Mixing
What is Einsteins ds?
$begingroup$
First off, sorry to trouble you with this, but it is really puzzling me.
(The 2s are powers, I don't know how to do that here.)
According to Professor Richard Conn Henry here Albert Einstein said:
Einstein's claim is that:
$dx^2 + dy^2 + dz^2 − dt^2 = ds^2 =dx^{'2} + dy^{'2} + dz^{'2} − dt^{′2}$ i.e, ds is
invariant.
Professor Henry goes on to show that photons know no time!
My hypothetical robot takes 1 step every second and can turn 90 degrees and move in any direction.
I am an observer. I send my robot a signal: take $3$ steps in $x$, $dx = 3$, $4$ steps in $y$, $dy = 4$, $12$ steps in $z$, $dz = 12$ together 19 steps, $dt = 19$ seconds. velocity $= 1$ step per second. Following professor Henry, I multiply $dt$ by $v$ to make the units fit.
In a straight line, my robot is now $13$ steps from its start position.
I will call $ds$ "$dE$".
$dE^2 = dx^2 + dy^2 + dz^2 - (vdt)^2$
$dE^2 = 9 + 16 + 144 - (1*19)^2 = 169 - 361 = -192$
and if we say $dx = 3, dy, dz = 0$
$dE^2 = 9 - (1*3)2 = 9 - 9 = 0$ My robot took $3$ steps and did not move.
and, if I say distance from origin $= 13$, $dt = 19$, $v = 13/19$
$dE^2 = 9 + 16 + 144 - (13/19*19)2 = 169 - 169 = 0$
Could you please tell me what $dE$ represents? I thought it should be a length as it is obtained from the sum of lengths squared.
linear-algebra
edited Jan 8 at 8:20
Zober
735
asked Jan 8 at 5:57
PedroskiPedroski
1051
$endgroup$
add a comment |
$begingroup$
First off, sorry to trouble you with this, but it is really puzzling me.
(The 2s are powers, I don't know how to do that here.)
According to Professor Richard Conn Henry here Albert Einstein said:
Einstein's claim is that:
$dx^2 + dy^2 + dz^2 − dt^2 = ds^2 =dx^{'2} + dy^{'2} + dz^{'2} − dt^{′2}$ i.e, ds is
invariant.
Professor Henry goes on to show that photons know no time!
My hypothetical robot takes 1 step every second and can turn 90 degrees and move in any direction.
I am an observer. I send my robot a signal: take $3$ steps in $x$, $dx = 3$, $4$ steps in $y$, $dy = 4$, $12$ steps in $z$, $dz = 12$ together 19 steps, $dt = 19$ seconds. velocity $= 1$ step per second. Following professor Henry, I multiply $dt$ by $v$ to make the units fit.
In a straight line, my robot is now $13$ steps from its start position.
I will call $ds$ "$dE$".
$dE^2 = dx^2 + dy^2 + dz^2 - (vdt)^2$
$dE^2 = 9 + 16 + 144 - (1*19)^2 = 169 - 361 = -192$
and if we say $dx = 3, dy, dz = 0$
$dE^2 = 9 - (1*3)2 = 9 - 9 = 0$ My robot took $3$ steps and did not move.
and, if I say distance from origin $= 13$, $dt = 19$, $v = 13/19$
$dE^2 = 9 + 16 + 144 - (13/19*19)2 = 169 - 169 = 0$
Could you please tell me what $dE$ represents? I thought it should be a length as it is obtained from the sum of lengths squared.
linear-algebra
edited Jan 8 at 8:20
Zober
735
asked Jan 8 at 5:57
PedroskiPedroski
1051
$endgroup$
$begingroup$
Please see math.meta.stackexchange.com/questions/5020
$endgroup$
– Lord Shark the Unknown
Jan 8 at 6:14
1
$begingroup$
I did some edits on your post to make it more comprehensible.
$endgroup$
– Zober
Jan 8 at 6:54
$begingroup$
Btw, if i understandly it correctly, $dE$ represent's spacetime displacement. in Euclidean geometry the metric (which represents infinitisemal spatial changes in the space) is given by $ds^2=dx^2+dy^2+dz^2$, in Einstein theory, the traditional euclidean space is replaced by the Minkowskian Space, which has a metric $ds^2$ of the form $c^2dt^2-dx^2-dy^2-dz^2$ ~because we use natural units, c is set to be $=1$.
$endgroup$
– Zober
Jan 8 at 6:59
1
$begingroup$
@Zober We seem to be using the other sign convention here, with the spatial $d$'s having positive sign and the temporal $d$ having negative sign
$endgroup$
– Arthur
Jan 8 at 7:46
add a comment |
$begingroup$
First off, sorry to trouble you with this, but it is really puzzling me.
(The 2s are powers, I don't know how to do that here.)
According to Professor Richard Conn Henry here Albert Einstein said:
Einstein's claim is that:
$dx^2 + dy^2 + dz^2 − dt^2 = ds^2 =dx^{'2} + dy^{'2} + dz^{'2} − dt^{′2}$ i.e, ds is
invariant.
Professor Henry goes on to show that photons know no time!
My hypothetical robot takes 1 step every second and can turn 90 degrees and move in any direction.
I am an observer. I send my robot a signal: take $3$ steps in $x$, $dx = 3$, $4$ steps in $y$, $dy = 4$, $12$ steps in $z$, $dz = 12$ together 19 steps, $dt = 19$ seconds. velocity $= 1$ step per second. Following professor Henry, I multiply $dt$ by $v$ to make the units fit.
In a straight line, my robot is now $13$ steps from its start position.
I will call $ds$ "$dE$".
$dE^2 = dx^2 + dy^2 + dz^2 - (vdt)^2$
$dE^2 = 9 + 16 + 144 - (1*19)^2 = 169 - 361 = -192$
and if we say $dx = 3, dy, dz = 0$
$dE^2 = 9 - (1*3)2 = 9 - 9 = 0$ My robot took $3$ steps and did not move.
and, if I say distance from origin $= 13$, $dt = 19$, $v = 13/19$
$dE^2 = 9 + 16 + 144 - (13/19*19)2 = 169 - 169 = 0$
Could you please tell me what $dE$ represents? I thought it should be a length as it is obtained from the sum of lengths squared.
linear-algebra
edited Jan 8 at 8:20
Zober
735
asked Jan 8 at 5:57
PedroskiPedroski
1051
$endgroup$
First off, sorry to trouble you with this, but it is really puzzling me.
(The 2s are powers, I don't know how to do that here.)
According to Professor Richard Conn Henry here Albert Einstein said:
Einstein's claim is that:
$dx^2 + dy^2 + dz^2 − dt^2 = ds^2 =dx^{'2} + dy^{'2} + dz^{'2} − dt^{′2}$ i.e, ds is
invariant.
Professor Henry goes on to show that photons know no time!
My hypothetical robot takes 1 step every second and can turn 90 degrees and move in any direction.
I am an observer. I send my robot a signal: take $3$ steps in $x$, $dx = 3$, $4$ steps in $y$, $dy = 4$, $12$ steps in $z$, $dz = 12$ together 19 steps, $dt = 19$ seconds. velocity $= 1$ step per second. Following professor Henry, I multiply $dt$ by $v$ to make the units fit.
In a straight line, my robot is now $13$ steps from its start position.
I will call $ds$ "$dE$".
$dE^2 = dx^2 + dy^2 + dz^2 - (vdt)^2$
$dE^2 = 9 + 16 + 144 - (1*19)^2 = 169 - 361 = -192$
and if we say $dx = 3, dy, dz = 0$
$dE^2 = 9 - (1*3)2 = 9 - 9 = 0$ My robot took $3$ steps and did not move.
and, if I say distance from origin $= 13$, $dt = 19$, $v = 13/19$
$dE^2 = 9 + 16 + 144 - (13/19*19)2 = 169 - 169 = 0$
Could you please tell me what $dE$ represents? I thought it should be a length as it is obtained from the sum of lengths squared.
linear-algebra
linear-algebra
edited Jan 8 at 8:20
Zober
735
asked Jan 8 at 5:57
PedroskiPedroski
1051
edited Jan 8 at 8:20
Zober
735
asked Jan 8 at 5:57
PedroskiPedroski
1051
edited Jan 8 at 8:20
Zober
735
edited Jan 8 at 8:20
Zober
735
edited Jan 8 at 8:20
Zober
735
735
asked Jan 8 at 5:57
PedroskiPedroski
1051
asked Jan 8 at 5:57
PedroskiPedroski
1051
asked Jan 8 at 5:57
PedroskiPedroski
1051
1051
$begingroup$
Please see math.meta.stackexchange.com/questions/5020
$endgroup$
– Lord Shark the Unknown
Jan 8 at 6:14
1
$begingroup$
I did some edits on your post to make it more comprehensible.
$endgroup$
– Zober
Jan 8 at 6:54
$begingroup$
Btw, if i understandly it correctly, $dE$ represent's spacetime displacement. in Euclidean geometry the metric (which represents infinitisemal spatial changes in the space) is given by $ds^2=dx^2+dy^2+dz^2$, in Einstein theory, the traditional euclidean space is replaced by the Minkowskian Space, which has a metric $ds^2$ of the form $c^2dt^2-dx^2-dy^2-dz^2$ ~because we use natural units, c is set to be $=1$.
$endgroup$
– Zober
Jan 8 at 6:59
1
$begingroup$
@Zober We seem to be using the other sign convention here, with the spatial $d$'s having positive sign and the temporal $d$ having negative sign
$endgroup$
– Arthur
Jan 8 at 7:46
add a comment |
$begingroup$
Please see math.meta.stackexchange.com/questions/5020
$endgroup$
– Lord Shark the Unknown
Jan 8 at 6:14
1
$begingroup$
I did some edits on your post to make it more comprehensible.
$endgroup$
– Zober
Jan 8 at 6:54
$begingroup$
Btw, if i understandly it correctly, $dE$ represent's spacetime displacement. in Euclidean geometry the metric (which represents infinitisemal spatial changes in the space) is given by $ds^2=dx^2+dy^2+dz^2$, in Einstein theory, the traditional euclidean space is replaced by the Minkowskian Space, which has a metric $ds^2$ of the form $c^2dt^2-dx^2-dy^2-dz^2$ ~because we use natural units, c is set to be $=1$.
$endgroup$
– Zober
Jan 8 at 6:59
1
$begingroup$
@Zober We seem to be using the other sign convention here, with the spatial $d$'s having positive sign and the temporal $d$ having negative sign
$endgroup$
– Arthur
Jan 8 at 7:46
$begingroup$
Please see math.meta.stackexchange.com/questions/5020
$endgroup$
– Lord Shark the Unknown
Jan 8 at 6:14
$begingroup$
Please see math.meta.stackexchange.com/questions/5020
$endgroup$
– Lord Shark the Unknown
Jan 8 at 6:14
1
1
$begingroup$
I did some edits on your post to make it more comprehensible.
$endgroup$
– Zober
Jan 8 at 6:54
$begingroup$
I did some edits on your post to make it more comprehensible.
$endgroup$
– Zober
Jan 8 at 6:54
$begingroup$
Btw, if i understandly it correctly, $dE$ represent's spacetime displacement. in Euclidean geometry the metric (which represents infinitisemal spatial changes in the space) is given by $ds^2=dx^2+dy^2+dz^2$, in Einstein theory, the traditional euclidean space is replaced by the Minkowskian Space, which has a metric $ds^2$ of the form $c^2dt^2-dx^2-dy^2-dz^2$ ~because we use natural units, c is set to be $=1$.
$endgroup$
– Zober
Jan 8 at 6:59
$begingroup$
Btw, if i understandly it correctly, $dE$ represent's spacetime displacement. in Euclidean geometry the metric (which represents infinitisemal spatial changes in the space) is given by $ds^2=dx^2+dy^2+dz^2$, in Einstein theory, the traditional euclidean space is replaced by the Minkowskian Space, which has a metric $ds^2$ of the form $c^2dt^2-dx^2-dy^2-dz^2$ ~because we use natural units, c is set to be $=1$.
$endgroup$
– Zober
Jan 8 at 6:59
1
1
$begingroup$
@Zober We seem to be using the other sign convention here, with the spatial $d$'s having positive sign and the temporal $d$ having negative sign
$endgroup$
– Arthur
Jan 8 at 7:46
$begingroup$
@Zober We seem to be using the other sign convention here, with the spatial $d$'s having positive sign and the temporal $d$ having negative sign
$endgroup$
– Arthur
Jan 8 at 7:46
add a comment |
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$begingroup$
Please see math.meta.stackexchange.com/questions/5020
$endgroup$
– Lord Shark the Unknown
Jan 8 at 6:14
1
$begingroup$
I did some edits on your post to make it more comprehensible.
$endgroup$
– Zober
Jan 8 at 6:54
$begingroup$
Btw, if i understandly it correctly, $dE$ represent's spacetime displacement. in Euclidean geometry the metric (which represents infinitisemal spatial changes in the space) is given by $ds^2=dx^2+dy^2+dz^2$, in Einstein theory, the traditional euclidean space is replaced by the Minkowskian Space, which has a metric $ds^2$ of the form $c^2dt^2-dx^2-dy^2-dz^2$ ~because we use natural units, c is set to be $=1$.
$endgroup$
– Zober
Jan 8 at 6:59
1
$begingroup$
@Zober We seem to be using the other sign convention here, with the spatial $d$'s having positive sign and the temporal $d$ having negative sign
$endgroup$
– Arthur
Jan 8 at 7:46