How does author derives $nabla_i V_{ij} = nabla_{ij} V_{ij} = - nabla_j V_{ji}$?
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In the book of Classical Mechanics by Golstein, at page 11, it is given that
When the forces are all conservative, the second term in Eq. (1.29) can be rewritten as a sum over pairs of particles, the terms for each pair being of the form
$$
-int_1^2(nabla_i V_{ij} cdot d{bf s}_i + nabla_j V_{ij} cdot d{bf s}_j)
$$
If the difference vector ${bf r}_i - {bf r}_j$ is denoted by ${bf r}_{ij}$ and $nabla_{ij}$ stands for the gradient with respect to ${bf r}_{ij}$, then
$$nabla_i V_{ij} = nabla_{ij} V_{ij} = - nabla_j V_{ji}$$
However, I cannot understand how does the author concludes the equality
$$nabla_i V_{ij} = nabla_{ij} V_{ij} = - nabla_j V_{ji}$$
real-analysis analysis partial-derivative classical-mechanics
edited Nov 24 at 10:39
caverac
12.8k21028
asked Nov 24 at 10:27
onurcanbektas
3,3251936
add a comment |
up vote
1
down vote
favorite
In the book of Classical Mechanics by Golstein, at page 11, it is given that
When the forces are all conservative, the second term in Eq. (1.29) can be rewritten as a sum over pairs of particles, the terms for each pair being of the form
$$
-int_1^2(nabla_i V_{ij} cdot d{bf s}_i + nabla_j V_{ij} cdot d{bf s}_j)
$$
If the difference vector ${bf r}_i - {bf r}_j$ is denoted by ${bf r}_{ij}$ and $nabla_{ij}$ stands for the gradient with respect to ${bf r}_{ij}$, then
$$nabla_i V_{ij} = nabla_{ij} V_{ij} = - nabla_j V_{ji}$$
However, I cannot understand how does the author concludes the equality
$$nabla_i V_{ij} = nabla_{ij} V_{ij} = - nabla_j V_{ji}$$
real-analysis analysis partial-derivative classical-mechanics
edited Nov 24 at 10:39
caverac
12.8k21028
asked Nov 24 at 10:27
onurcanbektas
3,3251936
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
In the book of Classical Mechanics by Golstein, at page 11, it is given that
When the forces are all conservative, the second term in Eq. (1.29) can be rewritten as a sum over pairs of particles, the terms for each pair being of the form
$$
-int_1^2(nabla_i V_{ij} cdot d{bf s}_i + nabla_j V_{ij} cdot d{bf s}_j)
$$
If the difference vector ${bf r}_i - {bf r}_j$ is denoted by ${bf r}_{ij}$ and $nabla_{ij}$ stands for the gradient with respect to ${bf r}_{ij}$, then
$$nabla_i V_{ij} = nabla_{ij} V_{ij} = - nabla_j V_{ji}$$
However, I cannot understand how does the author concludes the equality
$$nabla_i V_{ij} = nabla_{ij} V_{ij} = - nabla_j V_{ji}$$
real-analysis analysis partial-derivative classical-mechanics
edited Nov 24 at 10:39
caverac
12.8k21028
asked Nov 24 at 10:27
onurcanbektas
3,3251936
In the book of Classical Mechanics by Golstein, at page 11, it is given that
When the forces are all conservative, the second term in Eq. (1.29) can be rewritten as a sum over pairs of particles, the terms for each pair being of the form
$$
-int_1^2(nabla_i V_{ij} cdot d{bf s}_i + nabla_j V_{ij} cdot d{bf s}_j)
$$
If the difference vector ${bf r}_i - {bf r}_j$ is denoted by ${bf r}_{ij}$ and $nabla_{ij}$ stands for the gradient with respect to ${bf r}_{ij}$, then
$$nabla_i V_{ij} = nabla_{ij} V_{ij} = - nabla_j V_{ji}$$
However, I cannot understand how does the author concludes the equality
$$nabla_i V_{ij} = nabla_{ij} V_{ij} = - nabla_j V_{ji}$$
real-analysis analysis partial-derivative classical-mechanics
real-analysis analysis partial-derivative classical-mechanics
edited Nov 24 at 10:39
caverac
12.8k21028
asked Nov 24 at 10:27
onurcanbektas
3,3251936
edited Nov 24 at 10:39
caverac
12.8k21028
asked Nov 24 at 10:27
onurcanbektas
3,3251936
edited Nov 24 at 10:39
caverac
12.8k21028
edited Nov 24 at 10:39
caverac
12.8k21028
edited Nov 24 at 10:39
caverac
12.8k21028
12.8k21028
asked Nov 24 at 10:27
onurcanbektas
3,3251936
asked Nov 24 at 10:27
onurcanbektas
3,3251936
asked Nov 24 at 10:27
onurcanbektas
3,3251936
3,3251936
add a comment |
add a comment |
1 Answer
1
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oldest
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0
down vote
accepted
Because the potential $V_{ij}$ depends on the difference ${bf x}_{ij} = {bf x}_i - {bf x}_j$, so
$$
frac{partial V_{ij}}{partial {bf x}_j} =
frac{partial {bf x}_{ij}}{partial {bf x}_j} frac{partial V_{ij}}{partial {bf x}_{ij}}
= frac{partial ({bf x}_i - {bf x}_j)}{partial {bf x}_j} frac{partial V_{ij}}{partial {bf x}_{ij}} =
-frac{partial V_{ij}}{partial {bf x}_{ij}}
$$
Similarly
$$
frac{partial V_{ij}}{partial {bf x}_i} = +frac{partial V_{ij}}{partial {bf x}_{ij}}
$$
Or in Goldstein's notation
$$
nabla_i V_{ij} = nabla_{ij} V_{ij} = -nabla_j V_{ij}
$$
answered Nov 24 at 10:32
caverac
12.8k21028
Thanks both for the answer and the edit to the question.
– onurcanbektas
Nov 24 at 10:40
@onurcanbektas Happy to help
– caverac
Nov 24 at 10:40
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Because the potential $V_{ij}$ depends on the difference ${bf x}_{ij} = {bf x}_i - {bf x}_j$, so
$$
frac{partial V_{ij}}{partial {bf x}_j} =
frac{partial {bf x}_{ij}}{partial {bf x}_j} frac{partial V_{ij}}{partial {bf x}_{ij}}
= frac{partial ({bf x}_i - {bf x}_j)}{partial {bf x}_j} frac{partial V_{ij}}{partial {bf x}_{ij}} =
-frac{partial V_{ij}}{partial {bf x}_{ij}}
$$
Similarly
$$
frac{partial V_{ij}}{partial {bf x}_i} = +frac{partial V_{ij}}{partial {bf x}_{ij}}
$$
Or in Goldstein's notation
$$
nabla_i V_{ij} = nabla_{ij} V_{ij} = -nabla_j V_{ij}
$$
answered Nov 24 at 10:32
caverac
12.8k21028
Thanks both for the answer and the edit to the question.
– onurcanbektas
Nov 24 at 10:40
@onurcanbektas Happy to help
– caverac
Nov 24 at 10:40
add a comment |
up vote
0
down vote
accepted
Because the potential $V_{ij}$ depends on the difference ${bf x}_{ij} = {bf x}_i - {bf x}_j$, so
$$
frac{partial V_{ij}}{partial {bf x}_j} =
frac{partial {bf x}_{ij}}{partial {bf x}_j} frac{partial V_{ij}}{partial {bf x}_{ij}}
= frac{partial ({bf x}_i - {bf x}_j)}{partial {bf x}_j} frac{partial V_{ij}}{partial {bf x}_{ij}} =
-frac{partial V_{ij}}{partial {bf x}_{ij}}
$$
Similarly
$$
frac{partial V_{ij}}{partial {bf x}_i} = +frac{partial V_{ij}}{partial {bf x}_{ij}}
$$
Or in Goldstein's notation
$$
nabla_i V_{ij} = nabla_{ij} V_{ij} = -nabla_j V_{ij}
$$
answered Nov 24 at 10:32
caverac
12.8k21028
Thanks both for the answer and the edit to the question.
– onurcanbektas
Nov 24 at 10:40
@onurcanbektas Happy to help
– caverac
Nov 24 at 10:40
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Because the potential $V_{ij}$ depends on the difference ${bf x}_{ij} = {bf x}_i - {bf x}_j$, so
$$
frac{partial V_{ij}}{partial {bf x}_j} =
frac{partial {bf x}_{ij}}{partial {bf x}_j} frac{partial V_{ij}}{partial {bf x}_{ij}}
= frac{partial ({bf x}_i - {bf x}_j)}{partial {bf x}_j} frac{partial V_{ij}}{partial {bf x}_{ij}} =
-frac{partial V_{ij}}{partial {bf x}_{ij}}
$$
Similarly
$$
frac{partial V_{ij}}{partial {bf x}_i} = +frac{partial V_{ij}}{partial {bf x}_{ij}}
$$
Or in Goldstein's notation
$$
nabla_i V_{ij} = nabla_{ij} V_{ij} = -nabla_j V_{ij}
$$
answered Nov 24 at 10:32
caverac
12.8k21028
Because the potential $V_{ij}$ depends on the difference ${bf x}_{ij} = {bf x}_i - {bf x}_j$, so
$$
frac{partial V_{ij}}{partial {bf x}_j} =
frac{partial {bf x}_{ij}}{partial {bf x}_j} frac{partial V_{ij}}{partial {bf x}_{ij}}
= frac{partial ({bf x}_i - {bf x}_j)}{partial {bf x}_j} frac{partial V_{ij}}{partial {bf x}_{ij}} =
-frac{partial V_{ij}}{partial {bf x}_{ij}}
$$
Similarly
$$
frac{partial V_{ij}}{partial {bf x}_i} = +frac{partial V_{ij}}{partial {bf x}_{ij}}
$$
Or in Goldstein's notation
$$
nabla_i V_{ij} = nabla_{ij} V_{ij} = -nabla_j V_{ij}
$$
answered Nov 24 at 10:32
caverac
12.8k21028
answered Nov 24 at 10:32
caverac
12.8k21028
answered Nov 24 at 10:32
caverac
12.8k21028
answered Nov 24 at 10:32
caverac
12.8k21028
12.8k21028
Thanks both for the answer and the edit to the question.
– onurcanbektas
Nov 24 at 10:40
@onurcanbektas Happy to help
– caverac
Nov 24 at 10:40
add a comment |
Thanks both for the answer and the edit to the question.
– onurcanbektas
Nov 24 at 10:40
@onurcanbektas Happy to help
– caverac
Nov 24 at 10:40
Thanks both for the answer and the edit to the question.
– onurcanbektas
Nov 24 at 10:40
Thanks both for the answer and the edit to the question.
– onurcanbektas
Nov 24 at 10:40
@onurcanbektas Happy to help
– caverac
Nov 24 at 10:40
@onurcanbektas Happy to help
– caverac
Nov 24 at 10:40
add a comment |
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