Is it permissible to use infinity, which is not defined in physics, to assume the impossibility of traveling at the speed of light?

[Z.10.46]

1 minute ago, Genady said:

It is very easy to eliminate it. Don't divide by 0.

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.

 

2 hours ago, John Cuthber said:

Choosing to get the wrong answer is also absurd.

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.

 

[swansont]

12 hours ago, Z.10.46 said:

So, why did the journal contact me to publish it.

How much are they going to charge you?

[Genady]

17 minutes ago, Z.10.46 said:

Taking a value close to 0 often solves the problem

Dividing by a value close to zero does not pose a problem and does not require a solution. 

[swansont]

14 hours ago, Z.10.46 said:

However, by attempting to define it, we arrive at -M(c-1)

“arrive at” implies it was derived. You’ve just written it down. 

[Z.10.46]

22 minutes ago, swansont said:

How much are they going to charge you?

I went to their website, but I didn't see any option to pay for publication. There are not free options for membership.

 

16 minutes ago, Genady said:

Dividing by a value close to zero does not pose a problem and does not require a solution. 

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.

 

15 minutes ago, swansont said:

“arrive at” implies it was derived. You’ve just written it down. 

I didn't understand. What did you mean?

 

Edited July 24, 2023 by Z.10.46

[Genady]

2 minutes ago, Z.10.46 said:

In physics, when we encounter a divergence (infinity=1/0)

There is no such a divergence. Take any v<c and you get a finite number. For v=c the formula you use is not applicable. You never get 1/0.

[Z.10.46]

8 minutes ago, Genady said:

There is no such a divergence. Take any v<c and you get a finite number. For v=c the formula you use is not applicable. You never get 1/0.

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.

[swansont]

6 minutes ago, Z.10.46 said:

I didn't understand. What did you mean?

You did not “arrive” at your equation from accepted physics principles. You made it up. 

 

4 minutes ago, Z.10.46 said:

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.

Particle accelerators test what happens when you accelerate particles all the time. 

Your formula can’t be tested because it’s not a valid equation.

[Genady]

10 minutes ago, Z.10.46 said:

18 minutes ago, Genady said:

There is no such a divergence. Take any v<c and you get a finite number. For v=c the formula you use is not applicable. You never get 1/0.

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.

Each time you get a comment that you cannot answer, you change the topic. This means that you are not discussing in good faith. This means that you violate rules of the forum. I am reporting you.

[Z.10.46]

9 minutes ago, swansont said:

Vous n'êtes pas "arrivé" à votre équation à partir des principes de physique acceptés. Vous l'avez inventé. 

 

Les accélérateurs de particules testent ce qui se passe lorsque vous accélérez des particules tout le temps. 

Votre formule ne peut pas être testée car ce n'est pas une équation valide.

 

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.

 

Edited July 24, 2023 by Z.10.46

[joigus]

25 minutes ago, Z.10.46 said:

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.

That's not what regularisation/renormalisation is about. 

And there is zero in physics, as they've repeatedly told you. The charge of the neutron is zero, for example. It's not a tiny little non-zero smidgen of a thing. It's zero.

[Z.10.46]

2 minutes ago, Genady said:

Each time you get a comment that you cannot answer, you change the topic. This means that you are not discussing in good faith. This means that you violate rules of the forum. I am reporting you.

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.

 

[joigus]

On 7/22/2023 at 4:25 PM, Z.10.46 said:

To resolve this issue, we applied regularization through the Riemann zeta function. Eventually, we obtain a finite value of $-1/2$ for the divergent series $1+2+3+…$, and this result provides a good explanation for the Casimir effect.

Also, you got this wrong --perhaps a typo. The famous regularisation of this infinite divergent sum by means of the eta function produces -1/12.

[Genady]

Just now, Z.10.46 said:

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.

No, you did not answer my comment. My comment was about divergence, infinity, and 1/0. It was not about particles. Not observing particles at v=c is well explained in the theory of relativity.

[Z.10.46]

39 minutes ago, joigus said:

That's not what regularisation/renormalisation is about. 

And there is zero in physics, as they've repeatedly told you. The charge of the neutron is zero, for example. It's not a tiny little non-zero smidgen of a thing. It's zero.

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."

 

37 minutes ago, joigus said:

Also, you got this wrong --perhaps a typo. The famous regularisation of this infinite divergent sum by means of the eta function produces -1/12.

Yes, sorry, it was a spelling mistake. There is no typo in the attached document.

35 minutes ago, Genady said:

No, you did not answer my comment. My comment was about divergence, infinity, and 1/0. It was not about particles. Not observing particles at v=c is well explained in the theory of relativity.

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. .

Edited July 24, 2023 by Z.10.46

[Genady]

13 minutes ago, Z.10.46 said:

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 I told you in my comment that there is no infinity in relativity. It has nothing to do with your theory. 

There is infinity in your misunderstanding though.

[Z.10.46]

5 minutes ago, Genady said:

And I told you in my comment that there is no infinity in relativity. It has nothing to do with your theory. 

There is infinity in your misunderstanding though.

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...

Edited July 24, 2023 by Z.10.46

[Genady]

Just now, Z.10.46 said:

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

Not true! I did not "refute that a particle can reach the speed of light by assuming that it cannot attain infinite mass/energy." This was what you keep saying, and I keep saying that this is wrong. Let me repeat again:

I refute that a particle can reach the speed of light because any amount of energy can only accelerate it to a speed less than speed of light.

Do you see a word "infinity" in what I say?

[Z.10.46]

 

12 minutes ago, Genady said:

Not true! I did not "refute that a particle can reach the speed of light by assuming that it cannot attain infinite mass/energy." This was what you keep saying, and I keep saying that this is wrong. Let me repeat again:

I refute that a particle can reach the speed of light because any amount of energy can only accelerate it to a speed less than speed of light.

Do you see a word "infinity" in what I say?

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?

Edited July 24, 2023 by Z.10.46

[Bufofrog]

1 hour ago, Z.10.46 said:

c-1=299,792,458-1 to an object to see if it transforms into dark matter or dark energy.

How do you think this makes any sense?  Do you mean 299,792,458-1 = 299,792,457?  Do those 2 different speeds have some sort of deep meaning to you? Do you realize that dark energy and dark matter are completely different things?

Your entire presentation seems devoid of logic and reason.

You keeps say things like, "The mathematical proof for having M(c)=-M(c-1) is correct".  However you have not shown the proof or derived it, you just keep saying you think it is correct, when in fact it make no mathematical sense.  

[Genady]

8 minutes ago, Z.10.46 said:

 

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?

The equation is:

v=sqrt(c²-m²c⁶/E²)

Set E to any value in this equation and get v<c.

No infinities anywhere! All values are finite!

I do not "keep the value of infinity" at all, because it is not there!

I don't need to "transform" anything "into a finite value", because all values are already finite!

[Z.10.46]

31 minutes ago, Z.10.46 said:

 

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?

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.

9 minutes ago, Genady said:

The equation is:

v=sqrt(c²-m²c⁶/E²)

Set E to any value in this equation and get v<c.

No infinities anywhere! All values are finite!

I do not "keep the value of infinity" at all, because it is not there!

I don't need to "transform" anything "into a finite value", because all values are already finite!

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.😄

 

[Bufofrog]

12 minutes ago, Z.10.46 said:

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.

That may be the most absurd/humorous thing in this whole thread and that is saying something!

[Genady]

15 minutes ago, Z.10.46 said:

In the case of v=c, E becomes infinite.

Did you see my equation? There is no v=c in it. For every E, v<c.

There is no infinity that needs to be made a sense of.

[Z.10.46]

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?😃

20 minutes ago, Bufofrog said:

That may be the most absurd/humorous thing in this whole thread and that is saying something!

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.

Edited July 24, 2023 by Z.10.46