How fast does the particle go through spacetime?

In summary: No. Straight lines that appear below 45° (not 45° and below) on a Minkowski diagram drawn the usual way represent spacelike slices. There is a Euclidean metric on such slices, true, but this fact is pretty much useless to you. As represented on a Minkowski diagram, the scale on non-parallel slices is different and there is an offset between equivalent points on different slices. You can't work out either the scale or the offset without understanding Minkowski geometry, and you need both to be able to think about velocities.In summary, angles 45 and under from the x axis are not Euclidean.
  • #1
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This is a very basic question, and I am not sure I have the answer.

A photon goes from point A to point B, only 1 meter distance apart from each other. A spacetime diagram would show a line connecting points A and B at a 45 degree angle. This can be a right triangle with equal sides, with point C being the corner. The photon seems to have travelled faster along the hypotenuse (2^(1/2) meters) than how we would normally calculate its speed for a distance of 1 meter.

Is this right, or am I missing something here?
 
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  • #2
student34 said:
Is this right
No.

student34 said:
or am I missing something here?
Several things:

(1) Spacetime is not Euclidean. The metric of spacetime is the Minkowski metric; the interval between two events is ##(\Delta t)^2 - (\Delta x)^2## (note the minus sign), assuming we are using an inertial frame (and that we are using units in which the speed of light ##c## is ##1##).

(2) The concept of "speed through spacetime" does not make sense for light. Light travels on null worldlines; that means the "arc length" along such a worldline, calculated using the Minkowski metric, is zero. A "speed through spacetime", mathematically, would be a unit vector that points along the worldline, but if arc length along the worldline is zero, the norm of a vector pointing along the worldline is also zero, so there's no way to normalize it to a unit vector.

(3) We can calculate a "speed" for a light pulse (this is a better term than "photon" since we are doing classical physics here, not quantum physics) using an inertial frame; looking at the metric above, we can see that the interval being zero means that ##\Delta t = \Delta x## along the worldline of light. That means the speed of the light, which is given by ##\Delta x / \Delta t##, is ##1## (or ##c## in conventional units). But this is a "speed through space" in an inertial frame, not a "speed through spacetime".
 
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  • #3
PeterDonis said:
(1) Spacetime is not Euclidean.
@student34 didn’t we already have this discussion multiple times?
 
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  • #4
I thought angles 45 and under from the x axis are Euclicean.
 
  • #5
Thanks Peter for commenting on my post and not trying to me me look like an idiot.
 
  • #6
student34 said:
I thought angles 45 and under from the x axis are Euclicean.
What does this even mean?

Minkowski spacetime and its metric are a definite mathematical model. That is the model you should be using.
 
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  • #7
student34 said:
Thanks Peter for commenting on my post and not trying to me me look like an idiot.
You're welcome. But be advised that my generosity is limited. @Dale has already made a valid point. Looking at your past threads on the general topic of SR, a common factor seems to me to be an unwillingness on your part to clearly understand and use the mathematical model of Minkowski spacetime and its metric. But if you are going to do SR, you have no choice.
 
  • #8
student34 said:
I thought angles 45 and under from the x axis are Euclicean.
Huhh?! Features like angle size don't determine the geometry. The geometry is non-Euclidean by design, so it makes no difference what size angles you consider, you are dealing with a geometry that's non-Euclidean.
 
  • #9
student34 said:
Thanks Peter for commenting on my post and not trying to me me look like an idiot.
Sorry, my apologies. That was indeed unnecessary on my part.
 
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  • #10
student34 said:
The photon seems to have travelled faster along the hypotenuse (2^(1/2) meters) than how we would normally calculate its speed for a distance of 1 meter.
Please actually calculate the spacetime interval using the Minkowski metric. Show your work.

student34 said:
I thought angles 45 and under from the x axis are Euclicean.
No. The metric in the ##t,x## plane is Minkowski
 
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  • #11
student34 said:
I thought angles 45 and under from the x axis are Euclicean.
No. Straight lines that appear below 45° (not 45° and below) on a Minkowski diagram drawn the usual way represent spacelike slices. There is a Euclidean metric on such slices, true, but this fact is pretty much useless to you. As represented on a Minkowski diagram, the scale on non-parallel slices is different and there is an offset between equivalent points on different slices. You can't work out either the scale or the offset without understanding Minkowski geometry, and you need both to be able to think about velocities.

As already pointed out above, you seem to struggle to let go of Euclidean geometry and keep trying to wedge it into places where it won't fit. Stop trying that. There are strong analogies between Euclidean and Minkowski geometry but they are not the same and trying to "think Euclidean" in a Minkowski spacetime without understanding Minkowski is just banging your head on a brick wall.
 
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  • #12
Ibix said:
student34 said:
I thought angles 45 and under from the x axis are Euclicean.
No. Straight lines that appear below 45° (not 45° and below) on a Minkowski diagram drawn the usual way represent spacelike slices. There is a Euclidean metric on such slices, true, but this fact is pretty much useless to you.
The lines above 45° also have a Euclidean metric on them.

For spacetime, as others have said, it's not Euclidean but Minkowskian geometry
(in spite of many attempts to avoid Minkowskian spacetime geometry).

As a bridge from Euclidean geometry to Minkowskian geometry,
I think it's best to focus on what a circle is
(as something that defines a Euclidean metric, alternatively defining its dot-product)
and realize that the hyperbola is the Minkowski analogue of a circle.

It would further be helpful to realize where
the various constructions and operations you do in Euclidean geometry make use of a circle
and then consider the analogous construction in Minkowskian geometry.

One way to start is to play around with the E-slider in my visualization:
spacetime diagrammer for relativity https://www.desmos.com/calculator/kv8szi3ic8

where E=1 is Minkowski, E=0 is Galilean, and E=-1 is Euclidean...

1671098688206.png
1671098701022.png
1671098725231.png


In the spacetime cases, their "unit circles" represent the set of "first tick" events
for inertial clocks traveling with various velocities from the origin event.
Spacetime observers measure with a clock,
analogous to a Euclidean surveyor measuring with an odometer.
 
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  • #13
robphy said:
The lines above 45° also have a Euclidean metric on them.
Careful. If by lines you literally mean lines, i.e., timelike worldlines, they have a "Euclidean metric" in the trivial sense that they have a well-defined arc length along them.

But if by "lines" you mean 3-dimensional "slices" of spacetime obtained by adding the other two dimensions that were left out of the diagram (the way @Ibix described for lines with less than 45 degree slope), no, those slices do not have a Euclidean metric, because they are not purely spacelike.
 
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  • #14
Ibix said:
No. Straight lines that appear below 45° (not 45° and below) on a Minkowski diagram drawn the usual way represent spacelike slices. There is a Euclidean metric on such slices, true, but this fact is pretty much useless to you.
Thank you very much. But I don't understand why the lightlike/45 degree angles aren't Euclidean also. I get a 0 Minkowski distance using the Minkowski metric on the 45 degrees. Should I be getting 0?
 
  • #15
student34 said:
But I don't understand why the lightlike/45 degree angles aren't Euclidean also. I get a 0 Minkowski distance using the Minkowski metric on the 45 degrees. Should I be getting 0?
As I said in post #2, the interval is zero along a lightlike worldline. So yes, you should get 0 if you compute the Minkowski metric along a 45 degree line in your 2-dimensional. Which is not Euclidean.

If you are asking about the 3-dimensional "slices" that are obtained by adding the other two dimensions to a 45 degree line in the diagram, the geometry of those slices is also not Euclidean, because the slices are not purely spacelike.
 
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  • #16
student34 said:
Thank you very much. But I don't understand why the lightlike/45 degree angles aren't Euclidean also.
For any spacelike slice you can Lorentz transform into a frame where the slice is parallel to that frame's x, y, and z axes - i.e., it is space (at some chosen time) in that frame. In that frame's coordinates, then, ##\Delta t=0## on that slice and therefore ##\Delta s^2## for intervals in that slice only becomes ##\Delta s^2=\Delta x^2+\Delta y^2+\Delta z^2##, which is Euclidean. Thus the metric in that slice is Euclidean. For lightlike or timelike slices there is no frame where ##\Delta t=0## so you will always have an opposite-sign term in the metric.
student34 said:
I get a 0 Minkowski distance using the Minkowski metric on the 45 degrees. Should I be getting 0?
Of course. On the 45° line ##\Delta x=\Delta t##. What are you ever going to get out of the interval except zero?
 
  • #17
student34 said:
I get a 0 Minkowski distance using the Minkowski metric on the 45 degrees. Should I be getting 0?
Yes. Lightlike lines are also called "null" lines for exactly this reason. The Minkowski interval between lightlike separated events is 0.

student34 said:
Thank you very much. But I don't understand why the lightlike/45 degree angles aren't Euclidean also. I get a 0 Minkowski distance using the Minkowski metric on the 45 degrees. Should I be getting 0?
Nothing in the ##t,x## plane is Euclidean. It is Minkowski. I am not sure why you thought otherwise.

The metrics in the ##x,y## plane, the ##y,z## plane, and the ##x,z## plane as well as the metric in the ##x,y,z## subspace are all Euclidean. This is why we say that ##x,y,z## are spatial directions. As soon as you include ##t## (so ##t,x##, or ##t,y##, or ##t,z##, or ##t,x,y##, or ##t,x,z##, or ##t,y,z##, or ##t,x,y,z##) then the metric is Minkowski.
 
  • #18
Update at the end.


A way to visualize the square-interval is to draw the "causal diamond"
joining the endpoints of the spacetime-displacement vector.
For a timelike displacement, it's the intersection of the "future cone of the past event" with the "past cone of the future event".
Its area is equal to the square-interval [itex]\Delta t^2-\Delta(x/c)^2[/itex].
In the figure below, the area is 64.
So, the magnitude of the diagonal of the diamond equals 8.

Why? (It's based on light-cone coordinates from the eigenvectors of the boost in (1+1)-Minkowski.)
The lightlike sides have sizes
[itex]\Delta u=\Delta t+\Delta(x/c)[/itex]
[itex]\Delta v=\Delta t-\Delta(x/c)[/itex]
The area is [itex]\Delta u\Delta v=\Delta t^2-\Delta(x/c)^2.[/itex].
Since the boost has determinant 1, the signed-area is invariant under boosts.

I just made this:
https://www.desmos.com/calculator/4jg0ipstya
1671135120504.png

You can interact with it.
Move the events A and Z.

Since the signed-area is [itex]\Delta u\Delta v=\Delta t^2-\Delta(x/c)^2.[/itex],
  • when area > 0, then AZ is a timelike vector
  • when area <0. then AZ is a spacelike vector
  • when area =0, the AZ is a lightlike vector
For more information, consult my PF Insight:
https://www.physicsforums.com/insights/relativity-rotated-graph-paper/

The Insight focuses on light-clock diamonds
(as seen in the folder option in entry 33 of https://www.desmos.com/calculator/kv8szi3ic8 only when E=1).
For something more recent that uses the causal diamonds,
see https://www.aapt.org/docdirectory/m...dGraphPaper-CalculatingWithCausalDiamonds.pdf
1671135914599.png

Here's a conservation of momentum problem from the above slides.
Find the magnitude of [itex]P_{2f}[/itex]. (Hint: draw its causal diamond.)
1671136019587.png



If you want the spatial-velocity of the timelike segment, you could use
[tex] v=\frac{\Delta x}{\Delta t} =
\frac{
\frac{1}{2}(\Delta u-\Delta v)
}{
\frac{1}{2}(\Delta u+\Delta v)
}
=
\frac{
\frac{\Delta u}{\Delta v} -1
}{
\frac{\Delta u}{\Delta v}+1
},
[/tex]
where the aspect ratio [itex]\frac{\Delta u}{\Delta v}[/itex] is equal to square of the Doppler k-factor.
In the first diamond with [itex]\Delta u=16 [/itex] and [itex]\Delta v=4 [/itex], so [itex] k^2= \frac{16}{4}=4[/itex] and thus [itex] k= 2[/itex] so that [itex] v=\frac{4-1}{4+1}=\frac{3}{5} [/itex].
 
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  • #19
* I'm hoping this is a trick question on purpose, because nobody seems to have noticed the trick.

* If a is at the origin, and B is on the unit circle, then the sides of the triangle aren't 1, they're sin(45 degrees) = .7

* While it is true that the metric is invalid because it's not hyperbolic, your confusion about the triangle side length doesn't have anything to do with that.
 
  • #20
paige turner said:
* I'm hoping this is a trick question on purpose, because nobody seems to have noticed the trick.

* If a is at the origin, and B is on the unit circle, then the sides of the triangle aren't 1, they're sin(45 degrees) = .7
Which post are you referring to? The OP was talking about a triangle with sides ##\Delta x=1## and ##\Delta t=1##, which he erroneously calculated to have a hypotenuse of ##\sqrt{2}##. Using the Minkowski metric the hypotenuse is length ##0##.
 
  • #21
> he erroneously calculated

yes.

> Using the Minkowski metric the hypotenuse is length 0.

Because It's on the null cone, yes. Two objects can have an absolute (invariant) distance of zero even though they are on different sides of the universe. That's the magic of pseudometrics.

A long time ago in galaxy far, far away has a zero absolute distance from you when you see its light or feel it's gravity waves.

The Minkowski metric has a lot of magic in it that I haven't been able to find where other people have thought about.
 
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