Why is the scalar product of two four-vectors Lorentz-invariant?
Clash Royale CLAN TAG#URR8PPP
up vote
4
down vote
favorite
Why is the scalar product of two four-vectors Lorentz-invariant?
For instance, given two four-vector $A^mu$ and $B^mu$, so their scalar product is $Acdot B=A^mu B_mu=A^mu g_munuB^nu$.
Why is $A^mu g_munuB^nu$ Lorentz-invariant?
special-relativity metric-tensor tensor-calculus lorentz-symmetry invariants
add a comment |
up vote
4
down vote
favorite
Why is the scalar product of two four-vectors Lorentz-invariant?
For instance, given two four-vector $A^mu$ and $B^mu$, so their scalar product is $Acdot B=A^mu B_mu=A^mu g_munuB^nu$.
Why is $A^mu g_munuB^nu$ Lorentz-invariant?
special-relativity metric-tensor tensor-calculus lorentz-symmetry invariants
3
Have a look at eranreches' answer to Scalar invariance under Lorentz-transformation. That should answer your question or at least help you understand it.
– John Rennie
Nov 20 at 11:25
That's the only way the speed of light can be the same in all frames. Any standard reference on special relativity should explain this. I personally like the book "Introduction to Special Relativity" by Rindler.
– Eric David Kramer
Nov 20 at 12:14
A scalar product like $A^mu B_mu$ is invariant under any smooth coordinate transformation. The only thing that's special about Lorentz transformations is that they leave the components of the metric invariant.
– Ben Crowell
Nov 20 at 15:00
Possible duplicate of Lorentz invariance of the Minkowski metric
– AccidentalFourierTransform
Nov 20 at 18:52
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Why is the scalar product of two four-vectors Lorentz-invariant?
For instance, given two four-vector $A^mu$ and $B^mu$, so their scalar product is $Acdot B=A^mu B_mu=A^mu g_munuB^nu$.
Why is $A^mu g_munuB^nu$ Lorentz-invariant?
special-relativity metric-tensor tensor-calculus lorentz-symmetry invariants
Why is the scalar product of two four-vectors Lorentz-invariant?
For instance, given two four-vector $A^mu$ and $B^mu$, so their scalar product is $Acdot B=A^mu B_mu=A^mu g_munuB^nu$.
Why is $A^mu g_munuB^nu$ Lorentz-invariant?
special-relativity metric-tensor tensor-calculus lorentz-symmetry invariants
special-relativity metric-tensor tensor-calculus lorentz-symmetry invariants
edited Nov 20 at 13:20
Emilio Pisanty
81k21193398
81k21193398
asked Nov 20 at 10:26
A.Luo
858
858
3
Have a look at eranreches' answer to Scalar invariance under Lorentz-transformation. That should answer your question or at least help you understand it.
– John Rennie
Nov 20 at 11:25
That's the only way the speed of light can be the same in all frames. Any standard reference on special relativity should explain this. I personally like the book "Introduction to Special Relativity" by Rindler.
– Eric David Kramer
Nov 20 at 12:14
A scalar product like $A^mu B_mu$ is invariant under any smooth coordinate transformation. The only thing that's special about Lorentz transformations is that they leave the components of the metric invariant.
– Ben Crowell
Nov 20 at 15:00
Possible duplicate of Lorentz invariance of the Minkowski metric
– AccidentalFourierTransform
Nov 20 at 18:52
add a comment |
3
Have a look at eranreches' answer to Scalar invariance under Lorentz-transformation. That should answer your question or at least help you understand it.
– John Rennie
Nov 20 at 11:25
That's the only way the speed of light can be the same in all frames. Any standard reference on special relativity should explain this. I personally like the book "Introduction to Special Relativity" by Rindler.
– Eric David Kramer
Nov 20 at 12:14
A scalar product like $A^mu B_mu$ is invariant under any smooth coordinate transformation. The only thing that's special about Lorentz transformations is that they leave the components of the metric invariant.
– Ben Crowell
Nov 20 at 15:00
Possible duplicate of Lorentz invariance of the Minkowski metric
– AccidentalFourierTransform
Nov 20 at 18:52
3
3
Have a look at eranreches' answer to Scalar invariance under Lorentz-transformation. That should answer your question or at least help you understand it.
– John Rennie
Nov 20 at 11:25
Have a look at eranreches' answer to Scalar invariance under Lorentz-transformation. That should answer your question or at least help you understand it.
– John Rennie
Nov 20 at 11:25
That's the only way the speed of light can be the same in all frames. Any standard reference on special relativity should explain this. I personally like the book "Introduction to Special Relativity" by Rindler.
– Eric David Kramer
Nov 20 at 12:14
That's the only way the speed of light can be the same in all frames. Any standard reference on special relativity should explain this. I personally like the book "Introduction to Special Relativity" by Rindler.
– Eric David Kramer
Nov 20 at 12:14
A scalar product like $A^mu B_mu$ is invariant under any smooth coordinate transformation. The only thing that's special about Lorentz transformations is that they leave the components of the metric invariant.
– Ben Crowell
Nov 20 at 15:00
A scalar product like $A^mu B_mu$ is invariant under any smooth coordinate transformation. The only thing that's special about Lorentz transformations is that they leave the components of the metric invariant.
– Ben Crowell
Nov 20 at 15:00
Possible duplicate of Lorentz invariance of the Minkowski metric
– AccidentalFourierTransform
Nov 20 at 18:52
Possible duplicate of Lorentz invariance of the Minkowski metric
– AccidentalFourierTransform
Nov 20 at 18:52
add a comment |
3 Answers
3
active
oldest
votes
up vote
12
down vote
accepted
Frankly, you're looking at this backwards.
Why is $A^mu g_munuB^nu$ Lorentz-invariant?
That's the wrong way around: $A^mu g_munuB^nu$ is Lorentz invariant because Lorentz transformations are defined as the class of transformations that leaves $A^mu g_munuB^nu$ invariant.
Generally, if you transform $A^mu$ and $B^mu$ by some linear transformation with the transformation matrix $Lambda^mu_ nu$, then their transformed values will be $tilde A^mu = Lambda^mu_ nuA^nu$ and $tilde B^mu = Lambda^mu_ nu B^nu$ (using Einstein summations). This means that the transformed inner product will be
beginalign
tilde A^mu g_munu tilde B^nu
& =
(Lambda^mu_ alphaA^alpha) g_munu (Lambda^nu_ betaB^beta)
\ & =
A^alpha (Lambda^mu_ alpha g_munu Lambda^nu_ beta )B^beta
\ & =
A^mu (Lambda^gamma_ mu g_gammadelta Lambda^delta_ nu )B^nu
qquadqquadtext(by re-labelling)
\ & stackreltextrequired=
A^mu g_munu B^nu.
endalign
Thus, for $A^mu g_munu B^nu$ to be invariant we require that
$$
A^mu (Lambda^gamma_ mu g_gammadelta Lambda^delta_ nu )B^nu
=
A^mu g_munu B^nu
$$
for all $A^mu$ and $B^mu$, and by judicious choices of those vectors (basically running each independently over the basis in use) that can only be the case if
$$
Lambda^gamma_ mu g_gammadelta Lambda^delta_ nu
=
g_munu,
$$
which forms the core requirement on $Lambda^mu_ nu$ for it to be a Lorentz transformation.
add a comment |
up vote
2
down vote
Here's the way to think about this -- why is the standard Euclidean dot product, $sum x_iy_i$ interesting? Well, it is interesting primarily from the perspective of rotations, due to the fact that rotations leave dot products invariant. The reason this is so is that this dot product can be written as $|x||y|cosDeltatheta$, and rotations leave magnitudes and relative angles invariant.
Is the standard Euclidean norm $|x|$ invariant under Lorentz transformations? Of course not -- for instance, $Delta t^2+Delta x^2$ is clearly not invariant, but $Delta t^2-Delta x^2$ is. Similarly, $E^2+p^2$ is not important, but $E^2-p^2$ is. The reason this is the case is that Lorentz boosts are fundamentally skew transformations, which means the invariant locus is a hyperbola, not a circle. So you have $cosh^2 xi - sinh^2 xi = 1$, and $x_0^2-x_1^2$ is the right way to think of the norm on Minkowski space.
Similarly, Lorentz boosts change the rapidity $xi$ by a simple displacement, so $Delta xi$ is invariant. From this point, it's a simple exercise to show that
$$|x||y|coshxi=x_0y_0-x_1y_1$$
(as for the remaining dimensions -- remember that the standard Euclidean dot product is still relevant in space, so you just need to write $x_0y_0-xcdot y=x_0y_0-x_1y_1-x_2y_2-x_3y_3$.)
add a comment |
up vote
0
down vote
A vector $mathbfv = v^i , mathbfe_i = q^j ,mathbfu_j$ has different vector components ($v^i$, $q^j$ in this case) in different bases ($mathbfe$,$mathbfu$, in our example) which we can interpret as different reference frames (different axes with different origins).
Physicists are lazy: they refer to the vector components $v^i$ as vectors, which is a misnomer! A true vector $mathbfv$ exists out width whichever basis you choose to work in but to know its entries you must reference these with respect to a given basis: this is just elementary linear algebra.
Now, the magnitude of a vector is independent of whichever bases you choose for its description (that is, geometrically speaking its length is fixed):
$$
v^2 = v^i ; v_i , (mathbfe^i cdot mathbfe_i) = q^j , q_j,(mathbfu_jcdot mathbfu^j)
.
tagassuming orthonormal bases
$$
Hence, scalars do not transform upon of change of basis. In fact it doesn't make much sense to talk about basis for scalars since intuitively these are just numbers.
However, one other way to look at this is to consider a scalar a special type of vector with only one entry and one orthonormal basis (the number 1): its "length" must also be fixed. Hence, this one-dimensional "vector" is the same independent of reference frame.
This is true for all vectors, including special relativistic four-vectors.
As a sanity check, one of the tenets of special relativity is that $c$, the speed of light and a scalar, is the same for all observers. This could not be so if it was somehow different in different frames.
New contributor
@EmilioPisanty Yes, indeed! Fixed.
– OldTomMorris
Nov 20 at 21:59
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
12
down vote
accepted
Frankly, you're looking at this backwards.
Why is $A^mu g_munuB^nu$ Lorentz-invariant?
That's the wrong way around: $A^mu g_munuB^nu$ is Lorentz invariant because Lorentz transformations are defined as the class of transformations that leaves $A^mu g_munuB^nu$ invariant.
Generally, if you transform $A^mu$ and $B^mu$ by some linear transformation with the transformation matrix $Lambda^mu_ nu$, then their transformed values will be $tilde A^mu = Lambda^mu_ nuA^nu$ and $tilde B^mu = Lambda^mu_ nu B^nu$ (using Einstein summations). This means that the transformed inner product will be
beginalign
tilde A^mu g_munu tilde B^nu
& =
(Lambda^mu_ alphaA^alpha) g_munu (Lambda^nu_ betaB^beta)
\ & =
A^alpha (Lambda^mu_ alpha g_munu Lambda^nu_ beta )B^beta
\ & =
A^mu (Lambda^gamma_ mu g_gammadelta Lambda^delta_ nu )B^nu
qquadqquadtext(by re-labelling)
\ & stackreltextrequired=
A^mu g_munu B^nu.
endalign
Thus, for $A^mu g_munu B^nu$ to be invariant we require that
$$
A^mu (Lambda^gamma_ mu g_gammadelta Lambda^delta_ nu )B^nu
=
A^mu g_munu B^nu
$$
for all $A^mu$ and $B^mu$, and by judicious choices of those vectors (basically running each independently over the basis in use) that can only be the case if
$$
Lambda^gamma_ mu g_gammadelta Lambda^delta_ nu
=
g_munu,
$$
which forms the core requirement on $Lambda^mu_ nu$ for it to be a Lorentz transformation.
add a comment |
up vote
12
down vote
accepted
Frankly, you're looking at this backwards.
Why is $A^mu g_munuB^nu$ Lorentz-invariant?
That's the wrong way around: $A^mu g_munuB^nu$ is Lorentz invariant because Lorentz transformations are defined as the class of transformations that leaves $A^mu g_munuB^nu$ invariant.
Generally, if you transform $A^mu$ and $B^mu$ by some linear transformation with the transformation matrix $Lambda^mu_ nu$, then their transformed values will be $tilde A^mu = Lambda^mu_ nuA^nu$ and $tilde B^mu = Lambda^mu_ nu B^nu$ (using Einstein summations). This means that the transformed inner product will be
beginalign
tilde A^mu g_munu tilde B^nu
& =
(Lambda^mu_ alphaA^alpha) g_munu (Lambda^nu_ betaB^beta)
\ & =
A^alpha (Lambda^mu_ alpha g_munu Lambda^nu_ beta )B^beta
\ & =
A^mu (Lambda^gamma_ mu g_gammadelta Lambda^delta_ nu )B^nu
qquadqquadtext(by re-labelling)
\ & stackreltextrequired=
A^mu g_munu B^nu.
endalign
Thus, for $A^mu g_munu B^nu$ to be invariant we require that
$$
A^mu (Lambda^gamma_ mu g_gammadelta Lambda^delta_ nu )B^nu
=
A^mu g_munu B^nu
$$
for all $A^mu$ and $B^mu$, and by judicious choices of those vectors (basically running each independently over the basis in use) that can only be the case if
$$
Lambda^gamma_ mu g_gammadelta Lambda^delta_ nu
=
g_munu,
$$
which forms the core requirement on $Lambda^mu_ nu$ for it to be a Lorentz transformation.
add a comment |
up vote
12
down vote
accepted
up vote
12
down vote
accepted
Frankly, you're looking at this backwards.
Why is $A^mu g_munuB^nu$ Lorentz-invariant?
That's the wrong way around: $A^mu g_munuB^nu$ is Lorentz invariant because Lorentz transformations are defined as the class of transformations that leaves $A^mu g_munuB^nu$ invariant.
Generally, if you transform $A^mu$ and $B^mu$ by some linear transformation with the transformation matrix $Lambda^mu_ nu$, then their transformed values will be $tilde A^mu = Lambda^mu_ nuA^nu$ and $tilde B^mu = Lambda^mu_ nu B^nu$ (using Einstein summations). This means that the transformed inner product will be
beginalign
tilde A^mu g_munu tilde B^nu
& =
(Lambda^mu_ alphaA^alpha) g_munu (Lambda^nu_ betaB^beta)
\ & =
A^alpha (Lambda^mu_ alpha g_munu Lambda^nu_ beta )B^beta
\ & =
A^mu (Lambda^gamma_ mu g_gammadelta Lambda^delta_ nu )B^nu
qquadqquadtext(by re-labelling)
\ & stackreltextrequired=
A^mu g_munu B^nu.
endalign
Thus, for $A^mu g_munu B^nu$ to be invariant we require that
$$
A^mu (Lambda^gamma_ mu g_gammadelta Lambda^delta_ nu )B^nu
=
A^mu g_munu B^nu
$$
for all $A^mu$ and $B^mu$, and by judicious choices of those vectors (basically running each independently over the basis in use) that can only be the case if
$$
Lambda^gamma_ mu g_gammadelta Lambda^delta_ nu
=
g_munu,
$$
which forms the core requirement on $Lambda^mu_ nu$ for it to be a Lorentz transformation.
Frankly, you're looking at this backwards.
Why is $A^mu g_munuB^nu$ Lorentz-invariant?
That's the wrong way around: $A^mu g_munuB^nu$ is Lorentz invariant because Lorentz transformations are defined as the class of transformations that leaves $A^mu g_munuB^nu$ invariant.
Generally, if you transform $A^mu$ and $B^mu$ by some linear transformation with the transformation matrix $Lambda^mu_ nu$, then their transformed values will be $tilde A^mu = Lambda^mu_ nuA^nu$ and $tilde B^mu = Lambda^mu_ nu B^nu$ (using Einstein summations). This means that the transformed inner product will be
beginalign
tilde A^mu g_munu tilde B^nu
& =
(Lambda^mu_ alphaA^alpha) g_munu (Lambda^nu_ betaB^beta)
\ & =
A^alpha (Lambda^mu_ alpha g_munu Lambda^nu_ beta )B^beta
\ & =
A^mu (Lambda^gamma_ mu g_gammadelta Lambda^delta_ nu )B^nu
qquadqquadtext(by re-labelling)
\ & stackreltextrequired=
A^mu g_munu B^nu.
endalign
Thus, for $A^mu g_munu B^nu$ to be invariant we require that
$$
A^mu (Lambda^gamma_ mu g_gammadelta Lambda^delta_ nu )B^nu
=
A^mu g_munu B^nu
$$
for all $A^mu$ and $B^mu$, and by judicious choices of those vectors (basically running each independently over the basis in use) that can only be the case if
$$
Lambda^gamma_ mu g_gammadelta Lambda^delta_ nu
=
g_munu,
$$
which forms the core requirement on $Lambda^mu_ nu$ for it to be a Lorentz transformation.
answered Nov 20 at 13:33
Emilio Pisanty
81k21193398
81k21193398
add a comment |
add a comment |
up vote
2
down vote
Here's the way to think about this -- why is the standard Euclidean dot product, $sum x_iy_i$ interesting? Well, it is interesting primarily from the perspective of rotations, due to the fact that rotations leave dot products invariant. The reason this is so is that this dot product can be written as $|x||y|cosDeltatheta$, and rotations leave magnitudes and relative angles invariant.
Is the standard Euclidean norm $|x|$ invariant under Lorentz transformations? Of course not -- for instance, $Delta t^2+Delta x^2$ is clearly not invariant, but $Delta t^2-Delta x^2$ is. Similarly, $E^2+p^2$ is not important, but $E^2-p^2$ is. The reason this is the case is that Lorentz boosts are fundamentally skew transformations, which means the invariant locus is a hyperbola, not a circle. So you have $cosh^2 xi - sinh^2 xi = 1$, and $x_0^2-x_1^2$ is the right way to think of the norm on Minkowski space.
Similarly, Lorentz boosts change the rapidity $xi$ by a simple displacement, so $Delta xi$ is invariant. From this point, it's a simple exercise to show that
$$|x||y|coshxi=x_0y_0-x_1y_1$$
(as for the remaining dimensions -- remember that the standard Euclidean dot product is still relevant in space, so you just need to write $x_0y_0-xcdot y=x_0y_0-x_1y_1-x_2y_2-x_3y_3$.)
add a comment |
up vote
2
down vote
Here's the way to think about this -- why is the standard Euclidean dot product, $sum x_iy_i$ interesting? Well, it is interesting primarily from the perspective of rotations, due to the fact that rotations leave dot products invariant. The reason this is so is that this dot product can be written as $|x||y|cosDeltatheta$, and rotations leave magnitudes and relative angles invariant.
Is the standard Euclidean norm $|x|$ invariant under Lorentz transformations? Of course not -- for instance, $Delta t^2+Delta x^2$ is clearly not invariant, but $Delta t^2-Delta x^2$ is. Similarly, $E^2+p^2$ is not important, but $E^2-p^2$ is. The reason this is the case is that Lorentz boosts are fundamentally skew transformations, which means the invariant locus is a hyperbola, not a circle. So you have $cosh^2 xi - sinh^2 xi = 1$, and $x_0^2-x_1^2$ is the right way to think of the norm on Minkowski space.
Similarly, Lorentz boosts change the rapidity $xi$ by a simple displacement, so $Delta xi$ is invariant. From this point, it's a simple exercise to show that
$$|x||y|coshxi=x_0y_0-x_1y_1$$
(as for the remaining dimensions -- remember that the standard Euclidean dot product is still relevant in space, so you just need to write $x_0y_0-xcdot y=x_0y_0-x_1y_1-x_2y_2-x_3y_3$.)
add a comment |
up vote
2
down vote
up vote
2
down vote
Here's the way to think about this -- why is the standard Euclidean dot product, $sum x_iy_i$ interesting? Well, it is interesting primarily from the perspective of rotations, due to the fact that rotations leave dot products invariant. The reason this is so is that this dot product can be written as $|x||y|cosDeltatheta$, and rotations leave magnitudes and relative angles invariant.
Is the standard Euclidean norm $|x|$ invariant under Lorentz transformations? Of course not -- for instance, $Delta t^2+Delta x^2$ is clearly not invariant, but $Delta t^2-Delta x^2$ is. Similarly, $E^2+p^2$ is not important, but $E^2-p^2$ is. The reason this is the case is that Lorentz boosts are fundamentally skew transformations, which means the invariant locus is a hyperbola, not a circle. So you have $cosh^2 xi - sinh^2 xi = 1$, and $x_0^2-x_1^2$ is the right way to think of the norm on Minkowski space.
Similarly, Lorentz boosts change the rapidity $xi$ by a simple displacement, so $Delta xi$ is invariant. From this point, it's a simple exercise to show that
$$|x||y|coshxi=x_0y_0-x_1y_1$$
(as for the remaining dimensions -- remember that the standard Euclidean dot product is still relevant in space, so you just need to write $x_0y_0-xcdot y=x_0y_0-x_1y_1-x_2y_2-x_3y_3$.)
Here's the way to think about this -- why is the standard Euclidean dot product, $sum x_iy_i$ interesting? Well, it is interesting primarily from the perspective of rotations, due to the fact that rotations leave dot products invariant. The reason this is so is that this dot product can be written as $|x||y|cosDeltatheta$, and rotations leave magnitudes and relative angles invariant.
Is the standard Euclidean norm $|x|$ invariant under Lorentz transformations? Of course not -- for instance, $Delta t^2+Delta x^2$ is clearly not invariant, but $Delta t^2-Delta x^2$ is. Similarly, $E^2+p^2$ is not important, but $E^2-p^2$ is. The reason this is the case is that Lorentz boosts are fundamentally skew transformations, which means the invariant locus is a hyperbola, not a circle. So you have $cosh^2 xi - sinh^2 xi = 1$, and $x_0^2-x_1^2$ is the right way to think of the norm on Minkowski space.
Similarly, Lorentz boosts change the rapidity $xi$ by a simple displacement, so $Delta xi$ is invariant. From this point, it's a simple exercise to show that
$$|x||y|coshxi=x_0y_0-x_1y_1$$
(as for the remaining dimensions -- remember that the standard Euclidean dot product is still relevant in space, so you just need to write $x_0y_0-xcdot y=x_0y_0-x_1y_1-x_2y_2-x_3y_3$.)
answered Nov 20 at 15:59
Abhimanyu Pallavi Sudhir
4,41642343
4,41642343
add a comment |
add a comment |
up vote
0
down vote
A vector $mathbfv = v^i , mathbfe_i = q^j ,mathbfu_j$ has different vector components ($v^i$, $q^j$ in this case) in different bases ($mathbfe$,$mathbfu$, in our example) which we can interpret as different reference frames (different axes with different origins).
Physicists are lazy: they refer to the vector components $v^i$ as vectors, which is a misnomer! A true vector $mathbfv$ exists out width whichever basis you choose to work in but to know its entries you must reference these with respect to a given basis: this is just elementary linear algebra.
Now, the magnitude of a vector is independent of whichever bases you choose for its description (that is, geometrically speaking its length is fixed):
$$
v^2 = v^i ; v_i , (mathbfe^i cdot mathbfe_i) = q^j , q_j,(mathbfu_jcdot mathbfu^j)
.
tagassuming orthonormal bases
$$
Hence, scalars do not transform upon of change of basis. In fact it doesn't make much sense to talk about basis for scalars since intuitively these are just numbers.
However, one other way to look at this is to consider a scalar a special type of vector with only one entry and one orthonormal basis (the number 1): its "length" must also be fixed. Hence, this one-dimensional "vector" is the same independent of reference frame.
This is true for all vectors, including special relativistic four-vectors.
As a sanity check, one of the tenets of special relativity is that $c$, the speed of light and a scalar, is the same for all observers. This could not be so if it was somehow different in different frames.
New contributor
@EmilioPisanty Yes, indeed! Fixed.
– OldTomMorris
Nov 20 at 21:59
add a comment |
up vote
0
down vote
A vector $mathbfv = v^i , mathbfe_i = q^j ,mathbfu_j$ has different vector components ($v^i$, $q^j$ in this case) in different bases ($mathbfe$,$mathbfu$, in our example) which we can interpret as different reference frames (different axes with different origins).
Physicists are lazy: they refer to the vector components $v^i$ as vectors, which is a misnomer! A true vector $mathbfv$ exists out width whichever basis you choose to work in but to know its entries you must reference these with respect to a given basis: this is just elementary linear algebra.
Now, the magnitude of a vector is independent of whichever bases you choose for its description (that is, geometrically speaking its length is fixed):
$$
v^2 = v^i ; v_i , (mathbfe^i cdot mathbfe_i) = q^j , q_j,(mathbfu_jcdot mathbfu^j)
.
tagassuming orthonormal bases
$$
Hence, scalars do not transform upon of change of basis. In fact it doesn't make much sense to talk about basis for scalars since intuitively these are just numbers.
However, one other way to look at this is to consider a scalar a special type of vector with only one entry and one orthonormal basis (the number 1): its "length" must also be fixed. Hence, this one-dimensional "vector" is the same independent of reference frame.
This is true for all vectors, including special relativistic four-vectors.
As a sanity check, one of the tenets of special relativity is that $c$, the speed of light and a scalar, is the same for all observers. This could not be so if it was somehow different in different frames.
New contributor
@EmilioPisanty Yes, indeed! Fixed.
– OldTomMorris
Nov 20 at 21:59
add a comment |
up vote
0
down vote
up vote
0
down vote
A vector $mathbfv = v^i , mathbfe_i = q^j ,mathbfu_j$ has different vector components ($v^i$, $q^j$ in this case) in different bases ($mathbfe$,$mathbfu$, in our example) which we can interpret as different reference frames (different axes with different origins).
Physicists are lazy: they refer to the vector components $v^i$ as vectors, which is a misnomer! A true vector $mathbfv$ exists out width whichever basis you choose to work in but to know its entries you must reference these with respect to a given basis: this is just elementary linear algebra.
Now, the magnitude of a vector is independent of whichever bases you choose for its description (that is, geometrically speaking its length is fixed):
$$
v^2 = v^i ; v_i , (mathbfe^i cdot mathbfe_i) = q^j , q_j,(mathbfu_jcdot mathbfu^j)
.
tagassuming orthonormal bases
$$
Hence, scalars do not transform upon of change of basis. In fact it doesn't make much sense to talk about basis for scalars since intuitively these are just numbers.
However, one other way to look at this is to consider a scalar a special type of vector with only one entry and one orthonormal basis (the number 1): its "length" must also be fixed. Hence, this one-dimensional "vector" is the same independent of reference frame.
This is true for all vectors, including special relativistic four-vectors.
As a sanity check, one of the tenets of special relativity is that $c$, the speed of light and a scalar, is the same for all observers. This could not be so if it was somehow different in different frames.
New contributor
A vector $mathbfv = v^i , mathbfe_i = q^j ,mathbfu_j$ has different vector components ($v^i$, $q^j$ in this case) in different bases ($mathbfe$,$mathbfu$, in our example) which we can interpret as different reference frames (different axes with different origins).
Physicists are lazy: they refer to the vector components $v^i$ as vectors, which is a misnomer! A true vector $mathbfv$ exists out width whichever basis you choose to work in but to know its entries you must reference these with respect to a given basis: this is just elementary linear algebra.
Now, the magnitude of a vector is independent of whichever bases you choose for its description (that is, geometrically speaking its length is fixed):
$$
v^2 = v^i ; v_i , (mathbfe^i cdot mathbfe_i) = q^j , q_j,(mathbfu_jcdot mathbfu^j)
.
tagassuming orthonormal bases
$$
Hence, scalars do not transform upon of change of basis. In fact it doesn't make much sense to talk about basis for scalars since intuitively these are just numbers.
However, one other way to look at this is to consider a scalar a special type of vector with only one entry and one orthonormal basis (the number 1): its "length" must also be fixed. Hence, this one-dimensional "vector" is the same independent of reference frame.
This is true for all vectors, including special relativistic four-vectors.
As a sanity check, one of the tenets of special relativity is that $c$, the speed of light and a scalar, is the same for all observers. This could not be so if it was somehow different in different frames.
New contributor
edited Nov 20 at 21:58
New contributor
answered Nov 20 at 15:33
OldTomMorris
11
11
New contributor
New contributor
@EmilioPisanty Yes, indeed! Fixed.
– OldTomMorris
Nov 20 at 21:59
add a comment |
@EmilioPisanty Yes, indeed! Fixed.
– OldTomMorris
Nov 20 at 21:59
@EmilioPisanty Yes, indeed! Fixed.
– OldTomMorris
Nov 20 at 21:59
@EmilioPisanty Yes, indeed! Fixed.
– OldTomMorris
Nov 20 at 21:59
add a comment |
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f442119%2fwhy-is-the-scalar-product-of-two-four-vectors-lorentz-invariant%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
3
Have a look at eranreches' answer to Scalar invariance under Lorentz-transformation. That should answer your question or at least help you understand it.
– John Rennie
Nov 20 at 11:25
That's the only way the speed of light can be the same in all frames. Any standard reference on special relativity should explain this. I personally like the book "Introduction to Special Relativity" by Rindler.
– Eric David Kramer
Nov 20 at 12:14
A scalar product like $A^mu B_mu$ is invariant under any smooth coordinate transformation. The only thing that's special about Lorentz transformations is that they leave the components of the metric invariant.
– Ben Crowell
Nov 20 at 15:00
Possible duplicate of Lorentz invariance of the Minkowski metric
– AccidentalFourierTransform
Nov 20 at 18:52