Mixing
Potential Gradient relation
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I was just wondering why exactly the force is the negative gradient of the potential, which is shown by the relation $$bf F=−bf∇V$$I know that this equation only holds for conservative forces, where the potential only depends on the position of an object. The only problem I have is why there is a minus sign/ what the minus sign means in a physical context.
newtonian-mechanics kinematics potential
edited 4 hours ago
Garf
1,009117
asked 4 hours ago
Tatjana Gobold
325
add a comment |Â
up vote
2
down vote
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I was just wondering why exactly the force is the negative gradient of the potential, which is shown by the relation $$bf F=−bf∇V$$I know that this equation only holds for conservative forces, where the potential only depends on the position of an object. The only problem I have is why there is a minus sign/ what the minus sign means in a physical context.
newtonian-mechanics kinematics potential
edited 4 hours ago
Garf
1,009117
asked 4 hours ago
Tatjana Gobold
325
See: physics.stackexchange.com/questions/16339/… and physics.stackexchange.com/questions/18874/…
– LonelyProf
4 hours ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I was just wondering why exactly the force is the negative gradient of the potential, which is shown by the relation $$bf F=−bf∇V$$I know that this equation only holds for conservative forces, where the potential only depends on the position of an object. The only problem I have is why there is a minus sign/ what the minus sign means in a physical context.
newtonian-mechanics kinematics potential
edited 4 hours ago
Garf
1,009117
asked 4 hours ago
Tatjana Gobold
325
I was just wondering why exactly the force is the negative gradient of the potential, which is shown by the relation $$bf F=−bf∇V$$I know that this equation only holds for conservative forces, where the potential only depends on the position of an object. The only problem I have is why there is a minus sign/ what the minus sign means in a physical context.
newtonian-mechanics kinematics potential
newtonian-mechanics kinematics potential
edited 4 hours ago
Garf
1,009117
asked 4 hours ago
Tatjana Gobold
325
edited 4 hours ago
Garf
1,009117
asked 4 hours ago
Tatjana Gobold
325
edited 4 hours ago
Garf
1,009117
edited 4 hours ago
Garf
1,009117
edited 4 hours ago
Garf
1,009117
1,009117
asked 4 hours ago
Tatjana Gobold
325
asked 4 hours ago
Tatjana Gobold
325
asked 4 hours ago
Tatjana Gobold
325
325
See: physics.stackexchange.com/questions/16339/… and physics.stackexchange.com/questions/18874/…
– LonelyProf
4 hours ago
add a comment |Â
See: physics.stackexchange.com/questions/16339/… and physics.stackexchange.com/questions/18874/…
– LonelyProf
4 hours ago
See: physics.stackexchange.com/questions/16339/… and physics.stackexchange.com/questions/18874/…
– LonelyProf
4 hours ago
See: physics.stackexchange.com/questions/16339/… and physics.stackexchange.com/questions/18874/…
– LonelyProf
4 hours ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
1
down vote
accepted
Well, despite the fact thet the other answers are interesting, I'd prefer going directly to the "origin" of this thing, rather than its useful physical meaning.
The key idea is that
The minus sign is just a convention.
Forces do exist, and we define the potential based on them, not vice-versa.
By the way, I am commiting the mistake I hate: we are talking about potentials. But actually, the relation is
$$ vecF=-vecnablaE_p$$
Where $E_p$ is the potential energy. You'll say: "that's what I wrote", yes, but we are lazy and we say "potential", but we write potential energy. I curse the person who decided to use $V$ for potential energy, when it was being used for "potential", as in Volts of electric potential. Volts $neq$ joules, you know.
Well, sorry, back to the topic. The thing is that, the only thin we know is:
- A Conservative force is such that the work done by it does not depend on the path.
- But that implies that we can write work as "one function", evaluated at the final point, minus the same function evaluated atthe initial point:
$$W_cons=int_vecx_0^vecx_F vecFcdot dvecx=B(vecx_F)-B(vecx_0)$$
That's the only thing we know.
And then, we decide to define a new function, called "potential energy", such that
$$E_p=-B$$
And we do that because
$$W_Total=Delta E_k$$
$$W_total=W_cons+W_not cons$$
So
$$Delta E_k = W_not cons+ Delta B$$
$$Delta E_k - Delta B= W_not cons$$
And that's it. We could just work with that formula. But, we prefer havinga possitive quantity, so we write
$$Delta E_k + Delta E_p= W_not cons$$
And so, we can think of potential energy in the same way as kinetic energy, in the sense of, "the more energy, the more work".
That is the actual reason for the minus sign. But, as I said, it's jsut a convention. We could just work with the opposite funcion, which I called $B$.
Hope I helped.
answered 3 hours ago
FGSUZ
3,3102521
Great thank you! In a physical context I already got it, so it's nice to have a small deduction of this formula. Actually you would need some vector calculus for F = -dV/dx because it is a gradient, but your answer makes sense, too. And yes I am aware that V stands for potential energy, but it is just convenient to say potential if you know that you actually mean potential energy :D
– Tatjana Gobold
3 hours ago
add a comment |Â
up vote
1
down vote
The gradient is a vector pointing to the direction where the change of $V$ is maximum. If a force $F$ is pointing in the direction of $-nabla V$, it will push you in the direction where $V$ decreases the fastest. Think of a little marble placed on the edge of a bowl - it will roll down, but in which way? It can go down in a helix or just roll down to the center, and it chooses the second - why is that? Because that is the direction of fastest decrease in height (and thus gravitational potential energy). That is the physical meaning of the minus sign. If it was a plus, the ball would roll up the potential.
answered 4 hours ago
Ofek Gillon
1,139417
add a comment |Â
up vote
1
down vote
One of the simplest ways to look at this is in 1 dimension:
$$F=-fractextdVtextdx$$
Imagine a ball in a hilly landscape. Plotting a graph of gravitational potential energy vs $x$ might look something like
If the ball is at position $A$, then the gradient of the potential energy with $x$ is negative, and the resulting force will be in the positive $x$-direction.
If the ball is at position $B$, then the gradient of potential energy with $x$ is positive, so the force is in the negative $x$ direction.
This overall point here is that things like to try and lower their (potential) energy if they can. This means that if in a certain direction the potential energy is going to increase, the ball is going to want to move $away$ from this, and in a direction where the ball is decreasing potential energy, until equilibrium is reached at a stationary point.
answered 4 hours ago
Garf
1,009117
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Well, despite the fact thet the other answers are interesting, I'd prefer going directly to the "origin" of this thing, rather than its useful physical meaning.
The key idea is that
The minus sign is just a convention.
Forces do exist, and we define the potential based on them, not vice-versa.
By the way, I am commiting the mistake I hate: we are talking about potentials. But actually, the relation is
$$ vecF=-vecnablaE_p$$
Where $E_p$ is the potential energy. You'll say: "that's what I wrote", yes, but we are lazy and we say "potential", but we write potential energy. I curse the person who decided to use $V$ for potential energy, when it was being used for "potential", as in Volts of electric potential. Volts $neq$ joules, you know.
Well, sorry, back to the topic. The thing is that, the only thin we know is:
- A Conservative force is such that the work done by it does not depend on the path.
- But that implies that we can write work as "one function", evaluated at the final point, minus the same function evaluated atthe initial point:
$$W_cons=int_vecx_0^vecx_F vecFcdot dvecx=B(vecx_F)-B(vecx_0)$$
That's the only thing we know.
And then, we decide to define a new function, called "potential energy", such that
$$E_p=-B$$
And we do that because
$$W_Total=Delta E_k$$
$$W_total=W_cons+W_not cons$$
So
$$Delta E_k = W_not cons+ Delta B$$
$$Delta E_k - Delta B= W_not cons$$
And that's it. We could just work with that formula. But, we prefer havinga possitive quantity, so we write
$$Delta E_k + Delta E_p= W_not cons$$
And so, we can think of potential energy in the same way as kinetic energy, in the sense of, "the more energy, the more work".
That is the actual reason for the minus sign. But, as I said, it's jsut a convention. We could just work with the opposite funcion, which I called $B$.
Hope I helped.
answered 3 hours ago
FGSUZ
3,3102521
Great thank you! In a physical context I already got it, so it's nice to have a small deduction of this formula. Actually you would need some vector calculus for F = -dV/dx because it is a gradient, but your answer makes sense, too. And yes I am aware that V stands for potential energy, but it is just convenient to say potential if you know that you actually mean potential energy :D
– Tatjana Gobold
3 hours ago
add a comment |Â
up vote
1
down vote
accepted
Well, despite the fact thet the other answers are interesting, I'd prefer going directly to the "origin" of this thing, rather than its useful physical meaning.
The key idea is that
The minus sign is just a convention.
Forces do exist, and we define the potential based on them, not vice-versa.
By the way, I am commiting the mistake I hate: we are talking about potentials. But actually, the relation is
$$ vecF=-vecnablaE_p$$
Where $E_p$ is the potential energy. You'll say: "that's what I wrote", yes, but we are lazy and we say "potential", but we write potential energy. I curse the person who decided to use $V$ for potential energy, when it was being used for "potential", as in Volts of electric potential. Volts $neq$ joules, you know.
Well, sorry, back to the topic. The thing is that, the only thin we know is:
- A Conservative force is such that the work done by it does not depend on the path.
- But that implies that we can write work as "one function", evaluated at the final point, minus the same function evaluated atthe initial point:
$$W_cons=int_vecx_0^vecx_F vecFcdot dvecx=B(vecx_F)-B(vecx_0)$$
That's the only thing we know.
And then, we decide to define a new function, called "potential energy", such that
$$E_p=-B$$
And we do that because
$$W_Total=Delta E_k$$
$$W_total=W_cons+W_not cons$$
So
$$Delta E_k = W_not cons+ Delta B$$
$$Delta E_k - Delta B= W_not cons$$
And that's it. We could just work with that formula. But, we prefer havinga possitive quantity, so we write
$$Delta E_k + Delta E_p= W_not cons$$
And so, we can think of potential energy in the same way as kinetic energy, in the sense of, "the more energy, the more work".
That is the actual reason for the minus sign. But, as I said, it's jsut a convention. We could just work with the opposite funcion, which I called $B$.
Hope I helped.
answered 3 hours ago
FGSUZ
3,3102521
Great thank you! In a physical context I already got it, so it's nice to have a small deduction of this formula. Actually you would need some vector calculus for F = -dV/dx because it is a gradient, but your answer makes sense, too. And yes I am aware that V stands for potential energy, but it is just convenient to say potential if you know that you actually mean potential energy :D
– Tatjana Gobold
3 hours ago
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Well, despite the fact thet the other answers are interesting, I'd prefer going directly to the "origin" of this thing, rather than its useful physical meaning.
The key idea is that
The minus sign is just a convention.
Forces do exist, and we define the potential based on them, not vice-versa.
By the way, I am commiting the mistake I hate: we are talking about potentials. But actually, the relation is
$$ vecF=-vecnablaE_p$$
Where $E_p$ is the potential energy. You'll say: "that's what I wrote", yes, but we are lazy and we say "potential", but we write potential energy. I curse the person who decided to use $V$ for potential energy, when it was being used for "potential", as in Volts of electric potential. Volts $neq$ joules, you know.
Well, sorry, back to the topic. The thing is that, the only thin we know is:
- A Conservative force is such that the work done by it does not depend on the path.
- But that implies that we can write work as "one function", evaluated at the final point, minus the same function evaluated atthe initial point:
$$W_cons=int_vecx_0^vecx_F vecFcdot dvecx=B(vecx_F)-B(vecx_0)$$
That's the only thing we know.
And then, we decide to define a new function, called "potential energy", such that
$$E_p=-B$$
And we do that because
$$W_Total=Delta E_k$$
$$W_total=W_cons+W_not cons$$
So
$$Delta E_k = W_not cons+ Delta B$$
$$Delta E_k - Delta B= W_not cons$$
And that's it. We could just work with that formula. But, we prefer havinga possitive quantity, so we write
$$Delta E_k + Delta E_p= W_not cons$$
And so, we can think of potential energy in the same way as kinetic energy, in the sense of, "the more energy, the more work".
That is the actual reason for the minus sign. But, as I said, it's jsut a convention. We could just work with the opposite funcion, which I called $B$.
Hope I helped.
answered 3 hours ago
FGSUZ
3,3102521
Well, despite the fact thet the other answers are interesting, I'd prefer going directly to the "origin" of this thing, rather than its useful physical meaning.
The key idea is that
The minus sign is just a convention.
Forces do exist, and we define the potential based on them, not vice-versa.
By the way, I am commiting the mistake I hate: we are talking about potentials. But actually, the relation is
$$ vecF=-vecnablaE_p$$
Where $E_p$ is the potential energy. You'll say: "that's what I wrote", yes, but we are lazy and we say "potential", but we write potential energy. I curse the person who decided to use $V$ for potential energy, when it was being used for "potential", as in Volts of electric potential. Volts $neq$ joules, you know.
Well, sorry, back to the topic. The thing is that, the only thin we know is:
- A Conservative force is such that the work done by it does not depend on the path.
- But that implies that we can write work as "one function", evaluated at the final point, minus the same function evaluated atthe initial point:
$$W_cons=int_vecx_0^vecx_F vecFcdot dvecx=B(vecx_F)-B(vecx_0)$$
That's the only thing we know.
And then, we decide to define a new function, called "potential energy", such that
$$E_p=-B$$
And we do that because
$$W_Total=Delta E_k$$
$$W_total=W_cons+W_not cons$$
So
$$Delta E_k = W_not cons+ Delta B$$
$$Delta E_k - Delta B= W_not cons$$
And that's it. We could just work with that formula. But, we prefer havinga possitive quantity, so we write
$$Delta E_k + Delta E_p= W_not cons$$
And so, we can think of potential energy in the same way as kinetic energy, in the sense of, "the more energy, the more work".
That is the actual reason for the minus sign. But, as I said, it's jsut a convention. We could just work with the opposite funcion, which I called $B$.
Hope I helped.
answered 3 hours ago
FGSUZ
3,3102521
answered 3 hours ago
FGSUZ
3,3102521
answered 3 hours ago
FGSUZ
3,3102521
answered 3 hours ago
FGSUZ
3,3102521
3,3102521
Great thank you! In a physical context I already got it, so it's nice to have a small deduction of this formula. Actually you would need some vector calculus for F = -dV/dx because it is a gradient, but your answer makes sense, too. And yes I am aware that V stands for potential energy, but it is just convenient to say potential if you know that you actually mean potential energy :D
– Tatjana Gobold
3 hours ago
add a comment |Â
Great thank you! In a physical context I already got it, so it's nice to have a small deduction of this formula. Actually you would need some vector calculus for F = -dV/dx because it is a gradient, but your answer makes sense, too. And yes I am aware that V stands for potential energy, but it is just convenient to say potential if you know that you actually mean potential energy :D
– Tatjana Gobold
3 hours ago
Great thank you! In a physical context I already got it, so it's nice to have a small deduction of this formula. Actually you would need some vector calculus for F = -dV/dx because it is a gradient, but your answer makes sense, too. And yes I am aware that V stands for potential energy, but it is just convenient to say potential if you know that you actually mean potential energy :D
– Tatjana Gobold
3 hours ago
Great thank you! In a physical context I already got it, so it's nice to have a small deduction of this formula. Actually you would need some vector calculus for F = -dV/dx because it is a gradient, but your answer makes sense, too. And yes I am aware that V stands for potential energy, but it is just convenient to say potential if you know that you actually mean potential energy :D
– Tatjana Gobold
3 hours ago
add a comment |Â
up vote
1
down vote
The gradient is a vector pointing to the direction where the change of $V$ is maximum. If a force $F$ is pointing in the direction of $-nabla V$, it will push you in the direction where $V$ decreases the fastest. Think of a little marble placed on the edge of a bowl - it will roll down, but in which way? It can go down in a helix or just roll down to the center, and it chooses the second - why is that? Because that is the direction of fastest decrease in height (and thus gravitational potential energy). That is the physical meaning of the minus sign. If it was a plus, the ball would roll up the potential.
answered 4 hours ago
Ofek Gillon
1,139417
add a comment |Â
up vote
1
down vote
The gradient is a vector pointing to the direction where the change of $V$ is maximum. If a force $F$ is pointing in the direction of $-nabla V$, it will push you in the direction where $V$ decreases the fastest. Think of a little marble placed on the edge of a bowl - it will roll down, but in which way? It can go down in a helix or just roll down to the center, and it chooses the second - why is that? Because that is the direction of fastest decrease in height (and thus gravitational potential energy). That is the physical meaning of the minus sign. If it was a plus, the ball would roll up the potential.
answered 4 hours ago
Ofek Gillon
1,139417
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The gradient is a vector pointing to the direction where the change of $V$ is maximum. If a force $F$ is pointing in the direction of $-nabla V$, it will push you in the direction where $V$ decreases the fastest. Think of a little marble placed on the edge of a bowl - it will roll down, but in which way? It can go down in a helix or just roll down to the center, and it chooses the second - why is that? Because that is the direction of fastest decrease in height (and thus gravitational potential energy). That is the physical meaning of the minus sign. If it was a plus, the ball would roll up the potential.
answered 4 hours ago
Ofek Gillon
1,139417
The gradient is a vector pointing to the direction where the change of $V$ is maximum. If a force $F$ is pointing in the direction of $-nabla V$, it will push you in the direction where $V$ decreases the fastest. Think of a little marble placed on the edge of a bowl - it will roll down, but in which way? It can go down in a helix or just roll down to the center, and it chooses the second - why is that? Because that is the direction of fastest decrease in height (and thus gravitational potential energy). That is the physical meaning of the minus sign. If it was a plus, the ball would roll up the potential.
answered 4 hours ago
Ofek Gillon
1,139417
answered 4 hours ago
Ofek Gillon
1,139417
answered 4 hours ago
Ofek Gillon
1,139417
answered 4 hours ago
Ofek Gillon
1,139417
1,139417
add a comment |Â
add a comment |Â
up vote
1
down vote
One of the simplest ways to look at this is in 1 dimension:
$$F=-fractextdVtextdx$$
Imagine a ball in a hilly landscape. Plotting a graph of gravitational potential energy vs $x$ might look something like
If the ball is at position $A$, then the gradient of the potential energy with $x$ is negative, and the resulting force will be in the positive $x$-direction.
If the ball is at position $B$, then the gradient of potential energy with $x$ is positive, so the force is in the negative $x$ direction.
This overall point here is that things like to try and lower their (potential) energy if they can. This means that if in a certain direction the potential energy is going to increase, the ball is going to want to move $away$ from this, and in a direction where the ball is decreasing potential energy, until equilibrium is reached at a stationary point.
answered 4 hours ago
Garf
1,009117
add a comment |Â
up vote
1
down vote
One of the simplest ways to look at this is in 1 dimension:
$$F=-fractextdVtextdx$$
Imagine a ball in a hilly landscape. Plotting a graph of gravitational potential energy vs $x$ might look something like
If the ball is at position $A$, then the gradient of the potential energy with $x$ is negative, and the resulting force will be in the positive $x$-direction.
If the ball is at position $B$, then the gradient of potential energy with $x$ is positive, so the force is in the negative $x$ direction.
This overall point here is that things like to try and lower their (potential) energy if they can. This means that if in a certain direction the potential energy is going to increase, the ball is going to want to move $away$ from this, and in a direction where the ball is decreasing potential energy, until equilibrium is reached at a stationary point.
answered 4 hours ago
Garf
1,009117
add a comment |Â
up vote
1
down vote
up vote
1
down vote
One of the simplest ways to look at this is in 1 dimension:
$$F=-fractextdVtextdx$$
Imagine a ball in a hilly landscape. Plotting a graph of gravitational potential energy vs $x$ might look something like
If the ball is at position $A$, then the gradient of the potential energy with $x$ is negative, and the resulting force will be in the positive $x$-direction.
If the ball is at position $B$, then the gradient of potential energy with $x$ is positive, so the force is in the negative $x$ direction.
This overall point here is that things like to try and lower their (potential) energy if they can. This means that if in a certain direction the potential energy is going to increase, the ball is going to want to move $away$ from this, and in a direction where the ball is decreasing potential energy, until equilibrium is reached at a stationary point.
answered 4 hours ago
Garf
1,009117
One of the simplest ways to look at this is in 1 dimension:
$$F=-fractextdVtextdx$$
Imagine a ball in a hilly landscape. Plotting a graph of gravitational potential energy vs $x$ might look something like
If the ball is at position $A$, then the gradient of the potential energy with $x$ is negative, and the resulting force will be in the positive $x$-direction.
If the ball is at position $B$, then the gradient of potential energy with $x$ is positive, so the force is in the negative $x$ direction.
This overall point here is that things like to try and lower their (potential) energy if they can. This means that if in a certain direction the potential energy is going to increase, the ball is going to want to move $away$ from this, and in a direction where the ball is decreasing potential energy, until equilibrium is reached at a stationary point.
answered 4 hours ago
Garf
1,009117
edited 4 hours ago
answered 4 hours ago
Garf
1,009117
answered 4 hours ago
Garf
1,009117
answered 4 hours ago
Garf
1,009117
1,009117
add a comment |Â
add a comment |Â
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See: physics.stackexchange.com/questions/16339/… and physics.stackexchange.com/questions/18874/…
– LonelyProf
4 hours ago