Conversation 2. Potential energy
Colleague, about potential energy, please, in more detail.
You, my friend, are perfectly interested in the most important component of the total energy.
It is considered that the potential energy is a part of the total energy of the system, depending on the relative position of the material particles that make up this system and on their positions in the external force field (gravitational, electric field).
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We call the force field that part of the space, at each point of which a force of magnitude and direction acts on the material particle placed there.
Numerically, the potential energy of a system in its given position is equal to the work that forces acting on the system will produce as it moves from this position to where the potential energy is zero.
Colleague, does the test body have only energy in a potential field, or a potential field too?
To answer your question, open the TSB (Big Soviet Encyclopedia) and in the "Fields of Physics" section read (literally):
“The fields are physical, a special form of matter; a physical system with an infinitely large number of degrees of freedom. Examples of physical fields are electromagnetic and gravitational fields ... ".
It follows that the potential field is a material medium. So, like any material medium, this field has energy (and mass, respectively). By the way, this is confirmed, for example, by the presence of electromagnetic waves in the field, which are oscillations of this material medium.
It is difficult to determine the specific boundaries of the field, so physicists have long been accustomed to operating on the energy contained in a unit volume, that is, the volume density of the potential field energy (measured in J / m 3 ). Take, for example, Zilberman's book “Electricity and Magnetism” (Nauka, Moscow, 1970) and on page 136 we read (literally):
" In a flat capacitor and in general in a uniform field, the energy density, that is, the energy contained in a unit of volume, is constant and equal to the total energy divided by the volume ."
A colleague, since a potential field is a material medium, it must be characterized by specific parameters that can be calculated and measured.
You are absolutely right. We have already found out that the electric (potential) field is characterized by such a parameter as the bulk energy density (hereinafter referred to as pressure, J / m 3 or N / m 2 ). In addition, the potential field is characterized by the potential and its gradient - field strength. Moreover, pressure, potential and intensity characterize a potential field at a given point, regardless of the presence of a test body at this point, because, as we already know, the field itself has energy and mass.
If the potential energy (W P , J) is attributed to a unit mass (m, kg) or to a unit electric charge (q, C), then we obtain gravitational (v 2 = W P / m, J / kg) or electric (U = W P / q, j / Cl) potentials.
The gradient of the potential at this point is the field strength:
- for the gravitational field: g = - grad v 2 ;
- for electrical: E = - grad U (the sign will be discussed below).
Gradient (from lat. Gradiens, genus. Case gradientis - walking), a vector showing the direction of the fastest change of a certain value from one point of space to another.
With the distance from the center of the field, not only the potential, but also the potential energy changes. And its gradient is the force we call the force of force.
Addition : We have already agreed that the gradient of the gravitational potential is the gravitational field strength g = - grad v 2 . Multiplying these two parameters by the mass, we obtain, respectively, the value of force (F = mg) and potential energy (W P = mv 2 ). Therefore, the force can also be considered as the energy gradient at a given point of the field (F = - grad W W).
Similarly, for an electric field: the electric field strength is E = - grad U, force F = qE, potential energy W = qU. Hence, here, too, F = - grad W.
The equation F = - grad W shows that the work of forces along a closed trajectory in a potential field is always zero.
Colleague, what units of measurement are most appropriate for the above parameters?
Very good question. POWER is measured in Newtons (N = kg * m / s 2 ) or in J / m. The second version of the record is more acceptable, for it immediately gives us an indication that the force is only a GRADIENT OF ENERGY (J / m). This is important because it simplifies further understanding of physical processes. By the way, this concerns not only strength, but also parameters such as pressure and potential.
PRESSURE is measured in N / m 2 or in J / m 3 . Here, too, the second variant of the record is more acceptable, because it immediately indicates to us the VOLUME DENSITY OF ENERGY (J / m 3 ).
POTENTIAL is measured in m 2 / s 2 or J / kg (for the gravitational field) and in (kg / C) * (m 2 / s 2 ) or J / C (for the electric field). And here the second variant of the record is more acceptable, for it immediately indicates the value of potential energy per unit mass (J / kg) for the gravitational field or per unit electric charge (J / C) for the electric field.
Finally, colleague, let's consider how the value of potential energy is determined.
Perhaps now we are ready to solve this problem. The value of potential energy is determined in two ways:
- simplified (approximate) - for a uniform field;
- common (true) - for a non-uniform field, which really surrounds us.
A potential field can be conditionally considered homogeneous if the intensity vector at all its points has the same value and direction. For example, for a gravitational field this rule can be applied only at the surface of the Earth on a small part of it (say, in a laboratory experiment). In this case, to simplify the calculations, the value of the potential energy of a test body on the surface of the Earth is conventionally taken equal to zero, and its value at any other point is determined from the equation:
W P = mgh, J,
where g is the gravitational field strength (N / kg), and h is the vertical distance (m) from the Earth’s surface to the test body of mass m (kg).
Here, the sign in front of the potential energy value does not matter in principle.
Colleague, but this is the most common way to determine potential energy.
Unfortunately, many physics textbooks on this and complete the definition of potential energy. But not all. Take, for example, the General Course of Physics Sivukhina (Moscow, MIPT, 2005) or the American Course of Physics in translation edited by Akhmatov (Moscow, Science, 1974).
Here is considered:
- a method already known to us for determining the potential energy of a test body in a uniform gravitational field near the surface of the Earth (Volume 1, pp. 144-145 of the first source and Part III, pp. 1552-157 of the second source);
- and the general method of determining the potential energy for an inhomogeneous field (volume 1, pp. 145-146 of the first source and part III, pp. 157-159 of the second source).
The general calculation method already gives a negative value of potential energy:
- equation (25.6) W (U) = - GMm / r in the first source and
- the equation W (Ur) = - GMm / r - in the second.
The negative value of the potential energy here is explained as follows:
- in the first source (quotation): “The attracting masses have the maximum energy at an infinite distance between them. In this position, the potential energy is considered to be zero. In any other position it is less, that is, negative ”;
- in the second source, the proof of the correctness of the equation W (Ur) = - GMm / r is given.
Indeed, a body freely falling to the center of the field loses its potential energy, which is converted into kinetic energy. Hence, the potential energy decreases with decreasing distance between the centers of mass (M and m) and, conversely, with increasing distance increases.
Considering that in the already known equation W P = - GMm / r the radius symbol is in the denominator, it is extremely clear that with increasing distance (radius value tends to infinity) the potential energy increases to ... zero. This is possible only if the potential energy in any other position is negative.
Conclusion: the potential energy for all material particles is negative.
From this it follows that the value of the gravitational potential v 2 = W / m = - GM / r is also negative. And this is confirmed by the equation (3) in the section "The Encyclopedic Dictionary of Physics or a similar section of the Great Soviet Encyclopedia.
The value of potential energy and electric potential in an electric field is determined similarly. And then we will make sure that the potential energy and its bulk density (pressure) are the same for both the gravitational and the electric fields.
Colleague, now try to write down your statement as a formula ...
Formulas are written by mathematicians, and physicists use equations. The necessary equations are already given here. However, we will try, nevertheless, to do without them for the time being, all the more so without the “formulas”.
To do this, we use household observations, which suggest: to evaporate the water boiling in the kettle, you need to burn some wood or gas. In other words, you need to do the work. Using a thermometer, you can make sure that the temperature of boiling water and the temperature of steam above it are the same. Consequently, the average energy of particle motion in boiling water and vapor is the same.
Conclusion: thermal energy transferred to boiling water from the fuel is converted into the interaction energy of particles of evaporating water. This means that the binding energy of particles in boiling water is less than in water vapor. But in a pair, this energy is almost zero, therefore, the interaction energy of particles in a liquid is less than zero, i.e. is negative.
Colleague, your arguments are convincing and examples you give irrefutable. However, not everyone thinks the same.
And here you are absolutely right. For physicists, there is no problem with understanding the essence and sign of potential energy, because they consider the gravitational field, including the field of the Earth, to be inhomogeneous. For physicists, the gravitational field strength varies with the square distance: g = Gm / r 2 .
However, mathematicians do not think so. For them, the gravitational field is HOMOGENEOUS with a constant gravitational field strength (like this parameter and does not depend on the radius). The value of potential energy is determined by the simplified formula W = mgh. They do not associate h with the radius of the field, but consider it a simple segment between two arbitrary points of this field. Therefore, for them, the potential energy can take a zero value at any point they like. Nonsense, but it happens.
But there is also a "physics and mathematics". Their opinion depends on whether they are physicists or mathematicians.
Colleague, why do you think that mathematicians "to a uniform field"?
In support of this, we open the Short Course of Mathematical Analysis (Bermant, Aramanovich, 2005) and on page 520 in the section “Field Theory” we read:
“ A vector field is called homogeneous if A (P) is a constant vector, i.e. Ah, Ay and Az are constant values.
An example of a uniform field is, for example, a gravity field . ”
Now you see for yourself that mathematicians call the gravitational field “the field of gravity” and “seriously” consider it homogeneous. And this is not just a harmless error, for it prevents us from realizing the Nature of gravity. However, we will talk about this a little later.
You, my friend, are perfectly interested in the most important component of the total energy.
It is considered that the potential energy is a part of the total energy of the system, depending on the relative position of the material particles that make up this system and on their positions in the external force field (gravitational, electric field).
')
We call the force field that part of the space, at each point of which a force of magnitude and direction acts on the material particle placed there.
Numerically, the potential energy of a system in its given position is equal to the work that forces acting on the system will produce as it moves from this position to where the potential energy is zero.
Colleague, does the test body have only energy in a potential field, or a potential field too?
To answer your question, open the TSB (Big Soviet Encyclopedia) and in the "Fields of Physics" section read (literally):
“The fields are physical, a special form of matter; a physical system with an infinitely large number of degrees of freedom. Examples of physical fields are electromagnetic and gravitational fields ... ".
It follows that the potential field is a material medium. So, like any material medium, this field has energy (and mass, respectively). By the way, this is confirmed, for example, by the presence of electromagnetic waves in the field, which are oscillations of this material medium.
It is difficult to determine the specific boundaries of the field, so physicists have long been accustomed to operating on the energy contained in a unit volume, that is, the volume density of the potential field energy (measured in J / m 3 ). Take, for example, Zilberman's book “Electricity and Magnetism” (Nauka, Moscow, 1970) and on page 136 we read (literally):
" In a flat capacitor and in general in a uniform field, the energy density, that is, the energy contained in a unit of volume, is constant and equal to the total energy divided by the volume ."
A colleague, since a potential field is a material medium, it must be characterized by specific parameters that can be calculated and measured.
You are absolutely right. We have already found out that the electric (potential) field is characterized by such a parameter as the bulk energy density (hereinafter referred to as pressure, J / m 3 or N / m 2 ). In addition, the potential field is characterized by the potential and its gradient - field strength. Moreover, pressure, potential and intensity characterize a potential field at a given point, regardless of the presence of a test body at this point, because, as we already know, the field itself has energy and mass.
If the potential energy (W P , J) is attributed to a unit mass (m, kg) or to a unit electric charge (q, C), then we obtain gravitational (v 2 = W P / m, J / kg) or electric (U = W P / q, j / Cl) potentials.
The gradient of the potential at this point is the field strength:
- for the gravitational field: g = - grad v 2 ;
- for electrical: E = - grad U (the sign will be discussed below).
Gradient (from lat. Gradiens, genus. Case gradientis - walking), a vector showing the direction of the fastest change of a certain value from one point of space to another.
With the distance from the center of the field, not only the potential, but also the potential energy changes. And its gradient is the force we call the force of force.
Addition : We have already agreed that the gradient of the gravitational potential is the gravitational field strength g = - grad v 2 . Multiplying these two parameters by the mass, we obtain, respectively, the value of force (F = mg) and potential energy (W P = mv 2 ). Therefore, the force can also be considered as the energy gradient at a given point of the field (F = - grad W W).
Similarly, for an electric field: the electric field strength is E = - grad U, force F = qE, potential energy W = qU. Hence, here, too, F = - grad W.
The equation F = - grad W shows that the work of forces along a closed trajectory in a potential field is always zero.
Colleague, what units of measurement are most appropriate for the above parameters?
Very good question. POWER is measured in Newtons (N = kg * m / s 2 ) or in J / m. The second version of the record is more acceptable, for it immediately gives us an indication that the force is only a GRADIENT OF ENERGY (J / m). This is important because it simplifies further understanding of physical processes. By the way, this concerns not only strength, but also parameters such as pressure and potential.
PRESSURE is measured in N / m 2 or in J / m 3 . Here, too, the second variant of the record is more acceptable, because it immediately indicates to us the VOLUME DENSITY OF ENERGY (J / m 3 ).
POTENTIAL is measured in m 2 / s 2 or J / kg (for the gravitational field) and in (kg / C) * (m 2 / s 2 ) or J / C (for the electric field). And here the second variant of the record is more acceptable, for it immediately indicates the value of potential energy per unit mass (J / kg) for the gravitational field or per unit electric charge (J / C) for the electric field.
Finally, colleague, let's consider how the value of potential energy is determined.
Perhaps now we are ready to solve this problem. The value of potential energy is determined in two ways:
- simplified (approximate) - for a uniform field;
- common (true) - for a non-uniform field, which really surrounds us.
A potential field can be conditionally considered homogeneous if the intensity vector at all its points has the same value and direction. For example, for a gravitational field this rule can be applied only at the surface of the Earth on a small part of it (say, in a laboratory experiment). In this case, to simplify the calculations, the value of the potential energy of a test body on the surface of the Earth is conventionally taken equal to zero, and its value at any other point is determined from the equation:
W P = mgh, J,
where g is the gravitational field strength (N / kg), and h is the vertical distance (m) from the Earth’s surface to the test body of mass m (kg).
Here, the sign in front of the potential energy value does not matter in principle.
Colleague, but this is the most common way to determine potential energy.
Unfortunately, many physics textbooks on this and complete the definition of potential energy. But not all. Take, for example, the General Course of Physics Sivukhina (Moscow, MIPT, 2005) or the American Course of Physics in translation edited by Akhmatov (Moscow, Science, 1974).
Here is considered:
- a method already known to us for determining the potential energy of a test body in a uniform gravitational field near the surface of the Earth (Volume 1, pp. 144-145 of the first source and Part III, pp. 1552-157 of the second source);
- and the general method of determining the potential energy for an inhomogeneous field (volume 1, pp. 145-146 of the first source and part III, pp. 157-159 of the second source).
The general calculation method already gives a negative value of potential energy:
- equation (25.6) W (U) = - GMm / r in the first source and
- the equation W (Ur) = - GMm / r - in the second.
The negative value of the potential energy here is explained as follows:
- in the first source (quotation): “The attracting masses have the maximum energy at an infinite distance between them. In this position, the potential energy is considered to be zero. In any other position it is less, that is, negative ”;
- in the second source, the proof of the correctness of the equation W (Ur) = - GMm / r is given.
Indeed, a body freely falling to the center of the field loses its potential energy, which is converted into kinetic energy. Hence, the potential energy decreases with decreasing distance between the centers of mass (M and m) and, conversely, with increasing distance increases.
Considering that in the already known equation W P = - GMm / r the radius symbol is in the denominator, it is extremely clear that with increasing distance (radius value tends to infinity) the potential energy increases to ... zero. This is possible only if the potential energy in any other position is negative.
Conclusion: the potential energy for all material particles is negative.
From this it follows that the value of the gravitational potential v 2 = W / m = - GM / r is also negative. And this is confirmed by the equation (3) in the section "The Encyclopedic Dictionary of Physics or a similar section of the Great Soviet Encyclopedia.
The value of potential energy and electric potential in an electric field is determined similarly. And then we will make sure that the potential energy and its bulk density (pressure) are the same for both the gravitational and the electric fields.
Colleague, now try to write down your statement as a formula ...
Formulas are written by mathematicians, and physicists use equations. The necessary equations are already given here. However, we will try, nevertheless, to do without them for the time being, all the more so without the “formulas”.
To do this, we use household observations, which suggest: to evaporate the water boiling in the kettle, you need to burn some wood or gas. In other words, you need to do the work. Using a thermometer, you can make sure that the temperature of boiling water and the temperature of steam above it are the same. Consequently, the average energy of particle motion in boiling water and vapor is the same.
Conclusion: thermal energy transferred to boiling water from the fuel is converted into the interaction energy of particles of evaporating water. This means that the binding energy of particles in boiling water is less than in water vapor. But in a pair, this energy is almost zero, therefore, the interaction energy of particles in a liquid is less than zero, i.e. is negative.
Colleague, your arguments are convincing and examples you give irrefutable. However, not everyone thinks the same.
And here you are absolutely right. For physicists, there is no problem with understanding the essence and sign of potential energy, because they consider the gravitational field, including the field of the Earth, to be inhomogeneous. For physicists, the gravitational field strength varies with the square distance: g = Gm / r 2 .
However, mathematicians do not think so. For them, the gravitational field is HOMOGENEOUS with a constant gravitational field strength (like this parameter and does not depend on the radius). The value of potential energy is determined by the simplified formula W = mgh. They do not associate h with the radius of the field, but consider it a simple segment between two arbitrary points of this field. Therefore, for them, the potential energy can take a zero value at any point they like. Nonsense, but it happens.
But there is also a "physics and mathematics". Their opinion depends on whether they are physicists or mathematicians.
Colleague, why do you think that mathematicians "to a uniform field"?
In support of this, we open the Short Course of Mathematical Analysis (Bermant, Aramanovich, 2005) and on page 520 in the section “Field Theory” we read:
“ A vector field is called homogeneous if A (P) is a constant vector, i.e. Ah, Ay and Az are constant values.
An example of a uniform field is, for example, a gravity field . ”
Now you see for yourself that mathematicians call the gravitational field “the field of gravity” and “seriously” consider it homogeneous. And this is not just a harmless error, for it prevents us from realizing the Nature of gravity. However, we will talk about this a little later.
Source: https://habr.com/ru/post/206676/
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