is lighting bolt suspended in air? or installed on train?

Dear Professor Abkenar,

It's suspended in the air, so the flash scorches both the carriage and the platform symmetrically. So it's in both systems. The moving one, and the stationary one. (Confer also, attached, the original AE train-embankment experiment and the description therein).

So the light reaches the person on the train in less time, but the light passes by both people at the same speed according to relativity.

Dear Professor Abkenar,

If the light propagates at the same speed "c" for both observers, as you note, and the distance from the issue of the rays are the same, then

t = distance from issue / speed of propagation = the same for both observers.

So according to your latter statement then the rays must arrive at both observers simultaneously.

Light travels at a constant speed relative to a reference point. If you consider a person riding a train as a reference point, light still travels at a constant speed, because nothing can travel faster than light relative to any observer. The metric is different for two observers, space and time are different for a stationary and a moving observer.

That's right, speed of light is Absolute, and equal to "c" always and under any circumstance.

Dear Professor Abkenar,

The Lorentz relations are already applied in the above problem. We have x = ct, and x' = c t' in each frame. ie. the speed of light in the dashed frame and the undashed frame is "c". The Lorentz relations connect these two frames.

Preprint Use of the classical speed (c - v) in the Wiki Figure purpor...

Dear Gary stephen,

If an object moves at speed, time dilates for it, and if light comes from behind it and then passes by it, it appears to the object to have passed at the speed of light because time is slow motion for it and it sees. other speeds more.

in continue of previous situation: (if light come from behind of the moving object) On the other hand, the length of the object shortens, so the light passing by many times more than the length of the object, as if the metric of the moving object has shrunk and the passing light passes by it at the same speed of light in its own metric.

Dear Professor Abkenar,

Are the changes you describe above so the the light propagates at "c" in the moving frame? Is that the reason for the dilation and contraction you suggest?

Light passes by the moving man at a speed of C-V, but is observed by the moving man at the same speed C because time passes slowly for the moving man and he observes whatever passes by him as being faster than it actually is.

There is a difference between reality and what the observer observes.

Dear Professor Abkenar,

So according to you the man on the carriage will see light approach him at "c" ?

Dear Gary Stephens,

Yes, exactly.

The standing man sees the light passing by the moving man at a relative speed c+v (toward light source) or c_v (backward light source), but the moving man sees the light passing by him at a constant speed *C*.

So if the moving man moves towards the light source, the light will reach him first.

Dear Professor Abkenar,

So light is propagating at faster than the speed of light on the carriage, as viewed by the stationary man?

yes exactly,

But notice this is relative speed.

Dear Professor Abkenar,

But according to the Einstein vel. addition formula for light rays, the speed of the light on the moving carriage as viewed by the stationary observer is "c" and not (c + v) as you say. Confer above.

So speeds of (c + v) and (c - v) in "going between frames" are ruled out mathematically by this formula. Light can only travel at "c" both in each separate frame, and also only travel at "c" in "going between frames," on the moving as viewed from the stationary, or on the stationary as viewed from the moving.

Absolute speed is different from relative speed.

Stationary man see the light pass from moving man, at absolute speed *C* and relative speed *C+V*.

if you see two light move in opposite site together, so you see one light pass from other light at *2C* relative speed. but you see each light propagate at *C* in absolute speed.

Dear Professor Abkenar,

what is this speed relative to?

Like all other authors, including Einstein, Wheeler & Tayler treat light in the manner of bullets or ball-bearings, and employ the classical addition of velocities ( c + v), ( c – v) in their demonstration. This is exposed in Figure 1, above.

Preprint On Wheeler & Tayler's (c - v), (c + v) in error figure, purp...

Dear Gary Stephens,

speed *C+V* is relative to the man on train that is observed by the man on platform.

Dear Professor Abkenar,

As you can see, a speed of c + v for the light rays, relative to the man on the carriage, as viewed by the man on the plaform, is ruled out mathematically by the Einstein vel. addition formula applied to light rays. Instead we have C = c. Confer above.

Dear Gary Stephens,

I have a question for you.

Two spaceships are approaching each other in opposite directions at speed 0.6C .

What is their relative speed?

Why are you trying to claim the speed of light is not Absolute and equal to "c" but is 2c or 1.2c ?

If you think this is the case, please furnish -any- article on Special Relativity, from its inception in 1905 which gives the speed of light as 2c or 1.2c or 0 or anything else except "c", in all circumstances.

And explain to me why you are ignoring the Einstein vel. addition formula above, with the proof that C = c in the case described above, ie. the above gedanken experiment.

If you can provide any article at all, in which the speed of light is given as anything other that "c" in all circumstances, much appreciated.

Looking forward to your references.

Dear Gary stephens,

Einstein's relativity is completely correct, but we should not ignore the point of who the observer is.

The formula you have posted is for an observer who is himself moving and looks at the light passing by him.

If the observer is standing still, his relative speed with respect to the light or the ship passing by him will never exceed the speed of light.

But if the observer is standing still and looks at a moving person as the light passes by him, the relative speed will never be less than the speed of light.

The point of reference of who the observer is should not be ignored.

Dear Professor Abkenar,

You claimed above that :

You also claimed that :

and you also claim that :

So you claim above, that light speed is not Absolute and equal to "c" in all cases.

With your claim

(i) , if we put V = C, then you have a light speed of zero.

With your claim

(ii) , if we put V = C, we have a light speed of 2C.

This matches your other claim :

As I mentioned, can you find any references in the literature to support your idea that light speed is not Absolute? or is this your own personal view?

1) It goes without saying that the man from the train sees the light pulse first. The relative velocity between the train and the wavefront of the light pulse is vt = c + v

2) The light pulse reaches the man on the platform a little later. The relative speed between the platform and the wavefront of the light pulse is vp = c.

3) NOTE! The relative velocities in 1) and 2) only apply if the absolute velocity of the platform in space is 0! Otherwise, vt = c + v ± w and vp = c ± w , where w is the absolute velocity of the platform in space in the direction in which the light pulse is moving.

This means the light ray is propagating at c + v relative to the train, which is impossible. The speed of light is Absolute and equal to "c," as evidenced by Maxwell's equations, the Michelson Morley experiment, and the Einstein vel. addition formula for light rays, proved, as Einstein noted, in the --optical experiment-- of Fizeau.

Gary Stephens

3) "This means the light ray is propagating at c + v relative to the train, which is impossible."

You are thinking wrong in the same way as most people. Gurcharn Singh Sandhu probably thinks the same way as me.

Light NEVER moves relative to an IRF. Light moves RELATIVELY ONLY RELATIVE VACUUM!

4) "The speed of light is Absolute and equal to "c," as evidenced by Maxwell's equations, the Michelson Morley experiment, and the Einstein vel. addition formula for light rays, proved, as Einstein noted, in the --optical experiment-- of Fizeau."

4a. Correct regarding Maxwell's equations. (c = Absolute Speed of LIGHT in Vacuum, c = ASLV)

4b. Incorrect regarding MME (c + v is wrong, where v is, for example, the speed of the train relative to the platform -->

You can add c ONLY with another absolute speed, so we must have the train's absolute speed relative to vacuum)

4c. Incorrect regarding Einstein (idem)

Gary Stephens

Dear Gary and fellow scholars,

A fascinating question that touches both the physical mechanics of light propagation and the nature of perception itself.

Let us first address it scientifically:

When a lightning bolt occurs, it produces both visible light (the flash) and electromagnetic radiation (radio waves, X-rays, even gamma rays) almost instantaneously.

• Light from the bolt travels at approximately 299,792 kilometers per second (the speed of light in vacuum; slightly slower through air).
• The visible rays (photons) emitted are received first by observers located closest to the source, assuming direct line of sight and no obstructions.
• Since light travels uniformly in all directions, everyone within the visual range receives the photons almost simultaneously, but those nearest experience slightly earlier reception—though the difference in time is typically nanoseconds to microseconds, practically imperceptible to humans.

Thus, the person closest to the origin of the bolt receives the rays first—but the margin is so narrow that under normal human experience, it appears instantaneous to all observers.

Moreover, from a relativistic viewpoint, all observers receive the lightning's photons in their own inertial frames, preserving the fundamental speed of light as a cosmic limit. This is one of the extraordinary harmonies of nature: the light from lightning obeys universal law regardless of who observes it, where they are standing, or how they are moving.

If we stretch the question a bit philosophically— One could say that Nature itself receives the rays first. All molecules, particles, and fields along the path of the emitted photons interact before any conscious being "receives" the sensory information.

Thus, Nature is the first witness to lightning's light, and we, as conscious observers, are the interpreters of that cosmic event.

Nearest observer receives the rays first (technically). Practically, all nearby observers see it almost simultaneously. Philosophically, Nature witnesses the rays first, even before conscious perception.

Thank you for this stimulating and multi-layered question, Gary.

Such inquiries not only illuminate physical phenomena but also invite us to reflect on our relationship with light, time, and observation itself.

With regards,

Sandeep S. Jaiswal

Sandeep Jaiswal Gary Stephens

5) Mr. Sandeep Jaiswal: "A fascinating question that touches both the physical mechanics of light propagation and the nature of perception itself."

One should not confuse physics with the nature of human perception!

6) "Let us first address it scientifically"

Just because you write that you are doing a scientific analysis does NOT necessarily mean that the analysis is scientific.

7) "Since light travels uniformly in all directions, everyone within the visual range receives the photons almost simultaneously"

The light pulse reaches all points that are at the same distance from the light source simultaneously.

8) "This is one of the extraordinary harmonies of nature: the light from lightning obeys universal law regardless of who observes it, where they are standing, or how they are moving."

Yes, light moves independently of the movements of both the source and the observer.

Sandeep Jaiswal

Dear Professor Jaiswal,

I agree with much of what you say, except :

You can't elbow in "motion" without elbowing in a forbidden speed "v". The "motion" must be with respect to something, and in this case it is with respect to the light ray.

So it's impossible for the observer on the carriage to become by this "motion" toward the light-ray unless he can travel at (c + v) relative to the light ray, which is impossible. He can only travel at "c" relative to the light-ray, and so, in effect, he is as static relative to the advancing light ray, as the man on the platform.

This is also proved by saying : the man on the carriage is "stationary" -- it is the embankment that is doing the moving. So the light propagates towards the "static" man at c, over the fixed distance d. This will take time t.

(This is the same time t, for the man on the embankment, when we deemed the embankment "stationary"). And follows from the logic of Relative Motion.

They will both receive the rays after the same duration : simultaneously.

Preprint On Einstein’s use of the classical addition of velocities (c...

Perhaps someone can explain to me how light propagating at (c + v) and (c - v) relative to the end of the rod is "constancy of the velocity of light" ?

https://www.researchgate.net/post/Perhaps_someone_can_explain_to_me_how_light_propagating_at_c_v_c-v_relative_to_the_end_of_the_rod_is_constancy_of_the_velocity_of_light2

I take it, that's a no--- no one can explain to me the unexplainable.

Gary Stephens

If the distance between the light source and the two men at the beginning of the experiment is d (say it is the length of the cart) then the light pulse first reaches the man who is ON the cart (who is moving towards the light source).

tt = d/(c+v) - the time after which the light pulse reaches the man on the train

tp = d/c - the time after which the light pulse reaches the man on the platform

tt < tp

tt = d/(c+v) - the time after which the light pulse reaches the man on the train

This is the classical addition of velocities (c + v) which is not valid for light rays. Light has an Absolute speed of "c" for uniform motion, and cannot travel faster than the speed of light (c + v). If we put v = c, you have light propagating toward the man on the train at 2c.

Confer above. Speeds of (c + v) or (c - v) are invalid.

Gary Stephens

"This is the classical addition of velocities (c + v) which is not valid for light rays."

We can argue about this endlessly if we (you and the others) don't think strictly about the quantities we use.

tt = d/(c+v) - the time after which the light pulse reaches the man on the train

tp = d/c - the time after which the light pulse reaches the man on the platform

Thought experiments within SRT with the train and the platform are as follows:

We have two IRFs, IRF1 is stationary and IRF2 is moving relative to IRF1 with the relative velocity v. I have asked the question many times here on RG but no one wants to answer it. Relatively what is IRF1 stationary?

My answer is: It must be stationary with respect to vacuum, the medium in which the light signal travels.

Then we have the following situation:

a) The platform is a station with respect to vacuum, vp = 0

b) The train moves with speed vt (= v) with respect to the platform, with respect to vacuum.

c) The light signal moves with speed c with respect to vacuum.

In this case c, vp, and vt are absolute velocities. In this case it is correct to write

tt = d/(c+vt) = d/(c+v)

tp = d/(c+vp) = d/c

In both of these expressions, the speed of the light signal is c.

c+vt och c+vp are relative speeds!!!

Most physicists believe that c + v is the speed of light! NO! The speed of light (relative to vacuum) is c, c + v is the relative speed between the observer and the wavefront of the light signal.

BUT: v should be absolute speed, the observer's, train's, speed in space, and this speed is NOT KNOWN.

This is proven false by the above. Bohm above explicitly says (c + v) is the speed of light (along the moving carriage)--- so you cannot wriggle out of it.

I can choose the carriage to be the "stationary frame"-- something which you do not appear to appreciate. That is why Bohm says speeds of (c + v) and (c - v) are invalid. Einstein says the duplicate in his 1916 tract.

Einstein above, also saying speeds of (c - v) ... or (c + v) are invalid.

Only.... you... say speed of (c + v) and (c - v) are perfectly fine.

(c + v) with v = c, mean light propagating at 2c.

(c - v) with v = c, means light propagating at zero 0.

That is why both Bohm and Einstein rule these light-speeds out, a priori.

Gary Stephens

"(c + v) with v = c, mean light propagating at 2c.

(c - v) with v = c, means light propagating at zero 0.

That is why both Bohm and Einstein rule these light-speeds out, a priori."

WRONG! You are thinking wrong like many others!

In your example: (c + v) with v = c, mean light propagating at 2c.

What is moving at speed c and what is moving at speed v (v = c)? These are two different light signals moving towards each other. You can NOT add their speeds and say that one of them is moving at speed 2c.

2c is their relative speed!

When two cars are driving on the motorway towards each other at a speed of 100 km/h each, YOU CANNOT SAY that the car is moving at a speed of 200 km/h. That's it. Simple. But people (including you) use speed concepts incorrectly!!!

Dear Slowak,

I've explained to you that both Einstein and Bohm rule out speeds of light of (c + v) and (c - v) a priori. You do not deny this, because both Bohm and Einstein explicitly call speeds (c - v) and (c + v) light speeds. (Einstein only talks about the (c - v) light speed, Bohm both (c - v) and (c + v) light speeds.

So, unless you are saying both Einstein and Bohm got this wrong, you have no argument.

2c is their relative speed!

(c + v) refers to one light signal only. Not two. (c - v) refers to one light signal only. Not two.

So, 2c refers to the speed of one light signal, in the moving frame, when the frame (say the carriage) moves with a speed v equal to c. (This is the fore-ray, if issued from the ends, or the left-hand ray if issued from the middle, as depicted above).

The right-hand ray, is shown above, propagating at (c - v).

Gary Stephens

"I've explained to you that both Einstein and Bohm rule out speeds of light of (c + v) and (c - v) a priori. You do not deny this, because both Bohm and Einstein explicitly call speeds (c - v) and (c + v) light speeds. (Einstein only talks about the (c - v) light speed, Bohm both (c - v) and (c + v) light speeds.

So, unless you are saying both Einstein and Bohm got this wrong, you have no argument."

Your way of arguing is not leading anywhere. You keep recounting what one or the other person has said. Tell us about the phenomenon itself.

You contradict yourself many times:

1) "both Einstein and Bohm rule out speeds of light of (c + v) and (c - v) a priori."

2) "because both Bohm and Einstein explicitly call speeds (c - v) and (c + v) light speeds"

I have asked you to explain whose speed c is and whose speed v is. Can you describe it in a simple way in your own words and not paste the same png.file dozens of times.

"(c + v) refers to one light signal only. Not two. (c - v) refers to one light signal only. Not two." AND

"So, 2c refers to the speed of one light signal, in the moving frame, when the frame (say the carriage) moves with a speed v equal to c."

I understand that you have here a traffic light and a cart.

You say that the cart moves with speed v = c. Answer my question, please. Relative to what is the cart moving with speed c?