Three Experiments Challenging SRT
- Thread starter Masterov
- Start date
Look at my [POST=2958193]response[/POST] again, to which you have never adequately replied.
Particularly this part:
At least get the man's name right. It's "Lorentz", ok?Masterov said:
Can you provide evidence to back up your claim that "Lorentz argues that both times will be the same" from the literature?
And can you clarify for what kind of observer (in which frame of reference) Lorentz claims this? Unless of course your literature reference makes this clear.
If you can do neither, then you're just waving your arms around and making strange noises.
3. Distances are different and may not be the same.Look again:
1. Speed of red dot is equal in both directions.
2. Times are different and may not be the same.
Because the mirrors/travellers are moving. Every schoolboy knows that too, yet apparently you don't. Shame!
Advance traveller/mirror: $$x_{2} \,=\, vt \,+\, L$$.
Dog/red dot: $$x_{\mathrm{d}} \,=\, ct \,.$$
They coincide where $$x_{2} \,=\, x_{\mathrm{d}}$$, or $$vt \,+\, L \,=\, ct$$. So $$\Delta t_{2} \,=\, \frac{L}{c - v}$$.
The dog/red dot's position at that time is $$ct_{2} \,=\, \frac{cL}{c - v}$$, so $$\Delta x_{2} \,=\, \frac{L}{1 - v/c}$$.
Similarly, for the dog running in the other direction, $$\Delta t_{1} \,=\, \frac{L}{c + v}$$ and $$\Delta x_{1} \,=\, - \frac{L}{1 + v/c}$$.
Since you insist on using it wrong anyway, put that in Minkowski's formula and smoke it.
Masterov is wrong, as demonstrated in many posts which he has simply ignored.
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Lorentz transformations are the result of the expression: $$(ct')^2 - x'^2 = (ct)^2 - x^2$$
From this expression it follows that for all speeds: $$t_1=t_2$$ and $$t'_1=t'_2$$
It is impossible. If $$v\neq 0$$ then $$t'_1\neq t'_2$$
This expression ($$(ct')^2 - x'^2 = (ct)^2 - x^2$$) is a mistake.
Hence the Lorentz transformations are a mistake.
Slowing down time (and for a speed of light was constant) can not to do so: $$t'_1=t'_2$$
You don't even understand the equation you're using.
$$
\begin{eqnarray}
x_{1} &=& - \frac{L}{1 + v/c} \,, \quad & t_{1} &=& \frac{L}{c + v} \,, \\
x_{2} &=& \frac{L}{1 - v/c} \,, \quad & t_{2} &=& \frac{L}{c - v} \,.
\end{eqnarray}
$$
\begin{eqnarray}
x_{1} &=& - \frac{L}{1 + v/c} \,, \quad & t_{1} &=& \frac{L}{c + v} \,, \\
x_{2} &=& \frac{L}{1 - v/c} \,, \quad & t_{2} &=& \frac{L}{c - v} \,.
\end{eqnarray}
$$
In a boosted frame moving at velocity v, $$L' = \gamma L$$ (from length contraction), and:
$$
\begin{eqnarray}
x'_{1} &=& - \gamma L \,, \quad & t'_{1} &=& \gamma L/c \,, \\
x'_{2} &=& \gamma L \,, \quad & t'_{2} &=& \gamma L/c \,.
\end{eqnarray}
$$
\begin{eqnarray}
x'_{1} &=& - \gamma L \,, \quad & t'_{1} &=& \gamma L/c \,, \\
x'_{2} &=& \gamma L \,, \quad & t'_{2} &=& \gamma L/c \,.
\end{eqnarray}
$$
So:
$$
\begin{eqnarray}
(ct'_{1})^{2} \,-\, (x'_{1})^{2} &=\, 0 \, =& (ct_{1})^{2} \,-\, (x_{1})^{2} \,, \\
(ct'_{2})^{2} \,-\, (x'_{2})^{2} &=\, 0 \, =& (ct_{2})^{2} \,-\, (x_{2})^{2} \,.
\end{eqnarray}
$$
\begin{eqnarray}
(ct'_{1})^{2} \,-\, (x'_{1})^{2} &=\, 0 \, =& (ct_{1})^{2} \,-\, (x_{1})^{2} \,, \\
(ct'_{2})^{2} \,-\, (x'_{2})^{2} &=\, 0 \, =& (ct_{2})^{2} \,-\, (x_{2})^{2} \,.
\end{eqnarray}
$$
It works out fine, if you apply the equation correctly.
$$t'_1=L/(c+v)$$
$$t'_2=L/(c-v)$$
$$x'=L(1-v^2/c^2)$$
_________________________________________
I admit that I could be wrong.
BUT: If $$v\neq 0$$ then $$t'_1\neq t'_2$$ is true.
Consequently: expression $$(ct')^2 - x'^2 = (ct)^2 - x^2$$ is a mistake and Lorentz transformations are a mistake also.
przyk has shown, line by line, how the maths works. Your inability or unwillingness to apply equations correctly doesn't mean relativity is wrong because of it. I can't speak Japanese, doesn't mean it isn't a valid language (and yes, for those of you who read most of my posts I do like using that example....).
Remember how you said that the internet means people can found out if scientists are lying, via the free exchange of information. That works both ways, people can find out when you're lying too and you're lying now.
Remember how you said that the internet means people can found out if scientists are lying, via the free exchange of information. That works both ways, people can find out when you're lying too and you're lying now.
I do not understand it.
Your English is strange.
No, my English is pretty good. I do, however, have a tendency to construct somewhat overly elaborate sentence structures through the use of commas, much as I'm doing now by adding this comment. If you struggle to understand what I'm saying then tough. Your comprehension issues go much deeper than my excessive use of commas.
This statement is in error.
In a boosted frame moving at velocity v, $$L' = \gamma L$$ (from length contraction),
The primed from is always the other/moving frame.
Length contraction is described as
$$L' = L/\gamma$$
We have $$\gamma = 1/sqrt{1-v^2/c^2}$$
Hence, $$L' = L sqrt{1-v^2/c^2}= L/\gamma$$
You can find the source of your error and my correction to your error in section 4.
http://www.fourmilab.ch/etexts/einstein/specrel/www/
I showed in post #112 where przyk is in error. Since, you claimed his math was valid line by line, you are in error.
If you cannot understand this simple math error, then how could you possibly think you know Cantor';s argument is correct? In fact, it is not.
Then you shut down a thread because of a math error you made like above.
Open it back up and prove you know how it all works using a superior intelligence. You are not afraid are you?
Ahem:
If: In a boosted frame moving at velocity v, $$L' = \gamma L$$ (from length contraction)
then: $$L = L' / \gamma $$.
If: In a boosted frame moving at velocity v, $$L' = \gamma L$$ (from length contraction)
then: $$L = L' / \gamma $$.
LOL
So, what is L' in the context of the stationary frame where L is measured?
In other words, write me an equation for L' since I am in L.
The primed frame is whatever I say it is when I'm the one who defined the frame.
In [POST=2958462]post #106[/POST], the primed frame is moving, and the travellers are moving, and the primed frame is moving along with the travellers. In that case, that makes L the contracted length and L' the rest length (contrary to the usual convention). So $$L = L'/\gamma$$ according to relativity, which rearranges to $$L' = \gamma L$$.
You should think more instead of copying equations without understanding what they mean.
That's a complete non-sequitor. Good logic.
No, it was shut down because you repeatedly behaved like a troll. You refused to answer direct questions. You refused to respond to line by line corrections. You refused to engage in honest discussion. You acted, and still act, like a troll.
Firstly, that's more trolling behaviour. And secondly, you've already shown that you couldn't even understand what people were saying to you. The "add it to the end of the list" argument fails and you had why explained to you. When you now recall the discussion you misrepresent what was said to you, showing you didn't understand and haven't made any attempt to. You obviously don't even know what 'proof by contradiction' is, which illustrates how poor your maths skills are. Until you can demonstrate you're able to discuss things honestly and coherently you're not going to be allowed in the main maths and physics forum. Your behaviour here only serves to show that policy is completely correct.
If you think you have a disproof of relativity and Cantor then send it to a journal. You obviously don't listen to anything anyone here says to you and you think we're all simpletons. So why are you bothering with us? Why aren't you sending your work to a journal? Are the posts you make here the pinnacle of your mathematics capability and contribution? Is the sum of all your contributions to the scientific discourse posts on this forum? It's pretty sad if that's the case. Remember, I come here for fun, in between doing proper maths and physics in my day job. There's more to my maths and physics knowledge/contributions than forum posts. The fact you've spent years whining about this stuff and haven't gotten your work into a journal speaks volumes.
REMINDER:
Master Theory identifies two types of coordinates: actual (real) and visible.
In this topic discussed visual coordinates only.
The actual coordinates of a subjects obey to Galilean transformations and are calculated by double integration of acceleration over time.
In Master Theory are absolute (not depending on the speed of the observer): time, acceleration , strength, physical dimensions of objects, the speed of light.
Master Theory did a small corrections to Newtonian physics, which do reconciliation Newton with Michelson-Morley experiment.
Master Theory relieve many minds of physicists from rubbish, which Einstein (and his followers) generate one hundred years.
Master Theory identifies two types of coordinates: actual (real) and visible.
In this topic discussed visual coordinates only.
The actual coordinates of a subjects obey to Galilean transformations and are calculated by double integration of acceleration over time.
In Master Theory are absolute (not depending on the speed of the observer): time, acceleration , strength, physical dimensions of objects, the speed of light.
Master Theory did a small corrections to Newtonian physics, which do reconciliation Newton with Michelson-Morley experiment.
Master Theory relieve many minds of physicists from rubbish, which Einstein (and his followers) generate one hundred years.
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Newtonian physics is refuted by a lot more than just the Michelson-Morley experiment. And relativity is supported by a lot more than just the Michelson-Morley experiment.
Well done for ignoring basically all of the last century's worth of experimental results.
Well, phone up the people maintaining the GPS, or who work at institutions like CERN and Fermilab and DESY, or even just anyone involved in designing or operating electron microscopes or medical accelerators, and you explain that to them.