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(I'm a mathematician actually. This is my first question on this site. So please go easy on me if it's a trivial matter or mistake)

Lately I've been thinking about special relativity.

The thought came to my mind.

First consider the following paradox:

Considering a mass oscillating with infinite velocity between two walls. Now we can see that it is present everywhere at the same time between the line of motion. i.e. we can treat it as if it is an rod.

Now this can be seen in special relativity setting:

We have a particle which is oscillating with velocity $c$ ( lights speed) in above configuration. (Our eye can't trace it's motion as due to it's limiting speed $c$.)

Now by special relativity the relativistic mass is

$$m=\frac{m_0}{\sqrt{1-v^2/c^2}}.$$

So if $v=c$ and for finite rest mass $m_0$ the relativistic mass $m$ diverges.

Now of $m_0=0$ (photon) and $v=c$ expression becomes $0/0$ which could be finite.

So, the rod could successfully thought of as a oscillating photon with different limit of $ 0/0 $ than that of light.

So, every mass could be replaced by such phenomenon.

Only problem is exact limit of $0/0$ as we don't know more variables.

Is this argument valid or I'm just missing something very trivial here which I don't know or imagined. Please help?

(Nicola Tesla's "everything is light")

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    $\begingroup$ The problem occurs because you're taking a relationship that's only valid for $v<c$ and trying to use it for $v=c$. In modern terms, the more general relationships are $E^2-(pc)^2=(Mc^2)^2$ and $v/c = pc/E$, where $E$ is the object's total energy, $p$ is its momentum, and $M$ is its mass (formerly called "rest mass"). Combine these eqns to get $(1-v^2/c^2)E^2=(Mc^2)^2$. If $v=c$, then $M=0$. If $v<c$, then we can divide both sides by $1-v^2/c^2$ to get your equation (with $m := E/c^2$ and $m_0 := M$). The point is that your equation is only valid for $v<c$, not for $v=c$. $\endgroup$ – Chiral Anomaly Sep 7 '19 at 19:33
  • $\begingroup$ @Chiral Anomaly thank you very much for clearing my doubt. $\endgroup$ – TPC Sep 7 '19 at 19:36
  • $\begingroup$ But I still hopes for such "unification" $\endgroup$ – TPC Sep 7 '19 at 21:19
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    $\begingroup$ you cannot have a mass oscillating with infinite velocity between two walls $\endgroup$ – Wolphram jonny Sep 7 '19 at 22:50
  • $\begingroup$ FWIW in most straightforward interpretation of string theory all observed elementary particles arise from massless excitations of strings. i.e. waves propagating along strings with the speed of light. $\endgroup$ – A.V.S. Sep 11 '19 at 17:49
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Light in a box indeed can be said to have mass. Mass is simply rest frame energy and in this case the rest frame is that of the box. As a model of elementary particle mass this brings more questions than it solves in my opinion.

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So, every mass could be replaced by such phenomenon

Photons don’t have charge. So you couldn’t explain electrons or quarks this way, and therefore most matter. They also have integer spin, so you couldn’t get any fermions this way.

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  • $\begingroup$ Already thought of that ! But it's just my curiousity I've not given a detailed thought to the question. $\endgroup$ – TPC Sep 8 '19 at 5:19
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It is mathematically a loosing game to look at problems in special relativity algebraically. One should use the four vector formalism and then everything falls in place.

Mass is defined only as the "length" of the four vector, and is invariant under Lorenz transformations to other frames.

Infinite velocity is physically not possible as far as all the data and observations we have. Maximum velocity is the speed of light. Your box of photons will have an invariant mass given by the length of the four vector derived by the sum of the four vectors of individual photons, and that is all.

As others have observed, at the basis of nature there exist elementary particles and some of them have charge, and spin 1/2 , so you cannot use your box model to model them, i.e. to assume that the elementary particles are made of boxes of photons.

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