Mixing
Origin of negative term $-ct^2$ in the Lorentz invariance of a 4-vector
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So I am taking an introductory course in Special Relativity. In my book the spacetime 4-vector is defined as:
$$X= begin{bmatrix}ct \ x \ y \ z end{bmatrix}$$.
Then the book proceeds to say that the squared length of a 4-vector is given by:
$$|X|^2 = x^2+y^2+z^2- c^2t^2$$
Now what I don't understand why there is a negative term $-c^2t^2$, isn't the length of a vector defined as the square root of the sum of each component?
vectors linear-transformations mathematical-physics special-relativity
asked Oct 11 '18 at 18:38
daljit97daljit97
178111
$endgroup$
add a comment |
$begingroup$
So I am taking an introductory course in Special Relativity. In my book the spacetime 4-vector is defined as:
$$X= begin{bmatrix}ct \ x \ y \ z end{bmatrix}$$.
Then the book proceeds to say that the squared length of a 4-vector is given by:
$$|X|^2 = x^2+y^2+z^2- c^2t^2$$
Now what I don't understand why there is a negative term $-c^2t^2$, isn't the length of a vector defined as the square root of the sum of each component?
vectors linear-transformations mathematical-physics special-relativity
asked Oct 11 '18 at 18:38
daljit97daljit97
178111
$endgroup$
$begingroup$
the spacelike part does do that $x^2 + y^2 + z^2.$ These are split up as having positive, negative, or zero length; called spacelike, timelike, finally the light cone.
$endgroup$
– Will Jagy
Oct 11 '18 at 19:20
$begingroup$
@amd I have updated the title
$endgroup$
– daljit97
Oct 12 '18 at 11:37
add a comment |
$begingroup$
So I am taking an introductory course in Special Relativity. In my book the spacetime 4-vector is defined as:
$$X= begin{bmatrix}ct \ x \ y \ z end{bmatrix}$$.
Then the book proceeds to say that the squared length of a 4-vector is given by:
$$|X|^2 = x^2+y^2+z^2- c^2t^2$$
Now what I don't understand why there is a negative term $-c^2t^2$, isn't the length of a vector defined as the square root of the sum of each component?
vectors linear-transformations mathematical-physics special-relativity
asked Oct 11 '18 at 18:38
daljit97daljit97
178111
$endgroup$
So I am taking an introductory course in Special Relativity. In my book the spacetime 4-vector is defined as:
$$X= begin{bmatrix}ct \ x \ y \ z end{bmatrix}$$.
Then the book proceeds to say that the squared length of a 4-vector is given by:
$$|X|^2 = x^2+y^2+z^2- c^2t^2$$
Now what I don't understand why there is a negative term $-c^2t^2$, isn't the length of a vector defined as the square root of the sum of each component?
vectors linear-transformations mathematical-physics special-relativity
vectors linear-transformations mathematical-physics special-relativity
asked Oct 11 '18 at 18:38
daljit97daljit97
178111
asked Oct 11 '18 at 18:38
daljit97daljit97
178111
edited Jan 15 at 0:52
daljit97
asked Oct 11 '18 at 18:38
daljit97daljit97
178111
asked Oct 11 '18 at 18:38
daljit97daljit97
178111
asked Oct 11 '18 at 18:38
daljit97daljit97
178111
178111
$begingroup$
the spacelike part does do that $x^2 + y^2 + z^2.$ These are split up as having positive, negative, or zero length; called spacelike, timelike, finally the light cone.
$endgroup$
– Will Jagy
Oct 11 '18 at 19:20
$begingroup$
@amd I have updated the title
$endgroup$
– daljit97
Oct 12 '18 at 11:37
add a comment |
$begingroup$
the spacelike part does do that $x^2 + y^2 + z^2.$ These are split up as having positive, negative, or zero length; called spacelike, timelike, finally the light cone.
$endgroup$
– Will Jagy
Oct 11 '18 at 19:20
$begingroup$
@amd I have updated the title
$endgroup$
– daljit97
Oct 12 '18 at 11:37
$begingroup$
the spacelike part does do that $x^2 + y^2 + z^2.$ These are split up as having positive, negative, or zero length; called spacelike, timelike, finally the light cone.
$endgroup$
– Will Jagy
Oct 11 '18 at 19:20
$begingroup$
the spacelike part does do that $x^2 + y^2 + z^2.$ These are split up as having positive, negative, or zero length; called spacelike, timelike, finally the light cone.
$endgroup$
– Will Jagy
Oct 11 '18 at 19:20
$begingroup$
@amd I have updated the title
$endgroup$
– daljit97
Oct 12 '18 at 11:37
$begingroup$
@amd I have updated the title
$endgroup$
– daljit97
Oct 12 '18 at 11:37
add a comment |
2 Answers
2
active
oldest
votes
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This metric is the result of works of a mathematician named Minkowski. He observed that the following metric$$Delta S^2=Delta x^2+Delta y^2+Delta z^2-c^2Delta t^2$$is invariant under Lorentz transformation which connects two inertial frames. Each event namely $P$ carries a light cone with itself which is given by$$Delta x^2+Delta y^2+Delta z^2=c^2Delta t^2$$The events right on the cone are those who can be connected to event $P$ through a light signal. Inside the cone, are those events that can be connected to event $P$ with less that speed of light. There is a very good explanation in Wolfgang Rindler's Relativity: Special, General and Cosmological (which I found it very useful). Hope it also helps you on 4-vectors and 4-tensors.
answered Oct 11 '18 at 19:28
Mostafa AyazMostafa Ayaz
15.6k3939
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add a comment |
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The length $|X|^2=x^2+y^2+z^2+c^2t^2$ that you’re familiar with is the vector’s length in a Euclidean geometry. The geometry of spacetime in S.R. is non-Euclidean, though, and is basically defined by the metric given in your book. (BTW, calling this a “length” seems a bit off to me since $x^2+y^2+z^2-c^2t^2$ can be negative.)
answered Oct 11 '18 at 21:06
amdamd
30.4k21050
$endgroup$
$begingroup$
Ok so you are you saying that this is more a definition rather than a derivation?
$endgroup$
– daljit97
Oct 12 '18 at 11:38
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@daljit97 Just so, at least in the way it’s being presented in your text. Historically, this metric arose from studying the way that Maxwell’s equations behaved under coordinate transformations. Mostafa Ayaz’ answer speaks to this background.
$endgroup$
– amd
Oct 12 '18 at 19:29
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This metric is the result of works of a mathematician named Minkowski. He observed that the following metric$$Delta S^2=Delta x^2+Delta y^2+Delta z^2-c^2Delta t^2$$is invariant under Lorentz transformation which connects two inertial frames. Each event namely $P$ carries a light cone with itself which is given by$$Delta x^2+Delta y^2+Delta z^2=c^2Delta t^2$$The events right on the cone are those who can be connected to event $P$ through a light signal. Inside the cone, are those events that can be connected to event $P$ with less that speed of light. There is a very good explanation in Wolfgang Rindler's Relativity: Special, General and Cosmological (which I found it very useful). Hope it also helps you on 4-vectors and 4-tensors.
answered Oct 11 '18 at 19:28
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
add a comment |
$begingroup$
This metric is the result of works of a mathematician named Minkowski. He observed that the following metric$$Delta S^2=Delta x^2+Delta y^2+Delta z^2-c^2Delta t^2$$is invariant under Lorentz transformation which connects two inertial frames. Each event namely $P$ carries a light cone with itself which is given by$$Delta x^2+Delta y^2+Delta z^2=c^2Delta t^2$$The events right on the cone are those who can be connected to event $P$ through a light signal. Inside the cone, are those events that can be connected to event $P$ with less that speed of light. There is a very good explanation in Wolfgang Rindler's Relativity: Special, General and Cosmological (which I found it very useful). Hope it also helps you on 4-vectors and 4-tensors.
answered Oct 11 '18 at 19:28
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
add a comment |
$begingroup$
This metric is the result of works of a mathematician named Minkowski. He observed that the following metric$$Delta S^2=Delta x^2+Delta y^2+Delta z^2-c^2Delta t^2$$is invariant under Lorentz transformation which connects two inertial frames. Each event namely $P$ carries a light cone with itself which is given by$$Delta x^2+Delta y^2+Delta z^2=c^2Delta t^2$$The events right on the cone are those who can be connected to event $P$ through a light signal. Inside the cone, are those events that can be connected to event $P$ with less that speed of light. There is a very good explanation in Wolfgang Rindler's Relativity: Special, General and Cosmological (which I found it very useful). Hope it also helps you on 4-vectors and 4-tensors.
answered Oct 11 '18 at 19:28
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
This metric is the result of works of a mathematician named Minkowski. He observed that the following metric$$Delta S^2=Delta x^2+Delta y^2+Delta z^2-c^2Delta t^2$$is invariant under Lorentz transformation which connects two inertial frames. Each event namely $P$ carries a light cone with itself which is given by$$Delta x^2+Delta y^2+Delta z^2=c^2Delta t^2$$The events right on the cone are those who can be connected to event $P$ through a light signal. Inside the cone, are those events that can be connected to event $P$ with less that speed of light. There is a very good explanation in Wolfgang Rindler's Relativity: Special, General and Cosmological (which I found it very useful). Hope it also helps you on 4-vectors and 4-tensors.
answered Oct 11 '18 at 19:28
Mostafa AyazMostafa Ayaz
15.6k3939
answered Oct 11 '18 at 19:28
Mostafa AyazMostafa Ayaz
15.6k3939
answered Oct 11 '18 at 19:28
Mostafa AyazMostafa Ayaz
15.6k3939
answered Oct 11 '18 at 19:28
Mostafa AyazMostafa Ayaz
15.6k3939
15.6k3939
add a comment |
add a comment |
$begingroup$
The length $|X|^2=x^2+y^2+z^2+c^2t^2$ that you’re familiar with is the vector’s length in a Euclidean geometry. The geometry of spacetime in S.R. is non-Euclidean, though, and is basically defined by the metric given in your book. (BTW, calling this a “length” seems a bit off to me since $x^2+y^2+z^2-c^2t^2$ can be negative.)
answered Oct 11 '18 at 21:06
amdamd
30.4k21050
$endgroup$
$begingroup$
Ok so you are you saying that this is more a definition rather than a derivation?
$endgroup$
– daljit97
Oct 12 '18 at 11:38
$begingroup$
@daljit97 Just so, at least in the way it’s being presented in your text. Historically, this metric arose from studying the way that Maxwell’s equations behaved under coordinate transformations. Mostafa Ayaz’ answer speaks to this background.
$endgroup$
– amd
Oct 12 '18 at 19:29
add a comment |
$begingroup$
The length $|X|^2=x^2+y^2+z^2+c^2t^2$ that you’re familiar with is the vector’s length in a Euclidean geometry. The geometry of spacetime in S.R. is non-Euclidean, though, and is basically defined by the metric given in your book. (BTW, calling this a “length” seems a bit off to me since $x^2+y^2+z^2-c^2t^2$ can be negative.)
answered Oct 11 '18 at 21:06
amdamd
30.4k21050
$endgroup$
$begingroup$
Ok so you are you saying that this is more a definition rather than a derivation?
$endgroup$
– daljit97
Oct 12 '18 at 11:38
$begingroup$
@daljit97 Just so, at least in the way it’s being presented in your text. Historically, this metric arose from studying the way that Maxwell’s equations behaved under coordinate transformations. Mostafa Ayaz’ answer speaks to this background.
$endgroup$
– amd
Oct 12 '18 at 19:29
add a comment |
$begingroup$
The length $|X|^2=x^2+y^2+z^2+c^2t^2$ that you’re familiar with is the vector’s length in a Euclidean geometry. The geometry of spacetime in S.R. is non-Euclidean, though, and is basically defined by the metric given in your book. (BTW, calling this a “length” seems a bit off to me since $x^2+y^2+z^2-c^2t^2$ can be negative.)
answered Oct 11 '18 at 21:06
amdamd
30.4k21050
$endgroup$
The length $|X|^2=x^2+y^2+z^2+c^2t^2$ that you’re familiar with is the vector’s length in a Euclidean geometry. The geometry of spacetime in S.R. is non-Euclidean, though, and is basically defined by the metric given in your book. (BTW, calling this a “length” seems a bit off to me since $x^2+y^2+z^2-c^2t^2$ can be negative.)
answered Oct 11 '18 at 21:06
amdamd
30.4k21050
answered Oct 11 '18 at 21:06
amdamd
30.4k21050
answered Oct 11 '18 at 21:06
amdamd
30.4k21050
answered Oct 11 '18 at 21:06
amdamd
30.4k21050
30.4k21050
$begingroup$
Ok so you are you saying that this is more a definition rather than a derivation?
$endgroup$
– daljit97
Oct 12 '18 at 11:38
$begingroup$
@daljit97 Just so, at least in the way it’s being presented in your text. Historically, this metric arose from studying the way that Maxwell’s equations behaved under coordinate transformations. Mostafa Ayaz’ answer speaks to this background.
$endgroup$
– amd
Oct 12 '18 at 19:29
add a comment |
$begingroup$
Ok so you are you saying that this is more a definition rather than a derivation?
$endgroup$
– daljit97
Oct 12 '18 at 11:38
$begingroup$
@daljit97 Just so, at least in the way it’s being presented in your text. Historically, this metric arose from studying the way that Maxwell’s equations behaved under coordinate transformations. Mostafa Ayaz’ answer speaks to this background.
$endgroup$
– amd
Oct 12 '18 at 19:29
$begingroup$
Ok so you are you saying that this is more a definition rather than a derivation?
$endgroup$
– daljit97
Oct 12 '18 at 11:38
$begingroup$
Ok so you are you saying that this is more a definition rather than a derivation?
$endgroup$
– daljit97
Oct 12 '18 at 11:38
$begingroup$
@daljit97 Just so, at least in the way it’s being presented in your text. Historically, this metric arose from studying the way that Maxwell’s equations behaved under coordinate transformations. Mostafa Ayaz’ answer speaks to this background.
$endgroup$
– amd
Oct 12 '18 at 19:29
$begingroup$
@daljit97 Just so, at least in the way it’s being presented in your text. Historically, this metric arose from studying the way that Maxwell’s equations behaved under coordinate transformations. Mostafa Ayaz’ answer speaks to this background.
$endgroup$
– amd
Oct 12 '18 at 19:29
add a comment |
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the spacelike part does do that $x^2 + y^2 + z^2.$ These are split up as having positive, negative, or zero length; called spacelike, timelike, finally the light cone.
$endgroup$
– Will Jagy
Oct 11 '18 at 19:20
$begingroup$
@amd I have updated the title
$endgroup$
– daljit97
Oct 12 '18 at 11:37