# Why is "time orthogonal to space" in inertial reference frames?

• B

## Main Question or Discussion Point

I'm reading about the geometry of spacetime in special relativity (ref. Core Principles of Special and General Relativity by Luscombe). Here's the relevant section:

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Minkowski space is a four-dimensional vector space (with points in one-to-one correspondence with those of ##\mathbb{R}^4##) spanned by one timelike basis vector, ##\vec e_t##, and three spacelike basis vectors, ##\vec e_x, \vec e_y, \vec e_z##. While any four linearly independent vectors can constitute a basis, in IRFs we require time to be orthogonal to space.
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I'm not clear what is meant by this. By "time", does the author mean the ##[1,0,0,0]## basis vector and by "space", does he mean the ##[0, 1, 0, 0], [0, 0, 1, 0], [0, 0, 0, 1]## basis vectors? If that's the case, this is no different from the Euclidean ##\mathbb{R}^4## case - I could've just said that "we require ##x## to be orthogonal to the other dimensions", where by "##x##" I mean the corresponding standard basis vector. So I'm guessing that's not what the author meant.

If by "time" and "space" the author isn't referring to corresponding standard basis vectors, then I'm not sure what's stopping me from taking an arbitrary basis of ##\mathbb{R}^4## as a basis of Minkowski Space as well. Why do we require "time" to be orthogonal to "space"? (whatever the quoted words mean in this context) What is even meant by the statement?

Is there a proper mathematical justification for this?

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Orodruin
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By "time", does the author mean the [1,0,0,0][1,0,0,0][1,0,0,0] basis vector and by "space", does he mean the [0,1,0,0],[0,0,1,0],[0,0,0,1][0,1,0,0],[0,0,1,0],[0,0,0,1][0, 1, 0, 0], [0, 0, 1, 0], [0, 0, 0, 1] basis vectors?
Yes.

If that's the case, this is no different from the Euclidean R4R4\mathbb{R}^4 case
No. The metric in Minkowski space is pseudo-Riemannian while the metric in Euclidean space is Riemannian.

If by "time" and "space" the author isn't referring to corresponding standard basis vectors, then I'm not sure what's stopping me from taking an arbitrary basis of R4R4\mathbb{R}^4 as a basis of Minkowski Space as well. Why do we require "time" to be orthogonal to "space"? (whatever the quoted words mean in this context) What is even meant by the statement?
Nothing is stopping you from choosing an arbitrary basis. It will just not be a coordinate system that is what we refer to as Minkowski coordinates. Just as you could pick any coordinate system in Euclidean space (such as spherical coordinates), they just will not be Cartesian coordinates.

• Demystifier, sysprog, etotheipi and 2 others
PeroK
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While any four linearly independent vectors can constitute a basis, in IRFs we require time to be orthogonal to space.
Note that, strictly speaking, there is no concept of "orthogonal" in a vector space. To define orthogonal you need an inner product (or a metric). The orthogonality of time and space (if we can put it like that) is really a statement about the metric you choose to impose on your vector space.

• Demystifier, sysprog, etotheipi and 1 other person
Ibix
Why do we require "time" to be orthogonal to "space"?
You can indeed use any set of basis vectors you like, and in general relativity we frequently do, but we only call ones with four orthogonal basis vectors "frames". So orthogonal basis vectors, one timelike and three spacelike, is a definition of a frame.

There's a natural reason for defining your time axis orthogonal to your spatial axes. It corresponds to using the worldlines of clocks at rest with respect to you as a timelike direction and defining "space" by Einstein-synchronising those clocks and using the set of events with equal times as "space at a given time". This gives you a homogeneous and isotropic notion of physics, which is nice and simple. Or as simple as it gets, anyway...

• • sysprog, Dale and etotheipi
Note that, strictly speaking, there is no concept of "orthogonal" in a vector space. To define orthogonal you need an inner product (or a metric). The orthogonality of time and space (if we can put it like that) is really a statement about the metric you choose to impose on your vector space.
Totally agree with you that orthogonality requires us to define inner product, which in turn requires us to specify a metric. But I don't see how "the orthogonality of time and space" (not sure if we should put it like that) is a statement about the Minkowski metric. All that the Minkowski metric specifies is that ##-c\Delta t^2+\Delta x^2+\Delta y^2+\Delta z^2## is a constant. Without additional assumptions or without adopting a certain convention, I don't see where orthogonality springs up from.

You can indeed use any set of basis vectors you like, and in general relativity we frequently do, but we only call ones with four orthogonal basis vectors "frames". So orthogonal basis vectors, one timelike and three spacelike, is a definition of a frame.

There's a natural reason for defining your time axis orthogonal to your spatial axes. It corresponds to using the worldlines of clocks at rest with respect to you as a timelike direction and defining "space" by Einstein-synchronising those clocks and using the set of events with equal times as "space at a given time". This gives you a homogeneous and isotropic notion of physics, which is nice and simple. Or as simple as it gets, anyway...
Do you mean that it's a matter of convention that we select time and space basis vectors (in Minkowski space) in such a way that the time basis vector is orthogonal to space basis vectors?

PeroK
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Totally agree with you that orthogonality requires us to define inner product, which in turn requires us to specify a metric. But I don't see how "the orthogonality of time and space" (not sure if we should put it like that) is a statement about the Minkowski metric. All that the Minkowski metric specifies is that ##-c\Delta t^2+\Delta x^2+\Delta y^2+\Delta z^2## is a constant. Without additional assumptions or without adopting a certain convention, I don't see where orthogonality springs up from.
If we take our spacetime vectors as ##(ct, x, y, z)##, then the inner product that defines the metric is:
$$(ct_1, x_1, y_1, z_1) \cdot (ct_2, x_2, y_2, z_2) = -c^2t_1t_2 + x_1x_2 + y_1y_2 + z_1z_2$$
That in turn implies that ##(ct, 0, 0, 0)## is orthogonal to ##(0, x, 0, 0)## etc.

• FactChecker and etotheipi
If we take our spacetime vectors as ##(ct, x, y, z)##, then the inner product that defines the metric is:
$$(ct_1, x_1, y_1, z_1) \cdot (ct_2, x_2, y_2, z_2) = -c^2t_1t_2 + x_1x_2 + y_1y_2 + z_1z_2$$
That in turn implies that ##(ct, 0, 0, 0)## is orthogonal to ##(0, x, 0, 0)## etc.
But would this not hold for any arbitrary diagonal metric? The reason I'm confused is that this orthogonality between time and space dimensions seems to have been presented as a feature distinguishing Minkowski metric from Euclidean. The above conclusion of ##(ct, 0, 0, 0)## being orthogonal to ##(0, x, 0, 0)## should apply in the case of Euclidean metric as well.

Maybe I'm overthinking and this "time orthogonal to space" phrase is ill-posed.

PeroK
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But would this not hold for any arbitrary diagonal metric? The reason I'm confused is that this orthogonality between time and space dimensions seems to have been presented as a feature distinguishing Minkowski metric from Euclidean. The above conclusion of ##(ct, 0, 0, 0)## being orthogonal to ##(0, x, 0, 0)## should apply in the case of Euclidean metric as well.
It would, but ##c^2dt^2 + dx^2 + dy^2 + dz^2## doesn't work out as a model of the spacetime we live in.

The difference between Minkowski space and Euclidean is the minus sign on the ##c^2dt^2##.

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PeterDonis
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Do you mean that it's a matter of convention that we select time and space basis vectors (in Minkowski space) in such a way that the time basis vector is orthogonal to space basis vectors?
Yes, it's a convention. Note that what you quoted in the OP says "in IRFs". What it's actually defining there is not what Minkowski spacetime is, but what an IRF is.

would this not hold for any arbitrary diagonal metric?
The metric being diagonal is dependent on your choice of coordinates. It's easy to find coordinates in which the metric of Minkowski spacetime is not diagonal. But those coordinates would not describe an IRF.

this orthogonality between time and space dimensions seems to have been presented as a feature distinguishing Minkowski metric from Euclidean
No, it's being presented as a feature distinguishing an IRF from other types of coordinate charts.

• • sysprog, etotheipi, Shirish and 2 others
Yes, it's a convention. Note that what you quoted in the OP says "in IRFs". What it's actually defining there is not what Minkowski spacetime is, but what an IRF is.
Thank you! But then why would "space" be orthogonal to "time" in IRFs? Per my understanding, the defining feature of IRFs is that free particles have straight worldlines, which isn't related to the metric.

I'm doubtful on how the structure of spacetime geometry (more specifically, the Minkowski metric) has anything to do with the fact that it's an IRF, OR how "space" not being orthogonal to "time" could indicate a non-inertial frame.

Is there a mathematical chain of reasoning that connects the two?

Nugatory
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Do you mean that it's a matter of convention that we select time and space basis vectors (in Minkowski space) in such a way that the time basis vector is orthogonal to space basis vectors?
Sort of. The choice of basis vectors is conventional; we can choose any four linearly independent vectors to form our basis and whether they are orthogonal or not clearly depends on our choice. So yes, it is conventional.

However, you have constrained the choice (probably without noticing) by assuming that we have a time basis vector and three space vectors. What you are calling "the time basis vector" is the tangent to the worldline of some inertially moving clock, and having made that choice it would be perverse (at least in flat spacetime) not to choose space vectors orthogonal to it and to each other. This is the point that @Ibix was making in post #4 above.

But you don't have to a time basis vector at all.... Consider the two-dimensional Minkowski space that we use for space-time diagrams; we naturally use ##x## and ##t## to define our basis and call one of the them space direction and the other the time direction. But we could define coordinates ##a=x+t## and ##b=x-t## and choose basis vectors parallel to the ##a## and ##b## axes; and now we have neither a time vector nor a space vector in our basis.

• sysprog, Ibix and Shirish
PeroK
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Thank you! But then why would "space" be orthogonal to "time" in IRFs? Per my understanding, the defining feature of IRFs is that free particles have straight worldlines, which isn't related to the metric.
Let's start there. You could also stop there and decide to proceed without ever trying to define any geometric relationship between space and time. As I understand it, Einstein was initially reluctant to accept Minkowski's idea that SR could be modelled as a 4D spacetime. Your first option is to proceed with SR without looking for any 4D geometrical basis for the theory.

Suppose, however, you want to try to encapsulate the laws of SR geometrically by considering a 4D spacetime. You must consider what you mean by the length of a spacetime interval. Using an IRF is the simplest, so you consider those first. If a particle moves from ##(t_1, x_1)## to ##(t_2, x_2)##, then how do you define the spacetime distance between those points? The thing you want is that this spacetime distance is the same in all IRF's. That's the defining characteristic of a geometrical theory of spacetime.

By whatever process, you discover that ##-c^2(t_2 - t_1)^2 + (x_2 - x_1)^2## is invariant. You may have already derived the Lorentz transformation and prove it using that. Or, you may do some experiments and measure spacetime coordinates between events in different frames.

In any case, ##-c^2(t_2 - t_1)^2 + (x_2 - x_1)^2## turns out to be an invariant quantity. And that, as far as I can see, is essentially the whole story. Whatever further mathematical terminology you use to describe what you've discovered is immaterial.

Per my understanding, the defining feature of IRFs is that free particles have straight worldlines, which isn't related to the metric.
It isn't related to the metric, but it is related to the coordinates. In Minkowski coordinates, a free particle has constant speed and also light travels with constant speed in any direction. But this is exactly what is found in an inertial frame, by definition. Therefore Minkowski coordinates describe inertial frames.

And if you are wondering, it can be shown that Minkowski coordinates are the only coordinates in which that is possible.

• PeterDonis
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You may have already derived the Lorentz transformation
You can only derive the Lorentz transformation in the form you give it (that it leaves ##dt^2 - dx^2## invariant) if you have already assumed you are using standard inertial coordinates. So you can't use this to argue that standard inertial coordinates are the only possible ones for an IRF.

PeroK
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So you can't use this to argue that standard inertial coordinates are the only possible ones for an IRF.
Who's suggesting that?

PeterDonis
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Who's suggesting that?
I thought that was the argument of your post #13.

PeroK
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I thought that was the argument of your post #13, that you need standard inertial coordinates to make ##dt^2 - dx^2## invariant under Lorentz transformations, therefore standard inertial coordinates are the only possible ones for an IRF.
My argument was simply that by considering those coordinates we get an invariant spacetime interval. Once you have that there should be no argument about "time being orthogonal to space" in an IRF. Although, it remains to be seen what happens under a general Lorentz Transformation. I guess that would still have to be shown.

PS Or, starting with the OP and the assumption that time is orthogonal to space, it remains to be seen whether that definition of an IRF matches any prior definition: like a free particle moving in a straight line.

It isn't related to the metric, but it is related to the coordinates. In Minkowski coordinates, a free particle has constant speed and also light travels with constant speed in any direction. But this is exactly what is found in an inertial frame, by definition. Therefore Minkowski coordinates describe inertial frames.

And if you are wondering, it can be shown that Minkowski coordinates are the only coordinates in which that is possible.
Since I don't mind looking stupid, I'll confess I'm still confused and I'll tell you my thought process. Let's just assume the ##x## and ##t## coordinates for now.

In Minkowski coordinates, a free particle has constant speed and also light travels with constant speed in any direction.
This is what's not obvious to me. I thought this should be "In IRFs" instead of "In Minkowski coordinates". I thought that a coordinate system is just a mathematical construction we use to describe whatever happens in an IRF. So even if I used some random coordinate system to describe IRFs - sure, the free particle trajectory could potentially look very different/weird and the math would be much more complicated, but just selecting a different coordinate system to describe the IRF wouldn't change the physical fact that free particle and light travel at constant speed in an IRF.

The physical fact shouldn't get altered based on what mathematical construction we use to define the IRF, right?

And if you are wondering, it can be shown that Minkowski coordinates are the only coordinates in which that is possible.
Could you give some guidance on how it can be shown? Could you prove it/give an example/provide link to some reference? I'd be really thankful.

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Ibix
just selecting a different coordinate system to describe the IRF wouldn't change the physical fact that free particle and light travel at constant speed in an IRF.
Can you define "constant speed" without reference to a coordinate system?

The point about an inertial coordinate system is that ##\partial \vec x/\partial t=\mathrm{const}## (a statement about rate of change of three coordinates with respect to the fourth along some worldline) corresponds to inertial motion (a physical, measurable state). You are not obligated to use them, but a lot of maths is simpler if you do.

robphy
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Do you mean that it's a matter of convention that we select time and space basis vectors (in Minkowski space) in such a way that the time basis vector is orthogonal to space basis vectors?
(Apologies that I did not read all of the replies since this post I am quoting from.)

Physics (or at least our mathematical formulation of it) defines
spacelike-vectors to be orthogonal to timelike-vectors.

Here's how Minkowski describes this...
Minkowski's 'Space and Time' said:
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 $x_1, y_1, z_1, t_1$ to
be normal to each other.
In other words,
locate the intersection of
an observer's 4-velocity with the unit-hyperbola (the Minkowksi circle) centered at the tail of the observer's 4-velocity.
The tangent line to that hyperbola is Minkowski-perpendicular to that observer's 4-velocity.
That observer's x-axis is drawn through the tail of her 4-velocity, parallel to that tangent line.

The "intuition" to have is that
the tangent to the "circle" in that geometry
is orthogonal to the radius vector.
(If you feel better, you can add pseudo- and other such adjectives.)
Physically,
"Space" is orthogonal to "Time".

This idea, motivated by Euclidean geometry, works in the Galilean spacetime geometry and Minkowski spacetime geometry.

(All three are among the nine Cayley-Klein geometries in two dimensions...
the others include hyperbolic and elliptic geometry and,
for relativity, their Lorentzian analogues (called the de Sitter spacetimes) and
their Galilean limits (called the Newton-Hooke spacetimes).)

You can play around with this idea in my visualization (time runs upwards)
https://www.desmos.com/calculator/r4eij6f9vw

- The unit-hyperbola (the "Minkowski circle") is in blue.
This figure is unchanged by a Lorentz boost.
- The red dotted line is the [timelike] observer-worldline.
- The red tangent line is the prototype for "simultaneous events according to
the red-observer".

- The red-observer's [spacelike] x-axis is drawn parallel to that tangent line.

Here is the E= +1, Minkowski case: Play around with the E-slider to change signature to see the Galilean and Euclidean analogues.

Here E is approaching 0, the Galilean case (displaying the opening of the light-cone):
(Note how the tangent lines associated with different radii will coincide when E=0.
That's "absolute simultaneity".
Further note, that:
Generally, the tangents do not coincide.
Our "common-sense" Galilean "absolute simultaneity" case is the exception... not the [general] rule.
) Here is E= -1, the Euclidean case: Last edited:
• Shirish
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My argument was simply that by considering those coordinates we get an invariant spacetime interval.
You can have an invariant spacetime interval in any coordinates.

The metric being diagonal is dependent on your choice of coordinates. It's easy to find coordinates in which the metric of Minkowski spacetime is not diagonal.
This much I completely understand.

But those coordinates would not describe an IRF.
I didn't get this. Any set of coordinates that we use can potentially be used to describe an IRF, can't it? The behavior of worldlines or the math might get complicated, but can't potentially any set of coordinates be used to describe an IRF?

No, it's (orthogonality between time and space dimensions) being presented as a feature distinguishing an IRF from other types of coordinate charts.
But IRF itself isn't a coordinate chart? By IRF in this case, do you mean the standard inertial coordinates (i.e. the typical ones we use in spacetime diagrams), or have I completely misunderstood this point?

PeterDonis
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starting with the OP and the assumption that time is orthogonal to space
But that's the wrong place to start from to address the OP's question. The right place to start from is to not assume that time is orthogonal to space, since that's precisely the question at issue.

PeterDonis
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Any set of coordinates that we use can potentially be used to describe an IRF, can't it?
No. For a counterexample, consider Rindler coordinates on Minkowski spacetime. Observers at rest in these coordinates are not inertial.

IRF itself isn't a coordinate chart?
It depends on what definition of "IRF" you are using. (But it's worth noting that if you define "IRF" to mean a particular kind of coordinate chart, then your previous claim that any set of coordinates can be used to describe an IRF would make no sense.)

So let's take a step back: what is your definition of "IRF"?