Minus sign in Minkovsky´s metric

[DaTario]

Hi,

Is there a simple explanation for the presence of a minus sign in the Minkovsky´s metric?

Best wishes,

DaTario

 

[trurle]

Minus sign allows light to propagate in vacuum without any propagation medium, and make the photon stable.
Because all points on the surface of light cone have zero interval between them.

This make a good agreement with the experimental data, and allows to phase out the troublesome "aether" concept.

 

[Nugatory]

Is there a simple explanation for the presence of a minus sign in the Minkovsky´s metric?

As @trurle says, it makes for a good agreement with the experimental data.

Physicists choose to use and teach math that accurately describes the universe that we live in, so when we find that the Minkowski metric describes the way that spacetime behaves in our experiments and the Euclidean metric does not... that's what we use.

 

[DaTario]

Ok, reasonable, but is there a clear reason why time enters with opposite sign wrt spatial coordinates?

 

[Dale]

If it didn’t then it would be another spatial dimension. The sign is what distinguishes time from space.

 

[Mister T]

Ok, reasonable, but is there a clear reason why time enters with opposite sign wrt spatial coordinates?

The apparent existence of an invariant speed.

 

[DaTario]

Thank you all, firstly.

If it didn’t then it would be another spatial dimension. The sign is what distinguishes time from space.

Dear Dale, shoudn´t this notion you presented above, for its great simplicity, be used to derive a crisp clear distinction between time and space in introductory physics books on relativity? I am not using irony here with this comment, it so happens that the distinction you made (the way you put it in words) made it almost binary.

It seems that there must be a more meaningful way to explain either the distinction between time and space or the minus sign in $ds^2$.

 

[Nugatory]

It seems that there must be a more meaningful way to explain either the distinction between time and space or the minus sign in $ds^2$.

Loosely speaking: you can move back and forth in the three spatial dimensions but not the one temporal dimension. That's one physical difference between space and time. Another way of stating this physical difference is that we measure intervals in space with rulers and intervals in time with clocks; experience and observation tells us that we need three ruler-measured intervals and one clock-measured interval to completely and uniquely specify an event. There are more precise ways of stating this important difference, but this should be sufficient to show that one dimension is different than the other three.

Thus it's to be expected that whatever form the metric takes, it will incorporate time differently than space. We choose to use specifically the Minkowski metric because it leads to a particularly convenient and elegant mathematical description of the actual physics (including the observational fact that the speed of light is the same in all inertial frames). But there's not a lot of "why?" going one here - we were looking for a metric that worked and we stopped looking when we found one.

 

[DaTario]

Loosely speaking: you can move back and forth in the three spatial dimensions but not the one temporal dimension. That's one physical difference between space and time. Another way of stating this physical difference is that we measure intervals in space with rulers and intervals in time with clocks; experience and observation tells us that we need three ruler-measured intervals and one clock-measured interval to completely and uniquely specify an event. There are more precise ways of stating this important difference, but this should be sufficient to show that one dimension is different than the other three.

Ok, these are standard ways of explaining the difference.

Thus it's to be expected that whatever form the metric takes, it will incorporate time differently than space. We choose to use specifically the Minkowski metric because it leads to a particularly convenient and elegant mathematical description of the actual physics (including the observational fact that the speed of light is the same in all inertial frames). But there's not a lot of "why?" going one here - we were looking for a metric that worked and we stopped looking when we found one.

Ok with the historical comment. My point may be put in asking why the difference in sign suffices to provide a correct distinction between these coordinates (space and time).

A coupled question: Does Minkovsky´s metric imply Lorentz Transformations?
I ask this for Lorentz transformation is where, IMHO, the constancy of c is more clearly represented and manifested.

 

[trurle]

A coupled question: Does Minkovsky´s metric imply Lorentz Transformations?
I ask this for Lorentz transformation is where, IMHO, the constancy of c is more clearly represented and manifested.

Lorentz transformation can be derived from Minkowsky`s metric.
If you rotate a section of line in Minkowsky`s space using hyperbolic sine and cosine (sinh and cosh) instead of trigonometric (sin and cos) in Euclidean space, the true length (interval between endpoints of line) will be preserved while projections of line section to x and t axis will produce same length reduction and time dilation as in Lorentz transformation.

 

[Nugatory]

My point may be put in asking why the difference in sign suffices to provide a correct distinction between these coordinates (space and time).

There may not be any really satisfactory answer to that question. There are many possible mathematical formulations, this one works for this problem. You might as well ask why multiplication is the operation that we use to calculate a distance when given a speed and a time, but addition is the operation we use to calculate the number of apples in two baskets when given the number of apples in each basket.

A coupled question: Does Minkovsky´s metric imply Lorentz Transformations?

Yes (and I'm tempted to say "yes, of course" - otherwise we wouldn't be using it, we'd use something else that did). The Lorentz transformations are a hyperbolic rotation of the coordinate axes in a Minkowski diagram.

 

[Dale]

Dear Dale, shoudn´t this notion you presented above, for its great simplicity, be used to derive a crisp clear distinction between time and space in introductory physics books on relativity?

I think so, but I am not an author of such a book

All of the differences between space and time are encapsulated in the fact that the signs are different and that there are three dimensions of space and only one of time.

Loosely speaking: you can move back and forth in the three spatial dimensions but not the one temporal dimension. That's one physical difference between space and time.

I would say that is mostly about the number of dimensions. If you had two dimensions of time then you could have closed timelike curves in flat spacetime.

 

[GrayGhost]

DaTario,

I'll refer you to a former post here which may help ... https://www.physicsforums.com/threads/distance-between-two-points-in-spacetime.491216/#post-3254542

Best Regards,
GrayGhost

 

[vanhees71]

Well, here's my attempt to introcuce special relativity right away with the (in my opinion everything very much simplifying) math of Minkowski spacetime:

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

Maybe it's for some use to answer the question about the "minus sign" (or in my case the three minus signs ;-)) in the Minkowski fundamental form.

 

[bobob]

It seems that there must be a more meaningful way to explain either the distinction between time and space or the minus sign in $ds^2$.

Sure. There is such a way. If that minus sign were a plus sign, then it would be like every other spatial dimension and in particular, you would be able to turn around in time just like you turn around in space. The fact that you have that minus sign is what describes a geometry different from that.

 

[Dale]

If that minus sign were a plus sign, then it would be like every other spatial dimension and in particular, you would be able to turn around in time just like you turn around in space.

That is more due to the fact that there is only one dimension of time rather than two or more instead of the minus sign. If you had two dimensions of time, both with the minus sign, then you could turn around in time too.

 

[m4r35n357]

I am winging it a bit here, but only one dimension is needed to reverse direction. I would say the "turning around" bit would need an extra dimension to cater for changing orientation, but I think most people would be happy to count reversing in time as time travel ;).

 

[Dale]

I am winging it a bit here, but only one dimension is needed to reverse direction.

This is not correct. With one dimension of time (minus sign in the metric) the surfaces of equal proper time form a hyperboloid of two sheets. There is no way to smoothly transform from one sheet to the other, i.e. there is no way to reverse time forward and backwards are fundamentally distinct since they reside on different sheets. With two dimensions of time the surfaces of equal proper time form a hyperboloid of one sheet. This can then be smoothly transformed across all directions in time.

 

[robphy]

This is taken from my recent answer for https://physics.stackexchange.com/questions/449583/invariance-of-the-relativistic-interval

Why is the invariant of the form $S^2=\Delta t^2-\Delta x^2$?

- A good motivation is a radar measurement of an event $P=(t_P,x_P)$ not on your worldline.

Suppose you are an inertial observer.
To measure event $P$ ,
imagine sending a light signal to $P$
and waiting for its echo, and
noting the times on your wristwatch
when you sent it $t_{send}$
and when you receive it $t_{rec}$.

From those two times, you would assign
event $P$ the following coordinates:
- time coordinate $t_P=\frac{1}{2}(t_{rec}+t_{send})$ [the midway time during the round trip]
- spatial coordinate $x_P=\frac{1}{2}(t_{rec}-t_{send})$ [half of the roundtrip time (multiplied by c)]

Note that $t_{rec}=(t_P+x_P)$ and $t_{send}=(t_P-x_P)$.

Consider another inertial observer who met you when your wristwatch read zero and they set their wristwatch to zero.
They would make analogous measurements of event $P$ .
Thus, note that
$t'_{rec}=(t'_P+x'_P)$ and $t'_{send}=(t'_P-x'_P)$.

Taking an image from Bondi's "Relativity and Common Sense"

It turns out for events joined by a future-directed light-signal
that $t'_{send}=K t_{send}$ (where $K$ is a proportionality constant
[which depends on the relative velocities of the observers]) and that
$t_{rec}=K t'_{rec}$ (the same proportionality constant).
(Each is a Doppler effect.
This pair of Doppler effects is actually the Lorentz Boost transformation... in radar coordinates, which are related to the eigenbasis of the Lorentz boost transformation.)

So, it turns out that while $t_P\neq t'_P$ and $x_P\neq x'_P$ ,
it turns out
that $$ t_{rec}t_{send}=t'_{rec}t'_{send}.$$
(This is the product of times formula [Robb, Geroch],
which is the invariance of the square-interval in Radar coordinates.)
[This encodes the hyperbola, the “circle” of Minkowski Spacetime geometry. This also suggests the “area of a causal diamond”-interpretation of the square-interval that I use in my Relativity on Rotated Graph Paper approach. See my PF Insight for details.]

Expressing this back in terms of the $t_P$s and $x_P$s,
this says that $$({t_P}^2-{x_P}^2)=({t'_P}^2-{x'_P}^2).$$
(This is the invariance of the square-interval in standard form.)

 

[m4r35n357]

This is not correct.

It's a fair cop, don't know what i was thinking there ;)

 

[Dale]

It's a fair cop, don't know what i was thinking there ;)

No worries. Because time has both the opposite sign and a single “entry” in the signature it can be difficult to discern what is due to the sign and what is due to the number.

 

[DaTario]

Sure. There is such a way. If that minus sign were a plus sign, then it would be like every other spatial dimension and in particular, you would be able to turn around in time just like you turn around in space. The fact that you have that minus sign is what describes a geometry different from that.

Ok, I am trying to grasp this idea. But what sounds strange to me is how a minus sign can be connected to the irreversibility of time.

 

[Ibix]

Ok, I am trying to grasp this idea. But what sounds strange to me is how a minus sign can be connected to the irreversibility of time.

It isn't, directly. Loosely, it defines a geometry that requires you to exceed a particular finite velocity (more formally, to follow a path that changes from timelike to spacelike) to be able to "turn around in time" and also makes it impossible to reach that velocity.

 

[PeterDonis]

how a minus sign can be connected to the irreversibility of time.

It's not the minus sign by itself that makes time "irreversible". It's the fact that there is only one dimension of time, i.e., only one dimension with the minus sign. @Dale explained why in post #18.

 

[Ibix]

It's the combination of the two - the minus sign and there only being one of it, isn't it?

The minus sign means that you can't change a timelike vector into a spacelike vector by a smooth transformation. The "only one minus" means you can't draw a future-directed timelike path that connects to a past-directed one without having a spacelike segment. The combination means you cannot follow a path that changes from future-directed to past-directed.

If you have no minus signs then there is no "different" dimension - you just have 4d Euclidean space. It has no notion of "time". If you have more than one minus sign (label one of them t and the other T) then you can rotate a vector pointing in the +t direction in the t-T plane until it points in the -t direction. Neither corresponds to how our universe behaves.

 

[vanhees71]

The point is that with Minkowski space (special relativity) with the signature (1,3) (west-coast convention, which I adopt here) or equivalently (3,1) (east-coast convention) of its fundamental form ("pseduo-metric") you can define a causality structure.

The reason is that the corresponding symmetry group of this space-time model is the Poincare group (i.e., the semidirect product of the group of space-time translations and the Lorentz group). If you have a time-like vector, there's no way to change the temporal sequence of the corresponding two events by any such transformation (corresponding physically to the change from one inertial frame to another inertial frame) that is continuously connected with the identity transformation. The corresponding subgroup is the proper orthochronous Lorentz group, and that's the symmetry group that to the best of our knowledge is realized in nature (as far as you neglect gravity). In other words for any observer the time order of any two time-like separated events is the same and thus these events can be causally connected (of course they don't need to be, but they can). This is in accordance with the strict "speedlimit", $c$ (the speed of light in vacuo), in special relativity. You can only send a signal of any kind between two events, if they are time-like (or space-like) separated. Concerning causality that means that you can only influence an event by a signal sent from any event that it time-like or light-like separated from it, i.e., the influence can be only with a speed that is lower or equal the speed of light.

 

[kent davidge]

My answer to why there's a minus sign is much more simpler, though this holds only for light and thus it's not a general explanation. It's because for light $ds^2 = 0 = c^2dt^2 - dx^2$ or $dx/ dt = c$. If there were a positive sign it would be $dx / dt = -c$.

I think the best answer is: Because it's what you get when you make an appropriate Lorentz transformation.

 

[PeterDonis]

The minus sign means that you can't change a timelike vector into a spacelike vector by a smooth transformation.

Yes, you can. It just won't be a Lorentz transformation.

However, a Lorentz transformation is much more restrictive than just not changing the sign of the vector (more precisely, of its squared length). It doesn't change the length of the vector either. So you can't even change a spacelike vector into another spacelike vector (or timelike to timelike) with a Lorentz transformation unless they both have the same length. It isn't just a matter of sign.

I would say that the key thing the minus sign does is to make null vectors possible--i.e., vectors of zero length. Those vectors then form a natural "boundary" between timelike and spacelike vectors, and also give a natural way to model a finite invariant speed like the speed of light (since a Lorentz transformation does not rotate null vectors, it dilates them). A positive definite metric can't do that.

The "only one minus" means you can't draw a future-directed timelike path that connects to a past-directed one without having a spacelike segment.

Yes, this is the point @Dale was making about a hyperboloid of two sheets. (Note that his explanation implicitly assumes that a Lorentz transformation is being used, since the hyperboloid is a set of vectors of equal length.)

 

[stevendaryl]

I hope this isn't too far off-topic, but does anyone know of an application of Riemannian geometry that uses multiple timelike dimensions? That is, that uses a metric which when diagonalized has multiple +1 entries and multiple -1 entries? For that matter, are metrics with both positive and negative entries when in diagonal form used anywhere outside of special/general relativity?

 

[Dale]

for light $ds^2 = 0 = c^2dt^2 - dx^2$ or $dx/ dt = c$. If there were a positive sign it would be $dx / dt = -c$.

Just a small correction. $dx / dt = -c$ is a perfectly valid solution for $ds^2 = 0 = c^2dt^2 - dx^2$ (which simplifies to $(dx/dt)^2=c^2$ for this purpose), it is one of the two roots the other being $dx/ dt = c$. If the sign were positive then you would get $(dx/dt)^2=-c^2$ which has no real roots.

 

[martinbn]

I hope this isn't too far off-topic, but does anyone know of an application of Riemannian geometry that uses multiple timelike dimensions? That is, that uses a metric which when diagonalized has multiple +1 entries and multiple -1 entries? For that matter, are metrics with both positive and negative entries when in diagonal form used anywhere outside of special/general relativity?

I don't know, but my uneducated guess is that it is very unlikely. The reason I think so is that it would be a lot harder. In the Riemannian (definite metric) case geodesics have minimizing properties that are used in many geometric applications, and the PDE's that typically appear are elliptic or parabolic. In the Lorentzian (one sign different from the others), in some cases geodesics have maximizing properties, which is used in a similar manner as the minimizing in the Riemann case, for example the singularity theorems. The PDE's that appear are usually hyperbolic. In the case of more than one minus and more than one plus sign in the signature the geodesics will have no optimizing properties and the PDE's will be a lot harder to analyze.

About the very last question, Lorentzian geometry is used in the analysis of hyperbolic equations, whether they are GR related or not.

In the same line of questioning it would be interesting to know if there are any applications (to physics) where the metric is more general and not quadratic as in Finsler geometry.

 

[bobob]

That is more due to the fact that there is only one dimension of time rather than two or more instead of the minus sign. If you had two dimensions of time, both with the minus sign, then you could turn around in time too.

I question the validity of that statement. In particular, if you have two time components (and two or more space components), you have no spacelike or timelike hypersurfaces upon which to specify initial data and the spacetime would not make physical sense. (At this point I still see nothing physical in Itzhak Bars' two time theory, so I don't see that as a counterexample). Three time dimensions and one space dimensions describes a spacetime containing only tachyons, which is unphysical.

 

[Dale]

the spacetime would not make physical sense

Agreed, but that doesn’t invalidate anything I said. Purely geometrically two timelike dimensions easily leads to closed timelike curves in flat spacetime.

 

[stevendaryl]

Three time dimensions and one space dimensions describes a spacetime containing only tachyons, which is unphysical.

Actually, no. There is no physical difference between (1) 3 time axes and 1 space axis and (2) 3 space axes and 1 time axis.

 

[PeterDonis]

There is no physical difference between (1) 3 time axes and 1 space axis and (2) 3 space axes and 1 time axis.

I would not put it this way. I would say that there is no physical difference between a metric with a (3, 1) signature and a metric with a (1, 3) signature. That's because either way you can physically interpret the "1" dimension in the signature as the "time" dimension, which means, heuristically, that arc lengths along that dimension are measured with clocks instead of rulers, while arc lengths along the other 3 dimensions are measured with rulers. So there's no physical difference between the two, just a different choice of signature.

But saying that you have 3 time axes and 1 space axis is very different physically: it's saying that you have three orthogonal "directions" along which you measure arc lengths with clocks, and only one along which you measure arc lengths with rulers. That's a physical difference--and again, you could (if experiments supported it) adopt such an interpretation for either a (3, 1) or a (1, 3) signature.