Deriving Invariant Distance in Space Time: Pythagorean Theorem

[moatasim23]

Why in deriving invariant distance in space time we use Pythagorean Theorem with a negative sign?

 

[granpa]

c^2 = a^2 + b^2 + t^2

if we think of time as being imaginary we get

c^2 = a^2 + b^2 - t^2

 

[Dale]

Why in deriving invariant distance in space time we use Pythagorean Theorem with a negative sign?

Because it matches the results of physics experiments.

Basically, if we take a good clock, and we calculate the "length" of the path that the clock takes through spacetime using the above formula then we find experimentally that the clock measures that "length" as the elapsed time on the clock.

 

[Naty1]

Like many things in physics, when one first come across it one thinks "How did he/she ever figure out THAT?" I don't think it's a simple intuition for most of us. Is the Pythagorean Theorem itself intuitive?? It didn't seem to be for several thousand years.

My impression is that some really smart dudes played with various mathematical ideas and interpretations in the late 1800's and early 1900's trying to figure out descriptions(models) of light, ether, and Maxwell's equations for electromagnetic radiation...they all seemed mixed up together somehow.

We don't usually hear about what are the many likely formulations that didn't work. Einstein sure didn't "invent" that formulation; he took what was some available mathematics and it was his physical insights that gave him the ability to use that math to describe observational results. He did the same thing in general relativity using Riemann [curvature] which a mathematician friend found for him)as a basis for explaining his physical insights.

[Now that I think of it, that was the basis for an early formulation of the strong nuclear force...by Gabriele Veneziano...he noticed the Euler beta-function had some characteristics which matched physical observations. It worked but nobody knew why until
string theory.]

Here is how Sean Carroll explains the space time interval:

http://ned.ipac.caltech.edu/level5/March01/Carroll3/Carroll1.html



For some more insights, check here:
http://en.wikipedia.org/wiki/Minkowski_metric#History

 

[elfmotat]

Try calculating $(\Delta x')^2+(c\Delta t')^2$ in the unprimed frame with the Lorentz Transformation. Does it give you $(\Delta x)^2+(c\Delta t)^2$? No, it gives you something long and hairy. Now try doing the same thing with $(\Delta x')^2-(c\Delta t')^2$. This time you'll find that $(\Delta x')^2-(c\Delta t')^2=(\Delta x)^2-(c\Delta t)^2$.

 

[Naty1]

By pure chance I came across a related historical note that might be of interest:

See also History of Lorentz transformations.
Many physicists, including George FitzGerald, Joseph Larmor, Hendrik Lorentz and Woldemar Voigt, had been discussing the physics behind these equations since 1887.[1][2] Larmor and Lorentz, who believed the luminiferous ether hypothesis, were seeking the transformation under which Maxwell's equations were invariant when transformed from the ether to a moving frame.

here:

http://en.wikipedia.org/wiki/Lorentz_transformation#History

 

[guss]

The -c² appended onto that term in that equation acts as a correction factor. The negative is there because time is imaginary. The c₂ is there to act as a correction factor for the way we measure time. If we were to measure things very fundamentally, time and space would be measured in the same way. But we don't, so we must add the c₂.

See http://www.quora.com/Physics/How-di...arated-by-time-are-also-seperated-by-distance and read the comments on the answer.

 

[Naty1]

Here is another perspective:

http://www.knowledgerush.com/kr/encyclopedia/Spacetime/

A spacetime interval between two events is the frame-invariant quantity analogous to distance in Euclidean space. The spacetime interval s along a curve is defined by:



where c is the speed of light (see sign convention). A basic assumption of relativity is that coordinate transformations have to leave intervals invariant. Intervals are invariant under Lorentz transformations.

The spacetime intervals on a manifold define a pseudo-metric called the Lorentz metric. This metric is very similar to distance in Euclidean space. However, note that whereas distances are always positive, intervals may be positive, zero, or negative. Events with a spacetime interval of zero are separated by the propagation of a light signal. Events with a positive spacetime interval are in each other's future or past, and the value of the interval defines the proper time measured by an observer traveling between them. Spacetime together with this pseudo-metric makes up a pseudo-Riemannian manifold.

 

[nitsuj]

The -c² appended onto that term in that equation acts as a correction factor. The negative is there because time is imaginary. The c₂ is there to act as a correction factor for the way we measure time. If we were to measure things very fundamentally, time and space would be measured in the same way. But we don't, so we must add the c₂.

I thought it is negative because time has an inverse relationship with length.

What does "time is imaginary" mean?

It's not the way "we" measure it. It's the way it is measured. The physics of measuring time includes distance doesn't it?

In any case we do measure time and space the same with intervals.

 

[guss]

I thought it is negative because time has an inverse relationship with length.

What does "time is imaginary" mean?

It's not the way "we" measure it. It's the way it is measured. The physics of measuring time includes distance doesn't it?

In any case we do measure time and space the same with intervals.

Since time is unlike any other dimension, that equation calls it imaginary to make it clear that the time dimension is different than space dimensions. There are probably deeper reasons than that, though.

Time is part of the distance between two events in space-time (the other parts being physical distance). We traditionally measure this distance with seconds, but it needs to be converted to meters for the space-time distance equation. The c*t in (ct)² can be viewed sort of like velocity*time = distance.

 

[nitsuj]

...imaginary to make it clear that the time dimension is different than space dimensions.

I see another poster used the term "imaginary". Is this used in math/physics? It seems worse then reffering to gravity as a "fictitious force". (seems like an oxymoron)

Is there a less vague definition of "imaginary" then the context above?

An which equation are you reffering too with the -c2?

 

[Dale]

Imaginary just means $\sqrt{-1}$. The imaginary time concept has been out of use for several decades now, similar to the concept of relativistic mass. Imaginary time works in inertial frames in SR, but doesn't generalize to non inertial frames or curved space times.

 

[nitsuj]

Ha, is it the transition to understanding what equivalent means in the context of time = length, energy = mass.

Why not just, "the other side of the coin" eh Dalespam



"Imaginary time works in inertial frames in SR, but doesn't generalize to non inertial frames or curved space times."

I don't know GR at all, does that statement mean equivalence isn't so simple in GR? Are there no inertial frames in GR (i interprut curved spacetime as acceleration of somesort)? Said differently does GR ruin this awsome symmetry thing? (specifically complicates it)

How neat & tidy with; the "space" is time=length, the "stuff" is energy=mass. So there are two coins, space and stuff

lol I actual played around with the square root of -1 to see what is meant. If I consider squared and square root literally (as in lengths of space). Speaking loosely, I can see the signs distinguish between (have the same relationship as) length and time. I haven't thought much of it yet, but a long shot from here it looks like there is no quanta of time, in the same sense as length. Perhaps in that sense it was termed "imaginary".

 

[nitsuj]

The c*t in (ct)² can be viewed sort of like velocity*time = distance.

Does the squared in c2 come into play because in velocity*time = distance, time is being used twice? Once in velocity and again in duration of velocity.

I guess I am asking the why is c squared there.

 

[guss]

Does the squared in c2 come into play because in velocity*time = distance, time is being used twice? Once in velocity and again in duration of velocity.

I guess I am asking the why is c squared there.

Instead of thinking of it like -c^2t^2 think of it as -(ct)^2. Here, you can see the c*t is referring to velocity*time, so we get a distance. So, just like the other terms in that equation, we need to square it.

 

[nitsuj]

Maybe I need to ask it differently,

Why does c have to be squared (why are the terms squared in the first place)?

I'm pretty sure I've heard it has to do with the units (physical units I assume). I am just trying to piece that together.

 

[guss]

Maybe I need to ask it differently,

Why does c have to be squared (why are the terms squared in the first place)?

I'm pretty sure I've heard it has to do with the units (physical units I assume). I am just trying to piece that together.

Nope, it's just the pythagorean theorem. The distance between any two events in a two-dimensional world is √(x² + y²). In a three dimensional world with an "imaginary" time dimension being measured in seconds, it's √(x² + y² + z² - (ct)²)

 

[nitsuj]

Cool stuff, thanks guss.

I still can't see it though.

lets say the value for cordinate x is 3 and y is 5. What is the "function" of this part; (3+3+3)+(5+5+5+5+5).

Asked differently,
Why does the value of length along a dimension have to be squared in order to calculate the distance to an other value of length along a different dimension. (oohhh triangulating )

Maybe if I play with right angle triangles looking for this it'd make it more clear.

 

[Dale]

"Imaginary time works in inertial frames in SR, but doesn't generalize to non inertial frames or curved space times."

I don't know GR at all, does that statement mean equivalence isn't so simple in GR? Are there no inertial frames in GR (i interprut curved spacetime as acceleration of somesort)? Said differently does GR ruin this awsome symmetry thing? (specifically complicates it).

In introductory SR you use only inertial coordinate systems and generally simple algebra. Once you are a little more advanced you combine space and time into a single geometric structure, spacetime. There are several ways to do this.

One is to use a Euclidean line element $ds^2 = dx_0^2 + dx_1^2 + dx_2^2 + dx_3^2$ with coordinates $(x_0,x_1,x_2,x_3)=(i c t, x, y, z)$ where i is the imaginary number $i=\sqrt{-1}$. This is the "imaginary time approach". It is nice because the coordinates all have units of length and it allows you to use the familiar Euclidean line element to measure distances.

If you expand the line element then you can see immediately that it is equal to $ds^2= -c^2 dt^2 + dx^2 + dy^2 + dz^2$. With this approach the coordinates can be stripped of their units and the units can be collected in an object called the metric.
$$g=\left( \begin{array}{cccc} -c^2 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right)$$
So now the line element can be written in terms of the metric $ds^2=g \cdot dx \cdot dx$ where $dx=(dx_0,dx_1,dx_2,dx_3)=(dt,dx,dy,dz)$. This seems a little more cumbersome, but it is clearly equivalent. It also gets rid of the imaginary time thing.

However, at this point, there is no solid reason to prefer one approach over another, it is only once you go to non-inertial coordinate systems that the metric approach becomes truly useful. For instance, suppose that you wanted to use a rotating coordinate system. Now, the line element is: $ds^2=-(1-r^2\omega^2)dt^2+2\omega r^2 dt d\theta + dr^2 + r^2 d\theta^2 + dz^2$. This fits easily into the metric approach with a metric of
$$g=\left( \begin{array}{cccc} r^2 \omega ^2-1 & 0 & r^2 \omega & 0 \\ 0 & 1 & 0 & 0 \\ r^2 \omega & 0 & r^2 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right)$$
and coordinates of $(x_0,x_1,x_2,x_3)=(t,r,\theta,z)$, but it cannot fit into the imaginary time approach without converting back to inertial coordinates.

Once you add gravity it becomes impossible to find a set of coordinates which cover the entire space and are everywhere inertial, so in that case you are forced to abandon the imaginary time approach.

That is what I meant about the imaginary time concept not generalizing.

 

[sorax123]

if you draw a spacetime diagram, with time (ct) with c being the speed of light "calibrating speed" on the y axis, and space (x) on the x-axis then you can display this. Pythagorean theorem would lead to s²=x²+(ct)². This gives a setup which gives (in the way of the axes) x²+y²=s². This, when drawn on a spacetime diagram, gives a circle, which would mean that the same event could happen in both the past and present of another. This would obviously violate causality. To explain this, imagine an event 0, which is situated at the origin, and an event A which is s distance in spacetime away, this could be anywhere on the circle, so it could be, as i say both in past and present, depending on how you are relative to it. So the only other option is the minus sign in the pythagorean theorem, as you had a triangle, god knows why that works, it just does. It allows hyperbolic lines on the diagram instead of a circle, if you draw 4 lines which go through the origin and bisect the right angle between the axes, then you find that if you drew the hyperbolas they would track these 45 degree lines. If you drew the 4 hyperbola in between the lines, you would see that 1 is in the past only, one is in the present only, and 2 cross the space axis, so still create the problem of viloation of causality; however, simple y=mx+c shows use that the gradient of the 45 degree lines are 1, and therefore represent the speed c in space. If any line is shallower than this, then it is traveling faster than c. So, if nothing can travel faster than c, the 2 hyperbola that cross the spaces axes are invalid, as no event on them can effect the event at the origin. This proves that hyperbolic spacetime (minkowski space time) with the amended pythagorean formula just works.
At the end of the day, if something works, it just does. This is just a theory remember, it could be that spacetime is some very odd geometry that we have no clue about, but for the moment this hyperbolic way seems viable and so we stick with it.
Hope this gives some insight as to why the minus sign is used. :)

 

[nitsuj]

Thanks Dalespam!

I can see where I'm at now as far as what I know of SR.

Still wrapping my head around dimensions and physical units.

So looks like from a math perspective I am in between Pythagorean theorem and intervals, from a concept perspective pretty sure I understand intervals. I'd like to understand what a metric is/does. That funky looking equation/metric doesn't seem as intimidating now that you explained what it represents. And rather clearly.

The non inertial one is a little over the top for me yet.

 

[elfmotat]

Cool stuff, thanks guss.

I still can't see it though.

lets say the value for cordinate x is 3 and y is 5. What is the "function" of this part; (3+3+3)+(5+5+5+5+5).

Asked differently,
Why does the value of length along a dimension have to be squared in order to calculate the distance to an other value of length along a different dimension. (oohhh triangulating )

Maybe if I play with right angle triangles looking for this it'd make it more clear.

This is just the Pythagorean Theorem (which I'm assuming you're probably familiar with).

In 2-dimensional space, the distance between two points is given by a 2-dimensional version of the Pythagorean Theorem:

If there are two points, (x₁,y₁) and (x₂,y₂), and we define Δx=x₂-x₁ and Δy=y₂-y₁, then the distance between the two points (which we'll call d) is given by:

$$d^2=(\Delta x)^2+(\Delta y)^2$$Similarly, in 3-D space the distance between two points is given by a 3-D version of the Theorem:

If we have two points, (x₁,y₁,z₁) and (x₂,y₂,z₂), then the distance between the two points is given by:

$$d^2=(\Delta x)^2+(\Delta y)^2+(\Delta z)^2$$If we have two events in 4-D spacetime, (t₁,x₁,y₁,z₁) and (t₂,x₂,y₂,z₂), then the interval between these two events (which we'll call s) is given by:

$$s^2=-(c\Delta t)^2+(\Delta x)^2+(\Delta y)^2+(\Delta z)^2$$

The reason there is a c in front of the Δt is because Δt has units of time (i.e. seconds, minutes, years, etc.) whereas everything else has units of distance (meters, feet, miles, etc.). The speed of light, c, acts as a conversion factor between distance and time. Multiplying by c therefore gets everything into units of distance.

For example, if something takes 1 second of time to occur then we could also say that it takes 1 "light-second" of distance to occur. (A light second is just 1 second multiplied by c.)If you choose to view things from the "time is imaginary" perspective (which I personally do not like), then you can think of time as a sort of imaginary distance. In this case, events would be measured with coordinates such that:

(x₀,x₁,x₂,x₃)=(ict,x,y,z)

**In this case the subscripts on the x's don't refer to individual points - they refer to the name of the coordinate itself.

The interval would then be given by:

$$s^2=(\Delta x_0)^2+(\Delta x_1)^2+(\Delta x_2)^2+(\Delta x_3)^2=-(c\Delta t)^2+(\Delta x)^2+(\Delta y)^2+(\Delta z)^2$$

In this sense you can view the interval as a "distance" between two events in 4-D spacetime.

 

[Dale]

That funky looking equation/metric doesn't seem as intimidating now that you explained what it represents. And rather clearly.

Glad to help. For me, the most intimidating part was just the weird tensor notation with the Einstein summation convention, so I tried to avoid that and just use more standard matrix multiplication. Eventually you cannot easily avoid it, but I didn't go that far.

 

[nitsuj]

Hey elfmotat,

I had learned about c with light clocks and Pythagorean Theorem. That changed my persepctive the most, understanding what speed is. From there I went to spacetime diagrams. somewhere inbetween I forgot (neglected the importance) it is still triangles that are used in simple spacetime diagrams. Said differently, I back tracked a little


I understand the concepts of spacetime in SR (upto intervals), but not at a fundamental level. I had an oversight, regarding observations of distance, velocity ect, being triangulation.

With dimensions as the "fixed baseline", the observation being the hypotenuse. (i think that is how it is).

Which makes sense in that those "fixed" baselines (dimensions) are, for lack of a better word, effected by motion. Measurements of (along) those dimensions are in turn effected, yielding different calculated values for the hypotenuse (observation).

To perform measurements for an observation, regardless of relative motion, the observer "sets up" the "fixed" baseline (dimensions) at right angles to each other, so the hypotenuse can be calculated. Is that right? ( i hope so, i like thinking of it this way)


I really like that 3D image showing triangulating. I am going to spend some time trying to understand that. From the perspective above (triangulating), the way the time coordinate is treated still eludes me (can't picture it). It maybe I am just not familiar enough with spacetime diagrams yet.

Right now, it looks like time is negative in that equation simply because once time is equated to length the last part of the relationship is it is opposite. Time of course is not length (and not measured the same), if I'm allowed to say it is equal to but opposite of length, then it looks like once the units are the same (equal), the only missing part of the relationship is it must be opposite. I hope that's not too goofy

I wonder if hypotenuse to SR is = curves to GR.

 

[Naty1]

I thought it is negative because time has an inverse relationship with length.

What do you mean?

To me it is not logical if one were starting out from scratch, but makes sense within the context of the necessary math...as the quotes I posted above imply.


The sign is negative because that's what keeps transforms Lorentz invarient.

 

[nitsuj]

What do you mean?

To me it is not logical if one were starting out from scratch, but makes sense within the context of the necessary math...as the quotes I posted above imply.


The sign is negative because that's what keeps transforms Lorentz invarient.

I don't understand what you are saying. From which context do you think my comment is from? I know little of the math, but have read a number of times that intervals are invarient. It is possible to have intervals invarient so simply because of the relationship between time and length.

below is from an ealier post of mine that explains what I mean.

"Right now, it looks like time is negative in that equation simply because once time is equated to length the last part of the relationship is it is opposite. Time of course is not length (and not measured the same), if I'm allowed to say it is equal to but opposite of length, then it looks like once the units are the same (equal), the only missing part of the relationship is it must be opposite. I hope that's not too goofy."

I understand the term "inverse" to be the same as "equal but opposite". This all comes from knowing that intervals are invarient. Looks like there is a root word in there.

If you read my posts in this thread, it should be clear that the comment "The sign is negative because that's what keeps transforms Lorentz invarient." is vague.

But at least we agree on the function, invariance.

 

[bobc2]

Why in deriving invariant distance in space time we use Pythagorean Theorem with a negative sign?

moatasim23, here is a little different slant on how you could get that negative sign. I've used a space-time sketch in a way that allows you to start with the Pythagorean theorem, then solve for a distance. The blue guy can compute the distance red has traveled along red's X4 world line, but blue must now use a negative sign as shown in the derivation.

One way of looking at it in this context, is that you are not computing the distance for the hypotenuse of a right triangle, but rather the distance for one leg of a right triangle.

 

[smodak]

Here is how Sean Carroll explains the space time interval:

http://ned.ipac.caltech.edu/level5/March01/Carroll3/Carroll1.html

Carroll says

An event is defined as a single moment in space and time, characterized uniquely by (t, x, y, z). Then, without any motivation for the moment, let us introduce the spacetime interval between two events:

Then he writes the spacetime interval equation. He does not really explain how he got there. Or do i not understand what he says?