Mixing
How to prove the Hubble law is the unique expansion law compatible with homogeneity and isotropy?
$begingroup$
In the book physical foundations of cosmology, it says that Hubble's Law is unique and a problem seems to be a hint of proving that.
In order for a general expansion law,v=f(r,t), to be the same for all observers, the function f must satisfy the relation
$$f(bf{r_{CA}}−bf{r_{BA}},t) = f(bf{r_{CA}},t)−f(bf{r_{BA}},t),$$
where ABC are three points in space. Show that the only solution of this equation is given by the Hubble law.
With a little help from a Taylor approximation, I can convince myself that $f$ should be a linear function without a constant. But it seems to me that this is not good enough for a proof. How can one prove it in a more mathematical way?
Thanks!
physics mathematical-astronomy
edited Jan 21 at 2:21
NolantheNerd
134
asked Aug 9 '14 at 5:54
GeorGeor
1
$endgroup$
add a comment |
$begingroup$
In the book physical foundations of cosmology, it says that Hubble's Law is unique and a problem seems to be a hint of proving that.
In order for a general expansion law,v=f(r,t), to be the same for all observers, the function f must satisfy the relation
$$f(bf{r_{CA}}−bf{r_{BA}},t) = f(bf{r_{CA}},t)−f(bf{r_{BA}},t),$$
where ABC are three points in space. Show that the only solution of this equation is given by the Hubble law.
With a little help from a Taylor approximation, I can convince myself that $f$ should be a linear function without a constant. But it seems to me that this is not good enough for a proof. How can one prove it in a more mathematical way?
Thanks!
physics mathematical-astronomy
edited Jan 21 at 2:21
NolantheNerd
134
asked Aug 9 '14 at 5:54
GeorGeor
1
$endgroup$
add a comment |
$begingroup$
In the book physical foundations of cosmology, it says that Hubble's Law is unique and a problem seems to be a hint of proving that.
In order for a general expansion law,v=f(r,t), to be the same for all observers, the function f must satisfy the relation
$$f(bf{r_{CA}}−bf{r_{BA}},t) = f(bf{r_{CA}},t)−f(bf{r_{BA}},t),$$
where ABC are three points in space. Show that the only solution of this equation is given by the Hubble law.
With a little help from a Taylor approximation, I can convince myself that $f$ should be a linear function without a constant. But it seems to me that this is not good enough for a proof. How can one prove it in a more mathematical way?
Thanks!
physics mathematical-astronomy
edited Jan 21 at 2:21
NolantheNerd
134
asked Aug 9 '14 at 5:54
GeorGeor
1
$endgroup$
In the book physical foundations of cosmology, it says that Hubble's Law is unique and a problem seems to be a hint of proving that.
In order for a general expansion law,v=f(r,t), to be the same for all observers, the function f must satisfy the relation
$$f(bf{r_{CA}}−bf{r_{BA}},t) = f(bf{r_{CA}},t)−f(bf{r_{BA}},t),$$
where ABC are three points in space. Show that the only solution of this equation is given by the Hubble law.
With a little help from a Taylor approximation, I can convince myself that $f$ should be a linear function without a constant. But it seems to me that this is not good enough for a proof. How can one prove it in a more mathematical way?
Thanks!
physics mathematical-astronomy
physics mathematical-astronomy
edited Jan 21 at 2:21
NolantheNerd
134
asked Aug 9 '14 at 5:54
GeorGeor
1
edited Jan 21 at 2:21
NolantheNerd
134
asked Aug 9 '14 at 5:54
GeorGeor
1
edited Jan 21 at 2:21
NolantheNerd
134
edited Jan 21 at 2:21
NolantheNerd
134
edited Jan 21 at 2:21
NolantheNerd
134
134
asked Aug 9 '14 at 5:54
GeorGeor
1
asked Aug 9 '14 at 5:54
GeorGeor
1
asked Aug 9 '14 at 5:54
GeorGeor
1
1
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I spent a lot of time reading about Hubble's Law only to find that in a math context this problem isn't too hard to explain.
This function says that the change in $r$ corresponds to a constant change in $f$. So, let $x=frac{partial f}{partial r}+frac{partial f}{partial t}frac{dt}{dr}$. Since we assume that $t$ is a constantly changing function, it's derivative with respect to $r$ is 0. this means that $x=frac{partial f}{partial r}$. We can integrate this function with respect to $r$
begin{gather}
int frac{partial f}{partial r},dr=int x,dr = xr + c
end{gather}
I am unsure of as to why $c$ would be set to 0. There's probably an argument in there somewhere but I'm physics'd out for the day.
answered Aug 9 '14 at 8:03
EoinEoin
4,5061926
$endgroup$
$begingroup$
I am not certain if the assertion "the change in r corresponds to a constant change in f" is valid directly from the relation. I am quite sure f must be continuous and then assume that it's also differentiable and analytic, which I find rather unsettling... As for $c$, if you are still interested, here is how I managed it. Say on A (B), I observe B (A), at $bf{r}$ ($-bf{r}$), moving away from me at volocity of $Hbf{r}+bf{c}$($-Hbf{r}+bf{c}$). So $c$ should go to 0.
$endgroup$
– Geor
Aug 9 '14 at 15:55
add a comment |
$begingroup$
There is now a very simple way to calculate Hubble’s Constant, by inputting to an equation, the numerical value of Pi and the speed of light (C) from Maxwell’s equations, and the value of a parsec. NO space probe measurements (with their inevitable small measuring / interpretation errors) are now required. Hubble’s Constant is ‘fixed’ at 70.98047 PRECISELY. This maths method removes the errors / tolerances that is always a part of attempting to measuring something as ‘elusive’ as Hubble’s Constant. This has very deep implications for theoretical cosmology.
The equation to perform this is :- 2 X a meg parsec X light speed (C). This total is then divided by Pi to the power of 21. This gives 70.98047 kilometres per sec per meg parsec.
The equation to perform this can also be found in ‘The Principle of Astrogeometry’ on Amazon Kindle Books. This also explains how the Hubble 70.98047 ‘fixing’ equation was found. David.
answered Sep 23 '18 at 20:24
David HineDavid Hine
1
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
I spent a lot of time reading about Hubble's Law only to find that in a math context this problem isn't too hard to explain.
This function says that the change in $r$ corresponds to a constant change in $f$. So, let $x=frac{partial f}{partial r}+frac{partial f}{partial t}frac{dt}{dr}$. Since we assume that $t$ is a constantly changing function, it's derivative with respect to $r$ is 0. this means that $x=frac{partial f}{partial r}$. We can integrate this function with respect to $r$
begin{gather}
int frac{partial f}{partial r},dr=int x,dr = xr + c
end{gather}
I am unsure of as to why $c$ would be set to 0. There's probably an argument in there somewhere but I'm physics'd out for the day.
answered Aug 9 '14 at 8:03
EoinEoin
4,5061926
$endgroup$
$begingroup$
I am not certain if the assertion "the change in r corresponds to a constant change in f" is valid directly from the relation. I am quite sure f must be continuous and then assume that it's also differentiable and analytic, which I find rather unsettling... As for $c$, if you are still interested, here is how I managed it. Say on A (B), I observe B (A), at $bf{r}$ ($-bf{r}$), moving away from me at volocity of $Hbf{r}+bf{c}$($-Hbf{r}+bf{c}$). So $c$ should go to 0.
$endgroup$
– Geor
Aug 9 '14 at 15:55
add a comment |
$begingroup$
I spent a lot of time reading about Hubble's Law only to find that in a math context this problem isn't too hard to explain.
This function says that the change in $r$ corresponds to a constant change in $f$. So, let $x=frac{partial f}{partial r}+frac{partial f}{partial t}frac{dt}{dr}$. Since we assume that $t$ is a constantly changing function, it's derivative with respect to $r$ is 0. this means that $x=frac{partial f}{partial r}$. We can integrate this function with respect to $r$
begin{gather}
int frac{partial f}{partial r},dr=int x,dr = xr + c
end{gather}
I am unsure of as to why $c$ would be set to 0. There's probably an argument in there somewhere but I'm physics'd out for the day.
answered Aug 9 '14 at 8:03
EoinEoin
4,5061926
$endgroup$
$begingroup$
I am not certain if the assertion "the change in r corresponds to a constant change in f" is valid directly from the relation. I am quite sure f must be continuous and then assume that it's also differentiable and analytic, which I find rather unsettling... As for $c$, if you are still interested, here is how I managed it. Say on A (B), I observe B (A), at $bf{r}$ ($-bf{r}$), moving away from me at volocity of $Hbf{r}+bf{c}$($-Hbf{r}+bf{c}$). So $c$ should go to 0.
$endgroup$
– Geor
Aug 9 '14 at 15:55
add a comment |
$begingroup$
I spent a lot of time reading about Hubble's Law only to find that in a math context this problem isn't too hard to explain.
This function says that the change in $r$ corresponds to a constant change in $f$. So, let $x=frac{partial f}{partial r}+frac{partial f}{partial t}frac{dt}{dr}$. Since we assume that $t$ is a constantly changing function, it's derivative with respect to $r$ is 0. this means that $x=frac{partial f}{partial r}$. We can integrate this function with respect to $r$
begin{gather}
int frac{partial f}{partial r},dr=int x,dr = xr + c
end{gather}
I am unsure of as to why $c$ would be set to 0. There's probably an argument in there somewhere but I'm physics'd out for the day.
answered Aug 9 '14 at 8:03
EoinEoin
4,5061926
$endgroup$
I spent a lot of time reading about Hubble's Law only to find that in a math context this problem isn't too hard to explain.
This function says that the change in $r$ corresponds to a constant change in $f$. So, let $x=frac{partial f}{partial r}+frac{partial f}{partial t}frac{dt}{dr}$. Since we assume that $t$ is a constantly changing function, it's derivative with respect to $r$ is 0. this means that $x=frac{partial f}{partial r}$. We can integrate this function with respect to $r$
begin{gather}
int frac{partial f}{partial r},dr=int x,dr = xr + c
end{gather}
I am unsure of as to why $c$ would be set to 0. There's probably an argument in there somewhere but I'm physics'd out for the day.
answered Aug 9 '14 at 8:03
EoinEoin
4,5061926
answered Aug 9 '14 at 8:03
EoinEoin
4,5061926
answered Aug 9 '14 at 8:03
EoinEoin
4,5061926
answered Aug 9 '14 at 8:03
EoinEoin
4,5061926
4,5061926
$begingroup$
I am not certain if the assertion "the change in r corresponds to a constant change in f" is valid directly from the relation. I am quite sure f must be continuous and then assume that it's also differentiable and analytic, which I find rather unsettling... As for $c$, if you are still interested, here is how I managed it. Say on A (B), I observe B (A), at $bf{r}$ ($-bf{r}$), moving away from me at volocity of $Hbf{r}+bf{c}$($-Hbf{r}+bf{c}$). So $c$ should go to 0.
$endgroup$
– Geor
Aug 9 '14 at 15:55
add a comment |
$begingroup$
I am not certain if the assertion "the change in r corresponds to a constant change in f" is valid directly from the relation. I am quite sure f must be continuous and then assume that it's also differentiable and analytic, which I find rather unsettling... As for $c$, if you are still interested, here is how I managed it. Say on A (B), I observe B (A), at $bf{r}$ ($-bf{r}$), moving away from me at volocity of $Hbf{r}+bf{c}$($-Hbf{r}+bf{c}$). So $c$ should go to 0.
$endgroup$
– Geor
Aug 9 '14 at 15:55
$begingroup$
I am not certain if the assertion "the change in r corresponds to a constant change in f" is valid directly from the relation. I am quite sure f must be continuous and then assume that it's also differentiable and analytic, which I find rather unsettling... As for $c$, if you are still interested, here is how I managed it. Say on A (B), I observe B (A), at $bf{r}$ ($-bf{r}$), moving away from me at volocity of $Hbf{r}+bf{c}$($-Hbf{r}+bf{c}$). So $c$ should go to 0.
$endgroup$
– Geor
Aug 9 '14 at 15:55
$begingroup$
I am not certain if the assertion "the change in r corresponds to a constant change in f" is valid directly from the relation. I am quite sure f must be continuous and then assume that it's also differentiable and analytic, which I find rather unsettling... As for $c$, if you are still interested, here is how I managed it. Say on A (B), I observe B (A), at $bf{r}$ ($-bf{r}$), moving away from me at volocity of $Hbf{r}+bf{c}$($-Hbf{r}+bf{c}$). So $c$ should go to 0.
$endgroup$
– Geor
Aug 9 '14 at 15:55
add a comment |
$begingroup$
There is now a very simple way to calculate Hubble’s Constant, by inputting to an equation, the numerical value of Pi and the speed of light (C) from Maxwell’s equations, and the value of a parsec. NO space probe measurements (with their inevitable small measuring / interpretation errors) are now required. Hubble’s Constant is ‘fixed’ at 70.98047 PRECISELY. This maths method removes the errors / tolerances that is always a part of attempting to measuring something as ‘elusive’ as Hubble’s Constant. This has very deep implications for theoretical cosmology.
The equation to perform this is :- 2 X a meg parsec X light speed (C). This total is then divided by Pi to the power of 21. This gives 70.98047 kilometres per sec per meg parsec.
The equation to perform this can also be found in ‘The Principle of Astrogeometry’ on Amazon Kindle Books. This also explains how the Hubble 70.98047 ‘fixing’ equation was found. David.
answered Sep 23 '18 at 20:24
David HineDavid Hine
1
$endgroup$
add a comment |
$begingroup$
There is now a very simple way to calculate Hubble’s Constant, by inputting to an equation, the numerical value of Pi and the speed of light (C) from Maxwell’s equations, and the value of a parsec. NO space probe measurements (with their inevitable small measuring / interpretation errors) are now required. Hubble’s Constant is ‘fixed’ at 70.98047 PRECISELY. This maths method removes the errors / tolerances that is always a part of attempting to measuring something as ‘elusive’ as Hubble’s Constant. This has very deep implications for theoretical cosmology.
The equation to perform this is :- 2 X a meg parsec X light speed (C). This total is then divided by Pi to the power of 21. This gives 70.98047 kilometres per sec per meg parsec.
The equation to perform this can also be found in ‘The Principle of Astrogeometry’ on Amazon Kindle Books. This also explains how the Hubble 70.98047 ‘fixing’ equation was found. David.
answered Sep 23 '18 at 20:24
David HineDavid Hine
1
$endgroup$
add a comment |
$begingroup$
There is now a very simple way to calculate Hubble’s Constant, by inputting to an equation, the numerical value of Pi and the speed of light (C) from Maxwell’s equations, and the value of a parsec. NO space probe measurements (with their inevitable small measuring / interpretation errors) are now required. Hubble’s Constant is ‘fixed’ at 70.98047 PRECISELY. This maths method removes the errors / tolerances that is always a part of attempting to measuring something as ‘elusive’ as Hubble’s Constant. This has very deep implications for theoretical cosmology.
The equation to perform this is :- 2 X a meg parsec X light speed (C). This total is then divided by Pi to the power of 21. This gives 70.98047 kilometres per sec per meg parsec.
The equation to perform this can also be found in ‘The Principle of Astrogeometry’ on Amazon Kindle Books. This also explains how the Hubble 70.98047 ‘fixing’ equation was found. David.
answered Sep 23 '18 at 20:24
David HineDavid Hine
1
$endgroup$
There is now a very simple way to calculate Hubble’s Constant, by inputting to an equation, the numerical value of Pi and the speed of light (C) from Maxwell’s equations, and the value of a parsec. NO space probe measurements (with their inevitable small measuring / interpretation errors) are now required. Hubble’s Constant is ‘fixed’ at 70.98047 PRECISELY. This maths method removes the errors / tolerances that is always a part of attempting to measuring something as ‘elusive’ as Hubble’s Constant. This has very deep implications for theoretical cosmology.
The equation to perform this is :- 2 X a meg parsec X light speed (C). This total is then divided by Pi to the power of 21. This gives 70.98047 kilometres per sec per meg parsec.
The equation to perform this can also be found in ‘The Principle of Astrogeometry’ on Amazon Kindle Books. This also explains how the Hubble 70.98047 ‘fixing’ equation was found. David.
answered Sep 23 '18 at 20:24
David HineDavid Hine
1
answered Sep 23 '18 at 20:24
David HineDavid Hine
1
answered Sep 23 '18 at 20:24
David HineDavid Hine
1
answered Sep 23 '18 at 20:24
David HineDavid Hine
1
1
add a comment |
add a comment |
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