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!
This is the problem with you not understanding basics, you dont even understand the words your reading..
The **speed** of light is a constant, no speed is not the same as meaning "motion", motion is a non technical word but in the way you are using it you are trying to imply the **direction** light travels is invarianet, that is not the case.
And yes light is effected in the forward direction, it is just that its **speed** isnt effected by it. Likewise light is effected int he sideway direction, yet again its **speed** is not
Also the the millionth time, the experimental evidence directly contradicts all the nonsense your spouting.
Funny how you keep ignoring the experiments, experiments you can do and see for yourself, but keep relying on your own fauly interpritionation of something you dont seem to understand.
BS. If you have speed, a measure of motion, then you have motion. speed is just a measure of motion. Last I heard light, a photon travels in a straight line, so yes its direction is invariant in a vacuum.
I did not say that lights speed is affected by changing its direction. Im saying that this is the only possibility if we believe your claims about light.That its able to be affected by the motion of the source, but only sideways, not in the direction of the photons travel. A weird non physical claim. Thats what you claimed.
You said that speed is motion. If that is the definition you want to use, then no speed/motion is not effected forward or sideways, the speed is always the same.
Only thing changing is the direction of the light, not ts speed/motion.
And again this is expermentally proven (and replicated by anyone who has bothered).
Remind me again what experiments you or anyone has ever conducted to confirm what you said? None, zero, yet plenty to contradict you.
@freemo @CCoinTradingIdeas
I did not say that speed IS motion. I said that speed is just the MEASUREMENT of motion.
If you have changed the direction of a moving object, then you have necessarily changed its motion, and as newtons third law indicates, if you add force too a moving object, (force requiring a mass in motion) to the motion of the first object having mass, then the first object must change direction and gain some portion of the second objects inertia.
So it will both change direction AND change speed.
So, if a photon has mass and it inherited any of the light sources inertia as you claim then the photon will simultaneously change direction as you claim BUT it must also increase speed!
Now here you are saying two opposite things about light.
As long as you insist that light has mass what I just describe MUST be correct according to physics.
You can go away now.
It has relativistic mass, it does not have rest mass. You are unable and unwilling to learn what that distinction means
I would happily teach you if you were trying to learn, but you are not. In fact you are actively working to remain ignorant. y/ou have made no attempt to learn or correct any of your previous erroneous statements despite being proven wrong in multiple ways, not the least of which being experimentation.
I just posted the math proof that for light, there is no difference at all between the rest mass and the relativistic mass. both are zero mass.
(because there is no such thing as gamma)
Lets try this another way.
Tell me an experiment, either one that has been done or capable of being done, that demonstrates your assertion...
I'll wait. Because right now they all prove you wrong.
@freemo @CCoinTradingIdeas
Lets not go there yet, we are still trying to prove with math, using einsteins gamma (lorentz's actually) that a photon has a different mass at reat that ti does at c.
We have the correct equation for mass from Dr lincon of Fermi lab, we plug in the numbers... we get the result that for a photon, rest mass is identical to relativistic mass, namely it has no mass, zero is the correct answer.
Because gamma equates to just 1 at light speed.
Unless you math says something else here?
What, no thats not what the math showed at all.. please show me the steps you used to go from Dr. Lincolns equation to showing relativistic mass is equal to rest mass, because the equation you posted shows the exact opposit of what you just stated.
Show your math please, dont just state it to be so.
@freemo @CCoinTradingIdeas
OK, later, those chickens call.. have a break, i may not get back to the computer till tomorrow
but i will show my working.
please do, because you will quickly find the equation proves the exact opposite of what you claim, not to mention reality itself as we can all see.
Proving that light, a photon has no mass.
No rest mass and no relativistic mass.
There are two camps here, one classical and the other relativistic.
Both agree that the photon has no mass at rest.
Relativists claim that there is another value for mass called Relativistic mass, which allows the photon to possess momentum and be affected by inertia, and be affected by gravity. (spacetime curvature)
In keeping with Einstein’s and Classical physics rules, it is agreed that light speed in the following equations is speed of light in a vacuum, or c. Relativity always uses c in Einstein’s hypothesis.
Gamma (Lorentz transformation equation) is used uniformly throughout Einstein’s theory of special relativity. Gamma equation is
As we are only concerned about the photon here, which always goes at c, we have no variables to consider.
Therefore the value of gamma for the photon at c is simply 1. Gamma is 1, so that anytime we see gamma either as a multiplier or divisor of any value, we can ignore it, as the result will be unchanged. (45 x 1 is still 45, and 84 divided by 1 is still 84.)
To prove that there is no difference between the rest mass of something and the relativistic mass of the object, IF it is moving at c, we just need to do the math, the result is that both rest mass and the same mass at speed are equal. Actually, we get a math error of trying to divide by zero, which is impossible to do in Physics. Anything can be done in the imaginary world of theoretical mathematics, but not in Physics. The actual result is infinity, which is not a number.
Basing a fundamental law of Physics on a quirk of theoretical mathematics is beyond unreasonable.
In Physics, we demonstrate, not speculate when we make laws. For instance we can demonstrate division of 40 apples and 8 boxes. It is demonstrable. But its not rational or sensible to try to demonstrate 40 apples put into zero boxes. We cant demonstrate how division can work physically without the denominators, the boxes. And my calculator gives either “undefined error” or the infinity symbol, because the operation is nonsense.
And even if this were somehow true in physics when it comes to the photon, that means that every single photon has infinite mass.! Not even worth considering in Physics.
Now of course they (the relativists) try to claim that mass is really momentum So its not infinite mass but it is some relativistic value of momentum.
Please note here that it would have been simpler to just use the relativistic equation for converting rest mass to relativistic mass directly, which we shall address later
Pretending that Einstein really meant to say “momentum” when he wrote “mass” is really not going to help unless you think trickery is a valid way to do physics and we will now see why its invalid as well.
Momentum is simply p=mv.
They claim that the mass here is not normal mass, but relativistic mass….so they modify a perfectly good tried and tested equation with a relativistic friendly one in order to do a slight of hand with the maths, to “Prove” mathematically that black is really white, but only if you go fast enough.
Her is the relativistic equation for momentum.
p=γ*m*v
The mass here is rest mass.
So lets plug in the known values.
Lets see, the velocity is c so lets call that 300million
We are talking about photons so that still gamma which as we previously discovered, for photons gamma is 1
How can it be only 1, which is the same value we get for gamma when the velocity is reduced to a dead stop? That’s easy, this equation does not mimic reality, and this problem becomes obvious when we run the numbers.
So now we have the relativistic version of momentum giving the exact same result as the classical version of momentum because for a photon the gamma is 1. One times a value, the value remains unchanged.
So the momentum for a photon looks like this:
Momentum of a photon = zero*300million
= exactly ZERO.
But wait, some say that the m in the above equation means Relativistic mass, not rest mass… so what happens if we change the equation to use relativistic mass?
This is where we need to use the direct equation for converting rest mass to relativistic mass.
m_(rel )=rest mass divided by √(1-v^2/c^2 )
So plug the know values for rest mass (zero) and velocity of the photon squared is the same as c squared, so that equals 1…
Rest mass = 0 / sq root of (1 minus 1) =
Rest mass = 0/0 is … cant be anything else but zero.
AGAIN we show that the mass of a photon at rest or at c is always zero.
With this unsurprising, intuitive and mathematically perfect result for momentum of a photon, that also matches our observations of light, we go now to the relativistic equation we are trying to solve:
That is Einstein’s Equivalence of energy and mass equation, expressed in Relativistic terms.
Now we need to address the claim that the Energy -Mass equivalence equation can prove that a photon can have Mass but its magically disguised as “momentum”.
The full equation is:
.
E^2=(pc)^2+(〖mc〗^2 )^2
Apparently the variable m is supposed to be in relativistic terms, but its still ZERO which ever way you look at it as proven above.
Momentum of photon we know is zero, relativistic mass of photon is also zero, so (pc) squared is just zero…. And as mass is also zero we have square(m*300million) is exactly ZERO.
So zero plus zero then square it? Its ZERO.
E squared is exactly ZERO
Energy of a photon is exactly ZERO according to Einstein’s own formulas.
Or, the other way of looking at it, the photon has no mass, has no momentum and cant be imbued with any inertial from another object. The photon wont go at a diagonal as explained in my video.
Why is this so?
Simply because the mistake was made early on, when they continued to try to do math using an equation that requires a physical object that has necessarily got mass and size, they are weirdly trying to do a physical operation of something that is not physical.
If you have something that has no mass at rest, and no size, then that is the very precise statement that you don’t have anything physical that is going to obey any sets of physical laws!
Einstein’s equations REQUIRE physicality because the variables are properties that are possessed ONLY by things that have SOME mass and SOME size. Then he also mixes into this purely physical equation, light, which is not physical.
Mass, momentum, inertia, gravity, is never to be associated with light.
Here’s a good one for you to answer: “How much mass does TIME have?” How much does Time curve as it passes the Sun during an eclipse? How much stupidity can you put into a 1 liter container?
These are questions for Einstein.
Please address my math and explanations item by item, and not just make blanket statement. Explain where im wrong, and show the correct answer please.
@zeccano
Maybe this helps you get on the right track
http://www.desy.de/user/projects/Physics/ParticleAndNuclear/photon_mass.html
@CCoinTradingIdeas
Im not by a co.puter which is the only reason i didnt show him the math yet. The relative mass is embeded in the variable p in his equation. So you just expand the p, get the relative mass element, then solve for it and see it is nonzero
@zeccano
@freemo @CCoinTradingIdeas
You did not read my explanation very well.
I DID USE the relativistic or "correct" calculation for momentum (p)
the equation is p=gamma*REST MASS*v.
Straight from Dr Don Lincon's detailed video entitled "Is relativistic Mass real?"
The mas here IS BEING CONVERTED to relativistic mass by the use of gamma.
So now what are you going to show me?
@zeccano @freemo https://www.physicsforums.com/threads/relativistic-mass-of-a-photon.65749/#post-478590
Can't copy past it here so this has to do
@CCoinTradingIdeas @freemo
This guy is saying that a photon has no mass of any sort? correct? is this supposed to back up your claim that it magically has momentum but no mass?
You are saying the the highly regarded Physicist Dr Don Lincon, from Fermilab, is LYING when he stated that practically EVERY respectable Physicist today says that there is no such thing as relativistic mass? there is only just mass, which is also called rest mass. and mass does not change when seen from different frames.
SO he is WRONG? and also LYING?
If the professor literally said "Photons do not have relativistic mass" then yes, he was wrong, Though I am skeptical he said this.
@freemo @CCoinTradingIdeas
watch it first, then comment, anything else is speculation, which is what you relativists do a lot.
from 7:30 to 7:50.
or watch it all, same difference.
@zeccano
I just watched it, he never said relativistic mass was 0 in that video,
No, he said that there was no such thing as relativistic mass at ALL.
Its developed ONLY as for illustrative purposes as an intuitive aid when trying to teach relativity to students.
So, as there is no such thing as relativistic mass, we cant very well use it as a variable in an equation, can we?
BUT even if you insist that we do, the result in the case of the photon is still that the photon has no mass of any kind, (even thought there is only one kind)
its ZERO always, with anyone''s math.
Show me how you get a non zero, being a real number, not calculation error result which is infinity, or undefined, or divide by zero error.
Not exactly what he said, only that it is a way of thinking about it that doesnt lead to good intuition. It is real and positive mathematically and works if you know how to use it. But it is not real mass it is perceived mass (it acts as if it has tat mass when in motion).
This is pretty much all that he was saying.
By the way you asked for the equation for relativistic mass that gives a real positive value, I already provided this equation to you:
@freemo @CCoinTradingIdeas
""Pretty much what he said" is using your opinion or interpretation of what he said.
What he ACTUALLY said, off his carefully worded script written as a physicist, was, “We (Physicists) say that an object has only has one mass, which is the mass you measure for an object when it isn’t moving with respect to you. Some people called that the “rest mass” but it’s REALLY the ONLY mass.
Here Dr Lincon is making an ambiguous statement, when he should know better, as he is making this statement in order to explain or clarify the term “mass” as a Physicist in a statement specifically about the meaning of mass.
Its ambiguous because of the phrase “when it isn’t moving with respect to you”.
Seemingly indicating that there will be “another different” result for the normally static measurement of the mass of an object if you measure it when it is moving relative to you.
However he just got through saying that there is “only one mass”, and there is no physicist today who teaches a mass increase with relative velocity, they only ever talk about the idea that its only momentum that increases with velocity, and NOT mass that increases.
Dr Lincon made this clear when he mentioned the claim made by “some confused people” that claim that a photon passing the earth would have an infinite mass, which he said was wrong, its only the momentum or energy that changes with the speed of the photon. (because its impossible that that every photon has infinite or any near infinite mass.
So they swap over to using einstein’s energy mass equivalence here, to get out of a logic hole.
And for that they start talking about inertia and then momentum.
He does this beginning at 1:50 in the video (https://www.youtube.com/watch?v=LTJauaefTZM)
Where ha says that inertia is the property of WHAT? Of MATTER… last I looked, light is NOT MATTER.
Then he says that inertia is only for low speed, and at low speeds inertia is equivalent to mass, you must talk about momentum for really fast things, aka a photon.
(We should not be talking about inertia or momentum for light, it’s not made of matter.)
So this is exactly what Dr Lincon and all his Physicist buddies understand, spelled out in their own words.
There is only one mass, and it doesn’t change with velocity, momentum does. And they later equate momentum to energy with Einstein’s energy matter equivalence equation, because STILL light cant have anything other than a zero momentum, but you can hide this problem by doing the two step from mass to energy. But the very name of the equation gives it away as a sleght of hand. Its called the ENERGY-MATTER Equivalence equation. Again light is not matter so this equation will give a nonsense result if the vales measured are in incompatible terms.
May as well plug into the equation the numerical value in Hexidecimal of the color red, expecting that it will give some meaningful result.
I would e-mail dr. licoln just so he can tell you that you are wrong (He will likely respond to a fellow scientist like myself). If he tells you relativity is real will you believe him.
Also waiting for you to respond to this:
@freemo @CCoinTradingIdeas
How could I believe Dr Lincon or any relativist physicist? He's your boy, Im just using his statements to point out the errors.
These guys always leave the back door open to duck and dodge criticism of relativity. Always.
So here has says one thing clearly, next time if challenged, he he will say the exact opposite. Or employ some distracting mechanism, as a dodge.
Its like talking to a Mormon about scriptural truth.
No one leaves any backdoors opens, it is simply that these ideas are hard to explain to lay people, physicists talk to each other in a more technical language.
The fault you see is because you refuse to learn the language to describe these things
@freemo @CCoinTradingIdeas
Just show me the math.
NOTE: for those who cant read the below equations (they aren't rendered) you can follow this link to see: http://mathb.in/38381
Here you go a full derivation from the math equations you started with to the one I provided you several times...
axioms (given by you, agreed by physicist):
\[E = {m}_{rel} \cdot {C}^{2}\]
\[{E}^{2} = {p}^{2} \cdot {C}^{2} + {{m}_{rest}}^{2} \cdot {C}^{4}\]
De Broglie's Equation for the momentum of a photon:
\[p = \frac{h}{\lambda}\]
Formula to convert wavelength \(\lambda\) into a frequency:
\[\lambda = \frac{C}{f}\]
Simplify given axioms:
\[{({m}_{rel} \cdot {C}^{2})}^{2} = {p}^{2} \cdot {C}^{2} + {0}^{2} \cdot {C}^{4}\]
\[{{m}_{rel}}^{2} \cdot {C}^{4} = {p}^{2} \cdot {C}^{2}\]
\[{{m}_{rel}}^{2} = \frac{{p}^{2} \cdot {C}^{2}}{{C}^{4}}\]
\[{{m}_{rel}}^{2} = \frac{{p}^{2}}{{C}^{2}}\]
\[{m}_{rel} = \sqrt{\frac{{p}^{2}}{{C}^{2}}}\]
then use de broglie's equation for the momentum...
\[{m}_{rel} = \sqrt{\frac{{(\frac{h}{\lambda})}^{2}}{{C}^{2}}}\]
\[{m}_{rel} = \sqrt{\frac{\frac{{h}^{2}}{{\lambda}^{2}}}{{C}^{2}}}\]
\[{m}_{rel} = \sqrt{\frac{{h}^{2}}{{C}^{2} \cdot {\lambda}^{2}}}\]
Now convert wavelength to frequency...
\[{m}_{rel} = \sqrt{\frac{{h}^{2}}{{C}^{2} \cdot {(\frac{C}{f})}^{2}}}\]
\[{m}_{rel} = \sqrt{\frac{{h}^{2}}{{C}^{2} \cdot \frac{C^2}{f^2}}}\]
\[{m}_{rel} = \sqrt{\frac{{h}^{2}}{\frac{C^4}{f^2}}}\]
\[{m}_{rel} = \sqrt{{h}^{2} \cdot \frac{f^2}{C^4}}\]
\[{m}_{rel} = \sqrt{\frac{{h}^{2} \cdot f^2}{C^4}}\]
and bam you arrive at the equation for the relativistic mass of a photon.
\[{m}_{rel} = \frac{h \cdot f}{c^2} \]
@freemo @CCoinTradingIdeas
When did we agree to drag into the mix, the theory that Max Planck proposes, that there is such a thing as a magic constant to fudge away the errors of relativity?
We already have all the necessary equations and theory to be able to state the mass and therefore the momentum of light.
The problem is that by inventing a theoretical constant, you CAN fudge in to any equation a positive result when the rational result is actually zero.
So, given the equations of einstein, and newton which dont mention the need of using plancks constant, we should not be needing to use it.
Simply put, we have the equation to convert rest mass to relativistic mass, which does not exist, but is used to calculate the momentum, which also dosent exist, but is used to calculate the energy, which relies on the mass of the matter , of which light has NONE.
So why would you want to drag plancks constant into all this, if not to fudge you way out of a dilemma or rather a contradiction?
In Einsteins E=mc2 theory, or any additional similar alterantives, is the planck length a contributiing factor.
The claim is simply made ad hoc that a body emitting light will loose mass, because the light is being made by the conversion of matter.
That is the logical conclusion IF you agree that light is able to reduce mass of the emitter.
So try again to prove that a photon has momentum but this time stick to the original terms, not the fudges that were submitted later by the addition of an imaginary universal magic number, the planck length.
(under "normal"physical processes, again they claim that this tiny number has no observable affect! How bloody convenient!
Same claim exactly for ALL of Special relativity claimed effects.
You are not allowed to use other theories that hang on einstein''s SR, until you show that the photon has mass, using the equations supplied at the time the theory was proposed.
If the photon has momentum at c velocity, then it still will have the same mass when its stationary if it could be, because we have already have the statement that there is only one type of mass.
Mass of the photon unlike everything else on the universe, is not going to become a variable to suit Einstein.
Told ya that even after the equation was derived, followed his rules, and disproved him wrong, he would still try to cry fowl and make up more excuses. Dogma only at this point.
Shouldn't be surprising he tries so hard to avoid the experimental evidence all of it disproves him and not a single one supports him. So he pretends they dont exist and just acts like a fool and thinks it will get it somewhere.
