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Sum of two velocities is smaller than the speed of light
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Using the Lorentz transformation from special relativity, we get that the sum of two velocities can be expressed as
$$u=fracu'+v1+fracu'vc^2.$$
Given that $|u'|,|v| le c$, I want to prove that $|u| le c$, ie. that the velocity never exceeds $c$. However I am struggling to produce this bound. I have tried to bound the denominator from above but this produces zero and have tried a case wise approach but this has got me no where either.
real-analysis special-relativity
asked 2 hours ago
user258521
328210
add a comment |Â
up vote
3
down vote
favorite
Using the Lorentz transformation from special relativity, we get that the sum of two velocities can be expressed as
$$u=fracu'+v1+fracu'vc^2.$$
Given that $|u'|,|v| le c$, I want to prove that $|u| le c$, ie. that the velocity never exceeds $c$. However I am struggling to produce this bound. I have tried to bound the denominator from above but this produces zero and have tried a case wise approach but this has got me no where either.
real-analysis special-relativity
asked 2 hours ago
user258521
328210
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Using the Lorentz transformation from special relativity, we get that the sum of two velocities can be expressed as
$$u=fracu'+v1+fracu'vc^2.$$
Given that $|u'|,|v| le c$, I want to prove that $|u| le c$, ie. that the velocity never exceeds $c$. However I am struggling to produce this bound. I have tried to bound the denominator from above but this produces zero and have tried a case wise approach but this has got me no where either.
real-analysis special-relativity
asked 2 hours ago
user258521
328210
Using the Lorentz transformation from special relativity, we get that the sum of two velocities can be expressed as
$$u=fracu'+v1+fracu'vc^2.$$
Given that $|u'|,|v| le c$, I want to prove that $|u| le c$, ie. that the velocity never exceeds $c$. However I am struggling to produce this bound. I have tried to bound the denominator from above but this produces zero and have tried a case wise approach but this has got me no where either.
real-analysis special-relativity
real-analysis special-relativity
asked 2 hours ago
user258521
328210
asked 2 hours ago
user258521
328210
asked 2 hours ago
user258521
328210
asked 2 hours ago
user258521
328210
asked 2 hours ago
user258521
328210
328210
add a comment |Â
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
4
down vote
accepted
With only basic analysis, you can do it by fixing one of the two variables.
You can prove that, for a fixed $u'<c$:
- The function $f_u'(v)=fracu'+v1+fracu'c^2v$ is a monotonically increasing function.
- $f_u'(c)=c$
From $1$, you can conclude that if $v<c$, $f_u'(v)<f_u'(c)$,and including $2$, that gives you $f_u'(v)<c$.
This proves that if $u',v$ are both smaller than $c$, that $fracu'+v1+fracu'vc^2$ will also be smaller than $c$.
answered 2 hours ago
5xum
86.2k388155
This is very nice !
– user258521
2 hours ago
@user258521 Thanks. It's a little strange in the sense that you "break the symmetry" of the original expression - you view one variable as a variable, and the other as a parameter, and then solve the problem for each value of the parameter. Some call it ugly, some like it, it depends. But so long as it works :)
– 5xum
2 hours ago
I think it is perfectly fine, nothing wrong with this. Also in 1. it should have $c^2$ on the denominator, a small edit.
– user258521
2 hours ago
add a comment |Â
up vote
1
down vote
Hint: If you're familiar with techniques in optimization, try and choose values of $u'$ and $v$ so as to optimize $u$. If you can prove that the optimal values for making $u$ large yield a value of $u$ at or approaching $c$, then you are done.
answered 2 hours ago
Cade Reinberger
1527
The only optimisation techniques I can think of are derivatives so not too sure if this will be useful? Relating to your answer, $u'=v=c$ is a possible value but not sure how the argument follows?
– user258521
2 hours ago
Well, you're optimizing over a square, which importantly is compact. Explicitly, first prove the problem on the boundary (that is, when at least one of $u'$ or $v$ is $c$). Then you know if there is any value above c, it must be on the interior, and thus must be a simultaneous zero of both partial derivatives.
– Cade Reinberger
2 hours ago
add a comment |Â
up vote
1
down vote
Here's an elementary general proof.
Letting $x=u'/c, y = v/c, z = u/c$ the equation for the Einstein sum becomes
$$z = fracx+y1+x ytag1$$
and we have to prove that
$$-1lt z lt 1tag2$$
for $-1lt x lt 1, -1 lt y lt 1$.
Now we let
$$xto frac1-r1+r,yto frac1-s1+stag3a$$
or
$$r to frac1-x1+x, sto frac1-y1+ytag3b$$
which transforms $xin (-1,+1)$ to $r in (infty, 0)$ and $yin (-1,+1)$ to $s in (infty, 0)$ .
In other words, we have parametrized $x$ and $y$ with positive parameters $r$ and $s$.
Substituting (3) in (1) gives
$$z = frac1-t1+ttag4$$
with $t = r s$. Hence we have $t in (infty, 0)$ and from (4) follows (2). QED.
answered 40 mins ago
Dr. Wolfgang Hintze
2,785515
add a comment |Â
up vote
0
down vote
Such relativistic sum of speeds (here denoted as $oplus$) can be written in terms of a simple pullback:
$$ uoplus vstackreltextdef=ccdottanhleft(textarctanhtfracuc+textarctanhtfracvcright)$$
since $tanh$ is an increasing function with range $(-1,1)$, $left|uoplus vright|< c$ immediately follows.
answered 29 mins ago
Jack D'Aurizio♦
278k33272646
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
With only basic analysis, you can do it by fixing one of the two variables.
You can prove that, for a fixed $u'<c$:
- The function $f_u'(v)=fracu'+v1+fracu'c^2v$ is a monotonically increasing function.
- $f_u'(c)=c$
From $1$, you can conclude that if $v<c$, $f_u'(v)<f_u'(c)$,and including $2$, that gives you $f_u'(v)<c$.
This proves that if $u',v$ are both smaller than $c$, that $fracu'+v1+fracu'vc^2$ will also be smaller than $c$.
answered 2 hours ago
5xum
86.2k388155
This is very nice !
– user258521
2 hours ago
@user258521 Thanks. It's a little strange in the sense that you "break the symmetry" of the original expression - you view one variable as a variable, and the other as a parameter, and then solve the problem for each value of the parameter. Some call it ugly, some like it, it depends. But so long as it works :)
– 5xum
2 hours ago
I think it is perfectly fine, nothing wrong with this. Also in 1. it should have $c^2$ on the denominator, a small edit.
– user258521
2 hours ago
add a comment |Â
up vote
4
down vote
accepted
With only basic analysis, you can do it by fixing one of the two variables.
You can prove that, for a fixed $u'<c$:
- The function $f_u'(v)=fracu'+v1+fracu'c^2v$ is a monotonically increasing function.
- $f_u'(c)=c$
From $1$, you can conclude that if $v<c$, $f_u'(v)<f_u'(c)$,and including $2$, that gives you $f_u'(v)<c$.
This proves that if $u',v$ are both smaller than $c$, that $fracu'+v1+fracu'vc^2$ will also be smaller than $c$.
answered 2 hours ago
5xum
86.2k388155
This is very nice !
– user258521
2 hours ago
@user258521 Thanks. It's a little strange in the sense that you "break the symmetry" of the original expression - you view one variable as a variable, and the other as a parameter, and then solve the problem for each value of the parameter. Some call it ugly, some like it, it depends. But so long as it works :)
– 5xum
2 hours ago
I think it is perfectly fine, nothing wrong with this. Also in 1. it should have $c^2$ on the denominator, a small edit.
– user258521
2 hours ago
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
With only basic analysis, you can do it by fixing one of the two variables.
You can prove that, for a fixed $u'<c$:
- The function $f_u'(v)=fracu'+v1+fracu'c^2v$ is a monotonically increasing function.
- $f_u'(c)=c$
From $1$, you can conclude that if $v<c$, $f_u'(v)<f_u'(c)$,and including $2$, that gives you $f_u'(v)<c$.
This proves that if $u',v$ are both smaller than $c$, that $fracu'+v1+fracu'vc^2$ will also be smaller than $c$.
answered 2 hours ago
5xum
86.2k388155
With only basic analysis, you can do it by fixing one of the two variables.
You can prove that, for a fixed $u'<c$:
- The function $f_u'(v)=fracu'+v1+fracu'c^2v$ is a monotonically increasing function.
- $f_u'(c)=c$
From $1$, you can conclude that if $v<c$, $f_u'(v)<f_u'(c)$,and including $2$, that gives you $f_u'(v)<c$.
This proves that if $u',v$ are both smaller than $c$, that $fracu'+v1+fracu'vc^2$ will also be smaller than $c$.
answered 2 hours ago
5xum
86.2k388155
edited 2 hours ago
answered 2 hours ago
5xum
86.2k388155
answered 2 hours ago
5xum
86.2k388155
answered 2 hours ago
5xum
86.2k388155
86.2k388155
This is very nice !
– user258521
2 hours ago
@user258521 Thanks. It's a little strange in the sense that you "break the symmetry" of the original expression - you view one variable as a variable, and the other as a parameter, and then solve the problem for each value of the parameter. Some call it ugly, some like it, it depends. But so long as it works :)
– 5xum
2 hours ago
I think it is perfectly fine, nothing wrong with this. Also in 1. it should have $c^2$ on the denominator, a small edit.
– user258521
2 hours ago
add a comment |Â
This is very nice !
– user258521
2 hours ago
@user258521 Thanks. It's a little strange in the sense that you "break the symmetry" of the original expression - you view one variable as a variable, and the other as a parameter, and then solve the problem for each value of the parameter. Some call it ugly, some like it, it depends. But so long as it works :)
– 5xum
2 hours ago
I think it is perfectly fine, nothing wrong with this. Also in 1. it should have $c^2$ on the denominator, a small edit.
– user258521
2 hours ago
This is very nice !
– user258521
2 hours ago
This is very nice !
– user258521
2 hours ago
@user258521 Thanks. It's a little strange in the sense that you "break the symmetry" of the original expression - you view one variable as a variable, and the other as a parameter, and then solve the problem for each value of the parameter. Some call it ugly, some like it, it depends. But so long as it works :)
– 5xum
2 hours ago
@user258521 Thanks. It's a little strange in the sense that you "break the symmetry" of the original expression - you view one variable as a variable, and the other as a parameter, and then solve the problem for each value of the parameter. Some call it ugly, some like it, it depends. But so long as it works :)
– 5xum
2 hours ago
I think it is perfectly fine, nothing wrong with this. Also in 1. it should have $c^2$ on the denominator, a small edit.
– user258521
2 hours ago
I think it is perfectly fine, nothing wrong with this. Also in 1. it should have $c^2$ on the denominator, a small edit.
– user258521
2 hours ago
add a comment |Â
up vote
1
down vote
Hint: If you're familiar with techniques in optimization, try and choose values of $u'$ and $v$ so as to optimize $u$. If you can prove that the optimal values for making $u$ large yield a value of $u$ at or approaching $c$, then you are done.
answered 2 hours ago
Cade Reinberger
1527
The only optimisation techniques I can think of are derivatives so not too sure if this will be useful? Relating to your answer, $u'=v=c$ is a possible value but not sure how the argument follows?
– user258521
2 hours ago
Well, you're optimizing over a square, which importantly is compact. Explicitly, first prove the problem on the boundary (that is, when at least one of $u'$ or $v$ is $c$). Then you know if there is any value above c, it must be on the interior, and thus must be a simultaneous zero of both partial derivatives.
– Cade Reinberger
2 hours ago
add a comment |Â
up vote
1
down vote
Hint: If you're familiar with techniques in optimization, try and choose values of $u'$ and $v$ so as to optimize $u$. If you can prove that the optimal values for making $u$ large yield a value of $u$ at or approaching $c$, then you are done.
answered 2 hours ago
Cade Reinberger
1527
The only optimisation techniques I can think of are derivatives so not too sure if this will be useful? Relating to your answer, $u'=v=c$ is a possible value but not sure how the argument follows?
– user258521
2 hours ago
Well, you're optimizing over a square, which importantly is compact. Explicitly, first prove the problem on the boundary (that is, when at least one of $u'$ or $v$ is $c$). Then you know if there is any value above c, it must be on the interior, and thus must be a simultaneous zero of both partial derivatives.
– Cade Reinberger
2 hours ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Hint: If you're familiar with techniques in optimization, try and choose values of $u'$ and $v$ so as to optimize $u$. If you can prove that the optimal values for making $u$ large yield a value of $u$ at or approaching $c$, then you are done.
answered 2 hours ago
Cade Reinberger
1527
Hint: If you're familiar with techniques in optimization, try and choose values of $u'$ and $v$ so as to optimize $u$. If you can prove that the optimal values for making $u$ large yield a value of $u$ at or approaching $c$, then you are done.
answered 2 hours ago
Cade Reinberger
1527
answered 2 hours ago
Cade Reinberger
1527
answered 2 hours ago
Cade Reinberger
1527
answered 2 hours ago
Cade Reinberger
1527
1527
The only optimisation techniques I can think of are derivatives so not too sure if this will be useful? Relating to your answer, $u'=v=c$ is a possible value but not sure how the argument follows?
– user258521
2 hours ago
Well, you're optimizing over a square, which importantly is compact. Explicitly, first prove the problem on the boundary (that is, when at least one of $u'$ or $v$ is $c$). Then you know if there is any value above c, it must be on the interior, and thus must be a simultaneous zero of both partial derivatives.
– Cade Reinberger
2 hours ago
add a comment |Â
The only optimisation techniques I can think of are derivatives so not too sure if this will be useful? Relating to your answer, $u'=v=c$ is a possible value but not sure how the argument follows?
– user258521
2 hours ago
Well, you're optimizing over a square, which importantly is compact. Explicitly, first prove the problem on the boundary (that is, when at least one of $u'$ or $v$ is $c$). Then you know if there is any value above c, it must be on the interior, and thus must be a simultaneous zero of both partial derivatives.
– Cade Reinberger
2 hours ago
The only optimisation techniques I can think of are derivatives so not too sure if this will be useful? Relating to your answer, $u'=v=c$ is a possible value but not sure how the argument follows?
– user258521
2 hours ago
The only optimisation techniques I can think of are derivatives so not too sure if this will be useful? Relating to your answer, $u'=v=c$ is a possible value but not sure how the argument follows?
– user258521
2 hours ago
Well, you're optimizing over a square, which importantly is compact. Explicitly, first prove the problem on the boundary (that is, when at least one of $u'$ or $v$ is $c$). Then you know if there is any value above c, it must be on the interior, and thus must be a simultaneous zero of both partial derivatives.
– Cade Reinberger
2 hours ago
Well, you're optimizing over a square, which importantly is compact. Explicitly, first prove the problem on the boundary (that is, when at least one of $u'$ or $v$ is $c$). Then you know if there is any value above c, it must be on the interior, and thus must be a simultaneous zero of both partial derivatives.
– Cade Reinberger
2 hours ago
add a comment |Â
up vote
1
down vote
Here's an elementary general proof.
Letting $x=u'/c, y = v/c, z = u/c$ the equation for the Einstein sum becomes
$$z = fracx+y1+x ytag1$$
and we have to prove that
$$-1lt z lt 1tag2$$
for $-1lt x lt 1, -1 lt y lt 1$.
Now we let
$$xto frac1-r1+r,yto frac1-s1+stag3a$$
or
$$r to frac1-x1+x, sto frac1-y1+ytag3b$$
which transforms $xin (-1,+1)$ to $r in (infty, 0)$ and $yin (-1,+1)$ to $s in (infty, 0)$ .
In other words, we have parametrized $x$ and $y$ with positive parameters $r$ and $s$.
Substituting (3) in (1) gives
$$z = frac1-t1+ttag4$$
with $t = r s$. Hence we have $t in (infty, 0)$ and from (4) follows (2). QED.
answered 40 mins ago
Dr. Wolfgang Hintze
2,785515
add a comment |Â
up vote
1
down vote
Here's an elementary general proof.
Letting $x=u'/c, y = v/c, z = u/c$ the equation for the Einstein sum becomes
$$z = fracx+y1+x ytag1$$
and we have to prove that
$$-1lt z lt 1tag2$$
for $-1lt x lt 1, -1 lt y lt 1$.
Now we let
$$xto frac1-r1+r,yto frac1-s1+stag3a$$
or
$$r to frac1-x1+x, sto frac1-y1+ytag3b$$
which transforms $xin (-1,+1)$ to $r in (infty, 0)$ and $yin (-1,+1)$ to $s in (infty, 0)$ .
In other words, we have parametrized $x$ and $y$ with positive parameters $r$ and $s$.
Substituting (3) in (1) gives
$$z = frac1-t1+ttag4$$
with $t = r s$. Hence we have $t in (infty, 0)$ and from (4) follows (2). QED.
answered 40 mins ago
Dr. Wolfgang Hintze
2,785515
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Here's an elementary general proof.
Letting $x=u'/c, y = v/c, z = u/c$ the equation for the Einstein sum becomes
$$z = fracx+y1+x ytag1$$
and we have to prove that
$$-1lt z lt 1tag2$$
for $-1lt x lt 1, -1 lt y lt 1$.
Now we let
$$xto frac1-r1+r,yto frac1-s1+stag3a$$
or
$$r to frac1-x1+x, sto frac1-y1+ytag3b$$
which transforms $xin (-1,+1)$ to $r in (infty, 0)$ and $yin (-1,+1)$ to $s in (infty, 0)$ .
In other words, we have parametrized $x$ and $y$ with positive parameters $r$ and $s$.
Substituting (3) in (1) gives
$$z = frac1-t1+ttag4$$
with $t = r s$. Hence we have $t in (infty, 0)$ and from (4) follows (2). QED.
answered 40 mins ago
Dr. Wolfgang Hintze
2,785515
Here's an elementary general proof.
Letting $x=u'/c, y = v/c, z = u/c$ the equation for the Einstein sum becomes
$$z = fracx+y1+x ytag1$$
and we have to prove that
$$-1lt z lt 1tag2$$
for $-1lt x lt 1, -1 lt y lt 1$.
Now we let
$$xto frac1-r1+r,yto frac1-s1+stag3a$$
or
$$r to frac1-x1+x, sto frac1-y1+ytag3b$$
which transforms $xin (-1,+1)$ to $r in (infty, 0)$ and $yin (-1,+1)$ to $s in (infty, 0)$ .
In other words, we have parametrized $x$ and $y$ with positive parameters $r$ and $s$.
Substituting (3) in (1) gives
$$z = frac1-t1+ttag4$$
with $t = r s$. Hence we have $t in (infty, 0)$ and from (4) follows (2). QED.
answered 40 mins ago
Dr. Wolfgang Hintze
2,785515
edited 17 mins ago
answered 40 mins ago
Dr. Wolfgang Hintze
2,785515
answered 40 mins ago
Dr. Wolfgang Hintze
2,785515
answered 40 mins ago
Dr. Wolfgang Hintze
2,785515
2,785515
add a comment |Â
add a comment |Â
up vote
0
down vote
Such relativistic sum of speeds (here denoted as $oplus$) can be written in terms of a simple pullback:
$$ uoplus vstackreltextdef=ccdottanhleft(textarctanhtfracuc+textarctanhtfracvcright)$$
since $tanh$ is an increasing function with range $(-1,1)$, $left|uoplus vright|< c$ immediately follows.
answered 29 mins ago
Jack D'Aurizio♦
278k33272646
add a comment |Â
up vote
0
down vote
Such relativistic sum of speeds (here denoted as $oplus$) can be written in terms of a simple pullback:
$$ uoplus vstackreltextdef=ccdottanhleft(textarctanhtfracuc+textarctanhtfracvcright)$$
since $tanh$ is an increasing function with range $(-1,1)$, $left|uoplus vright|< c$ immediately follows.
answered 29 mins ago
Jack D'Aurizio♦
278k33272646
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Such relativistic sum of speeds (here denoted as $oplus$) can be written in terms of a simple pullback:
$$ uoplus vstackreltextdef=ccdottanhleft(textarctanhtfracuc+textarctanhtfracvcright)$$
since $tanh$ is an increasing function with range $(-1,1)$, $left|uoplus vright|< c$ immediately follows.
answered 29 mins ago
Jack D'Aurizio♦
278k33272646
Such relativistic sum of speeds (here denoted as $oplus$) can be written in terms of a simple pullback:
$$ uoplus vstackreltextdef=ccdottanhleft(textarctanhtfracuc+textarctanhtfracvcright)$$
since $tanh$ is an increasing function with range $(-1,1)$, $left|uoplus vright|< c$ immediately follows.
answered 29 mins ago
Jack D'Aurizio♦
278k33272646
answered 29 mins ago
Jack D'Aurizio♦
278k33272646
answered 29 mins ago
Jack D'Aurizio♦
278k33272646
answered 29 mins ago
Jack D'Aurizio♦
278k33272646
278k33272646
add a comment |Â
add a comment |Â
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