last day (16 days later) » 

23:04
-7
Q: Can we find a regularization of $M = \frac{m_{0}}{ \sqrt{1 - v^{2}/c^{2}}}$ at $v = c$ that gives $M = m_{0}$?

z.10.46Is there a mathematical regularization of the expression for the relativistic mass, $$M= \frac{m_{0}}{ \sqrt{1 - v^{2}/c^{2}}},$$ for a particle moving at the speed of light, $v=c$, to get $M=m_{0}$? This would be analogous to Ramanujan summation for divergent infinite series, which gives results...

Is this Homework?
No, just a question, and I added another question. Does this regularization of the infinite sum have a physical meaning?
@Ghoster Yes sorry copy paste problem fixed
the infinite sum What infinite sum? The Taylor series? If so, you should explain that in the question.
I have M = m0 / √(1 - v^2/c^2). Let's transform the function M into a series for v = c, then find an infinite sum regularization like Ramanujan's so that M = m0.
So that you can conclude that the enormous increase in relativistic mass that occurs at 0.999999999c completely disappears at 1.000000000c? Why would you want to think that there is that kind of bizarre discontinuity?
2
There is no need to do weird summation when you have the formula you start with.
23:04
No I want to find this infinite sum regulation and experiment with it to observe if v=c or not...
This will probably get closed as non-mainstream physics.
It's experimental physics then I already propose the experiment to do, just find this infinite sum regulation and test it expremientally to see if v=c or not
You are tossing around Ramanujan's name, I believe in bluff. Write down his identity you envision replicating. For luxe's, this is strictly 0/0. How do you propose to resolve it?
If the energy of the vacuum were not measured experimentally, we would believe that this energy is infinite and we would not use the regularization of the Ramanujan sum... Why bluff since I proposed the experiment to do?
In modern notation, your target expression is the requirement that E=0. Explain what you are hoping for.
23:04
I have a function M I can transform it into series, this series normally it diverges and M=infinity but I want an infinite sum regularization such as M=m0 and physically test this regulation to see if v=c... In the casmir effect this series 1+2+3+...=-1/12 and not infinity and this series is physically feasible...
There is a genuine singularity in the formula; that's why I call it a bluff. You do know that E is indefinite for luxons such as photons, and infinite for massive particles: the reason those cannot really achieve the speed of light. What is your point?
You cannot physically test this as it would require infinite energy. It's a ridiculous idea.
Infinity is just a mathematical notion and not a physical one. If why am I looking for a series in v=c which normally diverges in mathematics but converges in physics
@ StephenG - Help Ukraine Well it is physically testable since in the casimir effect can make such a sum which is normally divergent but physically convergent and testable
Where does the Casimir effect enter into this? You claim this is not a bluff? $E\geq m$. You saturate this for zero energy for massless particles like light, and stationary massive ones.
affirms that the series 1+2+3+4...=-1/12 and not 1+2+3+4..=infinity and it is experimentally testable
Here I am looking from the equation M when v=c a M=M0+M1+M2...=m0 (infinite sum regulation) Where M0+M1+M2.. is mathematically divergent but physically it is an infinite sum regulation which gives M=m0 and i want to test this infinite sum regulation to see if v=c
23:04
You can always raise this issue on Mathematics is you want to. I have no idea why you think this is testable. We cannot make thinks with non-zero mass reach the speed of light - that's established physics which no observations contradict.
There is not really a proof that a particle with mass cannot go to the speed of light without using the notion of infinity which makes no sense in physics...

  last day (16 days later) »