Why Does the Angle Between Coordinate Axes in Different Frames Equal atan(v/c)?

[olgerm]

according to this wikipedia page says that angle between coordinalines(and basevectors) in different frames of reference is atan(v/c), but I tried to derive it and got that these angles are $arcsin(v/c)$ and $\frac{2\pi}{4}-arccos(v/c)$. Derivation is on the picture that on the link. Where is the mistake in my derivation? $v_t$ is speed between the frsmes of references.
$\frac{\partial x´}{\partial t}=v_t$ because an object that is in rest in one frame of refence must move with speed $v_t$ in another frame of reference.

Also would that in 1. frame of reference c*t and x are crosswise but not in another frame of reference(x') make first one preferred or different?

 

[DEvens]

Your diagram is pretty unclear. But look at the section of the wiki page https://en.wikipedia.org/wiki/Minkowski_diagram#Loedel_diagram
and be careful about the following.
- Are you identifying the correct angles?
- Are you being careful to remember that in the Loedel diagram, one frame is going left at -v, the other to right at +v?
- Are you making good use of symmetry? Because they are going +v and -v the axis angles must be symmetric.

 

[Pencilvester]

angle between coordinalines(and basevectors) in different frames of reference is atan(v/c)

I think you’re making it way more complicated than it is. On a Minkowsi diagram, the worldline of a particle ($x’=0$) moving with speed $v \equiv \frac{dx}{dt}$ makes an angle $\alpha$ with the $t$ axis ($x=0$). The fact that $\tan{\alpha} = \frac{dx}{dt}$ should be obvious, so clearly, the inverse tangent function of $v$ gives $\alpha$. Pretty much the same argument can be used to get the same angle between $t’=0$ and $t=0$.

would that in 1. frame of reference c*t and x are crosswise but not in another frame of reference(x') make first one preferred or different?

I’m assuming by “crosswise” you mean orthogonal, in which case the $t’$ and $x’$ basis vectors are orthogonal (using the Minkowski metric).

 

[olgerm]

Your diagram is pretty unclear.

I improved it now.

- Are you identifying the correct angles?

I noted the $\alpha$ from wikipedia page with $\beta$.

- Are you being careful to remember that in the Loedel diagram, one frame is going left at -v, the other to right at +v?

0-frame is in rest; '-frame is moving left and -frame is moving right. But in my question only 2 frames were involved. $\alpha$ is angle from this picture.

- Are you making good use of symmetry? Because they are going +v and -v the axis angles must be symmetric.

In my question only 2 frames are involved. It is not about diagrams from Loedel diagram section in this wikipedia page.

 

[PeterDonis]

Derivation is on the picture that on the link.

Please post equations and derivations directly in the post using the LaTeX feature.

 

[olgerm]

Please post equations and derivations directly in the post using the LaTeX feature.

I can't post diagrams that way.

 

[BvU]

Diagrams can be drawn with a decent program and posted -- or drawn clearly and legibly and then scanned and posted

Your 'picture in the link' to me demonstrates a severe lack of consideration for potential helpers.

 

[PeterDonis]

I can't post diagrams that way.

Yes, you can; post them as images. But they need to just be images of diagrams. You need to post everything else directly.

 

[pervect]

according to this wikipedia page says that angle between coordinalines(and basevectors) in different frames of reference is atan(v/c)

I'm not sure what you're trying to do (lacking any diagrams), but I do recognize the formula you give:

$$\tan \alpha = \frac{v}{c} \quad \alpha = \tan^{-1}\frac{v}{c}$$

In this formula, $\alpha$ is an expression for the rapidity. (More specifically, the rapidity in a 2 d spacetime with one spatial dimension and 1 time dimension).

Rapidity can be called an angle in the Lorentzian geometry of space-time, which is hyperbolic, because, unlike velocities, rapdities add. So if we view velocites as angles, we want some sort of function of velocity that adds just like angles do. This function of velocity is the rapidity.

However, the rapidity is unbounded, in can have any value from -infinity to +infinity, so it's not quite analogous to an angle in Euclidean space which only has a range from $-\pi$ to $\pi$.

The wiki article on Rapidity might be of some help.

 

[olgerm]

lacking any diagrams

Diagram is on the link.

 

[Michael Price]

I'm not sure what you're trying to do (lacking any diagrams), but I do recognize the formula you give:

$$\tan \alpha = \frac{v}{c} \quad \alpha = \tan^{-1}\frac{v}{c}$$

In this formula, $\alpha$ is an expression for the rapidity. (More specifically, the rapidity in a 2 d spacetime with one spatial dimension and 1 time dimension).

Rapidity can be called an angle in the Lorentzian geometry of space-time, which is hyperbolic, because, unlike velocities, rapdities add. So if we view velocites as angles, we want some sort of function of velocity that adds just like angles do. This function of velocity is the rapidity.

However, the rapidity is unbounded, in can have any value from -infinity to +infinity, so it's not quite analogous to an angle in Euclidean space which only has a range from $-\pi$ to $\pi$.

The wiki article on Rapidity might be of some help.

Do you mean tanh, not tan?

 

[pervect]

Do you mean tanh, not tan?

Ooops, yes.

 

[Pencilvester]

Diagram is on the link.

I’m not sure if you didn’t see post #3, or if you’re having trouble understanding what I said, but this is just a very simple trig problem. Maybe you just need a diagram to help you understand, so I drew a quick picture. But again, I think you’re making this way more complicated than it really is.

 

[vanhees71]

I'm really puzzled, why still some textbooks and obviously also Wikipedia work in this highly confusing way.

As soon as you draw a Minkowski plane (i.e., a time axis and one of the three spatial axes of a Minkowski 4-bein) you should forget all about Euclidean geometry. The Minkowski plane is an affine space with a different fundamental form, defined by
$$\mathrm{d} s^2=\mathrm{d} t^2-\mathrm{d}x ^2$$
(setting $c=1$ and using the west-coast-convention of signature). There are no angles anymore!

This ingenious insight by Minkowski makes relativity much simpler than old-fashioned mixtures with Euclidean geometry which later have to unlearnt anyway!

 

[olgerm]

I renewed the diagrams in the link.
I now realize, that i made a mistake $\frac{\partial x'}{\partial t}\not = v_t$ in SR, but
$cos(\alpha_{t'-x})=\frac{ _\Delta x}{ _\Delta t'}=\frac{\partial x}{\partial t'}=\frac{v_t}{c*\sqrt{1-\frac{v_t^2}{c^2}}}$
and
$cos(\alpha_{t-x'})=\frac{ _\Delta x'}{ _\Delta t}=\frac{\partial x'}{\partial t}=-\frac{v_t}{c*\sqrt{1-\frac{v_t^2}{c^2}}}$ .
$\alpha_{t-x}=\frac{2\pi}{4}$ !
$\alpha_{t-t'}+\alpha_{t'-x}=\alpha_{t-x}$
$\alpha_{x-x'}+\alpha_{t-x'}=\alpha_{t-x}$
,but the result I get by finding $\alpha_{t-t'}$ and $\alpha_{x-x'}$ is still not $arctan(\frac{v}{c})$, but
$\alpha_{t-t'}=\frac{2\pi}{4}-\alpha_{t'-x}=\frac{2\pi}{4}-arccos(\frac{v_t}{c*\sqrt{1-\frac{v_t^2}{c^2}}})$
$\alpha_{x-x'}=\frac{2\pi}{4}-\alpha_{t-x'}=\frac{2\pi}{4}-arccos(-\frac{v_t}{c*\sqrt{1-\frac{v_t^2}{c^2}}})$

 

[Pencilvester]

but still $α_{t−t′}=2π4−αt′−x=2π4−arccos(vtc∗√1−v2tc2)αt−t′=2π4−αt′−x=2π4−arccos(vtc∗1−vt2c2)\alpha_{t-t'}=\frac{2\pi}{4}-\alpha_{t'-x}=\frac{2\pi}{4}-arccos(\frac{v_t}{c*\sqrt{1-\frac{v_t^2}{c^2}}})$ not $αt−t′=atan(vc)$

How exactly are you defining the term $\alpha_{t-t’}$? I think you may be inventing notation that is causing you to confuse yourself. In the Wikipedia article, $\alpha$ is simply the trigonometric angle between the corresponding basis vectors of two frames moving with speed $v$ relative to each other.
Also, there is no need to bring in the Lorentz transformation to understand why $\alpha = \arctan{v}$. Please describe what issues you are having in understanding what I’ve said in post #3 and #13.

 

[olgerm]

How exactly are you defining the term $\alpha_{t-t’}$?

$\alpha_{t-t’}$ is angle between t'-axis and t-axis.

 

[Pencilvester]

$\alpha_{t-t’}$ is angle between t'-axis and t-axis.

Okay, for the third time, you are making this way more complicated than it really is. Look at post #3. Look at the picture in post #13 if you need to. Tell me what exactly you’re having trouble understanding. Or explain to me why you think it’s necessary to use components of the Lorentz transformation to find the relation between $\alpha$ and $v$. Or re-read what @vanhees71 said in post #14, and stop worrying about trigonometric angles on a Minkowski diagram, since they have little to no use in learning SR.

 

[olgerm]

Look at post #3. Look at the picture in post #13 if you need to.

different ways of calculating should give the same result. If euclidean geometry does not work in minkowsky spacetime, like vanhees71 said, then why should your tangent work?
Maybe my assumation, that $\alpha_{t-x}=\frac{2\pi}{4}$ was wrong, but if it was wrong: what $\alpha_{t-x}$ is?

 

[Pencilvester]

different ways of calculating should give the same result.

Yes, but your way of calculating the angle is extremely convoluted and is using the Lorentz transformation for some reason, so it’s no surprise to me that you’re not getting the correct answer. So again, why are you using the Lorentz transformation in your derivation?

If euclidean geometry does not work in minkowsky spacetime, like vanhees71 said, then why should your tangent work?

Euclidean geometry is not useful on a Minkowsi diagram. But ultimately, the diagram is drawn on a piece of paper, and paper lends itself particularly well to Euclidean geometry since it’s a good approximation to a two dimensional Euclidean space. This fact is incidental and does not mean that Minkowsi diagrams should be interpreted using Euclidean geometry.

what αt−xαt−x\alpha_{t-x} is?

This appears to be your notation for the angle between the $t$ and $x$ axes, which, in the context of your original question and the Wikipedia article, is $\pi / 2$. If you really want to derive $\alpha$ using its complementary angle, then you could start with $\arctan{(1/v)} = \pi/2 - \alpha$.

 

[PeterDonis]

The wiki article on Rapidity might be of some help.

Not really, since, as you realized in response to @Michael Price, the rapidity is the inverse hyperbolic tangent of the relative velocity $v$. Whereas the angle the OP is looking at is the regular inverse tangent of the relative velocity $v$.

Which means, of course, that the angle the OP is looking at, the one described in the Wikipedia article, is actually not used at all in relativity. Which makes this another example of Wikipedia not being a good source to learn from.

 

[olgerm]

This appears to be your notation for the angle between the $t$ and $x$ axes,

Correct.

in the context of your original question and the Wikipedia article, is $\pi / 2$

Are you sure?

 

[olgerm]

Which means, of course, that the angle the OP is looking at, the one described in the Wikipedia article, is actually not used at all in relativity. Which makes this another example of Wikipedia not being a good source to learn from.

I think it is important to think of things in different terms and solve such seeming paradoxes to study any subject. So it is very useful to see things that are "not actually used".

 

[PeterDonis]

I think it is important to think of things in different terms and solve such seeming paradoxes to study any subject

But there is no "seeming paradox" and the angle in question has nothing whatever to do with the subject, namely special relativity. All of your "derivation" is looking at the wrong things: @Pencilvester has already explained where the $\text{atan} (v / c)$ comes from, and, as has been noted, it has nothing to do with relativity physics, it's just an observation about how the diagram is drawn on a Euclidean plane.

 

[olgerm]

there is no "seeming paradox"

The seeming paradox to me is that my derivation gave different result to $\alpha$, than wikipedia page.
Where exactly is mistake in my derivation in post#15?

 

[PeterDonis]

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]

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]

What is the correct value of these angles ($\alpha_{t'-x}$ , $\alpha_{t-x'}$) and $\alpha_{t-x}$?
As I understood $\alpha_{t'-t}=\alpha_{x-x'}=arctan(\frac{v}{c})$

 

[Pencilvester]

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.

 

[PeterDonis]

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.