Understanding Pseudo-Angles in Minkowski Space: A Geometric Interpretation

[Will Flannery]

TL;DR Summary

Does the concept of an angle between two vectors make sense in Minkowski space?

Does the concept of the angle between two vectors make sense in Minkowski space?

Does the concept of orthogonal basis for Minkowski space make sense? If it does, how is it defined?

When we start with the usual (time, distance) basis for 2-D Minkowski space, the axes as drawn make a right angle. However, when we add to the graph the basis vectors for frame moving with velocity v, these vectors don't form a right angle, but they must be 'orthogonal' as they are the (time, distance) basis for the moving frame - so ...

 

[Ibix]

Does the concept of the angle between two vectors make sense in Minkowski space?

Kind of. The equivalent of angle is usually called "rapidity" for timelike vectors. I'm not sure if there's an equivalent term for spacelike vectors that don't lie in a purely spacelike plane. The inner product of two unit timelike vectors is $\cosh\psi$, where $\psi$ is the rapidity, rather than $\cos\theta$, where $\theta$ is the angle.

these vectors don't form a right angle, but they must be 'orthogonal' as they are the (time, distance) basis for the moving frame - so ...

A Minkowski diagram is a Euclidean representation of a Minkowski space. It's not perfect. As you note, angles cannot be read easily, and neither do intervals ($\sqrt{\Delta t^2-\Delta x^2}$) correspond to distances ($\sqrt{\Delta t^2+\Delta x^2}$) on the diagram except in certain special cases. A Minkowski plane can't be represented perfectly on a Euclidean plane, or it would be a Euclidean plane.

 

[robphy]

Summary:: Does the concept of an angle between two vectors make sense in Minkowski space?

Does the concept of the angle between two vectors make sense in Minkowski space?

For two future-timelike vectors, the natural notion of angle in Minkowski spacetime is called the "rapidity" $\psi$, where $v/c=\tanh\psi$ and $\gamma =\cosh\psi$ ( https://en.wikipedia.org/wiki/Rapidity )

Does the concept of orthogonal basis for Minkowski space make sense? If it does, how is it defined?

Yes, using the dot product with Minkowski metric:
$\vec a \cdot \vec b \equiv a_t b_t - a_x b_x - a_y b_y -a_z b_z$,
two vectors orthogonal when their dot-product is zero.

When we start with the usual (time, distance) basis for 2-D Minkowski space, the axes as drawn make a right angle. However, when we add to the graph the basis vectors for frame moving with velocity v, these vectors don't form a right angle, but they must be 'orthogonal' as they are the (time, distance) basis for the moving frame - so ...

You can't use your Euclidean notion of "90 degees" to define perpendicularity (orthogonality) in Minkowski spacetime. Instead, you must use the dot product (algebraically) or use the geometric notion that the tangent to the Minkowski-circle (the hyperbola) is orthogonal to the radius (as Minkowski defined).

We decompose any vector, such as that from O to x, y, z, t into four
components x, y, z, t. If the directions of two vectors are, respectively, that
of a radius vector OR from O to one of the surfaces $\mp F=1$, and that of
a tangent RS at the point R on the same surface, the vectors are called
normal to each other. Accordingly,
$c^2tt_1 − xx_1 − yy_1 − zz_1 = 0$
is the condition for the vectors with components x, y, z, t and $x_1$, $y_1$, $z_1$, and $t_1$ to be normal to each other.

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[robphy]

Curiously, I have not been able to find a general definition of an angle between two arbitrary vectors in Minkowski spacetime...
just between
two future-timelike (or two past-timelike) [rapidity]
or two "spacelike vectors that are coplanar with two timelike vectors" [spacelike-rapidity]
or two "spacelike vectors that are othogonal to the same timelike vector" [spatial angle].

 

[Ibix]

Curiously, I have not been able to find a general definition of an angle between two arbitrary vectors in Minkowski spacetime...
just between
two future-timelike (or two past-timelike) [rapidity]
or two "spacelike vectors that are coplanar with two timelike vectors" [spacelike-rapidity]
or two "spacelike vectors that are othogonal to the same timelike vector" [spatial angle].

Doesn't that cover all options, except a future-directed timelike vector with a past-directed timelike vector?

 

[robphy]

Doesn't that cover all options, except a future-directed timelike vector with a past-directed timelike vector?

How would the "angle" be defined between a timelike and spacelike vector ?
For orthogonal vectors, one might define it as infinity.
But suppose they aren't orthogonal?
Presumably, with the definition given, the resulting angles could be used in the boosts and rotations to map one direction into the other.

I haven't mentioned null vectors so far.

 

[Ibix]

How would the "angle" be defined between a timelike and spacelike vector ?
For orthogonal vectors, one might define it as infinity.
But suppose they aren't orthogonal?
Presumably, with the definition given, the resulting angles could be used in the boosts and rotations to map one direction into the other.

I haven't mentioned null vectors so far.

Of course - I got hung up thinking about the possible combinations of spacelike and spacelike and then forgot the rest. I think the event of my bedtime must be in my past lightcone...

 

[PeterDonis]

How would the "angle" be defined between a timelike and spacelike vector ?

The generalization of "angle" for vectors in a vector space is "inner product". More precisely, the inner product of two vectors will be a trig function (either normal or hyperbolic) of the "angle in spacetime" between them (possibly with a minus sign). The inner product $a \cdot b$ is simply $\eta_{\mu \nu} a^\mu b^\nu$.

 

[robphy]

The generalization of "angle" for vectors in a vector space is "inner product". More precisely, the inner product of two vectors will be a trig function (either normal or hyperbolic) of the "angle in spacetime" between them (possibly with a minus sign). The inner product $a \cdot b$ is simply $\eta_{\mu \nu} a^\mu b^\nu$.

Since the dot product of those vectors may be negative,
such an angle would be complex-valued. It would be peculiar (but interesting if necessary) to use a past-timelike vector to get a positive dot product.
I went down that route a while back, but couldn't see an easy definition.
(I was also trying to define the complement and supplement of an angle... but got nowhere.)
I would hope to see properties of the "additivity of angles" and some geometrical interpretation,
not merely an inverse-cosine-like-function of the dot-product and not a definition depending on various cases.
When I last looked at it, I think the most promising plan of attack may be to use a generalization of the Gauss-Bonnet formula.

 

[pervect]

Summary:: Does the concept of an angle between two vectors make sense in Minkowski space?

Does the concept of the angle between two vectors make sense in Minkowski space?

Does the concept of orthogonal basis for Minkowski space make sense? If it does, how is it defined?

When we start with the usual (time, distance) basis for 2-D Minkowski space, the axes as drawn make a right angle. However, when we add to the graph the basis vectors for frame moving with velocity v, these vectors don't form a right angle, but they must be 'orthogonal' as they are the (time, distance) basis for the moving frame - so ...

Certainly the dot product of two vectors makes sense in Minkowskii space. And the dot product being zero defines orthogonality, so the concept of orthogonality in Minkowskii space makes sense.

As far as angles go - my take on this is that in Euclidean space, the angle between two orthogonal spatial vectors adds up to 90 degrees, or more formally, pi/2 radians.

However, the equivalent of angle between two time-like vectors, rapidity, is unbounded. Rapidities are known as hyperbolic angles, and, like angles, they add. There's a brief discussion of them in wiki.

Two time-like vectors can never be orthogonal. Their dot product will never be zero. I believe it's always greater than or equal to 1, for future directed vectors, though I'd have to check some references to make sure I'm not making a bonehead mistake.

The additive property of rapidies is why they are called hyperbolic angles. Velocity, the more familiar relationship between two time-like vectors, is related to the concept of angle. However, velocities don't add (and rapdities do), so rapidity is more closely related to the concept of angle than velocity is.

There is a relationship between velocity and rapidity, the relationship being that rapidity is the hyperbolic tangent of v/c. There's some discussion of this in wiki if you look up "rapidity". As you can see from the math, the rapidites add like angles do, but the behavior is different because an angular rotation of 2 pi radians is a full circle. There is no analogous concept in the hyperbolic geometry of a hyperbolic "rotation" being a full circle, as the geometry is hyperbolic, not circular.

 

[PeterDonis]

the dot product of those vectors may be negative

Yes, but if one of the vectors is timelike or null, you can always decompose that into a positive dot product (with an ordinary hyperbolic "angle", i.e., rapidity) and a sign change due to changing from future to past light cone or vice versa. And if both vectors are spacelike, you are using the ordinary cosine, so negative dot products just mean angles greater than 90 degrees.

 

[vanhees71]

Curiously, I have not been able to find a general definition of an angle between two arbitrary vectors in Minkowski spacetime...
just between
two future-timelike (or two past-timelike) [rapidity]
or two "spacelike vectors that are coplanar with two timelike vectors" [spacelike-rapidity]
or two "spacelike vectors that are othogonal to the same timelike vector" [spatial angle].

That's because there is no use of the notion of "angle" in Minkowski space. Also the Minkowski diagrams are NOT to be misunderstood as a Euclidean plane. Minkowski diagrams are hard to read, because you have to unlearn all Euclidean ideas we are used and trained with from elementary school on!

The rapidity is not an angle but an area. That's why the inverse functions of cosh and sinh, arcosh and arsinh are called area functions to begin with!

 

[Will Flannery]

For two future-timelike vectors, the natural notion of angle in Minkowski spacetime is called the "rapidity" $\psi$, where $v/c=\tanh\psi$ and $\gamma =\cosh\psi$ ( https://en.wikipedia.org/wiki/Rapidity )
Yes, using the dot product with Minkowski metric:
$\vec a \cdot \vec b \equiv a_t b_t - a_x b_x - a_y b_y -a_z b_z$,
two vectors orthogonal when their dot-product is zero.
You can't use your Euclidean notion of "90 degees" to define perpendicularity (orthogonality) in Minkowski spacetime. Instead, you must use the dot product (algebraically) or use the geometric notion that the tangent to the Minkowski-circle (the hyperbola) is orthogonal to the radius (as Minkowski defined).
Check out https://www.desmos.com/calculator/r4eij6f9vw
View attachment 265967

For Euclidean, move the E-slider to -1
View attachment 265968

Minkowski discusses this on page 46 in https://www.minkowskiinstitute.org/mip/MinkowskiFreemiumMIP2012.pdf
So, he's making an analogy between a radius vector for a circle in Euclidean space and the tangent to the circle, and a radius vector to the surface c2tt - xx - yy - zz = 1 in Minkowski space and the tangent to the surface, and saying that the tangent is orthogonal to the radius, and he writes ..

"We decompose any vector, such as that from O to x; y; z; t into four
components x; y; z; t. If the directions of two vectors are, respectively, that
of a radius vector OR from O to one of the surfaces F = 1, and that of
a tangent RS at the point R on the same surface, the vectors are called
normal to each other. Accordingly,
c2tt1 􀀀 xx1 􀀀 yy1 􀀀 zz1 = 0"

It's the 'accordingly' in the above that I don't get ... he seems to think it is obvious ?

 

[romsofia]

You can generalize the geometric formula of the dot product to one that is for infinitesimal vectors and brings in the metric in the following way:
$$ \cos \theta = \frac{g_{\alpha \beta}d^{(1)}x^{\alpha}d^{(2)}x^{\beta}}{\sqrt{g_{\nu \rho} d^{(1)}d^{\nu}d^{(1)}d^{\rho}}\sqrt{g_{\gamma \lambda}d^{(2)}x^{\gamma}d^{(2)}x^{\lambda}}} $$

where $d^{(1)}x^{\alpha},d^{(2)}x^{\alpha}$ are your infinitesimal vectors where $\theta$ is the angle between your two vectors. Now, this is just math, it is a way to "brute force" the notation of angles as it makes use of your metric. Trying to make sense of your result in a physical notion may be trickier, and I haven't really explored that to make much comment on it!

 

[vanhees71]

Well, in my opinion this doesn't make sense for an indefinite fundamental form. Angles are well defined only in Euclidean spaces, where the fundamental form is a true positive definite scalar product.

 

[Will Flannery]

While pondering the 'accordingly' in Minkowski's paper, I noted that he referred to the spacetime manifold, rather than vector space. But in any case, at each point in a manifold (I think) there is an associated vector space where the tangents to curves in the manifold live. And for Euclidean space the tangent vector space at a point is Euclidean. What about the vector space associated with a point in a spacetime manifold, is it a 4-D Euclidean space?

 

[PeterDonis]

What about the vector space associated with a point in a spacetime manifold, is it a 4-D Euclidean space?

No, it's 4-D Minkowski space. In other words, in special relativity, the spacetime manifold itself is identical to the tangent space at each point in the manifold.

However, this is only true in special relativity, where spacetime is flat and gravity is not present. In the presence of gravity, spacetime is curved and the spacetime manifold is no longer identical to the tangent space at each point in the manifold (the latter is still Minkowski space). In fact a general curved spacetime manifold is not even a vector space.

 

[emilija]

Maybe this reference can help
https://www.researchgate.net/publication/312256782_On_geometric_interpretation_of_pseudo-angle_in_Minkowski_plane
E. Nesovic , On geometric interpretation of pseudo angle in Minkowski plane.

 

[robphy]

Maybe this reference can help
https://www.researchgate.net/publication/312256782_On_geometric_interpretation_of_pseudo-angle_in_Minkowski_plane
E. Nesovic , On geometric interpretation of pseudo angle in Minkowski plane.

Yes, this is along the lines of the Gauss-Bonnet approach I mentioned in my post above.