How much is Special Relativity a needed foundation of General Relativity

[Deepak Kapur]

Nothing you've said is worth refuting because you don't understand what you are talking about.

For instance



That is not what GR predicts. Again you base your remarks on misunderstandings.

Not be able to understand the other person is indeed lack of understanding.

I never said 'it' predicts. It can be one of the implications. A clock so slow (as it apperas to an observer) that all the processes virtually coming to a stand still.

 

[Rasalhague]

Why is it same in all the inertial frames when there is no absolute definition of time and distance (time dialtion, length contraction) even in the inertial frames. As a matter of fact how can (or should) speed be determined with so much of relativity around (even in inertial frames of reference).

Speed is a ratio between a distance (how far travelled) and a time (how long it takes). Distances and times change when we switch from measuring them in one inertial reference frame to another, but the ratio between the distance light travels and the time it takes stays the same.

Suppose we measure the speed of light through a vacuum (I'll just call it "the speed of light" from now on) in one inertial reference frame (a non-accelerating spacetime coordinate system, with three coordinates for space and one for time, covering a region small enough and brief enough that the effects of gravity are negligible) and find it to be a certain value c.

I'll use the letter v to stand for the speed of some other inertial reference frame moving parallel to the pulse of light, as measured in our original reference frame. I'll use another variable, defined like this:

\gamma = \frac{1}{\sqrt{1-\left ( \frac{v}{c} \right )^2}}.

The Greek letter \gamma, called "gamma", is just a handy symbol, conventionally used to simplify the equations below. Relativity predicts that intervals of space and time will differ in our new inertial reference frame according to the following equations,

\Delta x' = \gamma \left ( \Delta x - v \; \Delta t \right )

and

\Delta t' = \gamma \left ( \Delta t - \frac{v}{c^2} \; \Delta x \right )

where \Delta x is some distance, for example the distance between the place at which a pulse of light is emitted and the place where it's received, measured according to our original reference frame (that is, as measured by rulers at rest in that reference frame), and \Delta x' is the distance between those same points measured according to our new reference frame. Similarly, \Delta t is an interval of time, for example the time between emission and reception of a pulse of light according to our original reference frame (that is, as measured by clocks at rest in that reference frame), and \Delta t' the time between these events according to our new reference frame.

The speed of light in our original reference frame is

c = \frac{\Delta x}{\Delta t},

the distance the light travels divided by the time it takes to travel that distance. The speed of light in our new reference frame is therefore

\frac{\Delta x'}{\Delta t'}=\frac{\gamma \left ( \Delta x - v \; \Delta t \right )}{\gamma \left ( \Delta t - \frac{v}{c^2} \; \Delta x \right )}=\frac{\frac{\Delta x}{\Delta t}-v}{1-\frac{v}{c^2}\frac{\Delta x}{\Delta t}}=\frac{c-v}{1-\frac{v}{c}}=\frac{c(c-v)}{c-v}=c

So, without paradox, the speed of light is the same in any inertial reference frame moving at some non-zero speed relative to our original inertial reference frame, even though neither the distance the light traveled nor the time it took are the same as they were in the original inertial reference frame.

 

[Fredrik]

Why is it same in all the inertial frames when there is no absolute definition of time and distance (time dialtion, length contraction) even in the inertial frames. As a matter of fact how can (or should) speed be determined with so much of relativity around (even in inertial frames of reference).

You should think of spacetime as an abstract set of points. Those points are called events. Inertial frames are functions that assign 4-tuples of real numbers (t,x,y,z) to events. Those numbers (the coordinates of events) are used to define velocity the same way as in pre-relativistic physics:

\vec r=(x,y,z)

\vec v=\frac{d\vec r}{dt}

Speed is the magnitude of the velocity:

v=|\vec v|=\sqrt{v_1^2+v_2^2+v_3^3}

It's hard to explain why this formula applied to a null geodesic gives us the result =1 in every inertial frame. To really understand it, you'd have to understand what a geodesic is, which requires that you know some differential geometry. So let me skip most of those technicalities and go directly to what we'd end up with if we went through with all those mathematical details. But before we get started, I should tell you that there's a very natural way to associate an inertial frame with the motion and spatial orientation of an observer that's moving with constant velocity forever. This allows us to identify "observers" with "frames".

We can pick an inertial frame (any inertial frame will do) and use it to identify Minkowski spacetime with \mathbb R^4. Because an inertial frame was used in this identification, we can think of this copy of \mathbb R^4, as representing the point of view of the observer associated with that inertial frame. We clearly need a formula that tells us how to calculate the coordinates that one observer assigns to an event, given the coordinates that another observer assigns to the same event. In 1+1 dimensions (let's keep it as simple as possible), it's a function

x\mapsto \Lambda x+a

where x and a are 2×1 matrices and

\Lambda=\gamma\begin{pmatrix}1 & -v\\ -v & 1\end{pmatrix}

\gamma=\frac{1}{\sqrt{1-v^2}}

This is in units such that c=1. Such a function is called a Poincaré transformation or a Lorentz transformation. (The term Lorentz transformation is often reserved for the case a=0). We can use this to calculate the speed of light in another inertial frame. The simplest possible scenario we can consider is when both inertial frames have the same origin, and the light we're considering is at x=0 when t=0 (that's in both frames, since they have the same origin). Now it's a really easy exercise to show that the curve that represents the motion of the light looks the same in both coordinate systems (a straight line through the origin with slope 1).

This may not be very satisfying unless you know why we're working with Lorentz transformations. As I said before, the requirement that inertial observers must describe each other's motion as straight lines is much stronger than it looks, and more or less forces us to only consider a spacetime where a coordinate change between inertial frames is done using a Lorentz transformation or a Galilei transformation. But we can actually ignore that completely and just say that using Minkowski spacetime (and therefore the Lorentz transformation) is an axiom of the theory, and is ultimately justified by the fact that theories of matter and interactions in Minkowski spacetime predict the results of experiments so well.

And anyway, you weren't really asking why the speed of light is the same in all inertial frames. You were just trying to find out how it's possible at all. I hope the above shed some light on it, even though I left out a lot. You might also want to have a look at pages 8-9 in Schutz. Light has speed 1 in all inertial frames because a Lorentz transformation tilts the x-axis by the same amount as the t axis. (Schutz's argument on those pages is actually that the invariance of the speed of light implies that tilting of the x axis, but the diagrams would have been the same even if he had been arguing for the converse statement).

 

[Deepak Kapur]

Speed is a ratio between a distance (how far travelled) and a time (how long it takes). Distances and times change when we switch from measuring them in one inertial reference frame to another, but the ratio between the distance light travels and the time it takes stays the same.

Suppose we measure the speed of light through a vacuum (I'll just call it "the speed of light" from now on) in one inertial reference frame (a non-accelerating spacetime coordinate system, with three coordinates for space and one for time, covering a region small enough and brief enough that the effects of gravity are negligible) and find it to be a certain value c.

I'll use the letter v to stand for the speed of some other inertial reference frame moving parallel to the pulse of light, as measured in our original reference frame. I'll use another variable, defined like this:

\gamma = \frac{1}{\sqrt{1-\left ( \frac{v}{c} \right )^2}}.

The Greek letter \gamma, called "gamma", is just a handy symbol, conventionally used to simplify the equations below. Relativity predicts that intervals of space and time will differ in our new inertial reference frame according to the following equations,

\Delta x' = \gamma \left ( \Delta x - v \; \Delta t \right )

and

\Delta t' = \gamma \left ( \Delta t - \frac{v}{c^2} \; \Delta x \right )

where \Delta x is some distance, for example the distance between the place at which a pulse of light is emitted and the place where it's received, measured according to our original reference frame (that is, as measured by rulers at rest in that reference frame), and \Delta x' is the distance between those same points measured according to our new reference frame. Similarly, \Delta t is an interval of time, for example the time between emission and reception of a pulse of light according to our original reference frame (that is, as measured by clocks at rest in that reference frame), and \Delta t' the time between these events according to our new reference frame.

The speed of light in our original reference frame is

c = \frac{\Delta x}{\Delta t},

the distance the light travels divided by the time it takes to travel that distance. The speed of light in our new reference frame is therefore

\frac{\Delta x'}{\Delta t'}=\frac{\gamma \left ( \Delta x - v \; \Delta t \right )}{\gamma \left ( \Delta t - \frac{v}{c^2} \; \Delta x \right )}=\frac{\frac{\Delta x}{\Delta t}-v}{1-\frac{v}{c^2}\frac{\Delta x}{\Delta t}}=\frac{c-v}{1-\frac{v}{c}}=\frac{c(c-v)}{c-v}=c

So, without paradox, the speed of light is the same in any inertial reference frame moving at some non-zero speed relative to our original inertial reference frame, even though neither the distance the light traveled nor the time it took are the same as they were in the original inertial reference frame.

Quite helpful! (though equations were not visible)

Since it's a forum, I presume that even if I ask (relply to) extremely large number of questions, I am not going to pester anybody.

1. Now coming to our galaxy (which is accelerating at tremendous speed). Is the speed of light same at every point in the galaxy (between galaxies for that matter). Is this speed equal to the speed of light in an inertial frame. Why?

2. Why does light itself not experience time-dialation. If you say its massless, it's at least energy that has probably resulted from mass annihilation. So, it's "something" after all and don't forget about the 'photons' of light that act like particles.

 

[Fredrik]

(though equations were not visible)

Search the feedback forum. I think others have had the same problem in the past. Maybe you're just using a really old browser and need to upgrade.

1. Now coming to our galaxy (which is accelerating at tremendous speed). Is the speed of light same at every point in the galaxy (between galaxies for that matter). Is this speed equal to the speed of light in an inertial frame. Why?

Our galaxy isn't accelerating significantly. Special relativity holds "locally", in small regions of spacetime. When we're talking about far away galaxies, we need general relativity. At any location in spacetime (in this galaxy or any other), we can consider coordinate systems that I'll call "local inertial frames" (unfortunately there doesn't seem to be a standard name for them). They are coordinate systems that can be associated with the motion of massive particles in a natural way, and the speed of light at the origin of the coordinate system is the same in any of them. However, things are not so simple when you compare things that are far away from each other.

Edit: Apparently it's complicated enough to confuse me too. I had to edit my post to rewrite this part.

The solutions of GR that describe homogeneous and isotropic universes are called FLRW solutions. When we're working with one of them, it's convenient to use a coordinate system in which the galaxies are more or less stationary (they'll have speeds of a few hundred km/s relative to a nearby objects that stay at constant position coordinates). In this coordinate system, it's convenient to define another kind of "speed" to be a measure of how fast distant objects are moving away from each other. Define d(t) to be the proper distance between the two objects in a hypersurface of constant coordinate time t, and define the new kind of speed to be d'(t). The expansion of space ensures that the speed is non-zero for two objects that stay at constant position coordinates, and for distant galaxies, it can even be much higher than c.

This doesn't contradict the invariance of the speed of light, because that's a statement about a different kind of speed in a different kind of coordinate system. If we apply that same definition of speed to the light emitted from a star in a distant galaxy, then it's speed clearly isn't going going to be c. But that light does move at c in a local inertial frame associated with the motion of the star that emits it, no matter how far away it is.

2. Why does light itself not experience time-dialation. If you say its massless, it's at least energy that has probably resulted from mass annihilation. So, it's "something" after all and don't forget about the 'photons' of light that act like particles.

One answer is that particles simply don't have experiences, but that's not the whole story, because we sometimes talk about a massive particle's point of view. When we do, we're specifically referring to the description of events in spacetime using the inertial frame we'd associate with the massive particle's motion. The concept of "the photon's point of view" doesn't quite make sense because there's no natural way to associate an inertial frame with its motion. The problem is that the method we'd like to use to determine which subset of spacetime to call "space, at time t" doesn't work for photons. (This method for massive particles is described in the part of Schutz's book that I linked to in my previous post).

 

[Rasalhague]

A quick fix if you're still having trouble with seeing the equations in earlier posts: you could try copying the LaTeX code (try left clicking on the equation; that should open a window with the code) and pasting it into an online LaTeX editor such as this one:

http://www.codecogs.com/components/equationeditor/equationeditor.php

Or if that doesn't work, there are lots of others.

 

[Deepak Kapur]

Search the feedback forum. I think others have had the same problem in the past. Maybe you're just using a really old browser and need to upgrade.


Our galaxy isn't accelerating significantly. Special relativity holds "locally", in small regions of spacetime. When we're talking about far away galaxies, we need general relativity. At any location in spacetime (in this galaxy or any other), we can consider coordinate systems that I'll call "local inertial frames" (unfortunately there doesn't seem to be a standard name for them). They are coordinate systems that can be associated with the motion of massive particles in a natural way, and the speed of light at the origin of the coordinate system is the same in any of them. However, things are not so simple when you compare things that are far away from each other.

Edit: Apparently it's complicated enough to confuse me too. I had to edit my post to rewrite this part.

The solutions of GR that describe homogeneous and isotropic universes are called FLRW solutions. When we're working with one of them, it's convenient to use a coordinate system in which the galaxies are more or less stationary (they'll have speeds of a few hundred km/s relative to a nearby objects that stay at constant position coordinates). In this coordinate system, it's convenient to define another kind of "speed" to be a measure of how fast distant objects are moving away from each other. Define d(t) to be the proper distance between the two objects in a hypersurface of constant coordinate time t, and define the new kind of speed to be d'(t). The expansion of space ensures that the speed is non-zero for two objects that stay at constant position coordinates, and for distant galaxies, it can even be much higher than c.

This doesn't contradict the invariance of the speed of light, because that's a statement about a different kind of speed in a different kind of coordinate system. If we apply that same definition of speed to the light emitted from a star in a distant galaxy, then it's speed clearly isn't going going to be c. But that light does move at c in a local inertial frame associated with the motion of the star that emits it, no matter how far away it is.


One answer is that particles simply don't have experiences, but that's not the whole story, because we sometimes talk about a massive particle's point of view. When we do, we're specifically referring to the description of events in spacetime using the inertial frame we'd associate with the massive particle's motion. The concept of "the photon's point of view" doesn't quite make sense because there's no natural way to associate an inertial frame with its motion. The problem is that the method we'd like to use to determine which subset of spacetime to call "space, at time t" doesn't work for photons. (This method for massive particles is described in the part of Schutz's book that I linked to in my previous post).

What would be the scenario if we consider that the galaxies are not moving at all but the space between them is expanding.

Mind you, this is not the opposite of what we have discussed before.

 

[Fredrik]

I'm not sure I understand the question (because it seems to me that the answer is in the text you quoted).