Z.10.46

Yes in math, but in physics, infinity can be handled through regularization and renormalization to give it a finite value. Many problems in physics are solved this way.

Je trouve E égal à l'infini comme solution quand v=c, pas vous ? Cependant, je ne suis pas sûr de la signification physique de cet infini mathématique en physique, et j'ai tenté de le trouver par régularisation. Je suis arrivé à E=-(M(c-1)+m0^c^2). N'est-ce pas intéressant ? En physique, l'infini n'existe pas, et une valeur finie peut résoudre une équation, contrairement aux mathématiques.

In physics, we are interested in energy to study motion.😁 How are you able to calculate the energy required to move an object?😁 And what is its value when v=c?😃 In physics, it is primarily an experimental science, nothing prohibits conducting this experiment since it originates from a physical equation at c=v, and we can explore it from all angles.

The proof for obtaining M(c)=-M(c-1) is in the attached document in my first message. What fascinates me about this mathematical relationship M(c)=-M(c-1) is how it can be interpreted in various ways. Firstly, the negative sign in the energy equation E=(y-1)m0*c^2=ym0-m0^c^2=-(M(c-1)+m0^c^2) could possibly hint at the existence of a particle, similar to the discovery of negative energy solutions that led to the existence of antimatter. Secondly, the fact that the number 1 has no dimension suggests that these particles could very well be related to dark energy and dark matter, making them undetectable. Thirdly, obtaining a result of 1 rather than something else could indicate the importance of using the international unit of velocity m/s for c-1=(3^10^8-1 )m/s when conducting experiments. Many physical discoveries and theories have been obtained through mathematical equations derived from physics, where these equations are manipulated and explored in various ways to draw conclusions and conduct experiments. In the case of v=c, E becomes infinite. Why don't you transform E = infinity into a finite value, To make physical sense when v=c instead of claiming that it is impossible.😄

And what is the physical equation that allows you to say that? Is it indeed M(v) = m0 / sqrt(1 - v^2 / c^2) and E=(y-1)m0^c^2=M(v)-m0^c^2 because at v = c, I get M(c) = infinity and E=infinity no? Why did you keep the value of infinity as it is without transforming it into a finite value in this equation to make this explanation?

Not true, as you refute the possibility of a particle reaching the speed of light by assuming that it cannot attain infinite mass/energy. Here, you are indeed providing explanations while keeping infinity as it is, without transforming it into a finite value...

But here, we are talking about an object that is characterized by other non-zero parameters. In regularization and renormalization, the process is carried out to give a finite value to a divergence like infinity=1/0. This is equivalent to treating 1/0 as a finite value, which would imply that 0 has a non-zero value." Yes, sorry, it was a spelling mistake. There is no typo in the attached document. In this new theory, its explanations of relativity, such as v=c, are criticized on the grounds that they retain infinity as it is, without transforming it into a finite value. and this new theory also provides explanations for why we do not observe a particle traveling at the speed of light. .

I have indeed answered your question. I simply stated that the fact of not observing this particle at v=c is well explained in this theory. It is possible that certain particles may travel at v=c without being noticed because they transform into undetectable dark energy or dark matter.

The formula was not invented; it is derived from a physical equation and provides an explanation for a physical observation: dark matter and dark energy. If it is possible to test it, I propose applying this negative energy (with E=-(M(c-1)+m0^c^2) and c-1=299,792,458-1) to an object to see if it transforms into dark matter or dark energy. Only after this experiment can we affirm or refute this theory.

Have you tested all particles with different energies, for example, by accelerating them using E=-(M(c-1)+m0^c^2) with c-1=299,792,458-1? The fact that we haven't observed such a particle is well explained here in M(c)=-M(c-1). The particle at v=c becomes undetectable.

I went to their website, but I didn't see any option to pay for publication. There are not free options for membership. In physics, when we encounter a divergence (infinity=1/0), we can use regularization and renormalization to obtain a finite value for this divergence. Essentially, it involves attributing a non-zero value to the mathematical zero to make sense of the divergence. I didn't understand. What did you mean?

In mathematics, yes, but in physics, zero doesn't really exist in many physical problems. Taking a value close to 0 often solves the problem and provides a physical meaning rather than sticking to the mathematical notion of 0 and declaring 1/0 as impossible. The mathematical proof for having M(c)=-M(c-1) is correct, and having a formula that can be explored in various ways to find a physical meaning is better than having an infinity.

It is rather a powerful but non-real mathematical trick used in physics to simplify calculations, but it poses a major problem in physics when we encounter 1/0. In such cases, we strive by all means to eliminate it to obtain a physical meaning.

What fascinates me about this mathematical relationship M(c)=-M(c-1) is how it can be interpreted in various ways. Firstly, the negative sign in the energy equation E=(y-1)m0*c^2=ym0-m0^c^2=-(M(c-1)+m0^c^2) could possibly hint at the existence of a particle, similar to the discovery of negative energy solutions that led to the existence of antimatter. Secondly, the fact that the number 1 has no dimension suggests that these particles could very well be related to dark energy and dark matter, making them undetectable. Thirdly, obtaining a result of 1 rather than something else could indicate the importance of using the international unit of velocity m/s for c-1=(3^10^8-1 )m/s when conducting experiments. Many physical discoveries and theories have been obtained through mathematical equations derived from physics, where these equations are manipulated and explored in various ways to draw conclusions and conduct experiments.

Yes, I agree with you. It's not very accurate to pose it mathematically; I did it just for simplicity. In your example, does it retain the notion of infinity as it is, or does it manipulate it to make it finite? My perspective on this approach is simple: just because something is not true in one context(c-1 not defin in physic) doesn't mean it can't open up new possibilities. So, why did the journal contact me to publish it. . Are you sure there could be wind or a small particle hitting it and making it move a little 😄?

0 as much as a measurable quantity does not exist, the ancient mathematicians exulate 0 as a number because it cannot imagine something with a unit and a quantity of measurement. The was added later by mathematicians just as an axiom even though it does not justify the eculide definition of a number since it has neither quantity nor unit of measure. "I would like to know if the Journal of Modern and Applied Physics is reputable, as there are experts who have shown interest in this theory?

Zero does not exist in physics, only in mathematics. Have you ever seen a quantity that has neither a unit of measurement nor a counterpart in nature? A divergence equals infinity, and 1/0 equals infinity

In physics, it is already done, and its formulas are well-established. However, mathematicians do not appreciate it very much. In physics, it is already considered that 1/0 is not equal to infinity to solve various physical problems, and it works very well. Nevertheless, if we keep this non-real abstraction of 1/0 in a physics equation, we cannot work with it without transforming it into a finite quantity, where 0 is not truly zero and can take different nonzero values depending on the physical problems to be solved... You can see this link for more information: https://en.m.wikipedia.org/wiki/Regularization_(physics)

M(v)=m0/sqrt(1-v^2/c^2)

there is no mistake I can do 1+i well even if 1 is a real number and i a complex number here c-1 is mathematically correct.

at v=c M(v)=infinity and E=infinity. by mathematically treating this expression we find M(c)=-M(c-1) yes 1 has no dimension but it can explain dark matter or dark energy if a particle reaches c=v, as for the casmir force where we have 1+2+3...=infinity but physically find 1+2+3...=-1/12 explain force effet casmir. @Bufofrog M is fonction of masse relative

Needing infinite amount of energy is a phrase which means that no finite amount of energy will do it. Say this: In all other physical theories when the equation comes out an infinity of energy makes it a regulization and renormalization and this infinity of energy is equal to a finite energy. why accept that in relativity the existence of an absurdity to make a conculsion?

And where does the conclusion come from that nothing can travel at the speed of light? As you mentioned, it is because we would need an infinite amount of energy, and, of course, this energy becomes from the equation. So, please answer these two questions.

Could you answer his two questions: Can you give an example in any physical theory or concussion when infinite energy comes out? Why do we agree to do this just in relativity? M(c)=M(c-1) can also have a physical meaning even if 1 has no dimension. It can explain dark matter and dark energy when a particle goes a speed v=c.

And how do we exclude the possibility of traveling at v=c, as it would require infinite energy/mass? We cannot precisely define what this infinite energy or mass is. However, by attempting to define it, we arrive at -M(c-1), where the 1 is dimensionless. Perhaps, it is due to this reason that we cannot observe dark matter and dark energy when a particle travels at v=c.