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The covariant equation of motion of a free particle, in flat Minkowski spacetime and Cartesian coordinates, reads $$ \frac{d^2x^\mu}{d\tau^2}=0, \tag{1} $$ with $\mu=0,1,2,3$, and has the solution $$ x^\mu=\alpha^\mu\tau+\beta^\mu \tag{2} $$ where $\alpha^\mu,\beta^\mu$ are (four-vectorial) integration constants. As far as the spatial coordinates, ($x^1=x,x^2=y,x^3=z$) are concerned, (1) represents a straight line trajectory in $3$-dimensional physical space. This is because the spatial coordinates satisfy (after eliminating the proper time, $\tau$, from (2)): $$ \frac{x-\beta^1}{\alpha^1}=\frac{y-\beta^2}{\alpha^2}=\frac{z-\beta^3}{\alpha^3}=\text{constant}. $$ My real question is, however, what is the nature of the curve (2) in spacetime and how to draw it? For this purpose, let us assume spacetime is $(2+1)$-dimensional (instead of $(3+1)$-dimensional, just for the simplicity of drawing). Now, if the $ct$ is plotted along $z$-axis, and the spatial coordinates, $x^1,x^2$ are plotted along $x$ and $y$ axes, the worldline will still appear as a straight line in spacetime because: $$ \frac{t-\beta^0}{\alpha^0}=\frac{x-\beta^1}{\alpha^1}=\frac{y-\beta^2}{\alpha^2}=\text{constant}. $$ Is it fair to say that the spacetime trajectory (i.e., the worldline) described by (2) is also a straight line? If not, what is that type of worldline referred to as?

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    $\begingroup$ Yes a straight line. Why the doubt? $\endgroup$
    – peek-a-boo
    Commented Apr 13 at 7:59
  • $\begingroup$ ...because spacetime is not Euclidean. A straight line is a strictly Euclidean concept. Isn't it? And spacetime diagram, with $ct$ and $x,y$ drawn "orthogonal" is just a convenient picturization. $\endgroup$
    – Solidification
    Commented Apr 13 at 8:02
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    $\begingroup$ “Straight” has a meaning if we decide to give it one. So unless you start with a definition of what a strsight line is, we can’t answer your question in the affirmative or negative. But anyway, here we have a real vector space equipped with the trivial connection, the (non-constant) autoparallels/geodesics of this connection coincide with the straight lines from affine/linear algebra (i.e a one-dimensional affine subspace). So in almost every sense of the word, it is justified to use the term “straight”. $\endgroup$
    – peek-a-boo
    Commented Apr 13 at 8:06
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    $\begingroup$ Your reasoning is circular. Straight lines are by definition geodesics. If you have some other definition you need to specify it clearly. Also it depends on how you embed your spacetime into $\mathbb{R}^3$. If you are using the standard embedding then geodesics in flat spacetime are indeed also straight lines in the Euclidean sense. $\endgroup$
    – Vincent Thacker
    Commented Apr 13 at 8:41
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    $\begingroup$ @Solidification the fact that you’re in Minkowski and not Euclidean space is irrelevant. I’ve given you two definitions of straightness which apply very generally (one is in the vector space setting, and the other is in the setting of a manifold with a connection. Then I said that in your special case, both notions coincide). So again, if you’re not satisfied with either of the definitions I gave, then you’re really going to have to dig deeper and tell us what notion of straight you mean (but since you said you’re unsure of the definition, I suggest you understand and accept the two I gave). $\endgroup$
    – peek-a-boo
    Commented Apr 13 at 23:16

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