Just an idea ...

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2. Nov 20, 2011

### atyy

When we say v ≤ c, v is the magnitude of a velocity, and c is the magnitude of a velocity in the same spatial direction as v, where "velocity" and "spatial direction" are defined using any particular inertial system of spacetime coordinates.

3. Nov 20, 2011

### digi99

For fun, let's suppose that the speed of light is indeed C in his own direction only and you could go faster (because you don't have equations now as Einstein had).

So when an object goes faster with e.g. speed = 2C, say a time dilation should occur of a half second (the word time dilation still exists), the distance is indeed C m (and C itself .5C m distance) in the objects rest frame. In the observers rest frame is C, Cm and the object 2C m further.

A muon still gives a time dilation. A neutrino gives a time dilation.
In CERN maybe a proton has to overwin the C limit, to go faster (ever read by suggestions of other scientists).
Everybody satisfied.

Only time is now different, you can go faster than time (if c presents time), maybe possible for a neutrino but never for humans I guess.

One thing is clear for me, Enstein did the most things alone I thought and on paper. Statistically it looks for me that it is impossible to analyse nature in such a way. Now we have not 1 Einstein anymore, but a group of thousends of scientists working together with highly sophisticated technics/equipment (and we can go into space), it seems for me that they must find out/analyse more than in the past was possible.

4. Nov 21, 2011

### Staff: Mentor

This is not correct. There is a lot of experimental evidence demonstrating that Jasper's light speed is c.

http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html

5. Nov 22, 2011

### digi99

Hi DaleSpam, an opportunity for me to find out more about light (Maxwell).

I have attached a document. Correct me what is wrong.

You see in A (a moving light source and many photons), with building light waves (vertically), so an em field like in B.

Do you tell me that on the hypotenusa also an em field is building, with a real light wave and light speed (so not an illusion) ?

And if yes, why is the horizontally line in D no em field ?

Am I right when I think em fields can have constant speeds like V (and other speeds), or is it the light source which can have a constant speed (and other speeds) and the em field is just following ?

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6. Nov 22, 2011

### Staff: Mentor

I am not sure what you are trying to describe with these drawings. If I am understanding the scenario correctly then I don't think that there is a device which can generate what you have drawn. Let me ask some questions about your intention.

Does the source produce collimated light like a laser or non collimated light like a light bulb? If collimated what is the direction?

Is the source producing light across a small aperture (approximately like a point) or across a large aperture (approximately like a line or plane)?

Is the light coherent with some particular phase relationship or incoherent?

Is the light pulsed or continuous?

7. Nov 22, 2011

### digi99

Hi DaleSpam,

I forgot to tell that C in my drawing belongs to the hypotenusa (my dictionary does not work anymore on Windows 7, so that's een problem in translations sometimes but I go the buy an update soon).

My drawings is the situation as Jaspers light .. so the hypotenusa in my drawings is the path of Japsers light ... and the light wave moving with speed V to right .. the many photons are part fom that moving light wave ..

I am very very curious, because I learn than a lot with one answer from you, how I must see this situation with em fields (so light) ...

8. Nov 22, 2011

### Staff: Mentor

OK, so the light is collimated along an axis perpendicular to the axis of relative motion in the source's rest frame, it uses a small aperture, coherence doesn't matter, and it is pulsed.

Yes.

I don't know what line you are refering to.

The EM wave will have a constant speed of c, regardless of the speed of the light source. That is the second postulate of relativity and is confirmed by lots of experimental evidence.

9. Nov 23, 2011

### digi99

It was the horizontally line on the level of the moving light source, D in my drawings.

Thanks for the answers, still hungry to answers and in fact difficult to find on the internet.

1) So an em field generates a light wave. The em field follows the moving light source. How many different light waves (= light signal) are generated by an moving light source with 1 em field, is that only one or many (could it be one in thought), so is it only one long row of photons or many parrallel rows ?

2) A light wave has always the speed c in its direction, can a light wave also have a speed V NOT in its direction (where the speed is c) (see it as a flying light wave) ? I think if you answer that's not possible, than a light wave can't move siteways ?

3) Last question. You agree that Japsers light (seen by Jasper) is formed by the tops of the vertical light waves from the moving light source (from Zoe). Why is that form also a light wave (I don't see that), so you can say the length of the hypotenuse is C.T and from the moving light source V.T ? (or is it no light wave but only the speed of the photons, crosspoints with that line or hypotenuse) With other words how is it possible that I don't see that ?

10. Nov 23, 2011

### Staff: Mentor

The light source is pulsed, so that is not an EM field.

I have no idea what this question means. I don't know how you would count the number of light waves.

Yes. Remember, speed is the magnitude of velocity. If you break a velocity up into different orthogonal components then each component can be less than or equal to the magnitude.

There is no vertical light wave in Jasper's frame, only the diagonal light wave. Remember, the light source is pulsed, so after it makes the pulse of light it shuts off and stops producing more EM. The empty boxes you drew in A don't have any fields or waves in them. That is why your drawing was confusing and I asked all of those clarifying questions.

11. Nov 23, 2011

### digi99

Again thanks for the answers, slowly I know want I wanted to know ..

Some questions more:

1) I mean generally, if a light source was moving to the right (no pulse) than it would generate many vertically waves (maybe there is 1 em field or many ?) , or if there was 1 em field the generated light wave is following this em field and a very long wave is generated until the em field is switched off, the light wave has a speed C up and a sideways speed V ? If am wrong can you tell me what is happening while a light source with vertically lightwaves is moving to the right (no pulse) and where the light waves are going if the light source is switch off, and what speeds are working on the light wave(s) ?

2) A light wave can only get a sideways speed from its em field, once "free" that sideways speeds stops (only a speed C in its own direction) or continues ? In vaccuum nothing can influence the light wave anymore with speeds NOT in its own direction (that's always C) ?

3) Exiciting to know how I must see this situation. I attached again a beautiful drawing.
See A. So I guess a pulse is a short lightwave (in fact maybe it answers partially questions 1/2), it has a speed C up and a sideways speed V (so keeps this speed after the em field is switched off) and bouncing between the mirrors.

See B. The pulse forms the diagonal light wave (or line) with his vertically short lightwave ? Why has the pulse on the diagonal line, speed C in the diagonal direction ( we can not speak about a light wave in the diagonal direction, because the short light wave is vertically) ?

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12. Nov 23, 2011

### Staff: Mentor

Your questions about numbers of waves are very confusing. As I already told you I don't know what it means and I don't know how you would count numbers of waves. Unless you can explain in detail what you mean there is no point in continuing to ask me about it.

The speed of a light wave is c. Speed is the magnitude of velocity by definition, not a component. So if you have a light wave with a horizontal component of velocity vx then the vertical component vy is given by
$$\sqrt{v_x^2+v_y^2}=c$$

You cannot get a component c up and a component v to the right because the speed would be greater than c.

Last edited: Nov 23, 2011
13. Nov 24, 2011

### digi99

Sorry DaleSpam, it's a language problem I think, in fact I was asking for my feelings many or 1 (not how much, of course you don't know).

I shall ask my questions (the lasts except the discussion topic of course, partially already answered with your calculation below) again, then I have a good view of light. You think maybe easy to find, but that behaviours oflight are difficult to find, and I want to have an exact view. I appreciates it very much to have the opportunity to ask experts (otherwise I have just to say to myself, this subject you should never known unless you follow a study, but thats not always easy in time near your own profession).

1) if a light source is moving to the right (or left) with speed V, and this light source has only 1 em field, I guess there is just one (long) light wave generated, on the location of this em field, and the Vx/Vy components of this light wave is given by the Vx/Vy component of the moving light source, if the em field (light source) is switched off, the light wave continues in some direction. Right ? What is the direction of that light wave, still calculated by the VX/Vy components ?

2) I don't know your answer from 1 already, but I try. Can a light wave only gets his velocity components by the components of an em field (light source), once "free" (em field switched off) in vaccuum are there other velocity components possible to change the direction of this light wave (by what caused) ?

3) If a light source should be accelerated by a force, what means this the generated light wave ?

So this means the component c up wins ?

Can you look to question (the discussion topic) 3 (Y 03:13 AM, I guess #11, not to see now during answering) ?

14. Nov 24, 2011

### Staff: Mentor

For the third time now: I don't know what it means to have only 1 em field. How are you counting the number of EM fields? What do you mean by 1 EM field? This is getting extremely irritating.

Light moves at a speed of c regardless of the velocity of the source. The velocity of the source can affect the direction or the frequency of the light, but not its speed.

I get the impression that you believe that light is not an EM field and that EM fields are somehow something that light is "stuck" to and then becomes released from.

Do you mean that the source is being accelerated during the emission of the light wave, or after?

Look at the equation I gave. If the component up, vy, is equal to c then:
$$\sqrt{v_x^2+v_y^2}=c$$$$\sqrt{v_x^2+c^2}=c$$$$v_x^2+c^2=c^2$$$$v_x^2=0$$$$v_x=0$$

15. Nov 25, 2011

### digi99

Sorry DaleSpam for the irritations, I have seen now it were misinterpretations from me (after reading some other articles about light). I know now what an em field is in terms of physics.

In generally, it is never to say when light waves are emitted, how many and how long, just a bundle of light waves ? E.g. when light arrives from a star in space, it's never to say how long the separted light waves are, not with todays stand of physics ?

In answer on your question, yes what is the effect on a light wave's direction before emitting and what is the effect after the emitting on a light wave's direction ?

Except in bending space, a light wave's direction can't be changed in space ?

For simplifying things, I work in mind (so with thought experiments) with only 1 light wave.

So my last drawing shows a vertically short light wave (Zoe's pulse).

Now I understand you say, because of the horizontal Vx component in Jasper's frame (was 0 in Zoe's frame) that's light wave direction is the diagonal line. Now I understand that a light wave's direction can be changed (or seen) as just for other objects (like a basket ball in Zoe's and Jasper's frames).

The diagonal line has a short light wave now, moving over that line with speed C.

Wihout knowing that a light wave's direction can be calculated in this way, you could not understand this example. Now I really understand .. thanks very much ..

But there is a difference, if a piramide object with his top above is moving up and down in Zoe's frame, the top will still be there in the same position in Jasper's frame only the moving direction of the piramide object is different. What is the difference with light ?

Last edited: Nov 25, 2011
16. Nov 26, 2011

### Staff: Mentor

Once a light wave has been emitted further acceleration of the source is irrelevant and will not affect the light in transit at all.

For a source which is undergoing acceleration during emission the effect will depend on the details of the construction of the device. If the device is sufficiently small and rigid and the acceleration sufficiently gentle and the pulse of light sufficiently brief then to a first order approximation you can consider only the inertial frame in which the source is momentarily at rest during the emission. The direction will be the usual unaccelerated direction in this frame, and the direction in any other frame can be obtained by a Lorentz transform.

Yes, the difference being that the speed of a basket ball is not frame-invariant.

I don't understand the question. What does the shape of the object have to do with anything? If it is light it moves at c and if it is matter it moves at some v<c. The shape is not important.

17. Nov 26, 2011

### digi99

The fact I asked you the question about how many waves, has to do what I learned 35 years ago about the electric current in a transistor, completely described by quantum phycics. So I was only curious if there is a model in quantum physics claiming how many waves there are under conditions of course. But I asked this earlier in unclear sentences. But I understand there is no such model for photons otherwise you had answered.

What you say about the transform is interesting because I did such transform also in another topic. But I thought it's only possible for the line light itselves, which other objects it is not possible. I thought you can use Lorenz only to see from the moving object itselves (standing still) how time and distances are. But is it also possible with a Lorenz transform that you have something moving X and a movement B in a rest frame A, and that you can transform moving X in a rest frame of B (by using the transformation for coordinates) ?

That's also interesting for me that you don't understand that.

If you look to my last drawing, you can interpretate the speed C on the diagonal line in 2 ways, the difference is very important for the result (I think).

The shape is very important now because that's the direction of the light wave. If the light wave's direction is not rotating like a compass needle (under pressure of own formulas), you can have 1) a component up = C or 2) the speed on the diagonal line = C (if ever real meassured like in this situation, that's different than meassuring the speed C in Zoe's frame where the light source is/was). If the speed is C on the diagonal line and the direction of the light wave is up, than the Vy component is < C (direction light wave). What is true (and really proved) ?

Besides normally you may not compare two frames in 1 view, and we do that now mathematically .. normally we work with clocks etc. to see it in 1 view and now it goes so easily (for my feelings strange, the chicken and the egg) ...

18. Nov 26, 2011

### Staff: Mentor

You can use the Lorentz transform to change coordinates for any object or path. It is not limited to light nor to paths of objects which are stationary in one of the frames.

1) is wrong. 2) is correct.

19. Nov 27, 2011

### digi99

Thanks DaleSpam for all the answers, I go to play with Lorenz the coming days/weeks (when I have time, it's an interesting subject) and to look carefully to its derivation because I want to understand the basics very well before to continue to next subjects in relativity.

Ok the postulate of Einstein says, the speed is always C but the light wave's direction is not really involved.

Maybe is it unlikely that the direction of a light wave is different than it's velocity direction, but it can be a leak if V > C will be proved maybe in the future by more experiments (something must be wrong in that case).

I let you know what I think for myselves after a while in this topic (maybe I fully agree with you, maybe not, who knows the real truth ..).

20. Dec 1, 2011

### digi99

I was trying with only a Lorenz transformation to find a function for the diagonal line, but that's difficult.

1) Can somebody shows with Lorenz that the velocity direction of the short/long lightwave is the diagonal line and the lightspeed is C on that line ?

2) If you look to the Lorenz derivation (transformation coordinates) only the speed of light (velocity direction) is involved and not the direction of the light wave itselves. This could mean in fact that speed (velocity component) of the light wave in its direction can be lower C if in a situation speed is C in it's velocity direction. This could be an error leading to another formula in the future (when V could be greater C possibly).

Can somebody proove (or is it proved), that the velocity direction is the same as the light wave direction ?