I have been following this conversation for a while now:
https://qoto.org/@zeccano/103205964102778422
@freemo, problem is that @zeccano does not understand what "relative to" and the "frame of reference" are or mean.
Without understanding this, it's all total waste of time.
It's also painfully obvious, when you bring up the bus and the ball.
I am not sure why zeccano cant (or refuses) to understand that if I send you a photon, while we both move, I see it travelling in a straight line from me to you and you see the same.
Now, for someone OUTSIDE of our frame, the photon moves diagonally relative to his reference point.
Exactly what happens if I was standing on a bridge, looking down at the bus and those 2 kids towing a ball to each other.
This is where people screw up. They mix the frames of reference where events occur and where they are - outside of it.
It's like they refuse to understand you can have a frame inside a frame.
This is also causes the confusion about why the laws of physics remain the same in all inertial frames of reference.
@CCoinTradingIdeas @freemo
Your illustration involving a ball tossed between two people, will be seen as a diagonal when viewed from a differnet moving perspective.
But its not like this with light.
According to every physicist including einstein, Light is the only thing that is absolute, its own self is the ONLY absolute frame of reference, which is why light is invariably always c.
Because physicists are saying that lights frame is the preferred frame. (the only absolute frame)
So in your scenario you have done the impossible, you have set the observer who sees the diagonal as if he were in the absolute preferred frame of light!
He cannot be in that frame of reference. Its absolute.
The two guys trying to toss the ball between each other wont have any problem if its a ball, which gains the inertia of the guys, as the ball has mass it CAN gain the inertia of the guys, but light cant, as its without mass.
If you try to reverse it, and claim that the guys are not moving, they are just tossing the ball back and forth, its the observer that is moving past, so he will see the diagonal, then still it only can work for a ball, not light? Why? Because in this scenerio, with the moving observer, you now have him AND light in the same absolute frame, again its not possible.
Anyway, what are you going to do with Einstein and every other physicist who say flat out, that light is NOT dependent on the motion of the source?
So move the guy who tosses the photon or the guy trying to catch it, and they will NOT stay in the same frame as the photon, as my video shows.
You guys are talking around in circles, contracting your own claims with weak logic.
A photon has no mass, therefore no inertia and cant have any momentum, relativistic or not.
🇳🇱
No thats not what einstein or physicists say, just you.
Its easy to make things up when you dont even understand the basics...
Not to mention what you are claiming directly contradicts experimental results like everything you said. The Michelson–Morley experiment directly contradicts your claim.
@freemo @CCoinTradingIdeas
Einstein said the the speed of light remains constant irrespective of the motion of the source. Its a basic postulate of SR.
speed is motion is it not? motion can be forward or sideways, so if light is not affected by motion forward then it cant be affected by the motion in in any other direction can it?
If you say it CAN be affected by sideways motion, then this is ADDING speed to light speed.
This is impossible.
Unless you think that light slows down in the original direction as you turn the source to face another direction????
If a ball is moving east at a set speed, and I add force in the north direction, then the result will be a new velocity, a new direction and an increase in speed!
At this point it is looking more and more like a psychiatric issue to be honest.
That isnt meant to attack, but you keep ignoring the things that easily dispute you (actual reality we see with our eyes through experiments) vs what you want to be true. Even when you admit you dont even know basic maths. That is the defining quality of delusion.
I seriously hope you can realize this (and why i am investing effort) and you can rise above it one day or seek professional help. It isnt healthy for you.
@freemo @CCoinTradingIdeas
Im ignoring the insanity claims, I have the exact same thinking about you, rest assured.
Now about a photon having mass of any sort, which is the crux of this discussion I think.
Without mass, that photon cant ever be going in the zigzag trajectory as in my video, so this is the bone of contention.
Here is the equation I wanted to see, which i took from Don Lincon of Fermilabs lecture.
He is discussing why a massless photon can have momentum specifically, so this is addressing the exact thing we are talking about.
I want to make a claim about the purpose of this equation. It is equating an amount of energy to what? The only possible answer is to A PHYSICAL OBJECT THAT HAS MASS.
The energy is in Joules, and a joule is equal to the energy transferred to an object when a force of one newton acts on that object in the direction of the force's motion through a distance of one metre.
And one Newton is the force needed to accelerate one kilogram of mass at the rate of one metre per second squared in the direction of the applied force.
So we have a big problem here.
If you have zero rest mass ( thats the m in the equation below) then you are not able to do anything with this equation. There is no measurable physical object to apply this equation to.
The definition of a physical object is something that has form and mass.
e=mc2 and its derivatives REQUIRE something to exist as physical matter having a measurable mass for the equation to be calculated.
There are TWO places in the equation that REQUIRE a positive value, the m and also the p, because here they are referring to newtonian physics, where p is equal to mv.
The chemical equivalent of what you propose by just setting mathematically all instances of m to the value o zero, is like trying to make water, H2O but only using H2 and having zero oxygen.
This an an abuse of math to try this sleight of hand to try to keep einstins stupid theories about light being able to take up inertia because it has mass at some speed but not another speed.
FYI, m in the equation you posted is not rest mass. Please learn about the equations you post before trying to talk about them.
those arent the same equations \( m_{0} \) is different than just m, that subscript of 0 is important.
No the 1 doesnt represent rest mass either.
Actually i take that back, you are correct that even in the original equation (Without the subscript 0) that the mass was specifically rest mass. Which is 0. The other term is the one that represents teh energy due to relatviistic mass, (which is contained within the p).
My apologies on that mistake.
Though your conclusion is still incorrect. Which as we covered has been experimentally proven (Something you like to just pretend doesnt exist in the conversations for some reason)
@freemo @CCoinTradingIdeas
I think you are needing to have a revision course on this topic before you find fault with my math.
In the equation I posted, m is rest mass OR relativistic mass, as they are BOTH identical values for a photon!
See the equation for Gamma below
The numeral "1" in the equation, the Numerator is what? Its the REST MASS. Because we are trying to determine what gamma is here, so we cant know the relativistic mass till we know gamma.
Anyway, in the case of the photon it equates to exactly the same result whether its rest or relativistic mass.!
Because the velocity of the photon is c.
so solving the equation, the ONLY correct result for light is that gamma = 1.
so multiplying anything by 1 achieves nothing!.
Therefore, claiming that I made a mistake by using rest mass instead of relativistic mass is just wrong.