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The spacetime interval in General Relativity


geordief

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I am familiar with the relationship between spatial and temporal measurements in Special Relativity.

s^2 =r^2 -(ct)^2

 

Does this relationship also apply in General Relativity or is there perhaps another relationship that applies when gravity applies rather than relative motion?

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46 minutes ago, geordief said:

I am familiar with the relationship between spatial and temporal measurements in Special Relativity.

s^2 =r^2 -(ct)^2

 

Does this relationship also apply in General Relativity or is there perhaps another relationship that applies when gravity applies rather than relative motion?

 

It still applies, but not in the same way.

In SR the expression is global, which is why we can use it for spacemen visiting Alpha Centauri etc.

In GR it is strictly local, a point function which varies from point to point as gravity varies.

s is the single scalar invariant for SR.

In GR there are up to 14 invariants here is a pdf about this.

https://www.researchgate.net/publication/243698965_Invariants_of_General_Relativity_and_the_Classification_of_Spaces

 

Edit:

On second thought it might be better if I worded this differently.

That is the trouble when you try to oversimplify a complicated matter.

I will add to this later today as it might give a false impression.

 

 

Edited by studiot
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On 4/21/2020 at 1:40 PM, geordief said:

Does this relationship also apply in General Relativity or is there perhaps another relationship that applies when gravity applies rather than relative motion?

As studiot correctly said, the spacetime element is of course also defined in GR (it is, in a sense, the fundamental quantity of the model), the main differences being that in GR it is a purely local quantity, and the coefficients are no longer constants, but functions of the coordinates. Very simply put, this means that, in the presence of gravity, the relationship between two given events will vary depending on where and/or when the events are located.

Note that SR is simply a special case of GR, for scenarios where gravitational effects are negligible. They are not two separate models.

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Let me see if I can really dumb it down ( not for you geordief ) for the layman.

In two dimensions, the general form of your relation is   ax^2 + by^2 = r^2  ( generalized Pythagoras ),
and in SR, the coefficients a = b = 1    as it is 'flat' ( Euclidian )spacetime.

In GR, we are no longer dealing with 'flat' or Euclidian spaces. So the coefficients a and b have values other than 1.
For positive curvature, the angles of a triangle add up to more than 180 deg. and the circumference of a circle is less than 2(pi)r, so the coefficients would be less than 1, and for negative curvature, where the angles of a triangle add to less than 180 deg. and the circumference of a circle is greater than 2(pi)r, the coefficients would be greater than 1.
And in the case where your graph paper has a compound curvature ( different in differing directions ), a and b are not equivalent.

IOW, the only difference between SR and GR is curvature, so you must use a more general form of the Pythagoras relation.

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2 minutes ago, MigL said:

IOW, the only difference between SR and GR is curvature, so you must use a more general form of the Pythagoras relation.

I bet Einstein started out thinking that. "How hard can it be, ja?"

After a year or so of work, he realised it was getting pretty complicated. "Sheisse! Grossman was right! I need to learn about tensors."

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Yes ( ja ).

 I could add that the distance between points ( r ) is a function of the local curvature of the space, and the metric tensor is then, the derivative of that function or the infinitesimal distance on that space.

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15 minutes ago, Strange said:

I bet Einstein started out thinking that. "How hard can it be, ja?"

After a year or so of work, he realised it was getting pretty complicated. "Sheisse! Grossman was right! I need to learn about tensors."

I was reading some stuff yesterday on what it takes to really understand GR, and tensors don't really get you there if you want to understand purely mathematically. There seems to be two types: those that can work with it and those that understand it.

Edited by StringJunky
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23 minutes ago, StringJunky said:

I was reading some stuff yesterday on what it takes to really understand GR, and tensors don't really get you there if you want to understand purely mathematically. There seems to be two types: those that can work with it and those that understand it.

Sounds interesting. Einstein had obviously realised what he needed to capture mathematically before he understood that tensors were the best tool for the job.

I have probably mentioned this before, but John Baez has written a great introduction to the concepts behind the Einstein equations using relatively simple math: http://math.ucr.edu/home/baez/einstein/

Quote

While there are many excellent expositions of general relativity, few adequately explain the geometrical meaning of the basic equation of the theory: Einstein's equation. Here we give a simple formulation of this equation in terms of the motion of freely falling test particles. We also sketch some of the consequences of this formulation and explain how it is equivalent to the usual one in terms of tensors. 

 

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Something which has always bothered me...

I understand that the 4dimensional line element

ds^2 = dx^2 + dy^2 + dz^2 - c^2dt^2

is invariant under a Lorentz transform, and that the 'time', on the 4dimensional space-time manifold,
needs to be expressed as a co-ordinate distance, hence    x, y, z, and ct.
But the fact that   c^2dt^2  is negative, implies the dimension is actually  ict , an imaginary quantity.

What exactly is the reason for making it imaginary ?
Is it as simple as the fact that the complex plane is orthogonal to the real plane ( of the x,y,z dimensions ) ?
Or is there a reason other than orthogonality requirements ?

 

 

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2 hours ago, MigL said:

Something which has always bothered me...

I understand that the 4dimensional line element

ds^2 = dx^2 + dy^2 + dz^2 - c^2dt^2

is invariant under a Lorentz transform, and that the 'time', on the 4dimensional space-time manifold,
needs to be expressed as a co-ordinate distance, hence    x, y, z, and ct.
But the fact that   c^2dt^2  is negative, implies the dimension is actually  ict , an imaginary quantity.

What exactly is the reason for making it imaginary ?
Is it as simple as the fact that the complex plane is orthogonal to the real plane ( of the x,y,z dimensions ) ?
Or is there a reason other than orthogonality requirements ?

 

 

 

18 hours ago, StringJunky said:

I was reading some stuff yesterday on what it takes to really understand GR, and tensors don't really get you there if you want to understand purely mathematically. There seems to be two types: those that can work with it and those that understand it.

Yes it is easy to loose sight of the Physics if you simply stick to mathematical formulae.

 

Yes it is necessary to write the equation in that form to express the fact that if a light pulse is emitted at the origin of the frame at t = 0 then there is a 3D spherical expanding wavefront of light with the equation  

x2 + y2 + z2 = r2.   for any given r.

That is the r2 is on the other side of the equation from the 3D spatial coordinates.

If we now take t as a parameter we recover the negative sign.

Of course to generate a line segment with the Euclidian norm we have to use ict not t.

 

It should be noted that the second postulate means that all inertial observers see an expanding sphere in their own frame.
This allows us to create the equation

 

x2 + y2 + z2 - c2t2 = x'2 + y'2 + z'2 - c2t'2

Where the primes on the coordinates denote the second inertial frame.

Leading to the Lorenz transformation and other consequences.

Does this help?

 

Edited by studiot
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But what, if any, is the result of making the coordinate distance in the time dimension, imaginary ?

Is there any Physical 'reality' associated with the  i in  ict ?

The mathematical formulae make sense with  i  in ict , but I am losing sight of the underlying Physics.
Does the  i  have any Physical meaning ?

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6 hours ago, MigL said:

But what, if any, is the result of making the coordinate distance in the time dimension, imaginary ?

Is there any Physical 'reality' associated with the  i in  ict ?

The mathematical formulae make sense with  i  in ict , but I am losing sight of the underlying Physics.
Does the  i  have any Physical meaning ?

 

The point I was trying to make is that (as Einstein said) we should get the Physics right first and adjust the maths to fit.

Ttwo observers, A and B with B moving past A at velocity v.
At the instant B is coincident with A,  a light pulse is generated.

The (postulated) Physics of Relativity says that :

Postulate (2) says.

Both A and B see the light pulse moving away in all directions at the same velocity c, as measured in their respective frames.

Since the velocity is the same in all directions, both A and B will see an expanding sphere of light surface.

Postulate (1) says both must see the same physics ie see the same expanding shape.

So all this is self consistent.

But it is not consistent with a Galilean transformation when you try to write the equations of a sphere in one frame and transform them into the other frame via a Galilean transformation.
This is incompatible with postulate (1).

 

Edited by studiot
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So this is just a case of using the complex number i  to make the mathematical model fit observational evidence and postulates ?
And it has no other Physical significance ?

Ok, thanks Studiot.

Edited by MigL
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20 hours ago, MigL said:

But the fact that   c^2dt^2  is negative, implies the dimension is actually  ict , an imaginary quantity.

What exactly is the reason for making it imaginary ?

Is that a fact? A spacetime interval can have a length that's positive, negative, or zero, so that value is not always negative.

Also, you're talking about a metric signature (+ + + -) which is only a convention, but so is (- - - +).

A measure of proper time uses the latter. Then, ds^2 is negative for spacelike intervals, and positive for timelike. As for physical significance, could you say that the proper time along a spacelike interval is imaginary?

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