Why is time orthogonal to space?

[jk22]

Basically how do we know that since it is not possible to see the 4th dimension, is it for simplicity ?

 

[Dale]

It is orthogonal because you can fix your position in space and measure time independently.

Similarly x is orthogonal to y and z because you can fix your position in y and z and measure x independently.

 

[jk22]

But SR treats moving frames then space and time are linked ?

 

[Dale]

But SR treats moving frames then space and time are linked ?

Yes. Similarly to how x and y are linked by rotations time and space are linked by boosts, which are a sort of 4-D rotations.

 

[DrGreg]

It is orthogonal because you can fix your position in space and measure time independently.

Similarly x is orthogonal to y and z because you can fix your position in y and z and measure x independently.

That's true, but there is more to it than that.

Einstein's second postulate says that the speed of light is the same in all directions. In particular, when you reflect light off a mirror, the time it takes to travel from source to mirror is the same time it takes to travel from mirror back to source. This gives you a procedure (called Einstein's syncrhonisation procedure) for assigning a time coordinate to a remote event. It also means that the "one-way speed of light" (on each leg of the journey) equals the "two-way speed of light" (the average for the round trip).

When you look at the mathematics behind this, it turns out that this condition (one-way speed = two-way speed) is equivalent to the time coordinate being orthogonal to all the space coordinates.

 

[Ibix]

Just to add to DrGreg's post, note that you can define coordinate systems in which the timelike and spacelike directions are not orthogonal. It just makes the maths nasty and hard to interpret and, consequently, rather error prone.

 

[Dale]

Just to add to DrGreg's post, note that you can define coordinate systems in which the timelike and spacelike directions are not orthogonal. It just makes the maths nasty and hard to interpret and, consequently, rather error prone.

Yes, just as you can define coordinate systems in which x and y are not orthogonal. The precise meaning of that must refer to the math.

 

[jk22]

I thought the inclination of the time axis could be seen as an extra parameter giving a slightly different Lorentz transform. So that any discrepancy in an accelerator experiment for example could be explained by that ?

 

[Dale]

I thought the inclination of the time axis could be seen as an extra parameter giving a slightly different Lorentz transform.

Similarly the angle of rotation is a parameter that gives a slightly different rotational transform.

 

[jk22]

What I mean is that the transform depends on $v$ the speed and $\alpha$ the angle of the time axis.

 

[FactChecker]

But SR treats moving frames then space and time are linked ?

In any given inertial frame, $I_1$, the measured position in space and time are independent of each other. An external inertial frame, $I_2$, might see a connection between $I_1$'s space and time, but that is irrelevant to the issue of orthogonality of $I_1$'s space and time.

 

[Dale]

What I mean is that the transform depends on v the speed and α the angle of the time axis.

The speed and the angle are essentially the same thing. (Technically the speed is the slope, but it is not much effort to convert a slope to an angle and each uniquely determines the other)

 

[jk22]

I thought they were independent.

 

[Ibix]

No. The (hyperbolic) angle the time axis makes with the axis of another frame is called the rapidity of that frame. You called this $\alpha$, and it is related to the frame velocity by $c\tanh\alpha=v$. There's no free parameter here.

 

[jk22]

Okay thanks, I calculated in a wrong way and obtained both of them in solving the equation to get the gamma factor.

If we do like

$\vec{x}=R(\alpha)\vec{x'}$

Then $G'(\alpha)=R^t(\alpha)GR(\alpha)$

And is sought $L(v)$ such that $x^tGx==x'^tG'(\alpha)x'==x'L(v)^tGL(v)x'$ ?

(In fact it comes to equations of a too high degree to solve)

 

[pervect]

I thought the inclination of the time axis could be seen as an extra parameter giving a slightly different Lorentz transform. So that any discrepancy in an accelerator experiment for example could be explained by that ?

Suppose one has a flat plane. One can parameterize it with coordinates so that x and y are orthogonal, or one could parameterize it so that they're not orthogonal.

Then someone, who is learning geometry, asks us "Why are x and y orthogonal? Why can't they be non-orthogonal". I would say the most accurate answer to this is that it's a matter of convention.

In this example, we don't even have to use x & y as coordinates. Sometimes, for instance, we find it convenient to use polar coordinates, r and $\phi$. Occasionally, even more exotic coordinates are used, for instance hyperbolic coordinates.

The important point here is that a plane is still a plane, regardless of what set of coordinates we use to label the points on it. We aren't changing anything significant about the geometry if we choose a different system of labeling points, we are only changing the description. Regardless of which set of coordinates we use, though, the plane is still a plane.

Using arbitrary coordinates is perfectly possible to do - but it's not really the way I'd suggest that someone learn geometry for the first time.

The basic point is that one is simple making geometry a bit easier and more familiar by making x orthogonal to y. A number of formula that one may have carefully memorized might no longer work properly if one does not assume that x is orthogonal to y, and unless one studies the topic at an advanced level, one is likely to run into problems by incorrectly applying a formula that one has learned that asssumes x and y are orthogonal if one works in coordinates where they are not.

With sufficient effort, one can do geometry where x is not orothogonal to y, but it's not a good approach for someone who is just learning geometry. It also doesn't make any actual difference. The shortest distance between two points is still a straight line, for instance, but the description of the straight line appears different if one uses different coordinates.

The situation is rather similar in physics. Techniques exist for using arbitrary coordinates correctly, but they are generally not taught at the introductory level. I would say the biggest issue that arises with using arbitrary coordinates is an issue of unlearning. One has simplified many formula by making certain key assumptions. Applying these simplified formula to cases where the assumptions are not met leads to incorrect answers. It takes a bit of learning to learn which formula are no longer valid in situations where the usual simplifying assumptions are not made.

The methods that allow one to use arbitrary coordinates correctly are basically called tensor methods, and are usually introduced in graduate school. As a practical matter, one usually learns the simpler techniques first, and then revisit them later.

Going back to physics, though, there is one well-regarded undergraduate text that at least briefly talks about special relativity in ways that are not focused on adopting a particular set of coordinates. This would be Taylor & Wheeler's "Space-time physics". In particular, the "parable of the surveyor" talks about why, in special relativity, we talk about a 4 dimensional space time rather than a 3 dimensional space and a separate one dimensional time. There's an old PF post of mine that talks a little more about this, https://www.physicsforums.com/threads/time-as-a-dimension.917639/#post-5784024

 

[PeroK]

Basically how do we know that since it is not possible to see the 4th dimension, is it for simplicity ?

I'd like to give you a slightly different answer. They are orthogonal by definition of the inner product. The same is true of the x and y axes. They are only orthogonal if you define the angle between them to be a right angle.

That the x, y, z and t axes are orthogonal is essentially the definition of "flat" spacetime.

That definition is then compatible with experimental results.

You could postulate any other geometry for gravity-free spacetime, but that wouldn't be compatible with experiment.

You could define a different inner product where the x, g and t axes are not orthogonal. That would, in general, give you curved spacetime.

 

[A.T.]

Basically how do we know that since it is not possible to see the 4th dimension, is it for simplicity ?

We don't directly see the 3 spatial dimensions either, just their 2D projection on our retinas. Our brains interpret it as 3 dimensions.

 

[FactChecker]

Basically how do we know that since it is not possible to see the 4th dimension, is it for simplicity ?

We are free to define coordinate systems that are most convenient for our use. So you are basically correct. It is for simplicity.

 

[sweet springs]

x coordinate and y coordinate are orthogonal and independent. After rotation on xy plane, x' coordinate and y' coordinate are orthogonal and independent. x' coordinate depends on x and y. y' coordinate depends on x and y.

Similarly space coordinate x and time coordinate t is independent, however, Galilean transformation
$$x'=x-vt$$
reveal us space x ant time t is mixed up to form space coordinate x'. x' is not orthogonal to both x and t.

 

[PeterDonis]

That the x, y, z and t axes are orthogonal is essentially the definition of "flat" spacetime.

This can't be right, because there are curved spacetimes that have a metric tensor which is everywhere diagonal (the best known example is Schwarzschild spacetime), which means that the four coordinate axes are orthogonal at every point.

 

[atyy]

I'd like to give you a slightly different answer. They are orthogonal by definition of the inner product. The same is true of the x and y axes. They are only orthogonal if you define the angle between them to be a right angle.

That the x, y, z and t axes are orthogonal is essentially the definition of "flat" spacetime.

That definition is then compatible with experimental results.

You could postulate any other geometry for gravity-free spacetime, but that wouldn't be compatible with experiment.

You could define a different inner product where the x, g and t axes are not orthogonal. That would, in general, give you curved spacetime.

In curved spacetime the timelike direction is also orthogonal to spacelike directions. We can see this in the tetrad formalism. https://en.wikipedia.org/wiki/Frame_fields_in_general_relativity

 

[PeterDonis]

In curved spacetime the timelike direction is also orthogonal to spacelike directions. We can see this in the tetrad formalism.

More precisely, in curved spacetime we can always pick, at a particular point, a set of one timelike direction (vector) and three spacelike directions (vectors) that are all mutually orthogonal; this is called a tetrad and a mapping of tetrads to points in a spacetime is called a frame field. But in a general curved spacetime there does not need to be any single global coordinate chart that matches up with a frame field (in the sense that at every point the vectors in the tetrad mapped to that point by the frame field also point in the same directions as the coordinate basis vectors at that point). The example I gave (Schwarzschild spacetime) is a curved spacetime in which there is a single global coordinate chart that matches up with a frame field.

 

[robphy]

"Time is orthogonal to space" in Minkowski Spacetime, as defined by Minkowski himself when he formulated "spacetime" (and "light cone" and "worldline" etc...).

From my post on StackExchange

From Minkowski's "Space and Time"... [bolding by me]
(e.g. http://www.minkowskiinstitute.org/mip/MinkowskiFreemiumMIP2012.pdf , p47)

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.

This orthogonality also holds in Galilean spacetime.
Both of these spacetime-orthogonalities are natural generalizations of Euclidean space.You can play around with this idea in my visualization [screencaptured below].
https://www.desmos.com/calculator/r4eij6f9vw
Play around with the E-slider to see the Galilean (E=0) and Euclidean (E=-1) analogues!

• The blue "circle" is the metric
• The dotted radial lines are worldlines of inertial observers that met at event O, representing time for each of them.
• The tangent lines are normal [orthogonal, perpendicular] to the radial lines.
  These correspond to "all of space at an instant of time according that observer".
• The line through the origin O parallel to the tangent line defines the "x-axis" of that observer at event O.

 

[jk22]

Using a classical Galileo transform also implies that time inclines over space by an angle $\arctan(v)$.

So if the equivalence principle were used that would also give a curved classical spacetime for gravitation and hence a local description of classical gravity ?

 

[PeterDonis]

Using a classical Galileo transform also implies that time inclines over space by an angle $\arctan(v)$.

No, with a Galilean transformation, the "direction" of time doesn't change at all, because the Galilean transform only transforms space, it does nothing to time.

if the equivalence principle were used that would also give a curved classical spacetime for gravitation and hence a local description of classical gravity ?

No. You can describe Newtonian gravity in geometric terms, as Cartan showed [1], but not the way you're describing here.

[1] https://en.wikipedia.org/wiki/Newton–Cartan_theory

 

[jk22]

In a Galilean transformation : $x'=x-vt$. If $x'=0$, hence the time primed axis, the equation is $x=vt$, which describe a straight line not orthogonal to x ?

 

[Ibix]

The Galilean transforms are$$\begin{eqnarray*}t'&=&t\\x'&=&x-vt\end{eqnarray*}$$So the $t'$ axis is not $x'=0$.

In relativity, because space and time are part of spacetime, changing your notion of "rest" changes your coordinates on spacetime, which includes changing your notion of time.

But Newtonian physics models time as something completely distinct from space. Changing your notion of rest doesn't change your notion of time - there is an absolute time parameter that is the same for all observers. So the time axis doesn't change when you change frames.

 

[jk22]

Yes took the line of the origin of the x-axis as the time axis, a misunderstanding of my self.

Or is it from differential geometry that $\vec{e}_{t'}=\frac{\partial\vec{r'}}{\partial t}$ ?

 

[FactChecker]

We naturally think of time and position as being independent and unrelated because, in our experience, positions do not generally move uniformly with time. Consider what would happen if you lived all your life floating in a moving stream of water. Then you would associate the passage of time with a uniform motion of all objects in the stream. Predicting the position of an object would always change as time passed. Your concept of time and space would not be orthogonal.
But that is not our experience. When one goes back home after a day of work, he goes to the same positional coordinates that he left home in the morning. Our concepts of time and position are orthogonal. Our mathematical definition of spacetime is made to match our natural concepts. That is practical, but not forced by mathematics.

[EDIT] I think we can see that in a Minkowski diagram, where the axes of a moving reference frame are not orthogonal.

 

[Dale]

In a Galilean transformation : $x'=x-vt$. If $x'=0$, hence the time primed axis, the equation is $x=vt$, which describe a straight line not orthogonal to x ?

It is still orthogonal to x. Orthogonality is based on some inner product. In relativity the metric that defines the inner product has a signature (-+++), but in Newtonian theory it is two degenerate metrics with signatures (+000) and (0+++) respectively. Thus the inner product of $\hat x’$ and $\hat t’$ is zero for both of the degenerate metrics.

 

[robphy]

In a Galilean transformation : $x'=x-vt$. If $x'=0$, hence the time primed axis, the equation is $x=vt$, which describe a straight line not orthogonal to x ?

Just adding to Dale's comment...
recall from my earlier post above that
"the tangent line to the circle [the metric]" is (following Minkowski) "orthogonal to the timelike radial direction".
That characterization is an extension of the Euclidean case and can be extended further to the Galilean case.
So, in a Galilean spacetime, the spatial direction [where t=constant] is Galilean-perpendicular (but not necessarily Euclidean-perpendicular) to the timelike-worldline. [This corresponds to the "temporal metric" with signature +000.]

 

[robphy]

Using a classical Galileo transform also implies that time inclines over space by an angle $\arctan(v)$.

No, with a Galilean transformation, the "direction" of time doesn't change at all, because the Galilean transform only transforms space, it does nothing to time.

In the Galilean case, it turns out that the "Galilean-rapidity" (the "Galilean-angle"= "arc-length [using the degenerate spatial metric mentioned by Dale] along the Galilean-circle) is equal to the ordinary velocity v. [There is a "Galilean-analogue" of the tangent function---called $tang(\theta)$ by the mathematician I.M Yaglom... and it's defined as $tang(\theta)\equiv \theta$... using "dimensionless units". Along these lines, $cosg(\theta)\equiv 1$.]

(The additivity of rapidity (angles = "spatial arc-lengths on unit circles) always holds...
but the corresponding slopes [physically, spatial-velocities] are not generally additive...
except in the Galilean case. Unfortunately, we hold on so tightly to the special-case Galilean additivity of spatial velocity that we have trouble with the more general non-additivity of slopes [velocities].)

Note that, in a boost, all timelike worldlines are transformed [in accordance with the principle of relativity... no preferred reference frame... no timelike eigenvectors].

• So, in this sense, the "direction of time" (direction of the observer's 4-velocity) does change [e.g. it can be transformed to being the "observer at rest" in the spacetime diagram]...
  however,
  in the Galilean case, the "time-coordinate" doesn't change: $t'=t$... that is "1 sec after event O" is still "1 sec after event O".
• Along these lines, in Galilean boost, the spatial hypersurfaces are mapped onto themselves... the set of events with "t=1 sec" are the same after a Galilean boost. However, the coordinate labeling of those events are different $x'=x-vt$.

 

[The Bill]

It is orthogonal because you can fix your position in space and measure time independently.

Similarly x is orthogonal to y and z because you can fix your position in y and z and measure x independently.

A coordinate system with linearly independent but not orthogonal x, y, and z axes would satisfy your condition while not being orthogonal.

 

[arydberg]

But we can go foward and backward in space. In time we can only go in one direction.