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How can velocity be a tensor?
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2
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I have just begun studying general relativity and have a question.
I know that if a tensor is zero in one coordinate system, it will be zero in all coordinate systems.
So how can the $mu$'th component of the velocity $dx^mu/dtau$ be a tensor?
There must be something I have misunderstood.
general-relativity special-relativity tensor-calculus
edited 3 hours ago
ZeroTheHero
17.3k52556
asked 4 hours ago
Christine
111
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Christine is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
2
down vote
favorite
I have just begun studying general relativity and have a question.
I know that if a tensor is zero in one coordinate system, it will be zero in all coordinate systems.
So how can the $mu$'th component of the velocity $dx^mu/dtau$ be a tensor?
There must be something I have misunderstood.
general-relativity special-relativity tensor-calculus
edited 3 hours ago
ZeroTheHero
17.3k52556
asked 4 hours ago
Christine
111
New contributor
Christine is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
The four velocity can't be zero in any coordinate system: In the rest system it is $(1, 0, 0, 0)$.
– Sebastian Riese
3 hours ago
5
Can you clarify what you are asking? You seem to be asking if the individual components of the four velocity are tensors, and of course they are not. But the four velocity as a whole is a tensor.
– John Rennie
3 hours ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I have just begun studying general relativity and have a question.
I know that if a tensor is zero in one coordinate system, it will be zero in all coordinate systems.
So how can the $mu$'th component of the velocity $dx^mu/dtau$ be a tensor?
There must be something I have misunderstood.
general-relativity special-relativity tensor-calculus
edited 3 hours ago
ZeroTheHero
17.3k52556
asked 4 hours ago
Christine
111
New contributor
Christine is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I have just begun studying general relativity and have a question.
I know that if a tensor is zero in one coordinate system, it will be zero in all coordinate systems.
So how can the $mu$'th component of the velocity $dx^mu/dtau$ be a tensor?
There must be something I have misunderstood.
general-relativity special-relativity tensor-calculus
general-relativity special-relativity tensor-calculus
edited 3 hours ago
ZeroTheHero
17.3k52556
asked 4 hours ago
Christine
111
New contributor
Christine is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 3 hours ago
ZeroTheHero
17.3k52556
asked 4 hours ago
Christine
111
New contributor
Christine is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 3 hours ago
ZeroTheHero
17.3k52556
edited 3 hours ago
ZeroTheHero
17.3k52556
edited 3 hours ago
ZeroTheHero
17.3k52556
17.3k52556
asked 4 hours ago
Christine
111
New contributor
Christine is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
Christine
111
asked 4 hours ago
Christine
111
111
New contributor
Christine is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Christine is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Christine is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
The four velocity can't be zero in any coordinate system: In the rest system it is $(1, 0, 0, 0)$.
– Sebastian Riese
3 hours ago
5
Can you clarify what you are asking? You seem to be asking if the individual components of the four velocity are tensors, and of course they are not. But the four velocity as a whole is a tensor.
– John Rennie
3 hours ago
add a comment |Â
2
The four velocity can't be zero in any coordinate system: In the rest system it is $(1, 0, 0, 0)$.
– Sebastian Riese
3 hours ago
5
Can you clarify what you are asking? You seem to be asking if the individual components of the four velocity are tensors, and of course they are not. But the four velocity as a whole is a tensor.
– John Rennie
3 hours ago
2
2
The four velocity can't be zero in any coordinate system: In the rest system it is $(1, 0, 0, 0)$.
– Sebastian Riese
3 hours ago
The four velocity can't be zero in any coordinate system: In the rest system it is $(1, 0, 0, 0)$.
– Sebastian Riese
3 hours ago
5
5
Can you clarify what you are asking? You seem to be asking if the individual components of the four velocity are tensors, and of course they are not. But the four velocity as a whole is a tensor.
– John Rennie
3 hours ago
Can you clarify what you are asking? You seem to be asking if the individual components of the four velocity are tensors, and of course they are not. But the four velocity as a whole is a tensor.
– John Rennie
3 hours ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
5
down vote
The three-velocity isn't a tensor - it's the four-velocity that is: this is a four-vector whose spatial components are the three-velocity of the particle (with times measured in the proper time of the particle), but which also has a zeroth (temporal) component
$$
fracmathrm d x^0mathrm dtau = fracmathrm d tmathrm dtau,
$$
which basically measures the rate of time dilation between your chosen frame of reference and the rest frame of the particle. No Lorentz transformation can transform a four-vector with a nonzero temporal component into one with a vanishing temporal component (and, moreover, no orthochronous Lorentz transformation can change the sign of the temporal component of a four-vector, so generally $fracmathrm d x^0mathrm dtau>0$).
This means that the four-velocity of any particle will always be nonzero: you can always set the spatial components to zero by transforming to the rest-frame of the particle, but if you do that then the temporal component will be
$$
fracmathrm d x^0mathrm dtau = fracmathrm d tmathrm dtau = fracmathrm d tau mathrm dtau = 1,
$$
and the four-vector will not vanish.
answered 3 hours ago
Emilio Pisanty
76k18181373
1
Based on OP's statement "how can the $mu$'th component of the velocity $dx^mu/dtau$ be a tensor?", I believe he may also have some misconceptions about the notation of a tensor vs components of a tensor and whether individual components of a velocity is a tensor or not. Perhaps you could include a little bit of explanation to clarify this point?
– enumaris
3 hours ago
@enumaris I'd rather get confirmation from Christine that that is indeed what the question was about. You're obviously welcome to add a supplementary / alternative / overencompassing answer.
– Emilio Pisanty
2 hours ago
Thank you for your response. I believe the problem is trouble understanding the notation.
– Christine
1 hour ago
Fair enough, but "I don't understand the notation" is very little to go on when explaining it. What is it about the notation that you don't understand?
– Emilio Pisanty
1 hour ago
1
@Christine you should explain in more detail why you think the four-velocity shouldn't be a tensor.
– Javier
12 mins ago
 |Â
show 1 more comment
up vote
0
down vote
velocity should not be a tensor
Why not? 4-velocity is a tensor: a $1 choose 0$ tensor. It is the tangent vector to the worldline of a material particle, parametrized by proper time.
Its components obey an identity:
$$g_munu u^mu u^nu = 1$$
($g_munu$ the metric tensor).
(I beg your pardon: there are two sign conventions here. I used the one I like better, but am not sure it is the one you are accustomed to.)
Also note that in GR you are widely free in the choice of coordinates (subject to some constraints I shall not dwell upon). In general you should not expect that coordinates themselves behave as components of a 4-vector. Still stranger as it may appear, you are not allowed (in general) to assign to one of the coordinates the character of a time and to the other three the one of space coordinates.
In other words, a coordinate line (e.g. $x^1=rm const.$, $x^2=rm const.$, $x^3=rm const.$) is not bound to have a timelike tangent vector and the other three spacelike. All that is required is that the four vectors are independent.
Things are easier with orthogonal coordinates: then the metric tensor has $ne0$ only the diagonal components, and in this case it is true that one of the coordinate lines is timelike and the remaining three spacelike. In many important cases such a choice of coordinates is possible, but not always. A counterexample is Kerr (rotating, uncharged) black hole.
answered 18 mins ago
Elio Fabri
862
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
The three-velocity isn't a tensor - it's the four-velocity that is: this is a four-vector whose spatial components are the three-velocity of the particle (with times measured in the proper time of the particle), but which also has a zeroth (temporal) component
$$
fracmathrm d x^0mathrm dtau = fracmathrm d tmathrm dtau,
$$
which basically measures the rate of time dilation between your chosen frame of reference and the rest frame of the particle. No Lorentz transformation can transform a four-vector with a nonzero temporal component into one with a vanishing temporal component (and, moreover, no orthochronous Lorentz transformation can change the sign of the temporal component of a four-vector, so generally $fracmathrm d x^0mathrm dtau>0$).
This means that the four-velocity of any particle will always be nonzero: you can always set the spatial components to zero by transforming to the rest-frame of the particle, but if you do that then the temporal component will be
$$
fracmathrm d x^0mathrm dtau = fracmathrm d tmathrm dtau = fracmathrm d tau mathrm dtau = 1,
$$
and the four-vector will not vanish.
answered 3 hours ago
Emilio Pisanty
76k18181373
1
Based on OP's statement "how can the $mu$'th component of the velocity $dx^mu/dtau$ be a tensor?", I believe he may also have some misconceptions about the notation of a tensor vs components of a tensor and whether individual components of a velocity is a tensor or not. Perhaps you could include a little bit of explanation to clarify this point?
– enumaris
3 hours ago
@enumaris I'd rather get confirmation from Christine that that is indeed what the question was about. You're obviously welcome to add a supplementary / alternative / overencompassing answer.
– Emilio Pisanty
2 hours ago
Thank you for your response. I believe the problem is trouble understanding the notation.
– Christine
1 hour ago
Fair enough, but "I don't understand the notation" is very little to go on when explaining it. What is it about the notation that you don't understand?
– Emilio Pisanty
1 hour ago
1
@Christine you should explain in more detail why you think the four-velocity shouldn't be a tensor.
– Javier
12 mins ago
 |Â
show 1 more comment
up vote
5
down vote
The three-velocity isn't a tensor - it's the four-velocity that is: this is a four-vector whose spatial components are the three-velocity of the particle (with times measured in the proper time of the particle), but which also has a zeroth (temporal) component
$$
fracmathrm d x^0mathrm dtau = fracmathrm d tmathrm dtau,
$$
which basically measures the rate of time dilation between your chosen frame of reference and the rest frame of the particle. No Lorentz transformation can transform a four-vector with a nonzero temporal component into one with a vanishing temporal component (and, moreover, no orthochronous Lorentz transformation can change the sign of the temporal component of a four-vector, so generally $fracmathrm d x^0mathrm dtau>0$).
This means that the four-velocity of any particle will always be nonzero: you can always set the spatial components to zero by transforming to the rest-frame of the particle, but if you do that then the temporal component will be
$$
fracmathrm d x^0mathrm dtau = fracmathrm d tmathrm dtau = fracmathrm d tau mathrm dtau = 1,
$$
and the four-vector will not vanish.
answered 3 hours ago
Emilio Pisanty
76k18181373
1
Based on OP's statement "how can the $mu$'th component of the velocity $dx^mu/dtau$ be a tensor?", I believe he may also have some misconceptions about the notation of a tensor vs components of a tensor and whether individual components of a velocity is a tensor or not. Perhaps you could include a little bit of explanation to clarify this point?
– enumaris
3 hours ago
@enumaris I'd rather get confirmation from Christine that that is indeed what the question was about. You're obviously welcome to add a supplementary / alternative / overencompassing answer.
– Emilio Pisanty
2 hours ago
Thank you for your response. I believe the problem is trouble understanding the notation.
– Christine
1 hour ago
Fair enough, but "I don't understand the notation" is very little to go on when explaining it. What is it about the notation that you don't understand?
– Emilio Pisanty
1 hour ago
1
@Christine you should explain in more detail why you think the four-velocity shouldn't be a tensor.
– Javier
12 mins ago
 |Â
show 1 more comment
up vote
5
down vote
up vote
5
down vote
The three-velocity isn't a tensor - it's the four-velocity that is: this is a four-vector whose spatial components are the three-velocity of the particle (with times measured in the proper time of the particle), but which also has a zeroth (temporal) component
$$
fracmathrm d x^0mathrm dtau = fracmathrm d tmathrm dtau,
$$
which basically measures the rate of time dilation between your chosen frame of reference and the rest frame of the particle. No Lorentz transformation can transform a four-vector with a nonzero temporal component into one with a vanishing temporal component (and, moreover, no orthochronous Lorentz transformation can change the sign of the temporal component of a four-vector, so generally $fracmathrm d x^0mathrm dtau>0$).
This means that the four-velocity of any particle will always be nonzero: you can always set the spatial components to zero by transforming to the rest-frame of the particle, but if you do that then the temporal component will be
$$
fracmathrm d x^0mathrm dtau = fracmathrm d tmathrm dtau = fracmathrm d tau mathrm dtau = 1,
$$
and the four-vector will not vanish.
answered 3 hours ago
Emilio Pisanty
76k18181373
The three-velocity isn't a tensor - it's the four-velocity that is: this is a four-vector whose spatial components are the three-velocity of the particle (with times measured in the proper time of the particle), but which also has a zeroth (temporal) component
$$
fracmathrm d x^0mathrm dtau = fracmathrm d tmathrm dtau,
$$
which basically measures the rate of time dilation between your chosen frame of reference and the rest frame of the particle. No Lorentz transformation can transform a four-vector with a nonzero temporal component into one with a vanishing temporal component (and, moreover, no orthochronous Lorentz transformation can change the sign of the temporal component of a four-vector, so generally $fracmathrm d x^0mathrm dtau>0$).
This means that the four-velocity of any particle will always be nonzero: you can always set the spatial components to zero by transforming to the rest-frame of the particle, but if you do that then the temporal component will be
$$
fracmathrm d x^0mathrm dtau = fracmathrm d tmathrm dtau = fracmathrm d tau mathrm dtau = 1,
$$
and the four-vector will not vanish.
answered 3 hours ago
Emilio Pisanty
76k18181373
answered 3 hours ago
Emilio Pisanty
76k18181373
answered 3 hours ago
Emilio Pisanty
76k18181373
answered 3 hours ago
Emilio Pisanty
76k18181373
76k18181373
1
Based on OP's statement "how can the $mu$'th component of the velocity $dx^mu/dtau$ be a tensor?", I believe he may also have some misconceptions about the notation of a tensor vs components of a tensor and whether individual components of a velocity is a tensor or not. Perhaps you could include a little bit of explanation to clarify this point?
– enumaris
3 hours ago
@enumaris I'd rather get confirmation from Christine that that is indeed what the question was about. You're obviously welcome to add a supplementary / alternative / overencompassing answer.
– Emilio Pisanty
2 hours ago
Thank you for your response. I believe the problem is trouble understanding the notation.
– Christine
1 hour ago
Fair enough, but "I don't understand the notation" is very little to go on when explaining it. What is it about the notation that you don't understand?
– Emilio Pisanty
1 hour ago
1
@Christine you should explain in more detail why you think the four-velocity shouldn't be a tensor.
– Javier
12 mins ago
 |Â
show 1 more comment
1
Based on OP's statement "how can the $mu$'th component of the velocity $dx^mu/dtau$ be a tensor?", I believe he may also have some misconceptions about the notation of a tensor vs components of a tensor and whether individual components of a velocity is a tensor or not. Perhaps you could include a little bit of explanation to clarify this point?
– enumaris
3 hours ago
@enumaris I'd rather get confirmation from Christine that that is indeed what the question was about. You're obviously welcome to add a supplementary / alternative / overencompassing answer.
– Emilio Pisanty
2 hours ago
Thank you for your response. I believe the problem is trouble understanding the notation.
– Christine
1 hour ago
Fair enough, but "I don't understand the notation" is very little to go on when explaining it. What is it about the notation that you don't understand?
– Emilio Pisanty
1 hour ago
1
@Christine you should explain in more detail why you think the four-velocity shouldn't be a tensor.
– Javier
12 mins ago
1
1
Based on OP's statement "how can the $mu$'th component of the velocity $dx^mu/dtau$ be a tensor?", I believe he may also have some misconceptions about the notation of a tensor vs components of a tensor and whether individual components of a velocity is a tensor or not. Perhaps you could include a little bit of explanation to clarify this point?
– enumaris
3 hours ago
Based on OP's statement "how can the $mu$'th component of the velocity $dx^mu/dtau$ be a tensor?", I believe he may also have some misconceptions about the notation of a tensor vs components of a tensor and whether individual components of a velocity is a tensor or not. Perhaps you could include a little bit of explanation to clarify this point?
– enumaris
3 hours ago
@enumaris I'd rather get confirmation from Christine that that is indeed what the question was about. You're obviously welcome to add a supplementary / alternative / overencompassing answer.
– Emilio Pisanty
2 hours ago
@enumaris I'd rather get confirmation from Christine that that is indeed what the question was about. You're obviously welcome to add a supplementary / alternative / overencompassing answer.
– Emilio Pisanty
2 hours ago
Thank you for your response. I believe the problem is trouble understanding the notation.
– Christine
1 hour ago
Thank you for your response. I believe the problem is trouble understanding the notation.
– Christine
1 hour ago
Fair enough, but "I don't understand the notation" is very little to go on when explaining it. What is it about the notation that you don't understand?
– Emilio Pisanty
1 hour ago
Fair enough, but "I don't understand the notation" is very little to go on when explaining it. What is it about the notation that you don't understand?
– Emilio Pisanty
1 hour ago
1
1
@Christine you should explain in more detail why you think the four-velocity shouldn't be a tensor.
– Javier
12 mins ago
@Christine you should explain in more detail why you think the four-velocity shouldn't be a tensor.
– Javier
12 mins ago
 |Â
show 1 more comment
up vote
0
down vote
velocity should not be a tensor
Why not? 4-velocity is a tensor: a $1 choose 0$ tensor. It is the tangent vector to the worldline of a material particle, parametrized by proper time.
Its components obey an identity:
$$g_munu u^mu u^nu = 1$$
($g_munu$ the metric tensor).
(I beg your pardon: there are two sign conventions here. I used the one I like better, but am not sure it is the one you are accustomed to.)
Also note that in GR you are widely free in the choice of coordinates (subject to some constraints I shall not dwell upon). In general you should not expect that coordinates themselves behave as components of a 4-vector. Still stranger as it may appear, you are not allowed (in general) to assign to one of the coordinates the character of a time and to the other three the one of space coordinates.
In other words, a coordinate line (e.g. $x^1=rm const.$, $x^2=rm const.$, $x^3=rm const.$) is not bound to have a timelike tangent vector and the other three spacelike. All that is required is that the four vectors are independent.
Things are easier with orthogonal coordinates: then the metric tensor has $ne0$ only the diagonal components, and in this case it is true that one of the coordinate lines is timelike and the remaining three spacelike. In many important cases such a choice of coordinates is possible, but not always. A counterexample is Kerr (rotating, uncharged) black hole.
answered 18 mins ago
Elio Fabri
862
add a comment |Â
up vote
0
down vote
velocity should not be a tensor
Why not? 4-velocity is a tensor: a $1 choose 0$ tensor. It is the tangent vector to the worldline of a material particle, parametrized by proper time.
Its components obey an identity:
$$g_munu u^mu u^nu = 1$$
($g_munu$ the metric tensor).
(I beg your pardon: there are two sign conventions here. I used the one I like better, but am not sure it is the one you are accustomed to.)
Also note that in GR you are widely free in the choice of coordinates (subject to some constraints I shall not dwell upon). In general you should not expect that coordinates themselves behave as components of a 4-vector. Still stranger as it may appear, you are not allowed (in general) to assign to one of the coordinates the character of a time and to the other three the one of space coordinates.
In other words, a coordinate line (e.g. $x^1=rm const.$, $x^2=rm const.$, $x^3=rm const.$) is not bound to have a timelike tangent vector and the other three spacelike. All that is required is that the four vectors are independent.
Things are easier with orthogonal coordinates: then the metric tensor has $ne0$ only the diagonal components, and in this case it is true that one of the coordinate lines is timelike and the remaining three spacelike. In many important cases such a choice of coordinates is possible, but not always. A counterexample is Kerr (rotating, uncharged) black hole.
answered 18 mins ago
Elio Fabri
862
add a comment |Â
up vote
0
down vote
up vote
0
down vote
velocity should not be a tensor
Why not? 4-velocity is a tensor: a $1 choose 0$ tensor. It is the tangent vector to the worldline of a material particle, parametrized by proper time.
Its components obey an identity:
$$g_munu u^mu u^nu = 1$$
($g_munu$ the metric tensor).
(I beg your pardon: there are two sign conventions here. I used the one I like better, but am not sure it is the one you are accustomed to.)
Also note that in GR you are widely free in the choice of coordinates (subject to some constraints I shall not dwell upon). In general you should not expect that coordinates themselves behave as components of a 4-vector. Still stranger as it may appear, you are not allowed (in general) to assign to one of the coordinates the character of a time and to the other three the one of space coordinates.
In other words, a coordinate line (e.g. $x^1=rm const.$, $x^2=rm const.$, $x^3=rm const.$) is not bound to have a timelike tangent vector and the other three spacelike. All that is required is that the four vectors are independent.
Things are easier with orthogonal coordinates: then the metric tensor has $ne0$ only the diagonal components, and in this case it is true that one of the coordinate lines is timelike and the remaining three spacelike. In many important cases such a choice of coordinates is possible, but not always. A counterexample is Kerr (rotating, uncharged) black hole.
answered 18 mins ago
Elio Fabri
862
velocity should not be a tensor
Why not? 4-velocity is a tensor: a $1 choose 0$ tensor. It is the tangent vector to the worldline of a material particle, parametrized by proper time.
Its components obey an identity:
$$g_munu u^mu u^nu = 1$$
($g_munu$ the metric tensor).
(I beg your pardon: there are two sign conventions here. I used the one I like better, but am not sure it is the one you are accustomed to.)
Also note that in GR you are widely free in the choice of coordinates (subject to some constraints I shall not dwell upon). In general you should not expect that coordinates themselves behave as components of a 4-vector. Still stranger as it may appear, you are not allowed (in general) to assign to one of the coordinates the character of a time and to the other three the one of space coordinates.
In other words, a coordinate line (e.g. $x^1=rm const.$, $x^2=rm const.$, $x^3=rm const.$) is not bound to have a timelike tangent vector and the other three spacelike. All that is required is that the four vectors are independent.
Things are easier with orthogonal coordinates: then the metric tensor has $ne0$ only the diagonal components, and in this case it is true that one of the coordinate lines is timelike and the remaining three spacelike. In many important cases such a choice of coordinates is possible, but not always. A counterexample is Kerr (rotating, uncharged) black hole.
answered 18 mins ago
Elio Fabri
862
edited 12 mins ago
answered 18 mins ago
Elio Fabri
862
answered 18 mins ago
Elio Fabri
862
answered 18 mins ago
Elio Fabri
862
862
add a comment |Â
add a comment |Â
Christine is a new contributor. Be nice, and check out our Code of Conduct.
Christine is a new contributor. Be nice, and check out our Code of Conduct.
Christine is a new contributor. Be nice, and check out our Code of Conduct.
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2
The four velocity can't be zero in any coordinate system: In the rest system it is $(1, 0, 0, 0)$.
– Sebastian Riese
3 hours ago
5
Can you clarify what you are asking? You seem to be asking if the individual components of the four velocity are tensors, and of course they are not. But the four velocity as a whole is a tensor.
– John Rennie
3 hours ago