Mixing
Is there a proof that a function which satisfies these axioms of relativity must be linear?
$begingroup$
Let us only consider the 2 dimensional case where the two frames of reference coincide at the origin. For each $vin (-c,c)$, let $f_v:mathbb{R}^2tomathbb{R}^2$ be a smooth function satisfying the following conditions (with the first coordinate representing time and the second representing space).
$(1),,f_v(0,0)=(0,0)$,
$(2),,forall tinmathbb{R},exists t'inmathbb{R}$ such that $f_v(t,vt)=(t',0)$,
$(3),,forall tinmathbb{R},exists t'inmathbb{R}$ such that $f_v(t,0)=(t',-vt')$,
$(4),,forall tinmathbb{R},forall xinmathbb{R},f_v(t,x)=((pi_1circ f_{-v})(t,-x),-(pi_2circ f_{-v})(t,-x))$, and
$(5),,forall tinmathbb{R},exists t'inmathbb{R}$ such that $f_v(t,ct)=(t',ct')$.
Are there other hypotheses needed in order to show that $f_v$ is linear? I can show that for all $vin (-c,c)$ if $f_v$ is a polynomial, then $f_v$ is linear; however, I cannot show that if $f_v$ is a limit of polynomial functions, then $f_v$ is linear.
Most of my attempts trying to figure this out have started with $"$let
$$f_v(t,x)=(sum_{i=0,, j=0}^{infty}a_{ij}t^i x^j, sum_{i=0,, j=0}^{infty}b_{ij}t^i x^j)."$$
I have then tried to show that some relationship between the coefficients implies that after a certain point, they are all $0$ (to no avail).
mathematical-physics
asked Jan 15 at 1:15
John BJohn B
1766
$endgroup$
add a comment |
$begingroup$
Let us only consider the 2 dimensional case where the two frames of reference coincide at the origin. For each $vin (-c,c)$, let $f_v:mathbb{R}^2tomathbb{R}^2$ be a smooth function satisfying the following conditions (with the first coordinate representing time and the second representing space).
$(1),,f_v(0,0)=(0,0)$,
$(2),,forall tinmathbb{R},exists t'inmathbb{R}$ such that $f_v(t,vt)=(t',0)$,
$(3),,forall tinmathbb{R},exists t'inmathbb{R}$ such that $f_v(t,0)=(t',-vt')$,
$(4),,forall tinmathbb{R},forall xinmathbb{R},f_v(t,x)=((pi_1circ f_{-v})(t,-x),-(pi_2circ f_{-v})(t,-x))$, and
$(5),,forall tinmathbb{R},exists t'inmathbb{R}$ such that $f_v(t,ct)=(t',ct')$.
Are there other hypotheses needed in order to show that $f_v$ is linear? I can show that for all $vin (-c,c)$ if $f_v$ is a polynomial, then $f_v$ is linear; however, I cannot show that if $f_v$ is a limit of polynomial functions, then $f_v$ is linear.
Most of my attempts trying to figure this out have started with $"$let
$$f_v(t,x)=(sum_{i=0,, j=0}^{infty}a_{ij}t^i x^j, sum_{i=0,, j=0}^{infty}b_{ij}t^i x^j)."$$
I have then tried to show that some relationship between the coefficients implies that after a certain point, they are all $0$ (to no avail).
mathematical-physics
asked Jan 15 at 1:15
John BJohn B
1766
$endgroup$
1
$begingroup$
It's hard to see how these conditions could imply linearity when there's no relation involving the sum of two vectors. It doesn't even seem like you can pull out scalars. What if $t'=t^2$?
$endgroup$
– Matt Samuel
Jan 15 at 1:54
$begingroup$
I suppose you are focusing on the Lorentz transformation for Special Relativity. If so, $f$ is required to be linear because it is part of an affine transformation. In fact, it is because $f$ is linear, you can then write down your conditions (1)-(5). Otherwise, the relation between $(t,x)$ and $(t',x')$ would not be as they are -- and that is why we need General Relativity when we lose the linearity condition.
$endgroup$
– hypernova
Jan 15 at 4:55
add a comment |
$begingroup$
Let us only consider the 2 dimensional case where the two frames of reference coincide at the origin. For each $vin (-c,c)$, let $f_v:mathbb{R}^2tomathbb{R}^2$ be a smooth function satisfying the following conditions (with the first coordinate representing time and the second representing space).
$(1),,f_v(0,0)=(0,0)$,
$(2),,forall tinmathbb{R},exists t'inmathbb{R}$ such that $f_v(t,vt)=(t',0)$,
$(3),,forall tinmathbb{R},exists t'inmathbb{R}$ such that $f_v(t,0)=(t',-vt')$,
$(4),,forall tinmathbb{R},forall xinmathbb{R},f_v(t,x)=((pi_1circ f_{-v})(t,-x),-(pi_2circ f_{-v})(t,-x))$, and
$(5),,forall tinmathbb{R},exists t'inmathbb{R}$ such that $f_v(t,ct)=(t',ct')$.
Are there other hypotheses needed in order to show that $f_v$ is linear? I can show that for all $vin (-c,c)$ if $f_v$ is a polynomial, then $f_v$ is linear; however, I cannot show that if $f_v$ is a limit of polynomial functions, then $f_v$ is linear.
Most of my attempts trying to figure this out have started with $"$let
$$f_v(t,x)=(sum_{i=0,, j=0}^{infty}a_{ij}t^i x^j, sum_{i=0,, j=0}^{infty}b_{ij}t^i x^j)."$$
I have then tried to show that some relationship between the coefficients implies that after a certain point, they are all $0$ (to no avail).
mathematical-physics
asked Jan 15 at 1:15
John BJohn B
1766
$endgroup$
Let us only consider the 2 dimensional case where the two frames of reference coincide at the origin. For each $vin (-c,c)$, let $f_v:mathbb{R}^2tomathbb{R}^2$ be a smooth function satisfying the following conditions (with the first coordinate representing time and the second representing space).
$(1),,f_v(0,0)=(0,0)$,
$(2),,forall tinmathbb{R},exists t'inmathbb{R}$ such that $f_v(t,vt)=(t',0)$,
$(3),,forall tinmathbb{R},exists t'inmathbb{R}$ such that $f_v(t,0)=(t',-vt')$,
$(4),,forall tinmathbb{R},forall xinmathbb{R},f_v(t,x)=((pi_1circ f_{-v})(t,-x),-(pi_2circ f_{-v})(t,-x))$, and
$(5),,forall tinmathbb{R},exists t'inmathbb{R}$ such that $f_v(t,ct)=(t',ct')$.
Are there other hypotheses needed in order to show that $f_v$ is linear? I can show that for all $vin (-c,c)$ if $f_v$ is a polynomial, then $f_v$ is linear; however, I cannot show that if $f_v$ is a limit of polynomial functions, then $f_v$ is linear.
Most of my attempts trying to figure this out have started with $"$let
$$f_v(t,x)=(sum_{i=0,, j=0}^{infty}a_{ij}t^i x^j, sum_{i=0,, j=0}^{infty}b_{ij}t^i x^j)."$$
I have then tried to show that some relationship between the coefficients implies that after a certain point, they are all $0$ (to no avail).
mathematical-physics
mathematical-physics
asked Jan 15 at 1:15
John BJohn B
1766
asked Jan 15 at 1:15
John BJohn B
1766
asked Jan 15 at 1:15
John BJohn B
1766
asked Jan 15 at 1:15
John BJohn B
1766
asked Jan 15 at 1:15
John BJohn B
1766
1766
1
$begingroup$
It's hard to see how these conditions could imply linearity when there's no relation involving the sum of two vectors. It doesn't even seem like you can pull out scalars. What if $t'=t^2$?
$endgroup$
– Matt Samuel
Jan 15 at 1:54
$begingroup$
I suppose you are focusing on the Lorentz transformation for Special Relativity. If so, $f$ is required to be linear because it is part of an affine transformation. In fact, it is because $f$ is linear, you can then write down your conditions (1)-(5). Otherwise, the relation between $(t,x)$ and $(t',x')$ would not be as they are -- and that is why we need General Relativity when we lose the linearity condition.
$endgroup$
– hypernova
Jan 15 at 4:55
add a comment |
1
$begingroup$
It's hard to see how these conditions could imply linearity when there's no relation involving the sum of two vectors. It doesn't even seem like you can pull out scalars. What if $t'=t^2$?
$endgroup$
– Matt Samuel
Jan 15 at 1:54
$begingroup$
I suppose you are focusing on the Lorentz transformation for Special Relativity. If so, $f$ is required to be linear because it is part of an affine transformation. In fact, it is because $f$ is linear, you can then write down your conditions (1)-(5). Otherwise, the relation between $(t,x)$ and $(t',x')$ would not be as they are -- and that is why we need General Relativity when we lose the linearity condition.
$endgroup$
– hypernova
Jan 15 at 4:55
1
1
$begingroup$
It's hard to see how these conditions could imply linearity when there's no relation involving the sum of two vectors. It doesn't even seem like you can pull out scalars. What if $t'=t^2$?
$endgroup$
– Matt Samuel
Jan 15 at 1:54
$begingroup$
It's hard to see how these conditions could imply linearity when there's no relation involving the sum of two vectors. It doesn't even seem like you can pull out scalars. What if $t'=t^2$?
$endgroup$
– Matt Samuel
Jan 15 at 1:54
$begingroup$
I suppose you are focusing on the Lorentz transformation for Special Relativity. If so, $f$ is required to be linear because it is part of an affine transformation. In fact, it is because $f$ is linear, you can then write down your conditions (1)-(5). Otherwise, the relation between $(t,x)$ and $(t',x')$ would not be as they are -- and that is why we need General Relativity when we lose the linearity condition.
$endgroup$
– hypernova
Jan 15 at 4:55
$begingroup$
I suppose you are focusing on the Lorentz transformation for Special Relativity. If so, $f$ is required to be linear because it is part of an affine transformation. In fact, it is because $f$ is linear, you can then write down your conditions (1)-(5). Otherwise, the relation between $(t,x)$ and $(t',x')$ would not be as they are -- and that is why we need General Relativity when we lose the linearity condition.
$endgroup$
– hypernova
Jan 15 at 4:55
add a comment |
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$begingroup$
It's hard to see how these conditions could imply linearity when there's no relation involving the sum of two vectors. It doesn't even seem like you can pull out scalars. What if $t'=t^2$?
$endgroup$
– Matt Samuel
Jan 15 at 1:54
$begingroup$
I suppose you are focusing on the Lorentz transformation for Special Relativity. If so, $f$ is required to be linear because it is part of an affine transformation. In fact, it is because $f$ is linear, you can then write down your conditions (1)-(5). Otherwise, the relation between $(t,x)$ and $(t',x')$ would not be as they are -- and that is why we need General Relativity when we lose the linearity condition.
$endgroup$
– hypernova
Jan 15 at 4:55