Science a GoGo's Home Page Forums General Discussion Physics Forum c-v

I have read in various groups that there is no such velocity as c-v however on the 9th. page of OEMB Einstein wrote "But the ray moves relatively to the initial point of k, when measured in the stationary system, with the velocity c-v...."

Einstein didn't write anything you read he writes in German, you are reading a translation :-)

Link me the part that's confusing you and I will see if I can work it out.

I have already provided 'the part that is confusing me' namely that physicists insist that there is no such thing as a velocity of c-v yet Einstein specifically referred to same (regardless of the fact that he wrote it in German).

Incidentally, I'm waiting on a response from you to my private message re superluminal neutrinos.

No full quote a few words like what you have said may easily be miss understood, you may be reading something that he did not mean.

For example he could be talking about a hypothetical zero reference frame such a frame does not exist it's just useful sometimes to imagine it.

The fact it's talking about a stationary system makes me think perhaps this is one of those situations.

It's a bit like talking about being outside space or the universe is sometimes conceptually useful although probably impossible.

Really I need much more of the text to work out what he is talking about and I am sure you are totally missing the point.

Yes I saw your message I have some background reading I need to do will be a few days.

Took me ages to find the reference and paragraph

No mystery he is talking about about a moving reference frame.

I have a torch and a person passes me at half the speed of light as they are 1 sec away from me I turn my light on when do I they see the light.

So lets follow the story my light beam takes off and takes 1 second to reach where the person was but they have moved 0.5 light seconds away in that time. My light beam will take 0.25 sec to close that distance but they have again move 0.125 light seconds away.


The solution is

Observer position = 1 light second + 0.5C * time
Chasing beam position = C * time.

When does Observer position = Chasing beam position when

1 light second + 0.5C * time = C * time

rearanging

1 light second = C * time - 0.5C time
1 light second = 0.5C time
time = 2 seconds.


Now Einstein is indeed talking about a static reference frame K like we have in the above.

Note in the above example the speed of light is still c when measured by any of us. It's just the closure speed of the photon appears to be c-v because of the static reference frame (K in Einsteins case).

The person moving, the person turning on the beam and me as a static observer would still agree the speed of light is c.

At no point is he saying the speed of light is c-v its the closure speed that is c-v.

In modern science speak we use the term "separation velocity".

Einstein wrote "...the ray moves relatively to the initial point of k, when measured in the stationary system, with the velocity c-v..."

He clearly wrote that the beam of light moves relatively to the initial point of k with the velocity c-v (when measured in the stationary system) and no denial of that fact will alter it.

"the ray moves relatively to the initial point of k"

That is the closure speed , otherwise he would simply say the ray moves at c-v

I just showed you what he meant look at my mathematics I had to even write the equation relative to the initial point in system k

>>> Observer position = 1 light second + 0.5C * time <<<

The situation he portraits means the closure speed of light would be c-v which is exactly what he says.

He said what he meant you can't interpret it any other way.

I am not denying anything I am showing you what he said why would I care one way or the other, I am reading what he said. Your comment that I an denying something is completely ridiculous what am I denying?

Do you want the truth or me to lie?

I am asuming you are trying to intimate that he said the speed of light to be something other than c for some reason.

If you go back up a bit in the article he states why he is doing this example and it is to show the effects of the constant speed of light so he is hardly then turning around and saying it does change.

You have taken that factor out of context! He wrote "..the ray moves relatively to the initial point of k WITH THE VELOCITY c-v!!!"

It makes no difference whatsoever that he is referring to the closing speed nor what your equations show I'm talking about what he WROTE!!!

He would NOT simply say 'the ray moves at c-v' because he is writing about the velocity of the ray relatively to a certain point!


YES!! The speed at which that light closes on k is c-v!


CORRECT!! Neither you nor I can interpret it any other way! Einstein wrote that the speed at which the light closes on k is c-v!!!!


You wrote, above, '..the closure speed would be c-v.' In other words - the speed at which the light is closing on k is c-v.

c-v is 'something other than' c.


If a light source is not moving away from k its emissions have a rate of closure relative to k of c; if a light source is moving away from k at v its emissions, according to Einstein, have a rate of closure relative to k of c-v!

If a beam of light has a rate of closure of c relatively to an observer who is located alongside the source that beam will strike the observer at c. If a beam of light has a rate of closure of c-v relatively to an observer (a moving source or a moving observer) that beam will strike the observer at c-v!

The speed at which that beam of light arrives at the observer's location and at which it strikes him is c-v NOT c!

Absolutely correct Bill S => beam will strike the observer at c-v flight time

BUT

the speed of light was c throught that whole thing it was never travelling at c-v anywhere, closure speed IS NOT the speed of light. No observer will ever see that c-v speed, he is showing the speed of light to the moving observer and everyone else is still c.

That was the point of this whole excercise.

Lets take the position of each observer and prove it by what we see.

The moving observer went past the stationary observer some 2 seconds later he sees the light from observer 1. He deduces at that time he is 2 light seconds from observer 1 ... he is we did that calculation. So for him the speed of light is still c.

For the staionary observer we need to do a trick we need the moving observer to have a mirror. 2 seconds after we turn on our light it hits the mirror and it will take 2 seconds for the light to get back to us. We see our light reflection after 4 seconds and we deduce the person was 2 seconds away from us because our beam travelled out and back. At the point of reflection the observer was 2 seconds away ... he was. To him the speed of light is still c.

The stationary observer story we covered they see the light beam chasing after the person and reaching the person after 2 seconds because we can see all the relative motions. The speed of light to us is still c.

All 3 observers will still deduce the speed of light to be c. The closure speed of the light to the moving observer is most definitely c-v but that is not the speed of light as measured by any observer.

When we say things can not exceed the speed of light we mean relative to a fixed point not relative to another moving point. Two light beams pointing in opposite directions move away from each other at 2c which is c+c in your terms.

Thats completely expected but it doesn't mean the speed of light as seen from one of the beams is suddenly c+c actually something quite strange happens in that case each beam would see the other beam as stationary because the last photon of the other beam they saw would travel with them and the beam would appear frozen. So again they would still see the speed of light as c.

It's weird isn't it even though the closure speed is most definitely c-v no observer would see light at that speed and that is the key point Einstein was making and why he says all observers see the speed of light as a constant.

He never was talking about light travelling at any other speed than c, he was showing that all observers see the speed of light as c no matter what there speed is.

BTW the effect is quite real and you don't need to be near the speed of light. The GPS system has a sagnac correction because the earth is moving relative to the satelite during the transit time. But again all the observers still see the speed of light as constant c but like my maths above the GPS system has to allow for sagnac to get the distance right.

Equally we can also put limits around closure speeds of -2c to 2c. At -2c we have two objects moving towards each other at the speed of light and they will collide without ever seeing each other. The other extreme 2c is two objects moving away from each other at the speed of light each other will never see each other or appear frozen in time if they started out together.

On the basis that the beam of light strikes the observer at c-v it must have previously approached him at c-v.


In his book 'Relativity' where he deals with the Relativity of Simultaneity Einstein writes that from the stationary observer's point of view the train passenger is hastening toward the beam of light coming from B whilst riding on ahead of the beam of light coming from A.

Contrary to your comment that no observer will ever see that c-v speed, the stationary observer of Einstein's RoS depiction 'sees' beam B as having a rate of closure of c+v relatively to the passenger and 'sees' beam A as having a rate of closure of c-v.

He 'sees' beam B moving toward (having a rate of closure with) and reaching/striking the passenger at c+v and beam A moving toward (having a rate of closure with) and reaching/striking the passenger at c-v.

And its the same above our stationary person in the above can see a c-v speed between the two objects but in there frame the speed of light is still c.

Noone says you can't observe two objects with relative speeds different from c as I said -2c to 2c is possible.

All einstein is saying is in any frame however c for you will appear to be c and we have shown that.


To make sure you finally get it try two observers flying towards each other at 0.9c start them 20 sec apart and tell me what they deduce the speed of light to be before they slam into each other with a combined speed of 1.8c .. 10 seconds later.


BTW in our current actual frame we can see both edges of the universe moving apart from each other at almost the speed of light... so we can actually see an almost 2c seperation event and still the speed for light for us is c as expected.


I should also say none of that proves you can't go faster than c thats a whole other story it just simply says you can't directly observe something faster than c.

Hence something like a tachyons are not possible for us to >>>DIRECTLY <<< observe and I do highlight the word directly meaning by using light or EM waves.

So lets look at a tachyon coming at us we would never see it coming because its moving faster than any light, any light emission coming of the front of it is actually falling behind it. The tachyon hits us and we go woah what was that. Many seconds later the light cone from the tachyon reaches us and we understand what happened. Sound and light from lightning storms work that way you see the lightning long before you hear the sound and so we would get hit by the tachyon long before we saw it assuming it emitted something in the light range at all.

Einstein 'is saying' (i.e. he wrote) "But the ray moves relatively to the initial point of k, when measured in the stationary system, with the velocity c-v..."

Where, in that comment, do you see him saying "in any frame however c for you will appear to be c."?


What has that got to do with whether or not Einstein used the term 'the velocity c-v.'?


According to Einstein, if I am looking at a person on a train who is hurtling toward a lightning flash the light from that flash will have a rate of closure with the passenger of c+v ergo I directly observe that beam (theoretically, I cannot actually see that beam) travel toward and arrive at the passenger's location at c+v!

I am observing something faster than c - a rate of closure of c+v!

I am observing that beam closing with and striking the passenger at c+v!

I am looking at a light source (o) that is moving across my line of vision to the right of screen and which projects beams of light in its direction of travel (beam A) and in the opposite direction (beam B).

According to special theory, beam B will, from my point of view, travel a greater distance away from the source in a given period of my time than will beam A

B<-------------o------->A

From my point of view - beam B has traveled AWAY FROM ITS SOURCE at c+v whilst beam B has traveled AWAY FROM ITS SOURCE at c-v!


This has nothing whatsoever to do with whether or not Einstein used the phrase 'the velocity c-v'.

Edit: Think I worked out your confusion so adjusted for next post ... I left a bit in about sound here.

Light is nothing special by the way.

There is only one speed of sound for given media conditions as well there isnt speed of sound + velocity either.

If you are standing on the wing of a plane doing mach 0.99 the speed of sound is still mach 1 its not mach 1.999 just because you are moving.

And if you used sound waves to deduce the speed of sound all observers would correctly deduce the speed of sound to be finite and constant for given media conditions.

We know that for example you can go faster than the speed of sound so this observation limitation does not mean you can't go faster. As with sound but once you pass through the sound barrier everything changes for sound.

re-reading I think I get your problem whats confusing you



That side on stationary observer position we are talking is the GOD view with no speed of light to us.

That is your so called zero frame in relativity and it does not exist for any actual observer because there is speed of light to us.

And I think thats may be where your coming unstuck now.

Einsteins K frame is the Zero reference frame because we have to be able to describe the movements on a common system because everyone will see something different. Thats not an actual observer like you or I, it's godmode so we can describe the movements.

If we were side of to that scene we wouldn't see it like that at all in real life.

WHY => There is speed of light to us as the side on observer we would actually see the whole scene different and thats why we have to create a zero frame.

Where your speeds where slow like with trains etc you put an observer there and they would be able to see almost what a zero frame looked like but not with stuff moving near the speed of light. Because the time of light to observer would change what they saw.

So what I should correctly say is you can deduce a c+v speed in a zero frame (Einstein K frame) but you could never directly observe it.

As I said we have deduced the edges of the universe must be moving away from each other faster than the speed of light and we have deduced two galaxies are moving away relative to each other faster than the speed of light.

As per post above speed of sound is no different it's only ever one speed. Stand on a moving car and yell and the sound still moves at the speed of sound but what happens is the pitch changes to people stationary on the ground.

I have attempted to resolve this discussion by asking you direct and relevant questions however you have conveniently deleted any reference to same thereby censoring my arguments and now I find that your grammar is becoming incomprehensible.

I am not sure I can help you Bill 6 .. English is not my native language .. prob ask someone with better language skills.

The effect being described is rather trivial it doesn't really prove light can or can not move faster than light it simply says you will never directly observe the effect unless you are god and can see things faster than light in Einsteins "K" reference frame.

In my mind there is not much to resolve it's all pretty trivial but you seem to get hung up on c+v closure speed which isnt real to any real observer.

As I have said think about sound it behaves exactly the same way ... there is also only one speed of sound as well just because you are moving in aircraft doesn't change the speed of sound. Changing the media conditions is the only way to change the speed of sound and that is the only way to change the speed of light we know of as well.