I Why ##\alpha=atan(v/c)##?

  • Thread starter olgerm
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Where exactly is mistake in my derivation in post#15?
The angle you are referring to has nothing to do with ##dx' / dt## or ##dx / dt'##. It's just ##\text{atan} (dx / dt)## in the "stationary" frame for an object that is at rest in the "moving" frame. Just as @Pencilvester has been telling you.
 

Nugatory

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Where exactly is mistake in my derivation in post#15?
....
##cos(\alpha_{t'-x})=\frac{ _\Delta x}{ _\Delta t'}=....##
and
##cos(\alpha_{t-x'})=\frac{ _\Delta x'}{ _\Delta t}=....##
Both of these expressions are wrong (or at least they have nothing to do with the angle you're looking for). The value of a trig function can be expressed as a ratio of lengths only if the lengths are both taken in the same frame. This should be more clear if you consider that lengths from different frames cannot both be drawn to the same scale on the Euclidean surface of the sheet of paper.
 

olgerm

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What is the correct value of these angles (##\alpha_{t'-x}## , ##\alpha_{t-x'}##) and ##\alpha_{t-x}##?
As I understood ##\alpha_{t'-t}=\alpha_{x-x'}=arctan(\frac{v}{c})##
 
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What is the correct value of these angles (##\alpha_{t'-x}## , ##\alpha_{t-x'}##) and ##\alpha_{t-x}##?
As I understood ##\alpha_{t'-t}=\alpha_{x-x'}=arctan(\frac{v}{c})##
Since these are cartesian coordinates, the angle between the x and t axes is ##\pi /2## by definition. That means ##\alpha_{t'-x}## and ##\alpha_{t-x'}## are just the complementary angles to ##\alpha_{t'-t}## and ##\alpha_{x-x'}##: (##\pi/2-\alpha_{t'-t}##). If you want that in terms of ##v##, that’s the same as ##\arctan(1/v)##.

But again, I feel I must reiterate—these are simple trig problems in 2D Euclidean space that don’t have anything to do with learning SR.
 
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As I understood ##\alpha_{t'-t}=\alpha_{x-x'}=arctan(\frac{v}{c})##
If by this you mean the Euclidean angles between the ##t'## and ##t## axes, and between the ##x'## and ##x## axes, on a spacetime diagram where both sets of axes are drawn with the same origin, then yes, this is true. But, as has already been commented, this angle is not useful for actually learning SR or solving SR problems.
 

olgerm

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If by this you mean the Euclidean angles between the ##t'## and ##t## axes, and between the ##x'## and ##x## axes, on a spacetime diagram where both sets of axes are drawn with the same origin, then yes, this is true. But, as has already been commented, this angle is not useful for actually learning SR or solving SR problems.
I mean the actual angle between coordinate lines.
 

vanhees71

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If by this you mean the Euclidean angles between the ##t'## and ##t## axes, and between the ##x'## and ##x## axes, on a spacetime diagram where both sets of axes are drawn with the same origin, then yes, this is true. But, as has already been commented, this angle is not useful for actually learning SR or solving SR problems.
It's even worse: It's misleading for students of SR. To the contrary, one should stress from the very beginning that Minkowski diagrams should not be read as if it were a Euclidean plane! The fundamental form is not th Euclidean one but the Minkowskian. The crucial point is the different signature of these fundamental forms of the "Minkowskian" rather than the "Euclidean" affine plane.
 
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I mean the actual angle between coordinate lines.
The coordinate lines are parallel to the axes, so it’s the same angle...
EDIT: Oops, just saw @Michael Price got there first. Over 4 hours ago.
 

olgerm

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I know that axes and coordinatelines are same thing, but I meant, that I want the angles between, coordinatelines not between representation of coordinatelines on euclidean sheet.
 

Ibix

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I know that axes and coordinatelines are same thing, but I meant, that I want the angles between, coordinatelines not between representation of coordinatelines on euclidean sheet.
You mean, you want the angle between the ##t## axis in spacetime and the ##t'## axis in spacetime, not the angle between the representation of the ##t## axis on the diagram and the representation of the ##t'## axis on the diagram? Then you need to use hyperbolic geometry, not trigonometry.
 

olgerm

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You mean, you want the angle between the ##t## axis in spacetime and the ##t'## axis in spacetime, not the angle between the representation of the ##t## axis on the diagram and the representation of the ##t'## axis on the diagram? Then you need to use hyperbolic geometry, not trigonometry.
Sorry I got confused. I did not realease that angels in spacetime and euclidean representation are diferent, when I started thread.
 
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Sorry I got confused. I did not realease that angels in spacetime and euclidean representation are diferent, when I started thread.
There is a nice relationship between the Euclidean angle ##\alpha## and the rapidity ##\rho##
tan(##\alpha##)=tanh(##\rho##)=v/c
where the axes are ct, x
 
Last edited:

vanhees71

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Well, may be, but why not using the right geometry from the very beginning. The Minkowski plane is not a Euclidean plane and thus should not be treated as such!
 

olgerm

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So on euclidean papersheet:
##\alpha_{t-t'}=arctan(v/c)##
##\alpha_{t-x}=\frac{2\pi}{4}##
##\alpha_{t-x'}=\frac{2\pi}{4}-arctan(v/c)##
##\alpha_{t'-x}=\frac{2\pi}{4}-arctan(v/c)##
##\alpha_{t'-x'}=\frac{2\pi}{4}-2*arctan(v/c)##
##\alpha_{x-x'}=arctan(v/c)##

And really in spacetime:
##\alpha_{t-t'}=arctanh(v/c)##
##\alpha_{t-x}=##
##\alpha_{t-x'}=##
##\alpha_{t'-x}=##
##\alpha_{t'-x'}=##
##\alpha_{x-x'}=arctanh(v/c)##

Can you confirm these are correct and fill the empty ones?
 

Ibix

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You can write the Lorentz transforms in terms of rapidity:
$$\begin{eqnarray*}
x'&=&\cosh(\alpha) x-\sinh(\alpha) t\\
t'&=&\cosh(\alpha) t-\sinh(\alpha) x
\end{eqnarray*}$$
That should let you solve for the various angles.
 

olgerm

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That should let you solve for the various angles.
I do not know how to do that.
Is ##\vec{v1}\cdot \vec{v2}=|\vec{v1}|∗|\vec{v2}|∗cos(\alpha)## still true in non-euclidean space?
Is it
##\alpha_{x−t}=arccos(-\vec{e_x}\cdot\vec{e_t})=\frac{2\pi}{4}##
##\alpha_{x'−t'}=arccos(-\vec{e_x'}\cdot\vec{e_t'})=\frac{2\pi}{4}##
##\alpha_{x−t′}=arccos(-\vec{e_x}\cdot\vec{e_t'})=arccos(-\vec{e_x}\cdot(\gamma∗\vec{e_t}+\beta∗\gamma∗\vec{e_x}))=arccos(-\frac{v}{\sqrt{c^2−v^2}})##
##\alpha_{x′−t}=arccos(-\vec{e_x'}\cdot\vec{e_t})=arccos(-(\gamma∗\beta∗\vec{e_t}+\gamma∗\vec{e_x})* \vec{e_t})=arccos(-\frac{v}{\sqrt{c^2−v^2}})##
 
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And really in spacetime
The quantities you get from the inverse hyperbolic functions are not angles. They are not limited to the range ##0## to ##2 \pi##. So thinking of them as the "angles in spacetime" between axes is not a good idea. The rapidity ##\alpha## is just a different way of parameterizing the relative velocity between frames, one which makes the math easier and more elegant in some respects. It is not an "angle between the axes in spacetime" in any useful sense.
 
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That should let you solve for the various angles.
I don't think rapidity can be correctly called an "angle". It certainly is not an "angle between axes in spacetime" in the sense the OP seems to want to interpret that; rapidities are not limited to the range ##0## to ##2 \pi##. See my previous post in response to the OP.
 

olgerm

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The quantities you get from the inverse hyperbolic functions
Should I use ##\vec{v_1}\cdot\vec{v_2}=|\vec{v_1}|*|\vec{v_2}|*cosh(\alpha)## instead of ##\vec{v_1}\cdot\vec{v_2}=|\vec{v_1}|*|\vec{v_2}|*cos(\alpha)##?
 
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Should I use ##\vec{v_1}\cdot\vec{v_2}=|\vec{v_1}|*|\vec{v_2}|*cosh(\alpha)## instead of ##\vec{v_1}\cdot\vec{v_2}=|\vec{v_1}|*|\vec{v_2}|*cos(\alpha)##?
What do you think ##\vec{v_1}\cdot\vec{v_2}## means, physically?

Note that in relativity, the dot product of 3-vectors is not an invariant; it changes when you change frames. Only the dot product of 4-vectors is invariant.
 

olgerm

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In euclidean space it has many properties I do not, know which one you mean. For example on vecotrs projection to another times another vectors length.
 
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In euclidean space
Which is, as has already been said a number of times in this thread, irrelevant since SR is not done in Euclidean space, it is done in Minkowski spacetime. Trying to think of the relative velocity between frames in SR as a Euclidean 3-vector is only going to confuse you.
 

olgerm

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Note that in relativity, the dot product of 3-vectors is not an invariant; it changes when you change frames. Only the dot product of 4-vectors is invariant.
I meant 4-vectors by ##\vec{v_1}## and ##\vec{v_2}##.
 

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