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Rigorous formulation of special relativity.
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I want to understand the theory of special relativity and I am reading from Resnick's Introduction to Special Relativity.
But I want to prove everything rigorously.
$newcommandRmathbf R$
$newcommandmcmathcal$
$newcommandvpvarphi$
Of course, this requires the definitions to be rigorous.
So here is an attempt, which is really lame right now.
I am hoping those who already understand the theory can add to the formalism which I describe below.
I am trying to understand the theory for the simplest case where there is one spatial dimension and one temporal dimension.
Definition.
Spacetime is a $2$-dimensional smooth manifold homeomorphic to $R^2$ whose elements are called events.
We will denote spacetime as $E$.
Definition.
A frame of reference is a global smooth chart $vp:Eto R^2$.
(The above definition of a frame of reference allows for all sorts of crazy things to be called frames of reference. So it needs more work.)
We postulate that spacetime comes equipped with certain spacial kind of subsets called photons, and the collection of all photons will be denoted by $mc P$.
Definition.
A frame of reference $vp:Eto R^2$ will be called inertial if for each $Pin mc P$, the set $vp(P)$ is a straight line with slope $1$.
(What we are tying to say is that light travels with unit speed in an inertial frame).
Now I want to formalize Einstein's postulate that light travels with the same speed in all inertial frames.
For this first I need to defined what is meant by the velocity of a frame with respect to another.
Let $vp$ and $psi$ be two frames.
We want to track the origin of $psi$ as seen by $vp$.
If $s(t)$ is the spatial location of the origin of $psi$ when $vp$'s clock reads $t$ (that is amongst the events which have their second coordinate equal to $t$ under $vp$, $s(t)$ is the first coordinate under $vp$ of that event whose first coordinate is zero under $psi$) the velocity of $psi$ with respect to $vp$ is $s'(t)$.
(Of course, this quantity may not even exist. or may not be well-defined because there could be multiple candidates for $s(t)$).
So we need to say some more words, which I am not sure about.
At any rate. Running with it, we have:
Any two inertial frames move at a constant speed with respect to each other.
Okay. So can somebody see as to how to develop this. Or perhaps there is a text where this approach has already been taken.
Just one more thing. I think one needs to insist that is $vp:Eto mathbf R^2$ is an inertial frame, and $ell$ is a line with slope $1$, then $vp^-1(ell)$ is a photon. This is ugly I know. Perhaps there could be given a more intrinsic axiom to the set of all the photons.
What are your thoughts? Thanks.
physics smooth-manifolds special-relativity
asked Jul 24 at 13:02
caffeinemachine
6,07221145
 |Â
show 1 more comment
up vote
1
down vote
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I want to understand the theory of special relativity and I am reading from Resnick's Introduction to Special Relativity.
But I want to prove everything rigorously.
$newcommandRmathbf R$
$newcommandmcmathcal$
$newcommandvpvarphi$
Of course, this requires the definitions to be rigorous.
So here is an attempt, which is really lame right now.
I am hoping those who already understand the theory can add to the formalism which I describe below.
I am trying to understand the theory for the simplest case where there is one spatial dimension and one temporal dimension.
Definition.
Spacetime is a $2$-dimensional smooth manifold homeomorphic to $R^2$ whose elements are called events.
We will denote spacetime as $E$.
Definition.
A frame of reference is a global smooth chart $vp:Eto R^2$.
(The above definition of a frame of reference allows for all sorts of crazy things to be called frames of reference. So it needs more work.)
We postulate that spacetime comes equipped with certain spacial kind of subsets called photons, and the collection of all photons will be denoted by $mc P$.
Definition.
A frame of reference $vp:Eto R^2$ will be called inertial if for each $Pin mc P$, the set $vp(P)$ is a straight line with slope $1$.
(What we are tying to say is that light travels with unit speed in an inertial frame).
Now I want to formalize Einstein's postulate that light travels with the same speed in all inertial frames.
For this first I need to defined what is meant by the velocity of a frame with respect to another.
Let $vp$ and $psi$ be two frames.
We want to track the origin of $psi$ as seen by $vp$.
If $s(t)$ is the spatial location of the origin of $psi$ when $vp$'s clock reads $t$ (that is amongst the events which have their second coordinate equal to $t$ under $vp$, $s(t)$ is the first coordinate under $vp$ of that event whose first coordinate is zero under $psi$) the velocity of $psi$ with respect to $vp$ is $s'(t)$.
(Of course, this quantity may not even exist. or may not be well-defined because there could be multiple candidates for $s(t)$).
So we need to say some more words, which I am not sure about.
At any rate. Running with it, we have:
Any two inertial frames move at a constant speed with respect to each other.
Okay. So can somebody see as to how to develop this. Or perhaps there is a text where this approach has already been taken.
Just one more thing. I think one needs to insist that is $vp:Eto mathbf R^2$ is an inertial frame, and $ell$ is a line with slope $1$, then $vp^-1(ell)$ is a photon. This is ugly I know. Perhaps there could be given a more intrinsic axiom to the set of all the photons.
What are your thoughts? Thanks.
physics smooth-manifolds special-relativity
asked Jul 24 at 13:02
caffeinemachine
6,07221145
For purposes of doing everything rigorously, let me mention what your question overlooks, namely the theory of symmetric bilinear forms, in particular forms of signature $(n-1,1)$. 2-dimensional spacetime should be more than just a 2-manifold homeomorphic to $mathbb R^2$, it should be a 2-dimensional vector space over $mathbb R$ equipped with a symmetric bilinear form of signature $(1,1)$.
– Lee Mosher
Jul 24 at 13:39
@LeeMosher Can you provide some motivation behind the choice of the inner product? Starting with a vector space structure is something I can accept. But a bilinear form of index 1 seems like a post-hoc artifice.
– caffeinemachine
Jul 24 at 16:12
Actually, a bilinear form of index $1$ is exactly what special relativity is all about. I'll write a little bit of an answer along those lines.
– Lee Mosher
Jul 24 at 16:35
Have you considered rigorous text on Special relativity ?, such as Geometry of Minkowski Spacetime by G. Naber (describe Minkowski space as 4-dim real vector space with nondegenerate sym. bilinear form of index 1) and Barett O'neill' Semi-Riemannian Geometry (more general than Naber, I have not done read this one yet).
– Sou
Jul 25 at 13:32
@Sou Yes I took a look at that book. But it seems to me (and I am pretty green) that there is no motivation given behind the formalism. What I am looking for is a formulation which takes "speed of light is same in all inertial frames" and makes this precise, and then deduce relativity of simultaneity. The text starts with the universe being a $4$-dimensional real vector space with a certain bilinear form. I am unable to see why would anyone think of this formulation. What is the, well, story?
– caffeinemachine
Jul 25 at 15:13
 |Â
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I want to understand the theory of special relativity and I am reading from Resnick's Introduction to Special Relativity.
But I want to prove everything rigorously.
$newcommandRmathbf R$
$newcommandmcmathcal$
$newcommandvpvarphi$
Of course, this requires the definitions to be rigorous.
So here is an attempt, which is really lame right now.
I am hoping those who already understand the theory can add to the formalism which I describe below.
I am trying to understand the theory for the simplest case where there is one spatial dimension and one temporal dimension.
Definition.
Spacetime is a $2$-dimensional smooth manifold homeomorphic to $R^2$ whose elements are called events.
We will denote spacetime as $E$.
Definition.
A frame of reference is a global smooth chart $vp:Eto R^2$.
(The above definition of a frame of reference allows for all sorts of crazy things to be called frames of reference. So it needs more work.)
We postulate that spacetime comes equipped with certain spacial kind of subsets called photons, and the collection of all photons will be denoted by $mc P$.
Definition.
A frame of reference $vp:Eto R^2$ will be called inertial if for each $Pin mc P$, the set $vp(P)$ is a straight line with slope $1$.
(What we are tying to say is that light travels with unit speed in an inertial frame).
Now I want to formalize Einstein's postulate that light travels with the same speed in all inertial frames.
For this first I need to defined what is meant by the velocity of a frame with respect to another.
Let $vp$ and $psi$ be two frames.
We want to track the origin of $psi$ as seen by $vp$.
If $s(t)$ is the spatial location of the origin of $psi$ when $vp$'s clock reads $t$ (that is amongst the events which have their second coordinate equal to $t$ under $vp$, $s(t)$ is the first coordinate under $vp$ of that event whose first coordinate is zero under $psi$) the velocity of $psi$ with respect to $vp$ is $s'(t)$.
(Of course, this quantity may not even exist. or may not be well-defined because there could be multiple candidates for $s(t)$).
So we need to say some more words, which I am not sure about.
At any rate. Running with it, we have:
Any two inertial frames move at a constant speed with respect to each other.
Okay. So can somebody see as to how to develop this. Or perhaps there is a text where this approach has already been taken.
Just one more thing. I think one needs to insist that is $vp:Eto mathbf R^2$ is an inertial frame, and $ell$ is a line with slope $1$, then $vp^-1(ell)$ is a photon. This is ugly I know. Perhaps there could be given a more intrinsic axiom to the set of all the photons.
What are your thoughts? Thanks.
physics smooth-manifolds special-relativity
asked Jul 24 at 13:02
caffeinemachine
6,07221145
I want to understand the theory of special relativity and I am reading from Resnick's Introduction to Special Relativity.
But I want to prove everything rigorously.
$newcommandRmathbf R$
$newcommandmcmathcal$
$newcommandvpvarphi$
Of course, this requires the definitions to be rigorous.
So here is an attempt, which is really lame right now.
I am hoping those who already understand the theory can add to the formalism which I describe below.
I am trying to understand the theory for the simplest case where there is one spatial dimension and one temporal dimension.
Definition.
Spacetime is a $2$-dimensional smooth manifold homeomorphic to $R^2$ whose elements are called events.
We will denote spacetime as $E$.
Definition.
A frame of reference is a global smooth chart $vp:Eto R^2$.
(The above definition of a frame of reference allows for all sorts of crazy things to be called frames of reference. So it needs more work.)
We postulate that spacetime comes equipped with certain spacial kind of subsets called photons, and the collection of all photons will be denoted by $mc P$.
Definition.
A frame of reference $vp:Eto R^2$ will be called inertial if for each $Pin mc P$, the set $vp(P)$ is a straight line with slope $1$.
(What we are tying to say is that light travels with unit speed in an inertial frame).
Now I want to formalize Einstein's postulate that light travels with the same speed in all inertial frames.
For this first I need to defined what is meant by the velocity of a frame with respect to another.
Let $vp$ and $psi$ be two frames.
We want to track the origin of $psi$ as seen by $vp$.
If $s(t)$ is the spatial location of the origin of $psi$ when $vp$'s clock reads $t$ (that is amongst the events which have their second coordinate equal to $t$ under $vp$, $s(t)$ is the first coordinate under $vp$ of that event whose first coordinate is zero under $psi$) the velocity of $psi$ with respect to $vp$ is $s'(t)$.
(Of course, this quantity may not even exist. or may not be well-defined because there could be multiple candidates for $s(t)$).
So we need to say some more words, which I am not sure about.
At any rate. Running with it, we have:
Any two inertial frames move at a constant speed with respect to each other.
Okay. So can somebody see as to how to develop this. Or perhaps there is a text where this approach has already been taken.
Just one more thing. I think one needs to insist that is $vp:Eto mathbf R^2$ is an inertial frame, and $ell$ is a line with slope $1$, then $vp^-1(ell)$ is a photon. This is ugly I know. Perhaps there could be given a more intrinsic axiom to the set of all the photons.
What are your thoughts? Thanks.
physics smooth-manifolds special-relativity
asked Jul 24 at 13:02
caffeinemachine
6,07221145
asked Jul 24 at 13:02
caffeinemachine
6,07221145
asked Jul 24 at 13:02
caffeinemachine
6,07221145
asked Jul 24 at 13:02
caffeinemachine
6,07221145
6,07221145
For purposes of doing everything rigorously, let me mention what your question overlooks, namely the theory of symmetric bilinear forms, in particular forms of signature $(n-1,1)$. 2-dimensional spacetime should be more than just a 2-manifold homeomorphic to $mathbb R^2$, it should be a 2-dimensional vector space over $mathbb R$ equipped with a symmetric bilinear form of signature $(1,1)$.
– Lee Mosher
Jul 24 at 13:39
@LeeMosher Can you provide some motivation behind the choice of the inner product? Starting with a vector space structure is something I can accept. But a bilinear form of index 1 seems like a post-hoc artifice.
– caffeinemachine
Jul 24 at 16:12
Actually, a bilinear form of index $1$ is exactly what special relativity is all about. I'll write a little bit of an answer along those lines.
– Lee Mosher
Jul 24 at 16:35
Have you considered rigorous text on Special relativity ?, such as Geometry of Minkowski Spacetime by G. Naber (describe Minkowski space as 4-dim real vector space with nondegenerate sym. bilinear form of index 1) and Barett O'neill' Semi-Riemannian Geometry (more general than Naber, I have not done read this one yet).
– Sou
Jul 25 at 13:32
@Sou Yes I took a look at that book. But it seems to me (and I am pretty green) that there is no motivation given behind the formalism. What I am looking for is a formulation which takes "speed of light is same in all inertial frames" and makes this precise, and then deduce relativity of simultaneity. The text starts with the universe being a $4$-dimensional real vector space with a certain bilinear form. I am unable to see why would anyone think of this formulation. What is the, well, story?
– caffeinemachine
Jul 25 at 15:13
 |Â
show 1 more comment
For purposes of doing everything rigorously, let me mention what your question overlooks, namely the theory of symmetric bilinear forms, in particular forms of signature $(n-1,1)$. 2-dimensional spacetime should be more than just a 2-manifold homeomorphic to $mathbb R^2$, it should be a 2-dimensional vector space over $mathbb R$ equipped with a symmetric bilinear form of signature $(1,1)$.
– Lee Mosher
Jul 24 at 13:39
@LeeMosher Can you provide some motivation behind the choice of the inner product? Starting with a vector space structure is something I can accept. But a bilinear form of index 1 seems like a post-hoc artifice.
– caffeinemachine
Jul 24 at 16:12
Actually, a bilinear form of index $1$ is exactly what special relativity is all about. I'll write a little bit of an answer along those lines.
– Lee Mosher
Jul 24 at 16:35
Have you considered rigorous text on Special relativity ?, such as Geometry of Minkowski Spacetime by G. Naber (describe Minkowski space as 4-dim real vector space with nondegenerate sym. bilinear form of index 1) and Barett O'neill' Semi-Riemannian Geometry (more general than Naber, I have not done read this one yet).
– Sou
Jul 25 at 13:32
@Sou Yes I took a look at that book. But it seems to me (and I am pretty green) that there is no motivation given behind the formalism. What I am looking for is a formulation which takes "speed of light is same in all inertial frames" and makes this precise, and then deduce relativity of simultaneity. The text starts with the universe being a $4$-dimensional real vector space with a certain bilinear form. I am unable to see why would anyone think of this formulation. What is the, well, story?
– caffeinemachine
Jul 25 at 15:13
For purposes of doing everything rigorously, let me mention what your question overlooks, namely the theory of symmetric bilinear forms, in particular forms of signature $(n-1,1)$. 2-dimensional spacetime should be more than just a 2-manifold homeomorphic to $mathbb R^2$, it should be a 2-dimensional vector space over $mathbb R$ equipped with a symmetric bilinear form of signature $(1,1)$.
– Lee Mosher
Jul 24 at 13:39
For purposes of doing everything rigorously, let me mention what your question overlooks, namely the theory of symmetric bilinear forms, in particular forms of signature $(n-1,1)$. 2-dimensional spacetime should be more than just a 2-manifold homeomorphic to $mathbb R^2$, it should be a 2-dimensional vector space over $mathbb R$ equipped with a symmetric bilinear form of signature $(1,1)$.
– Lee Mosher
Jul 24 at 13:39
@LeeMosher Can you provide some motivation behind the choice of the inner product? Starting with a vector space structure is something I can accept. But a bilinear form of index 1 seems like a post-hoc artifice.
– caffeinemachine
Jul 24 at 16:12
@LeeMosher Can you provide some motivation behind the choice of the inner product? Starting with a vector space structure is something I can accept. But a bilinear form of index 1 seems like a post-hoc artifice.
– caffeinemachine
Jul 24 at 16:12
Actually, a bilinear form of index $1$ is exactly what special relativity is all about. I'll write a little bit of an answer along those lines.
– Lee Mosher
Jul 24 at 16:35
Actually, a bilinear form of index $1$ is exactly what special relativity is all about. I'll write a little bit of an answer along those lines.
– Lee Mosher
Jul 24 at 16:35
Have you considered rigorous text on Special relativity ?, such as Geometry of Minkowski Spacetime by G. Naber (describe Minkowski space as 4-dim real vector space with nondegenerate sym. bilinear form of index 1) and Barett O'neill' Semi-Riemannian Geometry (more general than Naber, I have not done read this one yet).
– Sou
Jul 25 at 13:32
Have you considered rigorous text on Special relativity ?, such as Geometry of Minkowski Spacetime by G. Naber (describe Minkowski space as 4-dim real vector space with nondegenerate sym. bilinear form of index 1) and Barett O'neill' Semi-Riemannian Geometry (more general than Naber, I have not done read this one yet).
– Sou
Jul 25 at 13:32
@Sou Yes I took a look at that book. But it seems to me (and I am pretty green) that there is no motivation given behind the formalism. What I am looking for is a formulation which takes "speed of light is same in all inertial frames" and makes this precise, and then deduce relativity of simultaneity. The text starts with the universe being a $4$-dimensional real vector space with a certain bilinear form. I am unable to see why would anyone think of this formulation. What is the, well, story?
– caffeinemachine
Jul 25 at 15:13
@Sou Yes I took a look at that book. But it seems to me (and I am pretty green) that there is no motivation given behind the formalism. What I am looking for is a formulation which takes "speed of light is same in all inertial frames" and makes this precise, and then deduce relativity of simultaneity. The text starts with the universe being a $4$-dimensional real vector space with a certain bilinear form. I am unable to see why would anyone think of this formulation. What is the, well, story?
– caffeinemachine
Jul 25 at 15:13
 |Â
show 1 more comment
2 Answers
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2
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accepted
Since you seem unfamiliar with the nature of special relativity as the geometry of index 1 symmetric bilinear forms, aka "Lorentz forms", I think the best I can do is to give you a little bit of intuition about how these forms are connected to the reality of physical space time. This might allow you to realize the direction that one must go in order to formalize special relativity.
Beyond that, the best I can suggest is to suggest that you read on. I started with Resnick too, when I was a freshman, oh so many years ago, and it took me a long time until I properly appreciated the linear algebra basis of special relativity.
The starting point for special relativity in 2-dimensional space time is a 2-dimensional vector space $V$ equipped with a symmetric bilinear form $langle v,w rangle$ defined for all $v,w in V$ such that the signature of this form is $(1,1)$. By definition, this means that $V$ has a basis in which the matrix of the form is $beginpmatrix1 & 0 \ 0 & -1endpmatrix$.
Using these coordinates, we can identify $V$ with $mathbb R^2$, and we can write the two coordinate functions as $x$, $t$. The first coordinate $x$ is usually called something like the "space-like" coordinate, and $t$ is called the "time-like coordinate".
Given $p,q in V$ representing two space-time events, it is useful to ponder the meaning of the quantity
$$|q-p|^2 = -langle q-p, q-p rangle
$$
This quantity can be zero, positive, or negative, and I'll discuss the physical meaning of these possibilities.
Understanding a space time interval $|q-p|^2=0$ is a good start. Mathematically, this just means that the straight line through $p$ and $q$ has slope $pm 1$. What $|q-p|^2=0$ means physically is that in order for a particle to move under inertia between the events $p$ and $q$, that particle must essentially be a photon: it must move at the speed of light (the coordinates have been normalized so that the speed of light equals $1$).
In general, the world line of a photon, or any particle moving in a straight line at the speed of light, is a path of the form
$$f(s) = p_0 + s cdot v
$$
where the vector $v$ satisfies $langle v,v rangle = 0$. The appearance of this world line in the space-time $V$ is that it is a line of slope $pm 1$.
The physical meaning of $|q-p|^2 > 0$ is that it is possible to move between the events $p,q$ by "real", inertial motion. Furthermore, if you move in that manner, and if your clock is ticking along, then the amount of time that your clock ticks off between $p$ and $q$ is equal to
$$|q-p| = sqrt
$$
Inertial motion from $p$ to $q$ is modelled by the parameterized path
$$f(s) = p + s u
$$
where $u = fracq-pq-p$ (in order to be moving "forward" in time, the $t$ coordinate of the vector $u$ should be positive; in other words, $p$ is in the "past" of $q$ and $q$ is in the "future" of $p$). For example, if an object starts at $p_0=(x_0,t_0)$ and does not "move" in this coordinate system, then $u = (0,1)$ and the world line is
$$f(s) = (x_0,t_0+s)
$$
The specific example of an unmoving particle sitting at the origin $x_0=0$ is $p(s)=(0,s)$.
The physical meaning assigned to $|q-p|^2 < 0$ is that communication between the space time events $q$ and $p$ is impossible, because moving between them would require faster than light motion. So, for example, it is impossible to see what is happening at this exact moment one inch in front of your face.
answered Jul 24 at 17:13
Lee Mosher
45.4k33478
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Any two inertial frames move at a constant speed with respect to each other.
There's an important subtlety you're in danger of overlooking. The composition of two Lorentz boosts is in general not a Lorentz boost, but rather a composition of one with a rotation. If you want to flesh out the details properly, you'll need the theory of gyrovector spaces or something equivalent.
Perhaps there could be given a more intrinsic axiom to the set of all the photons.
I assume you mean luxons, since in theory photons aren't the only speed-$c$ particle. An alternative approach is to define speed-$c$ trajectories by $ds^2=eta_munudx^mu dx^nu=0$.
answered Jul 24 at 13:07
J.G.
13.2k11424
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Since you seem unfamiliar with the nature of special relativity as the geometry of index 1 symmetric bilinear forms, aka "Lorentz forms", I think the best I can do is to give you a little bit of intuition about how these forms are connected to the reality of physical space time. This might allow you to realize the direction that one must go in order to formalize special relativity.
Beyond that, the best I can suggest is to suggest that you read on. I started with Resnick too, when I was a freshman, oh so many years ago, and it took me a long time until I properly appreciated the linear algebra basis of special relativity.
The starting point for special relativity in 2-dimensional space time is a 2-dimensional vector space $V$ equipped with a symmetric bilinear form $langle v,w rangle$ defined for all $v,w in V$ such that the signature of this form is $(1,1)$. By definition, this means that $V$ has a basis in which the matrix of the form is $beginpmatrix1 & 0 \ 0 & -1endpmatrix$.
Using these coordinates, we can identify $V$ with $mathbb R^2$, and we can write the two coordinate functions as $x$, $t$. The first coordinate $x$ is usually called something like the "space-like" coordinate, and $t$ is called the "time-like coordinate".
Given $p,q in V$ representing two space-time events, it is useful to ponder the meaning of the quantity
$$|q-p|^2 = -langle q-p, q-p rangle
$$
This quantity can be zero, positive, or negative, and I'll discuss the physical meaning of these possibilities.
Understanding a space time interval $|q-p|^2=0$ is a good start. Mathematically, this just means that the straight line through $p$ and $q$ has slope $pm 1$. What $|q-p|^2=0$ means physically is that in order for a particle to move under inertia between the events $p$ and $q$, that particle must essentially be a photon: it must move at the speed of light (the coordinates have been normalized so that the speed of light equals $1$).
In general, the world line of a photon, or any particle moving in a straight line at the speed of light, is a path of the form
$$f(s) = p_0 + s cdot v
$$
where the vector $v$ satisfies $langle v,v rangle = 0$. The appearance of this world line in the space-time $V$ is that it is a line of slope $pm 1$.
The physical meaning of $|q-p|^2 > 0$ is that it is possible to move between the events $p,q$ by "real", inertial motion. Furthermore, if you move in that manner, and if your clock is ticking along, then the amount of time that your clock ticks off between $p$ and $q$ is equal to
$$|q-p| = sqrt
$$
Inertial motion from $p$ to $q$ is modelled by the parameterized path
$$f(s) = p + s u
$$
where $u = fracq-pq-p$ (in order to be moving "forward" in time, the $t$ coordinate of the vector $u$ should be positive; in other words, $p$ is in the "past" of $q$ and $q$ is in the "future" of $p$). For example, if an object starts at $p_0=(x_0,t_0)$ and does not "move" in this coordinate system, then $u = (0,1)$ and the world line is
$$f(s) = (x_0,t_0+s)
$$
The specific example of an unmoving particle sitting at the origin $x_0=0$ is $p(s)=(0,s)$.
The physical meaning assigned to $|q-p|^2 < 0$ is that communication between the space time events $q$ and $p$ is impossible, because moving between them would require faster than light motion. So, for example, it is impossible to see what is happening at this exact moment one inch in front of your face.
answered Jul 24 at 17:13
Lee Mosher
45.4k33478
add a comment |Â
up vote
2
down vote
accepted
Since you seem unfamiliar with the nature of special relativity as the geometry of index 1 symmetric bilinear forms, aka "Lorentz forms", I think the best I can do is to give you a little bit of intuition about how these forms are connected to the reality of physical space time. This might allow you to realize the direction that one must go in order to formalize special relativity.
Beyond that, the best I can suggest is to suggest that you read on. I started with Resnick too, when I was a freshman, oh so many years ago, and it took me a long time until I properly appreciated the linear algebra basis of special relativity.
The starting point for special relativity in 2-dimensional space time is a 2-dimensional vector space $V$ equipped with a symmetric bilinear form $langle v,w rangle$ defined for all $v,w in V$ such that the signature of this form is $(1,1)$. By definition, this means that $V$ has a basis in which the matrix of the form is $beginpmatrix1 & 0 \ 0 & -1endpmatrix$.
Using these coordinates, we can identify $V$ with $mathbb R^2$, and we can write the two coordinate functions as $x$, $t$. The first coordinate $x$ is usually called something like the "space-like" coordinate, and $t$ is called the "time-like coordinate".
Given $p,q in V$ representing two space-time events, it is useful to ponder the meaning of the quantity
$$|q-p|^2 = -langle q-p, q-p rangle
$$
This quantity can be zero, positive, or negative, and I'll discuss the physical meaning of these possibilities.
Understanding a space time interval $|q-p|^2=0$ is a good start. Mathematically, this just means that the straight line through $p$ and $q$ has slope $pm 1$. What $|q-p|^2=0$ means physically is that in order for a particle to move under inertia between the events $p$ and $q$, that particle must essentially be a photon: it must move at the speed of light (the coordinates have been normalized so that the speed of light equals $1$).
In general, the world line of a photon, or any particle moving in a straight line at the speed of light, is a path of the form
$$f(s) = p_0 + s cdot v
$$
where the vector $v$ satisfies $langle v,v rangle = 0$. The appearance of this world line in the space-time $V$ is that it is a line of slope $pm 1$.
The physical meaning of $|q-p|^2 > 0$ is that it is possible to move between the events $p,q$ by "real", inertial motion. Furthermore, if you move in that manner, and if your clock is ticking along, then the amount of time that your clock ticks off between $p$ and $q$ is equal to
$$|q-p| = sqrt
$$
Inertial motion from $p$ to $q$ is modelled by the parameterized path
$$f(s) = p + s u
$$
where $u = fracq-pq-p$ (in order to be moving "forward" in time, the $t$ coordinate of the vector $u$ should be positive; in other words, $p$ is in the "past" of $q$ and $q$ is in the "future" of $p$). For example, if an object starts at $p_0=(x_0,t_0)$ and does not "move" in this coordinate system, then $u = (0,1)$ and the world line is
$$f(s) = (x_0,t_0+s)
$$
The specific example of an unmoving particle sitting at the origin $x_0=0$ is $p(s)=(0,s)$.
The physical meaning assigned to $|q-p|^2 < 0$ is that communication between the space time events $q$ and $p$ is impossible, because moving between them would require faster than light motion. So, for example, it is impossible to see what is happening at this exact moment one inch in front of your face.
answered Jul 24 at 17:13
Lee Mosher
45.4k33478
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Since you seem unfamiliar with the nature of special relativity as the geometry of index 1 symmetric bilinear forms, aka "Lorentz forms", I think the best I can do is to give you a little bit of intuition about how these forms are connected to the reality of physical space time. This might allow you to realize the direction that one must go in order to formalize special relativity.
Beyond that, the best I can suggest is to suggest that you read on. I started with Resnick too, when I was a freshman, oh so many years ago, and it took me a long time until I properly appreciated the linear algebra basis of special relativity.
The starting point for special relativity in 2-dimensional space time is a 2-dimensional vector space $V$ equipped with a symmetric bilinear form $langle v,w rangle$ defined for all $v,w in V$ such that the signature of this form is $(1,1)$. By definition, this means that $V$ has a basis in which the matrix of the form is $beginpmatrix1 & 0 \ 0 & -1endpmatrix$.
Using these coordinates, we can identify $V$ with $mathbb R^2$, and we can write the two coordinate functions as $x$, $t$. The first coordinate $x$ is usually called something like the "space-like" coordinate, and $t$ is called the "time-like coordinate".
Given $p,q in V$ representing two space-time events, it is useful to ponder the meaning of the quantity
$$|q-p|^2 = -langle q-p, q-p rangle
$$
This quantity can be zero, positive, or negative, and I'll discuss the physical meaning of these possibilities.
Understanding a space time interval $|q-p|^2=0$ is a good start. Mathematically, this just means that the straight line through $p$ and $q$ has slope $pm 1$. What $|q-p|^2=0$ means physically is that in order for a particle to move under inertia between the events $p$ and $q$, that particle must essentially be a photon: it must move at the speed of light (the coordinates have been normalized so that the speed of light equals $1$).
In general, the world line of a photon, or any particle moving in a straight line at the speed of light, is a path of the form
$$f(s) = p_0 + s cdot v
$$
where the vector $v$ satisfies $langle v,v rangle = 0$. The appearance of this world line in the space-time $V$ is that it is a line of slope $pm 1$.
The physical meaning of $|q-p|^2 > 0$ is that it is possible to move between the events $p,q$ by "real", inertial motion. Furthermore, if you move in that manner, and if your clock is ticking along, then the amount of time that your clock ticks off between $p$ and $q$ is equal to
$$|q-p| = sqrt
$$
Inertial motion from $p$ to $q$ is modelled by the parameterized path
$$f(s) = p + s u
$$
where $u = fracq-pq-p$ (in order to be moving "forward" in time, the $t$ coordinate of the vector $u$ should be positive; in other words, $p$ is in the "past" of $q$ and $q$ is in the "future" of $p$). For example, if an object starts at $p_0=(x_0,t_0)$ and does not "move" in this coordinate system, then $u = (0,1)$ and the world line is
$$f(s) = (x_0,t_0+s)
$$
The specific example of an unmoving particle sitting at the origin $x_0=0$ is $p(s)=(0,s)$.
The physical meaning assigned to $|q-p|^2 < 0$ is that communication between the space time events $q$ and $p$ is impossible, because moving between them would require faster than light motion. So, for example, it is impossible to see what is happening at this exact moment one inch in front of your face.
answered Jul 24 at 17:13
Lee Mosher
45.4k33478
Since you seem unfamiliar with the nature of special relativity as the geometry of index 1 symmetric bilinear forms, aka "Lorentz forms", I think the best I can do is to give you a little bit of intuition about how these forms are connected to the reality of physical space time. This might allow you to realize the direction that one must go in order to formalize special relativity.
Beyond that, the best I can suggest is to suggest that you read on. I started with Resnick too, when I was a freshman, oh so many years ago, and it took me a long time until I properly appreciated the linear algebra basis of special relativity.
The starting point for special relativity in 2-dimensional space time is a 2-dimensional vector space $V$ equipped with a symmetric bilinear form $langle v,w rangle$ defined for all $v,w in V$ such that the signature of this form is $(1,1)$. By definition, this means that $V$ has a basis in which the matrix of the form is $beginpmatrix1 & 0 \ 0 & -1endpmatrix$.
Using these coordinates, we can identify $V$ with $mathbb R^2$, and we can write the two coordinate functions as $x$, $t$. The first coordinate $x$ is usually called something like the "space-like" coordinate, and $t$ is called the "time-like coordinate".
Given $p,q in V$ representing two space-time events, it is useful to ponder the meaning of the quantity
$$|q-p|^2 = -langle q-p, q-p rangle
$$
This quantity can be zero, positive, or negative, and I'll discuss the physical meaning of these possibilities.
Understanding a space time interval $|q-p|^2=0$ is a good start. Mathematically, this just means that the straight line through $p$ and $q$ has slope $pm 1$. What $|q-p|^2=0$ means physically is that in order for a particle to move under inertia between the events $p$ and $q$, that particle must essentially be a photon: it must move at the speed of light (the coordinates have been normalized so that the speed of light equals $1$).
In general, the world line of a photon, or any particle moving in a straight line at the speed of light, is a path of the form
$$f(s) = p_0 + s cdot v
$$
where the vector $v$ satisfies $langle v,v rangle = 0$. The appearance of this world line in the space-time $V$ is that it is a line of slope $pm 1$.
The physical meaning of $|q-p|^2 > 0$ is that it is possible to move between the events $p,q$ by "real", inertial motion. Furthermore, if you move in that manner, and if your clock is ticking along, then the amount of time that your clock ticks off between $p$ and $q$ is equal to
$$|q-p| = sqrt
$$
Inertial motion from $p$ to $q$ is modelled by the parameterized path
$$f(s) = p + s u
$$
where $u = fracq-pq-p$ (in order to be moving "forward" in time, the $t$ coordinate of the vector $u$ should be positive; in other words, $p$ is in the "past" of $q$ and $q$ is in the "future" of $p$). For example, if an object starts at $p_0=(x_0,t_0)$ and does not "move" in this coordinate system, then $u = (0,1)$ and the world line is
$$f(s) = (x_0,t_0+s)
$$
The specific example of an unmoving particle sitting at the origin $x_0=0$ is $p(s)=(0,s)$.
The physical meaning assigned to $|q-p|^2 < 0$ is that communication between the space time events $q$ and $p$ is impossible, because moving between them would require faster than light motion. So, for example, it is impossible to see what is happening at this exact moment one inch in front of your face.
answered Jul 24 at 17:13
Lee Mosher
45.4k33478
edited Jul 25 at 12:57
answered Jul 24 at 17:13
Lee Mosher
45.4k33478
answered Jul 24 at 17:13
Lee Mosher
45.4k33478
answered Jul 24 at 17:13
Lee Mosher
45.4k33478
45.4k33478
add a comment |Â
add a comment |Â
up vote
0
down vote
Any two inertial frames move at a constant speed with respect to each other.
There's an important subtlety you're in danger of overlooking. The composition of two Lorentz boosts is in general not a Lorentz boost, but rather a composition of one with a rotation. If you want to flesh out the details properly, you'll need the theory of gyrovector spaces or something equivalent.
Perhaps there could be given a more intrinsic axiom to the set of all the photons.
I assume you mean luxons, since in theory photons aren't the only speed-$c$ particle. An alternative approach is to define speed-$c$ trajectories by $ds^2=eta_munudx^mu dx^nu=0$.
answered Jul 24 at 13:07
J.G.
13.2k11424
add a comment |Â
up vote
0
down vote
Any two inertial frames move at a constant speed with respect to each other.
There's an important subtlety you're in danger of overlooking. The composition of two Lorentz boosts is in general not a Lorentz boost, but rather a composition of one with a rotation. If you want to flesh out the details properly, you'll need the theory of gyrovector spaces or something equivalent.
Perhaps there could be given a more intrinsic axiom to the set of all the photons.
I assume you mean luxons, since in theory photons aren't the only speed-$c$ particle. An alternative approach is to define speed-$c$ trajectories by $ds^2=eta_munudx^mu dx^nu=0$.
answered Jul 24 at 13:07
J.G.
13.2k11424
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Any two inertial frames move at a constant speed with respect to each other.
There's an important subtlety you're in danger of overlooking. The composition of two Lorentz boosts is in general not a Lorentz boost, but rather a composition of one with a rotation. If you want to flesh out the details properly, you'll need the theory of gyrovector spaces or something equivalent.
Perhaps there could be given a more intrinsic axiom to the set of all the photons.
I assume you mean luxons, since in theory photons aren't the only speed-$c$ particle. An alternative approach is to define speed-$c$ trajectories by $ds^2=eta_munudx^mu dx^nu=0$.
answered Jul 24 at 13:07
J.G.
13.2k11424
Any two inertial frames move at a constant speed with respect to each other.
There's an important subtlety you're in danger of overlooking. The composition of two Lorentz boosts is in general not a Lorentz boost, but rather a composition of one with a rotation. If you want to flesh out the details properly, you'll need the theory of gyrovector spaces or something equivalent.
Perhaps there could be given a more intrinsic axiom to the set of all the photons.
I assume you mean luxons, since in theory photons aren't the only speed-$c$ particle. An alternative approach is to define speed-$c$ trajectories by $ds^2=eta_munudx^mu dx^nu=0$.
answered Jul 24 at 13:07
J.G.
13.2k11424
answered Jul 24 at 13:07
J.G.
13.2k11424
answered Jul 24 at 13:07
J.G.
13.2k11424
answered Jul 24 at 13:07
J.G.
13.2k11424
13.2k11424
add a comment |Â
add a comment |Â
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For purposes of doing everything rigorously, let me mention what your question overlooks, namely the theory of symmetric bilinear forms, in particular forms of signature $(n-1,1)$. 2-dimensional spacetime should be more than just a 2-manifold homeomorphic to $mathbb R^2$, it should be a 2-dimensional vector space over $mathbb R$ equipped with a symmetric bilinear form of signature $(1,1)$.
– Lee Mosher
Jul 24 at 13:39
@LeeMosher Can you provide some motivation behind the choice of the inner product? Starting with a vector space structure is something I can accept. But a bilinear form of index 1 seems like a post-hoc artifice.
– caffeinemachine
Jul 24 at 16:12
Actually, a bilinear form of index $1$ is exactly what special relativity is all about. I'll write a little bit of an answer along those lines.
– Lee Mosher
Jul 24 at 16:35
Have you considered rigorous text on Special relativity ?, such as Geometry of Minkowski Spacetime by G. Naber (describe Minkowski space as 4-dim real vector space with nondegenerate sym. bilinear form of index 1) and Barett O'neill' Semi-Riemannian Geometry (more general than Naber, I have not done read this one yet).
– Sou
Jul 25 at 13:32
@Sou Yes I took a look at that book. But it seems to me (and I am pretty green) that there is no motivation given behind the formalism. What I am looking for is a formulation which takes "speed of light is same in all inertial frames" and makes this precise, and then deduce relativity of simultaneity. The text starts with the universe being a $4$-dimensional real vector space with a certain bilinear form. I am unable to see why would anyone think of this formulation. What is the, well, story?
– caffeinemachine
Jul 25 at 15:13