Why is the flat spacetime of special relativity, not a real linear vector space? It seems to satisfy all the axioms for a set to form a vector space. I mean adding to spacetime points $(t_1,\vec{r}_1)$ and $(t_2,\vec{r}_2)$, is yet another spacetime point $(t_1+t_2,\vec{r}_1+\vec{r}_2)$. So I don't see a problem with closure, or vector addition being commutative or associative. The null vector is the spacetime point $(0,\vec{0})$. I also hope that it is also closed under scalar multiplication.
$\begingroup$
$\endgroup$
5
asked Feb 12, 2023 at 16:45
-
2$\begingroup$ Who says it's not? Also, what do you gain by giving it a vector space structure? How does it help? $\endgroup$– PraharCommented Feb 12, 2023 at 17:04
-
2$\begingroup$ It is a vector space since it is really just $\mathbb{R}^4$ endowed with the Lorentzian metric $\eta = \operatorname{diag}(-1,1,1,1)$, so you can of course define the operation of addition and multiplication by scalar as usual and gain a vector space structure. But being able to does not mean it is useful. What adding events would actually mean? In fact, when you view $\mathbb{R}^4$ with this Lorentzian product as a spacetime you are more interested in viewing $\mathbb{R}^4$ as a smooth manifold than as a vector space. $\endgroup$– GoldCommented Feb 12, 2023 at 17:37
-
$\begingroup$ When you view $\mathbb{R}^4$ with this Lorentzian product as a space of four-vectors as four-momenta, or tangent vectors at some point in spacetime like four-velocities, then you are in a situation in which the vector space structure plays a role. $\endgroup$– GoldCommented Feb 12, 2023 at 17:38
-
1$\begingroup$ In this context, a “null vector” is a vector with zero norm using the metric. While the zero vector has 0 norm using the minkowski metric (and the Euclidean metric), there are non-zero null vectors in special relativity (unlike in Euclidean geometry). $\endgroup$– robphyCommented Feb 12, 2023 at 17:42
-
$\begingroup$ Possible duplicates: physics.stackexchange.com/q/158946/2451 and links therein. $\endgroup$– Qmechanic ♦Commented Feb 13, 2023 at 1:39
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
13
Flat spacetime is more naturally described as an affine space, not a vector space. An affine space is basically a vector space without an origin. There is no unique natural event in flat spacetime which is naturally distinguished as "the" origin.
If we neglect curvature and take two events in spacetime say A is the supernova SN 2003fg and B is the supernova SN 2006gy then without choosing a third event and designating it as an origin, what event in spacetime is A+B? Indeed, it doesn't make sense to add A and B, any more than it makes sense to add Paris to Caracas.
In contrast, without choosing an origin we can subtract B-A to get a vector. This is what an affine space does.
-
2
-
$\begingroup$ By "vector space without an origin", do you mean "a vector space without a null vector"? But why is there no natural null vector? When we talk about flat spacetime, isn't it the collection of events w.r.t a given inertial observer? Suppose the observer monitors a clock kept at rest in the origin of his or her coordinate system. When that clock reads t=5 s, can't we take that event as the "origin"? Please correct me. $\endgroup$ Commented Feb 12, 2023 at 17:28
-
2$\begingroup$ @Solidification Yes, by origin I mean the unique vector $\vec 0$. The point is that the observer is something in addition to spacetime. Yes, you can use a given event on a given observer’s worldline as an origin, but the same is true of any other event on any other observer’s worldline. There is nothing about spacetime that distinguishes any of the infinite possible origins $\endgroup$– DaleCommented Feb 12, 2023 at 17:32
-
$\begingroup$ Thank you very much for clarifying this to me. Let me paraphrase your point. Observers merely assign coordinate labels to events but events exist independently of the observers. Is that right? If so, I have a last question. When you say spacetime as a mathematical space, does each point must have to correspond to a physical event? In other words, is spacetime a collection of all possible spacetime locations or spacetime locations of real events (supernova explosion, turning on a light bulb, an apple falling from the tree etc)? Thanks in advance. $\endgroup$ Commented Feb 12, 2023 at 17:39
-
1$\begingroup$ @Dale I fully agree with your assessment. My impression was that Solidification was using "physical event" to mean what you describe as something interesting happening, but perhaps I'm mistaken in that sense. Regardless, perhaps this back-and-forth will be helpful to clarify the issue (assuming it's not deleted first). $\endgroup$ Commented Feb 13, 2023 at 18:47