Master Theory (edition 2)

Relativity of truth will justify everything. This isn't a matter of interpretation, it's fact.

 

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Relativity of truth will justify everything. This isn't a matter of interpretation, it's fact. It:

\(x^2-(ct)^2=(x')^2-(ct')^2=0\)

for example.

Notwithstanding the fact that the correct formula is as follows:

\(x^2-(ct)^2=(x'-vt')^2-(ct')^2=0\)

 

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Lorentz transforms are used when modelling accelerator physics. If the Lorentz boost was not as SR says but as you say then the models wouldn't match the observations. It's literally a homework problem to calculate results of particle scatterings in different frames related by Lorentz transforms. They match with experiments. Basically if your claims were right then the very first particle accelerator would have invalided special relativity. Instead they only help validate it's accuracy.

This isn't a matter of "Show me the experimental results!" because if an accelerator produced a result which didn't agree with SR we'd all know about it. In fact that's precisely why there's been all this discussion about neutrinos from the OPERA experiment, it's the first one to seriously seem to not square with relativity. At least at first glance.

 

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Is it? Confirm whether the expression:

\(E = mc^2=\frac{m_o}{\sqrt{1-v^2/c^2}}c^2\)

a direct (in the calorimeter) measurement of the energy of relativistic particles?

Could you give me a link to these (which published) experiments?

No. Master Theory is one of the solutions of the same problems that are solved by Einstein. In some experiments, both theory (MT and SR) give the same prediction. The difference in interpretation.

But exist experiments which clearly show justice of only one of theories.

You suggest I to believe your words. But I need the facts.

 

If the transform rules were \(x^{2} - (ct)^{2} = (x'-vt')^{2} - (ct')^{2}\) then Lorentz transforms which say \(x^{2} - (ct)^{2} = (x')^{2} - (ct')^{2}\) would be wrong. Then predictions about ALL particle accelerators would be wrong. Basically every test of quantum field theory would have failed to be accurate.

This isn't a matter of providing a specific experiment which tests your specific expression but that if your claims were true then every accelerator experiment would have contradicted predictions. Quantum field theory would never have made it past the 50s if there were the case.

That's flat out not true. In computing differential cross sections in quantum field theory Lorentz transforms are used all the time. You claim that \(x \to x'-vt'\) and \(t \to t'\). Applying these transforms instead of the Lorentz transforms lead to different predictions, predictions for experiments we've done and found QFT to be accurate.

Your claims are experimentally excluded. The problem is you're too stupid to understand that. If you were right no QFT experiment would be accurate, yet not only do we have one, we have all the experiments which bear out its accuracy.

Then you should learn how quantum field theory makes its predictions and how they are tested. Then you'll see that if you were right QFT would always be wrong. Since that isn't the case your claimed transformations are false. Job done.

 

Sure. So I suggest you show the experimental results that confirm Einstein's formula:

\(E = mc^2=\frac{m_o}{\sqrt{1-v^2/c^2}}c^2\)

It suffices to prove that the temperature of the calorimeter is proportional to:
\(T\sim\frac{1}{\sqrt{1-v^2/c^2}}\)

 

You seem to be failing to grasp some pretty simple logic here. I'm saying that your claims (ie the formulae I mentioned a few posts ago) are false, categorically false. If they were even close to right then QFT would never work. As such I don't need to show the experimental data to prove QFT right in order to prove you wrong.

The fact QFT wasn't immediately falsified decades ago shows you're incorrect. Case closed, your 'master theory' is dead.

How so? When dealing with individual scattering processes you don't talk about the temperature of a collider calorimeter but the energy levels of the individual particles, which can then be turned into a temperature for a particular collision (ie mean velocities in the centre of mass frame).

 

\(E = mc^2=\frac{m_o}{\sqrt{1-v^2/c^2}}c^2\) - are this formulae categorically false?
If this formula is true, then publish her experimental support.

 

You keep comming back to this.

Just what is it you believe it means? No one could really answer you any better than several have tried without some further clarification.

There are some differences in how "scientists" interpret the \(m\) in \(E = mc^2\). Some interpretations are that \(m\) = rest mass and others that \(m\) = inertial or relativistic mass. In some sense it may be that both could be accurate while, at the same time not be referring to the same conditions. Rest mass and relativistic or inertial mass are not the same.

So, what does this formula mean to you, \(E = mc^2=\frac{m_o}{\sqrt{1-v^2/c^2}}c^2\)."

 

Experimental support

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I assume that you at least then consider \(m\) to be rest mass? And in the test above it is converted to \(E\) energy?

 

This formula:

\(E = mc^2=\frac{m_o}{\sqrt{1-v^2/c^2}}c^2\)

means that the temperature of the calorimeter in an experiment Bertozzi continues to grow, while the rate remains almost unchanged.

I have info that monstrously huge energies that (as if) manage to get at accelerators exist only on paper. These huge energies are derived from this formula. In a calorimeter this monstrously huge energy generate a ridiculous zilch. It is this circumstance is the reason for the lack of results of these experiments in a public press.

I am confident that in an accelerator of an elementary particles get speed up faster the light speed does not allow the Doppler effect, but not the unlimited growth of energy and mass.

 

To me what is significant about this formula, \(E = mc^2=\frac{m_o}{\sqrt{1-v^2/c^2}}c^2\), is that it is defining the \(m\) in \(E = mc^2\) as inertial or relativistic mass, rather than rest mass.

I have heard a number of prominent physists over the years explain \(E = mc^2\). Some have explained it such that \(m\) should be interpreted as relativistic mass and others as rest mass. Personally, I am partial to the rest mass interpretation, but I can understand the other side of the picture also.

The first or rest mass approach I believe is likely more consistent with how Einstein saw it, but that is really more my impression than anything else. The second or relativistic mass approach is consistent with special relativity, in that the formula for relativistic mass, incorporates the Lorentz Transformation. Neither, is contrary to relativity. They are just different ways of looking at things.

It seems that portion you are most opposed to, and I may be taking some liberty here, is that both versions include \(c^2\). Doing so, both sets a limit to velocities, which cannot exceed the speed of light, and at the same time removes limits from energy, which can essentially become infinite in principle. (i.e. No matter how much energy you put in you cannot accelerate mass to the speed of light...) This is very similar to what your statement above is about,

the energy continues to increase while the rate or velocity remains almost unchanged. That is what the formula says, or at least one interpretation, and it is completely consistent with both special and general relativity.

I don't see that either version is in conflict with relativity. And yet the question you keep asking, when you ask for experimental proof.., I am not sure there is a clear answer.

A few posts up AlexG posted a picture of an atomic bomb blast as an answer and yet though there is no argument that there is a great deal of energy released, we cannot measure it directly. We cannot measure even what happens at a much smaller scale in particle accellerators directly. Everything that occurs at both scales, we can only evaluate through the theories and models that we have. Those theories have been very successful. Will they be the same theories we rely on in a hundred years? Perhaps not. We seem currently to be running up against some difficult obstacles. But I am fairly certain that it is more likely that what the future holds will incorporate or be built upon what we know today, rather than completely overturn it.

 

This formula:

\(E = mc^2=\frac{m_o}{\sqrt{1-v^2/c^2}}c^2\)

is the foundation of modern physics, and so is very important for science. We should not take on trust this formula. The direct experimental (of a calorimeter) confirmation of this formula are need long ago.

 

Gee, it seems like I might have seen someone ask something like this before, hmmm....

I think the best course of action is to post this same idea over and over and most importantly; ignore all responses, just keep on posting.

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Published experiment (in which to did a direct (of a calorimeter) measurement of the energy of relativistic particles) by an answer may be only .

Words (which not proved by experiment) to serve by an answer and a proof can not.

 

I language know english not your first. But understand talk you say if can't know how you say.

Answers questions to you ask tell have you been given. Not sure barrier language is problem of it or brain function problem is cause.

 

Origin,

Mocking a language translation issue this way is out of line. It comes very close to insulting the person, rather than addressing the argument or discussion.

 

Yeah, I know, it was a cheap shot.

 

Passed another month, but experimental evidence for Einstein's theory, there is still no.

\(E = mc^2=\frac{m_o}{\sqrt{1-v^2/c^2}}c^2\) - is a mess?