# Einstein got it all wrong?

Discussion in 'Physics & Math' started by scifes, Mar 26, 2011.

1. ### OnlyMeValued Senior Member

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Anyone who is currently involved in undergraduate work in physics should take the following post with a large grain of salt, it steps outside of the boundaries of currently accepted theory...

Radioactive decay, is attributed to the weak nuclear force (the electro-weak force). Once our understanding of the weak nuclear force evolved to to include an electromagnetic component, it does not seem reasonable to me, to assume that just because stable isotopes do not decay through electro-weak interaction(s), that electro-weak interaction(s) do not have any affect on stable isotopes.

The questions are:
• Can electro-weak interactions affect electron transitions in stable isotopes?
• If so, can a solar event and/or variations in the the earth/sun distance, that affect the decay rate of unstable isotopes, have an influence, either directly or indirectly, on the electron transitions in a stable isotope?
• And.., if the conditions above do affect electron transitions in stable isotopes, is there a measurable difference in that effect, when measured on earth as opposed to in an orbit around the earth?

Should electro-weak interactions, be found to influence electron transitions and vary under the influence of solar activity and distance, it could represent a serious challenge to the application of the Lorentz transformations, to length contraction and time dilation of an object in motion, as well as requiring some remodeling of general relativity. This would not affect the use of the Lorentz transformation as applied to observations made from frames of reference in relative motion to one another, only to their application to objects in motion.

If what is suggested above were to be confirmed, it could lead to significant re-evaluations of both quantum theory and general relativity. It also has some potential to resolve one of the conflicts between the two.., the definition and nature of space itself.

3. ### RJBeeryNatural PhilosopherValued Senior Member

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What's the difference? And why would, say, neutrinos altering decay rates affect the use of Lorentz transforms in any situation?
Again, I'm not making the connection why decay having a partial externally-caused explanation would be such an unsettling discovery for QM or GR. Could you elaborate?

5. ### OnlyMeValued Senior Member

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The Lorentz transformations are involved in reconciling differences in the way events are observed to occur from differing frames of reference in uniform motion relative to one another. This is observer oriented. i.e. If I am at rest and you are moving at a constant velocity relative to me, we see the same events occurring at different times and even perhaps in different orders. This is a perception of events caused by the finite speed of light and our different frames of reference. Here, the transformations do not define what is length contracted they help to reconcile the observations of an event or series of events that appears length contracted with a frame of reference in which there is no length contraction observed.

The Lorentz transformations themselves when applied to objects in motion with a known velocity, define the amount that the object's velocity results in both length contraction and time dilation.

It is, once again the twin paradox. Both twins see the other as moving, length contracted and time dilated, while only one twin is moving and aging at a slower rate.

Neutrinos, don't have anything (known) to do with this. I at one point made reference to the muon decay resulting from cosmic radiation and neutrino experiments only as an example that cosmic rays interact with the atmosphere all the way to the ground. One reason neutrino detectors are underground or ice, shielding them from most cosmic radiation.

It is not the decay event that is at issue. My assumption is that the electro-weak interaction/force, which is involved with radio active decay, must also have some affect on atoms, isotopes that are stable. In which case a decay event would not be the result. So the electro-weak force affects both stable and unstable isotopes, resulting in the decay of the unstable isotope.

The electromagnetic nature of electrons and the electromagnetic aspect of the electro-weak force imply the potential for interaction between the two. If variations in solar activity affects the electro-weak interaction and the electro-weak interaction contributes to electron transitions, variations in solar activity could affect the accuracy of atomic clocks, which depend on electron transitions.

How this affects length contraction and general relativity is that currently, variations in electron transitions between atomic clocks on earth and in orbit are based on time dilation, as defined by Lorentz transformations for moving objects. Should it be found that another mechanism is at work or also involved, a great deal of reworking the dynamics of a portion of both special and general relativity would be necessary. They both relay on the accuracy of the Lorentz transformations to some extent.

This should be testable. However, it involves both quantum mechanics and general relativistic components. Quantum mechanics to work out what affect the electro-weak force might have on electron transitions, if any. And relativity to incorporate that information and determine its impact and implications, on related observed phenomena.

I fear I am not doing a good job of relating this. Though it has been on my list of topics of interest and exploration for some time, it has been on a back burner.

7. ### EmilValued Senior Member

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2,789
I follow with interest the discussion between you and I'm not going to interfere.I just want to draw attention to the following fact.

If the logic used in the muons experiment is correct ie muons lifetime value τ = 2 × 10 ^ -6 has no practical value because this value is according to muons. According to the Earth the value is τ = 1.4 ×10^-4, under Lorentz transform and the muons speed.
Then the particles lifetime from this table do not have a practical use.The values should be recalculated under Lorentz transformation and particle velocity.

8. ### Janus58Valued Senior Member

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1,882
You know, it is one thing to be mistaken. It is quite another to stubbornly hold on to a mistaken belief after it has carefully explained to you that it is wrong.

What good would be a chart of values recalculated under Lorentz transformation be when particles, even of the same type, do not have fixed speeds relative to the Earth? Even the Muons generated by cosmic rays have individual variations of velocity. And the Lorentz factor difference between even 0.9999c and 0.9998c is the difference between 70 and 50.

The only type of chart that would be practical at all is one that gives the "proper"(at rest) values for the half-lifes of these particles.

9. ### AlphaNumericFully ionizedRegistered Senior Member

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6,697
How many times are you going to avoid facing up to the fact you've been demonstrated, repeatedly, to be wrong and ignorant on a multitude of issues brought up in this thread?

A table of half lifes from the point of view of particle rest frames is of practical application. From that you can easily calculate what any inertial observer would measure them to be via simple Lorentz transforms.

This is something built into quantum field theory, where the notion of a 'decay width' exists. Since quantum field theory has Lorentz transformations built into it by design decay widths correctly behaviour under said transformations.

These are experimentally tested, they are used in the design of experiments, they are well understood. You simply don't know, don't understand and by the looks of it have no wish to understand.

10. ### EmilValued Senior Member

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2,789

Not you I meant when I posted "I follow with interest the discussion between you".
Your posts do not follow a logic.They are quoted from a book memorized,and if not exactly the same words.
I stopped trying to have a discussion with you, time ago.
But I try again.

Following the logic used here
and applied to the photon (β =1), distance becomes 0.
What that means, according to logic, the photon travels instantly.
In other words, after the photons are emitted and leaves the local conditions around the body which issued,it travels through space instantaneously.
This is not true, follows that the logic used is not correct.

11. ### AlphaNumericFully ionizedRegistered Senior Member

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6,697
Ah, the "You read a book therefore I will ignore you" defence. Farsight uses that one too.

The fact there's agreement in the scientific community doesn't mean we're all parroting books mindlessly, it means there's reason and evidence behind what is said.

The fact you haven't read any book on the subject doesn't somehow elevate your position.

You simply ignored my posts almost from the get go, despite them being directly relevant to things you've said. You whined about muons and Sun-Earth travel after I explained them. The fact you talk about good faith makes you a hypocrite.

It doesn't make sense to put v=c into Lorentz transforms, since they have singularities there. However, if you consider the effect of $v \to c$ in them then yes, the amount of time a photon experiences in its transit from emitters to detector is zero. The amount of distance it views between those two places is zero.

You're whining about relativity you know nothing about. You bring up these points as if its stuff no one has heard of. This is stuff covered in first courses on relativity, taught to 1st years, if not high school students. You aren't presenting or raising novel issues, you're dredging up stuff everyone who has studied relativity has considered.

Jack_ had the same issue, thinking that because he didn't understand it then points he raised must not have been considered by anyone else. In his case, as in yours, every single issue was covered in introductory courses on relativity. As with him, all you do by raising these points as if they are problems or unconsidered is show how flat out ignorant and dishonest you are. If you had got a book on relativity when this thread started and spent some time reading it you'd have been able to answer all your own posts. Instead you post stuff with a "Ah ha! Look at what I've found!" attitude, only for it to be so basic I'd have been embarrassed if one of my first term first year relativity students had done as you have.

Why are you continuing? You know you don't know any relativity. You know some of us do. You know you're just Googling for something, anything, to throw at us. Are you so detached from reality you don't know when to fess up and walk away?

12. ### scifesIn withdrawal.Valued Senior Member

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2,573
ok, now that the "dark physics" part of my brain is up and running again;

is matter to one observer, light[or energy] to another observer, and vice versa?

if matter turns into energy at the speed of light, then what is matter to a relatively stationary observer is actually going the speed of light to another observer going the speed of light relative to the first observer..

so your cup of coffee is matter to you, but to an alien spying on you while zooming around at the speed of light it's a blotch of light [and so are you, then why spy in the first place:huh

]

13. ### originTrump is the best argument against a democracy.Valued Senior Member

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10,296
No need to worry about the rest because the above statement is not correct. Matter does not turn into energy at the speed of light, as a matter of fact, it is impossible for matter to even travel at the speed of light

14. ### Fraggle RockerStaff Member

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24,690
Only particles without mass can travel at the speed of light, such as photons.

You can see this by doing the math. The energy required to accelerate an object with non-zero mass to the speed of light is infinite.

15. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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23,198
I don't intend to challenge general relativity but may have unintentionally done so here: http://www.sciforums.com/showpost.php?p=2764618&postcount=52

I would appreciate it if some of the more knowledgeable posters I see are active here would read that post, which attempts to defend the "pushing version of gravity" by postulating existence of quantized unit of momentum can exist with neither energy nor mass. Here is compressed version of the summary of that link's post:

"... SUMMARY: What would make my defense fail is some fundamental reason why the postulated mass-less particle must carry energy if it carries momentum. As I noted above they {can reproduce gravity's laws and} need not violate either conservation of momentum nor energy if this separation is conceptually possible. ...

I know that E^2 = m^2 c^4 + p^2 c^4 but do not know this MUST apply to a yet unknown "particle" which is only a quantized unit of momentum. Showing that "must" is what is required. Simply citing this fact, known to be true for all KNOWN particles, is not a proof of "MUST." As Hamlet said to Horatio, "There are more things in heaven and earth than are dreamt of in your philosophy." At the level of this question, physics is philosophy, not known equations about currently known "particles". ..."

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I believe it was Janus58 who previously pointed out to me that a very massive object with true mass M, such as black hole at center of a galaxy, would not produce all the gravity normally associated with mass M, (due to internal self shadowing) but we only know its computed mass, M', by the effect of its gravity, so how could we know that M' < M?

Last edited by a moderator: Jun 8, 2011