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

[PeterDonis]

I want the angles between, coordinatelines not between representation of coordinatelines on euclidean sheet

What's the difference?

 

[Ibix]

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]

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.

 

[Michael Price]

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

 

[vanhees71]

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]

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]

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]

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
$\alpha_{x−t}=arccos(-\vec{e_x}\cdot\vec{e_t})=\frac{2\pi}{4}$
$\alpha_{x'−t'}=arccos(-\vec{e_x'}\cdot\vec{e_t'})=\frac{2\pi}{4}$
$\alpha_{x−t′}=arccos(-\vec{e_x}\cdot\vec{e_t'})=arccos(-\vec{e_x}\cdot(\gamma∗\vec{e_t}+\beta∗\gamma∗\vec{e_x}))=arccos(-\frac{v}{\sqrt{c^2−v^2}})$
$\alpha_{x′−t}=arccos(-\vec{e_x'}\cdot\vec{e_t})=arccos(-(\gamma∗\beta∗\vec{e_t}+\gamma∗\vec{e_x})* \vec{e_t})=arccos(-\frac{v}{\sqrt{c^2−v^2}})$

 

[PeterDonis]

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.

 

[PeterDonis]

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]

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 $\vec{v_1}\cdot\vec{v_2}=|\vec{v_1}|*|\vec{v_2}|*cos(\alpha)$?

 

[PeterDonis]

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]

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.

 

[PeterDonis]

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]

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 $\vec{v_2}$.

 

[PeterDonis]

I meant 4-vctirs by $\vec{v_1}$ and $\vec{v_2}$.

Ok, then what do you think their 4-vector dot product means, physically? Or, for that matter, what do you think each 4-vector by itself means, physically? What do these 4-vectors represent?

 

[olgerm]

Now I am quite sure, that it is $\vec{v_1}\cdot\vec{v_2}=|\vec{v_1}|*|\vec{v_2}|*cos(\alpha)$, because otherwise would be that
$\alpha_{t-t'}=arccosh(\frac{\vec{e_t}\cdot \vec{e_t'}}{|\vec{e_t}|*|\vec{e_t'}|})= arccosh(\frac{\vec{e_t}\cdot (\gamma*\vec{e_t}+\beta*\gamma*\vec{e_x})}{(-1)*(-1)})=arccosh(\frac{1}{\sqrt{1-v^2/c^2}})$
, but I was told that $\alpha_{t-t'}=arccos(\frac{\vec{e_t}\cdot \vec{e_t'}}{|\vec{e_t}|*|\vec{e_t'}|})= arccos(\frac{\vec{e_t}\cdot (\gamma*\vec{e_t}+\beta*\gamma*\vec{e_x})}{(-1)*(-1)})=arccos(\frac{1}{\sqrt{1-v^2/c^2}})=i*sgn(v)*arctanh(v/c)$.

 

[PeterDonis]

Now I am quite sure, that it is $\vec{v_1}\cdot\vec{v_2}=|\vec{v_1}|*|\vec{v_2}|*cos(\alpha)$,

If they are 4-vectors, you are wrong. You haven't answered the questions I asked about what these vectors mean, physically. If you don't know the answer to that you aren't going to understand anything else.

Can you answer the question, what do the 4-vectors $\vec{v_1}$ and $\vec{v_2}$ mean physically? Or don't you know? If you don't know, then you need to figure that out first before even trying to guess about what their dot product and its relationship with the rapidity $\alpha$ might be.

 

[olgerm]

If they are 4-vectors, you are wrong. You haven't answered the questions I asked about what these vectors mean, physically. If you don't know the answer to that you aren't going to understand anything else.

Can you answer the question, what do the 4-vectors $\vec{v_1}$ and $\vec{v_2}$ mean physically? Or don't you know? If you don't know, then you need to figure that out first before even trying to guess about what their dot product and its relationship with the rapidity $\alpha$ might be.

These are quantities that describe location in SR-spacetime. I am not sure what kind of answer you espect.

 

[PeterDonis]

These are quantities that describe location in SR-spacetime.

No, they aren't. We are not dealing with spacetime position vectors here--if we were, everything that has been written in this thread would be nonsense.

I am not sure what kind of answer you espect.

In other words, you don't know.

The standard math of SR describes the worldline of a particle with a 4-velocity vector, usually denoted $u$, which is a unit vector tangent to the worldline at a particular event. If we have two particles in relative motion, we can choose the event to be the point in spacetime where they pass each other, and then we have two 4-velocity vectors $u_1$ and $u_2$. Their dot product is $u_1 \cdot u_2 = \gamma$, where $\gamma$ is the relativistic gamma factor corresponding to the ordinary "relative velocity" between the particles, i.e., $\gamma = 1 / \sqrt{1 - v^2}$.

In terms of rapidity, we have $\gamma = \cosh \alpha$, and also $v = \tanh \alpha$. From this it is easy to calculate that $\gamma v = \sinh \alpha$. So we have $u_1 \cdot u_2 = \cosh \alpha$. Note that, since $u_1$ and $u_2$ are both unit vectors, we could indeed write the dot product as $u_1 \cdot u_2 = |u_1| |u_2| \cosh \alpha$, since both magnitudes are $1$. But $\alpha$ here, as has been said, is the rapidity; it is not an "angle in spacetime between the worldlines" in any useful sense, since it is not limited to the range $0$ to $2 \pi$.

 

[PeterDonis]

But $\alpha$ here, as has been said, is the rapidity; it is not an "angle in spacetime between the worldlines" in any useful sense, since it is not limited to the range $0$ to $2 \pi$.

And just to repeat once more, this means that the rapidity $\alpha$ is not the angle $\alpha$ that appears in the Wikipedia article linked to in the OP. The angle that appears in the Wikipedia article has no useful physical meaning.

 

[Ibix]

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.

Fair point. But I think it is fair to say that rapidity is analogous to angle in the same way that interval is analogous to distance. Like the angle in Euclidean geometry, rapidity is a measure of how far off parallel two lines are and is additive. But it can take any value, not just $0-2\pi$, fundamentally because it is not periodic. As rapidity, I think it's also only usually defined for timelike vectors since they are the only things that correspond to something moving. It's certainly possible to write the inner product of a timelike and a spacelike vector, but if you try to derive a rapidity from it you'll get a complex number. Not necessarily a bad thing, but maybe goes some way to explaining why it doesn't get used so much.

Anyway, since there are all these differences you cannot easily read the rapidity off a Minkowski diagram.

 

[olgerm]

I meant that
4-vectors indicate direction and magnitude in spacetime(by direction I do not mean spatial direction, but direction in spacetime that includes time).
Spacetime position 4-vectors are quantities that describe displacement from coordinate 0-point in spacetime(by displacement I do not mean spatial displacement, but displacement in spacetime that includes time).

 

[olgerm]

it is poosible to choose base metric is euclidean just $\vec{e_t'}=\sqrt{-1}*\vec{e_t}$ and $\vec{e_x'}=\vec{e_x}$. Then $\vec{v_1}\cdot\vec{v_2}=|\vec{v_1}|*|\vec{v_2}|*cos(\alpha)$

then:
$\frac{e_t'}{i}\cdot e_x'=|e_t'/i|*|e_x'|*cos(\alpha_{t'-x'})$
$e_t\cdot e_x=|e_t'/i|*|e_x|*cos(\alpha_{t-x})$
$0=|e_t'/i|*|e_x|*cos(\alpha_{t-x})$
$\alpha_{t-x}=arccos(0)$
$\alpha_{t-x}=\frac{2\pi}{4}$

 

[vanhees71]

Should I use $\vec{v_1}\cdot\vec{v_2}=|\vec{v_1}|*|\vec{v_2}|*cosh(\alpha)$ instead of $\vec{v_1}\cdot\vec{v_2}=|\vec{v_1}|*|\vec{v_2}|*cos(\alpha)$?

No! No! No! You should use the correct math. You are completely mislead by whatever book. A Minkowski plane is not a euclidean plane! For a simple introduction, see

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

 

[Nugatory]

it is poosible to choose base metric is euclidean just $\vec{e_t'}=\sqrt{-1}*\vec{e_t}$ and $\vec{e_x'}=\vec{e_x}$.

No. Introducing that factor of $i$ allows many equations to take on the same form as they do for Euclidean space, but the similarity is superficial and just in the mathematical formalism. The space is still non-Euclidean in ways that cannot be avoided - for example, a straight line is not, in general, the shortest distance between two points.

There's some history here. My old copy of Goldstein and many other textbooks of that vintage used the $ict$ formalism. However, by the mid-1970's MTW had a short section stating that our old friend $ict$ was to be "put to the sword", something that had to be unlearned.

 

[Ibix]

it is poosible to choose base metric is euclidean just $\vec{e_t'}=\sqrt{-1}*\vec{e_t}$ and $\vec{e_x'}=\vec{e_x}$. Then $\vec{v_1}\cdot\vec{v_2}=|\vec{v_1}|*|\vec{v_2}|*cos(\alpha)$

then:
$\frac{e_t'}{i}\cdot e_x'=|e_t'/i|*|e_x'|*cos(\alpha_{t'-x'})$
$e_t\cdot e_x=|e_t'/i|*|e_x|*cos(\alpha_{t-x})$
$0=|e_t'/i|*|e_x|*cos(\alpha_{t-x})$
$\alpha_{t-x}=arccos(0)$
$\alpha_{t-x}=\frac{2\pi}{4}$

As @Nugatory says, not really. It's a mistake to try in my opinion. One of the key points for clear thinking is to make assumptions and differences between concepts as clear as possible. The $ict$ approach has always struck me as attempting to bury something quite subtle, and it comes back to bite when you move on to GR.

Honestly, I think the best approach is to explain that the inner product of two vectors, usually written $\vec v^T\vec v$ (in matrix notation, assuming $\vec v$ is a column vector), is more generally written as $\vec v^T\mathbf g\vec v$, where $\mathbf g$ is the metric tensor, which can be written as a square matrix. To get Euclidean geometry in Cartesian coordinates you set $\mathbf g$ to the identity matrix (which is why you don't normally see it written). To get Minkowski geometry in Einstein coordinates you make one of the diagonal elements negative (all of special relativity and our notions of cause and effect spring from that one change!). To get general relativity you let it get more complicated still.

This way you can see that the rule for taking inner products is the same in Minkowski and Euclidean spaces - but the spaces are fundamentally different in ways encoded in the metric tensor. And you are laying the groundwork for general relativity at the same time. My 2p, anyway.

 

[vanhees71]

To make the concepts even clearer, rather call $\mathbf{g}$ the components of the fundamental form, since in general it's not a metric, because it's not positive definite (but nondegenerate). Sometimes one calls it a pseudometric in this case. It's simply a non-degenerate bilinear form on a vector space.

 

[olgerm]

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:
$\alpha_{t-t'}=arccos(\frac{\vec{e_t}\cdot \vec{e_t'}}{|\vec{e_t}|*|\vec{e_t'}|})= arccos(\frac{\vec{e_t}\cdot (\gamma*\vec{e_t}+\beta*\gamma*\vec{e_x})}{(-1)*(-1)})=arccos(\frac{1}{\sqrt{1-v^2/c^2}})=i*sgn(v)*arctanh(v/c)$
$\alpha_{t-x}=arccos(-\vec{e_x}\cdot\vec{e_t})=\frac{2\pi}{4}$
$\alpha_{t-x'}=arccos(-\vec{e_x'}\cdot\vec{e_t})=arccos(-(\gamma∗\beta∗\vec{e_t}+\gamma∗\vec{e_x})* \vec{e_t})=arccos(-\frac{v}{\sqrt{c^2−v^2}})$
$\alpha_{t'-x}=arccos(-\vec{e_x}\cdot\vec{e_t'})=arccos(-\vec{e_x}\cdot(\gamma∗\vec{e_t}+\beta∗\gamma∗\vec{e_x}))=arccos(-\frac{v}{\sqrt{c^2−v^2}})$
$\alpha_{t'-x'}arccos(-\vec{e_x'}\cdot\vec{e_t'})=\frac{2\pi}{4}$
$\alpha_{x-x'}=arccos(\frac{\vec{e_x}\cdot \vec{e_x'}}{|\vec{e_x}|*|\vec{e_x'}|})= arccos(\frac{\vec{e_x}\cdot (\beta*\gamma*\vec{e_t}+\gamma*\vec{e_x})}{(1)*(1)})=arccos(\frac{1}{\sqrt{1-v^2/c^2}})=i*sgn(v)*arctanh(v/c)$

Can you confirm these are correct?

 

[PeterDonis]

Can you confirm these are correct?

The "Euclidean papersheet" values look fine, but have no physical meaning.

The "really in spacetime" values are all nonsense except for the first one, if the first one is interpreted as the rapidity. Besides the rapidity $\alpha$, which you appear to be calling $\alpha_{t-t'}$, the only other meaningful quantities expressible in terms of trig functions (hyperbolic or otherwise) are $\gamma = \cosh (v / c)$ and $\gamma v = \sinh (v / c)$; but neither of those appear on your list.

 

[olgerm]

I had to edit $\alpha_{t-t'}$ and $\alpha_{x-x'}$.

 

[PeterDonis]

I had to edit $\alpha_{t-t'}$ and $\alpha_{x-x'}$.

That doesn't change anything I said in post #65.

At this point you have been given the correct information multiple times. There is no point in simply continuing to repeat the same corrective information if you aren't going to accept it.

Thread closed.