How do you draw a spacetime diagram and what equation describes it?

[runningc]

According to Minkowski, spacetime for non-accelerated systems is a four dimensional continuum which has four axes (three space, one time) set mutually at right angles to each other. The velocity of a body moving in a straight line will be the resultant of the component velocities along the three space axes. Its direction of motion will clearly not be at right angles to any of those axes. This would seem to imply that the direction of motion of the body through space cannot be at right angles to the time axis either. Any comments?
RC

 

[Andrew Mason]

Trying to represent 4 dimensions in 3 dimensional space will make your head hurt. It can't be done. That is why most space-time diagrams use only one spatial dimension + the time dimension. But using a 4 dimensional vectorspace, it is easy to express 4 mutually orthogonal basis vectors mathematically - just not geometrically. If you do that, there is no problem: any vector in the 3 dimensional physical space is perpendicular to the time axis.

AM

 

[Orodruin]

What do you mean by "angles" here? The geometry of Minkowski space is not Euclidean (nor Riemannian) and works differently. The 4-velocity is certainly not in the spatial plane (that would make it space-like and not time-like).

 

[Ibix]

Its direction of motion will clearly not be at right angles to any of those axes.

It will if the velocity is zero in this frame. Or if it has no component in the direction of a chosen axis (e.g. motion in the x-y plane is perpendicular to the z axis).

This would seem to imply that the direction of motion of the body through space cannot be at right angles to the time axis either.

I'm not sure this makes any sense. Did you mean "the direction of motion of the body through spacetime"?

 

[Andrew Mason]

This would seem to imply that the direction of motion of the body through space cannot be at right angles to the time axis either. Any comments?

I may have misinterpreted your question. If you meant to say that "the direction of the path of a body through space-time" cannot be at right angles to the time axis, you would be correct. That is because it takes time for a body to change its position in space as measured in any inertial reference frame. So the path of a body through space-time will always have a time component.

AM

 

[Dale]

According to Minkowski, spacetime for non-accelerated systems is a four dimensional continuum which has four axes (three space, one time) set mutually at right angles to each other.

Yes.

The velocity of a body moving in a straight line will be the resultant of the component velocities along the three space axes.

Not if you are using Minkowski spacetime. In spacetime the four-velocity is a four dimensional vector with (potentially) components on all four axes. At a minimum it will have a component on the time axis.

Its direction of motion will clearly not be at right angles to any of those axes.

It cannot be at right angles with the time axis, but it could be at right angles with all of the space axes. This would be an object at rest in that frame.

This would seem to imply that the direction of motion of the body through space cannot be at right angles to the time axis either.

No, this implication does not follow at all.

 

[Orodruin]

It cannot be at right angles with the time axis

I am very uncomfortable with the use of the word "right angle" here. It may be fine for the projection onto a simultaneity, where there is a Riemannian notion of angles, but certainly not in Minkowski space, where the relevant tilt from the time axis would instead be given by a hyperbolic angle. I would be fine with the nomenclature "orthogonal", i.e., "it cannot be orthogonal to the time axis" as this is standard nomenclature regardless of the signature of the metric. The word "angle" to me is intimately tied to a positive definite metric.

 

[Dale]

I would be fine with the nomenclature "orthogonal",

I agree that would be better terminology.

 

[runningc]

Yes.

It cannot be at right angles with the time axis,

How, therefore, should a spacetime diagram showing the three vectors,: velocity through space, velocity through spacetime, and velocity through time be drawn and what equation would describe it?

<Mentor's note, edited to get the quotes right>

 

[Ibix]

Note: please don't reply inside the quote tag. It makes it hard to quote you and looks like you are falsely attributing your own comments to Dale.

How, therefore, should a spacetime diagram showing the three vectors,: velocity through space, velocity through spacetime, and velocity through time be drawn?

Velocity through spacetime is a somewhat sloppy notation for the four-velocity. Once you choose that and pick a frame, any meaning of "velocity through space" and "velocity through time" is completely constrained. Usually you'd interpret "velocity through space" as the spacelike components of the four vector (give or take a factor of gamma) and "velocity through time" as the timelike component (I have seen more complicated interpretations).

So what that means is that you draw a Minkowski diagram. The four-velocity is tangent to the object's worldline, and its spatial and timelike components give you everything else you asked for.

 

[Dale]

How, therefore, should a spacetime diagram showing the three vectors,: velocity through space, velocity through spacetime, and velocity through time be drawn and what equation would describe it?

You wouldn't show three vectors, you would show one four-vector. You would draw it at less than a 45 degree angle from the time axis and with the arrow pointing towards the future.

See here for the equations/math: https://en.wikipedia.org/wiki/Four-velocity

 

[robphy]

How, therefore, should a spacetime diagram showing the three vectors,: velocity through space, velocity through spacetime, and velocity through time be drawn and what equation would describe it?

Quoting myself from a post earlier this month
https://www.physicsforums.com/threads/velocity-through-spacetime.957038/#post-6068213

Here's a spacetime diagram
View attachment 231852
In trigonometric form,
v/c\quad\ =\tanh\theta=\frac{opp}{adj}
\gamma\qquad \ =\cosh\theta =\frac{adj}{hyp}
\gamma(v/c)=\sinh\theta =\frac{opp}{hyp}

(I updated the diagram by indicating the rapidity \theta (the Minkowski-analogue of the angle).
By the way, the rapidity of the Minkowski-right-angle is infinity.
I also added the interpretation of "tangent" as "opposite over adjacent", etc...

Also note: the tangent-line to the Minkowski-circle is Minkowski-perpendicular to the radius.)