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

[Genady]

5 minutes ago, Z.10.46 said:

And what is its value when v=c?

None. The equation does not have a solution for v=c. Just as it does not have a solution for v=2c, v=3c, etc.

[Z.10.46]

il y a 7 minutes, Genady a dit :

Aucun. L'équation n'a pas de solution pour v=c. De même qu'il n'a pas de solution pour v=2c, v=3c, etc.

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.

Edited July 24, 2023 by Z.10.46

[Genady]

Just now, Z.10.46 said:

I find E equal to infinity as a solution when v=c, don't you?

No, I don't. E is a real number value. Infinity is not.

[Z.10.46]

3 minutes ago, Genady said:

No, I don't. E is a real number value. Infinity is not.

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.

 

[Genady]

3 minutes ago, Z.10.46 said:

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.

 

However, there is no such problem in this case. YOU make the problem up.

[Z.10.46]

16 minutes ago, Genady said:

However, there is no such problem in this case. YOU make the problem up.

and why not do it since it is possible to open up new avenues in  theoretical physics.?

Edited July 24, 2023 by Z.10.46

[swansont]

2 hours ago, Z.10.46 said:

The formula was not invented; it is derived from a physical equation

Then provide this derivation.

[Z.10.46]

4 minutes ago, swansont said:

Then provide this derivation.

derived=obtained

 

[swansont]

Just now, Z.10.46 said:

derived=obtained

 

x+2 = 1/(c-v) isn’t valid, because the units don’t match.

[Z.10.46]

(x+2) and 1/c-v have the same units because I've established an equality.

I can even set x+2/c=1/(c-v) and I will still have M(c)=M(c-1)

[swansont]

1 minute ago, Z.10.46 said:

(x+2) and 1/c-v have the same units because I've established an equality.

It doesn’t work that way.

1 minute ago, Z.10.46 said:

I can even set x+2/c=1/(c-v) and I will still have M(c)=M(c-1)

The notion that you can multiply one side of the equation by some factor and think it remains an equality underscores how ridiculous this assertion is.

[Z.10.46]

1 hour ago, swansont said:

Cela ne fonctionne pas ainsi.

L'idée que vous pouvez multiplier un côté de l'équation par un facteur et penser que cela reste une égalité souligne à quel point cette affirmation est ridicule.

I don't see anything ridiculous about x+2=1/c-v, as x, 2, and 1/c-v all have the same unit. Even if I write x+a=1/c-v, I still obtain M(c)=-M(c-1), even if the unit of c changes. So, regardless of the units used, it doesn't pose any problem, and I always get M(c)=-M(c-1).

In essence, x+a=1/c-v is valid for any value of a, so even if the unit of c changes, it doesn't pose any problem because it corresponds to a specific value of a where this equality is always true x+a=x+2=1/c-v at v=c .

here exemple c=3^10^8 and c=1   x+2=1/c-v with x et 2 and 1/c-v have the same unit is true.😁 becouse infinty=infinity=a*infinity=infinity+a 😁

Edited July 24, 2023 by Z.10.46

[Phi for All]

1 hour ago, Z.10.46 said:

(x+2) and 1/c-v have the same units because I've established an equality.

!

Moderator Note

No. They don't have the same units.

You've had 5 pages to persuade us that your idea has merit, and this isn't the first mistake you've made. You aren't going to be able to convince anyone unless you show more rigor in your explanation. If you can't do that, we'll have to close this thread.

 

[Z.10.46]

30 minutes ago, Phi for All said:

!

Remarque du modérateur

Non. Ils n'ont pas les mêmes unités.

Vous avez eu 5 pages pour nous persuader que votre idée a du mérite, et ce n'est pas la première erreur que vous commettez. Vous n'arriverez à convaincre personne que si vous faites preuve de plus de rigueur dans vos explications. Si vous ne pouvez pas le faire, nous devrons fermer ce fil.

 

The problem with units is that if I change them, for example, to (c-1), I will obtain a different value each time. But here, when I state x+2=1/c-v, it poses no problem because regardless of the unit of c, this formula x+2=1/c-v will always hold true since infinity=infinity=a*infinity. Here's an example, let c=3.10^8 m/s, and I have x+2=1/c-v. Now, if I change the unit to c=3.10^5 m/s or any other unit, I will still have this relation true, x+2=1/c-v. Thus, this relationship does not depend on any unit of measurement as it is an equation involving divergences or infinity=infinity=a*infinity, unlike (c-1) where its value changes depending on the unit of c And it depends on the unit of c that I choose.

Here is a very solid proof to set x+2=1/c-v because this equality relationship between infinities does not depend on any unit of c, it is true for any unit of c.

 

 

.

.

Edited July 24, 2023 by Z.10.46

[Bufofrog]

2 minutes ago, Z.10.46 said:

let c=3.10^8 m/s, and I have x+2=1/c-v. Now, if I change the unit to c=3.10^5 m/s or any other unit

I see part of the problem.  In the above you did not change the units.

The units are m/s, the exponents are 8 and 5.

[Z.10.46]

29 minutes ago, Bufofrog said:

I see part of the problem.  In the above you did not change the units.

The units are m/s, the exponents are 8 and 5.

Go ahead, you can choose any unit for c. Does the relationship remain true even if I change the unit for c?😁

If the relationship is true for any unit of c, that means it does not depend on the unit of c to pose x+2=1/c-v at v=c and x tends to infinity .😀

Why we say, for example in this equation 1=c/c, it does not depend on any unit of c, is because regardless of the chosen unit, you always obtain 1. Similarly, here in the equation x+2=1/(c-v), when c=v and x tends to infinity, this relationship holds true. Hence, it does not depend on any unit of c, just like c/c=1.😇

Edited July 24, 2023 by Z.10.46

[Genady]

2 hours ago, Z.10.46 said:

derived=obtained

 

What are the units on the left and on the right sides of the first equation,

?

[MJ kihara]

Where there is smoke...their is fire ....keep following the smoke...when did you start developing this concept?

[Z.10.46]

7 minutes ago, Genady said:

What are the units on the left and on the right sides of the first equation,

?

It is a mathematical calculation for example:

 M = m0 / √(1 - v^2/c^2) ~ m0 / √(c^2 - v^2)/c^2 ~ m0 / √(c + v)√(c - v)/c^2 ~ m0c / √(c + v)√(c - v)=√(c + c)√(c - v) ~ m0c / √(c + c)√(c - v) ~ m0c / √2c√(c - v) ~ m0√c / √2√(c - v)~ m0√c / √2√n

[Genady]

1 minute ago, Z.10.46 said:

It is a mathematical calculation for example:

 M = m0 / √(1 - v^2/c^2) ~ m0 / √(c^2 - v^2)/c^2 ~ m0 / √(c + v)√(c - v)/c^2 ~ m0c / √(c + v)√(c - v)=√(c + c)√(c - v) ~ m0c / √(c + c)√(c - v) ~ m0c / √2c√(c - v) ~ m0√c / √2√(c - v)~ m0√c / √2√n

I don't think your mathematical calculation is correct. To make it readable, break it down line by line. Then I think you will see an error.

[Z.10.46]

5 minutes ago, Genady said:

I don't think your mathematical calculation is correct. To make it readable, break it down line by line. Then I think you will see an error.

no it's correct you can show it to any mathematician, I'm not going to give a theory either where there is a calculation error and without verification by mathematicians😁

And the journal wouldn't have offered me the possibility to publish this theory if it were mathematically invalid.😄

13 minutes ago, MJ kihara said:

Where there is smoke...their is fire ....keep following the smoke...when did you start developing this concept?

several years I thought about this idea but I just reformulated it mathematically less than a year ago thanks to the help of online mathematicians.

[Genady]

3 minutes ago, Z.10.46 said:

no it's correct you can show it to any mathematician, I'm not going to give a theory either where there is a calculation error and without verification by mathematicians😁

And the journal wouldn't have offered me the possibility to publish this theory if it were mathematically invalid.😄

I've asked a mathematician and he wanted to clarify, if this is what you are saying:

1. M = m0 / √(1 - v^2/c^2) =

2. = m0 / √(c^2 - v^2)/c^2 =

3. = m0 / √(c + v)√(c - v)/c^2 =

4. = m0c / √(c + v)√(c - v) =

5. = √(c + c)√(c - v) =

6. = m0c / √(c + c)√(c - v) =

7. = m0c / √2c√(c - v) =

8. = m0√c / √2√(c - v) =

9. = m0√c / √2√n

?

[Z.10.46]

8 minutes ago, Genady said:

I've asked a mathematician and he wanted to clarify, if this is what you are saying:

1.  M = m0 / √(1 - v^2/c^2) =

2. = m0 / √(c^2 - v^2)/c^2 =

3. = m0 / √(c + v)√(c - v)/c^2 =

4. = m0c / √(c + v)√(c - v) =

5. = √(c + c)√(c - v) =

6. = m0c / √(c + c)√(c - v) =

7. = m0c / √2c√(c - v) =

8. = m0√c / √2√(c - v) =

9. = m0√c / √2√n

?

 

1. M = m0 / √(1 - v^2/c^2) =

2. = m0 / √(c^2 - v^2)/c^2 =

3. = m0 / √(c + v)√(c - v)/c^2 =

4. = m0c / √(c + v)√(c - v) =

 

5.c=v  so (c+v)=c+c=2c

 

6. = m0c / √(c + c)√(c - v) =

7. = m0c / √2c√(c - v) =

8. = m0√c / √2√(c - v) =

9. = m0√c / √2√n

Edited July 24, 2023 by Z.10.46

[Genady]

4 minutes ago, Z.10.46 said:

 

1.  M = m0 / √(1 - v^2/c^2) =

2. = m0 / √(c^2 - v^2)/c^2 =

3. = m0 / √(c + v)√(c - v)/c^2 =

 

4. = m0c / √(c + v)√(c - v) =

5.c=v  (c+v)=c+c=2c

6. = m0c / √(c + c)√(c - v) =

7. = m0c / √2c√(c - v) =

8. = m0√c / √2√(c - v) =

9. = m0√c / √2√n

Yes, there is an error.

After you set c=v on line 5, you get m₀c/0 on line 6. Anything after that is meaningless, because there is no such thing as dividing by 0.

[Z.10.46]

12 minutes ago, Genady said:

Yes, there is an error.

After you set c=v on line 5, you get m₀c/0 on line 6. Anything after that is meaningless, because there is no such thing as dividing by 0.

I haven't done it. The purpose of this is to let the divergence appear in order to apply regularization using the zeta function.

in other words, Md is a simplification of M in the vicinity of v=c that is equivalent.

Edited July 24, 2023 by Z.10.46