I was unable to translate this phrase from the English language.
I could not translate this phrase from the English language.
Three Experiments Challenging SRT
- Thread starter Masterov
- Start date
I could not translate this phrase from the English language.
L is the distance between the travellers in the reference frame in which they are moving with velocity v.
L' is the distance between the travellers in the reference frame in which they are at rest (i.e. in the reference frame moving along with the travellers).
I don't speak Russian, so I don't see how I can make it clearer than that.
L' is the distance between the travellers in the reference frame in which they are at rest (i.e. in the reference frame moving along with the travellers).
I don't speak Russian, so I don't see how I can make it clearer than that.
In the problem of the two travelers and the dog: L '= L.
This little problem only shows that dog's times (in different directions) will be different:
$$t'_1=L/(c+v)$$
$$t'_2=L/(c-v)$$
In the problem (with two mirrors, with a photon and with a moving observer) times will be different too, but to this added to the visual relativistic effect: the distance between the mirrors will become smaller (visual). Therefore:
$$L'=L(1-v^2/c^2)$$
$$t'_1=L'/(c+v)=L(1+v/c)/c$$
$$t'_2=L'/(c-v)=L(1-v/c)/c$$
TEST:
$$t'_1+t'_2=L(1+v/c)/c+L(1-v/c)/c=2L/c=t_1+t_2$$ - is true!
My English is not good really.
I use Google's translator, that to read and write English.
Long sentences is very difficult to understand for me.
Me is difficult to understand those posts of an authors of which use the English language as a foreign language.
I ask to write simple and short sentences.
Then I can respond to your posts.
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You've always had a problem doing frame changes. That was clear from your touting of Andrew Banks's work.
You must be aware you've not read enough relativity, that you've never been able to do simple homework problems expected of undergraduates learning special relativity (likewise for logic and countability). Yet you presume, without evidence and even contrary to explicit evidence, you have capabilities in such things. You know you've never done any relativity yet you continue with this laughable attitude.
Your notation is confusing. Why are you putting primes (') on your variables here? Why t' instead of t?
I'll use your earlier notation. The times are different, like you said:
$$
\begin{eqnarray}
T_{1} &=& \frac{L}{c + v} \,,
T_{2} &=& \frac{L}{c - v} \,.
\end{eqnarray}
$$
\begin{eqnarray}
T_{1} &=& \frac{L}{c + v} \,,
T_{2} &=& \frac{L}{c - v} \,.
\end{eqnarray}
$$
But the distances travelled by the dog are also different:
$$
\begin{eqnarray}
X_{1} &=& \frac{L}{1 + v/c} \,,
X_{2} &=& \frac{L}{1 - v/c} \,.
\end{eqnarray}
$$
\begin{eqnarray}
X_{1} &=& \frac{L}{1 + v/c} \,,
X_{2} &=& \frac{L}{1 - v/c} \,.
\end{eqnarray}
$$
in both cases, $$(X_{i})^{2} \,-\, (cT_{i})^{2} \,=\, 0$$.
It's hard to believe you had to write that down. What's been your main area of scientific research? If you wouldn't mind saying.
No, excuse me, your friend measured his own L' from the context of measurements of the moving frame.
How exactly would you know measurements in the moving frame?
Are you going to use ESP?
So, as we can see, you are in error.
You do not seem to understand.
You do not measure your frame in units of the moving frame.
You measure them in units of your own frame.
That is the error you made.
You claimed your measurements are a result of measurements of the moving frame. That is not SR.
You use your measurements and construct measurements in the moving frame.
So, you and AN are in error.
Correction, you're the one who doesn't understand.chinglu said:
Which frame is moving, how do you tell?
The answer is, you can't tell. That's relativity.
So, therefore, which frame is primed (marked, if you will, as being in uniform motion) and which isn't is an arbitrary choice made by an observer.
Yes.
Quantum information. Mostly I've been doing theoretical work related to quantum cryptography in the last 18 months or so.
I have no idea what you're talking about. If you know a quantity in one reference frame, then with enough information you can work out the quantity in another frame. That's what the Lorentz transformation is for. In this case, $$L$$ is given in the first frame I'm considering, and you can work out $$L'$$ in a boosted frame that matches the velocity of the travellers. You get $$L' = \gamma L$$. I gave a [POST=2959149]full, systematic derivation from the Lorentz transform here[/POST]. If you think the result is wrong, you are welcome to point out an error in that.
I don't think you're qualified to be telling anyone here what SR is and isn't.
No, I'm going to use knowledge and ability. Concepts you seem unfamiliar with.
Thanks for your reply.
Please look at LT and prove your case.
A frame measures its own context and then LT translates this to the other frame coordinates.
In short, you are frame mixing.
You must use your frame measurements and then use LT to translate.
So, your equations are all wrong. You cannot use the other frames information for your own measurements.
In practice that may be accurate. The case here is a hypothetical where the values in the "other" frame are given. In that situation the "other" frame can be treated, as known and the LT used to translate the other way.
Yes that was said very awkwardly. The point is przyk was working with the hypothetical as it had been presented. What it seems you are saying is that the hypotheical was set up in the wrong way or direction. Seems przyk began with a similar point of view and then applied the math to the situation as it was presented to him.
IOW The LT can be used to transform between any two inertial farms of reference, so long as one is fully known and the relative velocity or velocities of the two are known. Neither need be the rest frame of the observer. There are layers upon layers....