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Is the principle of energy a tautology ?

  1. Oct 30, 2011 #1
    In classical mechanics, if we consider a force field (uniform or non-uniform) in which the acceleration [itex]\vec{a}_A[/itex] of a particle A is constant, then

    [tex]{\vphantom{\int}} \vec{a}_A = \vec{a}_A[/tex][tex]\int \vec{a}_A \cdot d\vec{r}_A = \int \vec{a}_A \cdot d\vec{r}_A[/tex][tex]{\vphantom{\int}} \Delta \; {\textstyle \frac{1}{2}} \; \vec{v}_A^{\;2} = \Delta \; \vec{a}_A \cdot \vec{r}_A[/tex][tex]{\vphantom{\int}} \Delta \; {\textstyle \frac{1}{2}} \; \vec{v}_A^{\;2} - \, \Delta \; \vec{a}_A^{\vphantom{^{\;2}}} \cdot \vec{r}_A^{\vphantom{^{\;2}}} = 0[/tex][tex]{\vphantom{\int}} m_A^{\vphantom{^{\;2}}} \left( \Delta \; {\textstyle \frac{1}{2}} \; \vec{v}_A^{\;2} - \, \Delta \; \vec{a}_A^{\vphantom{^{\;2}}} \cdot \vec{r}_A^{\vphantom{^{\;2}}} \right) = 0[/tex][tex]{\vphantom{\int}} \Delta \; T_A + \, \Delta \; V_A = 0[/tex][tex]{\vphantom{\int}} T_A + \, V_A = constant[/tex]
    where

    [tex]T_A = {\textstyle \frac{1}{2}} \; m_A^{\vphantom{^{\;2}}}\vec{v}_A^{\;2}[/tex][tex]V_A = - \; m_A^{\vphantom{^{\;2}}} \; \vec{a}_A^{\vphantom{^{\;2}}} \cdot \vec{r}_A^{\vphantom{^{\;2}}}[/tex]

    If [itex]\vec{a}_A[/itex] is not constant but [itex]\vec{a}_A[/itex] is function of [itex]\vec{r}_A[/itex] then the same result is obtained, even if Newton's second law were not valid.
    [itex]{\vphantom{aat}}[/itex]
     
  2. jcsd
  3. Oct 30, 2011 #2

    BruceW

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    Why? if acceleration were not constant, it looks like the energy is not conserved.

    Speaking more generally, energy in classical mechanics is conserved if the sum of the potential energies is not explicitly time dependent. For example, for an object falling in a constant gravitational field, energy is conserved.

    Now speaking even more generally, energy is simply a symmetry that arises due to invariance of the system under time translation. (I think its to do with Noether's theorem, or something like that).
     
  4. Oct 30, 2011 #3

    Dale

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    Where do you get the left hand side of this?
     
  5. Oct 30, 2011 #4

    BruceW

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    In the 1d case:
    [tex]\frac{d \frac{1}{2} \dot{x}^2}{dx}[/tex]
    Is equal to:
    [tex] \dot{x} \frac{d \dot{x}}{dx}[/tex]
    Which is equal to:
    [tex] \frac{dx}{dt} \frac{d \dot{x}}{dx}[/tex]
    And using the chain rule is equal to:
    [tex] \frac{d \dot{x}}{dt}[/tex]
    Which is simply [itex]\ddot{x}[/itex]. So the integral of [itex]\ddot{x}[/itex] with respect to x gives [itex]\frac{1}{2} \dot{x}^2[/itex]. Now, I'm not sure if this applies for the 3d case..
     
  6. Oct 31, 2011 #5

    Dale

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    This is essentially called the work energy theorem. If you assume the work energy theorem then energy conservation is indeed a tautology. More to the point, energy is a defined quantity and that definition is constructed so that it is conserved, so in that sense it is, by design, tautological that energy is conserved.
     
  7. Nov 13, 2011 #6
    For example, if we consider only the vertical motion of a particle in the Earth's gravitational field (one dimension), then
    [tex]{\vphantom{\int}} a = a[/tex][tex]\int_a^b a \cdot dr = \int_a^b a \cdot dr[/tex][tex]\int_a^b a \cdot dr = \int_a^b \frac{dv}{dt} \cdot {v}\;{dt} = \Delta\:{\textstyle \frac{1}{2}}\:v^{2}[/tex][tex]\int_a^b a \cdot dr = a \int_a^b dr = a\:\Delta\:r = \Delta\:a\:r [/tex][tex]{\vphantom{\int}} \Delta\:{\textstyle \frac{1}{2}}\:v^{2} = \Delta\:a\:r [/tex][tex]{\vphantom{\int}} \Delta\:{\textstyle \frac{1}{2}}\:v^{2} - \Delta\:a\:r= 0 [/tex][tex]{\vphantom{\int}} m \; \left( \Delta\:{\textstyle \frac{1}{2}}\:v^{2} - \Delta\:a\:r \right) = 0 [/tex][tex]{\vphantom{\int}} \Delta\:{\textstyle \frac{1}{2}}\:m\:v^{2} - \Delta\:m\:a\:r = 0 [/tex][tex]{\vphantom{\int}} \Delta\:T + \Delta\:V = 0 [/tex][tex]{\vphantom{\int}} T + V = constant [/tex]where

    [tex]T = {\textstyle \frac{1}{2}}\:m\:v^{2}[/tex][tex]V = - m\:a\:r[/tex][tex]a=g=constant[/tex][tex]r=y[/tex]
    BruceW: It is possible that this is related to Noether's theorem.
     
  8. Nov 13, 2011 #7

    BruceW

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    I like your working. It is very concise. You've essentially created a definition of energy which is constant with time for this system.

    An interesting point is that we could say that horizontal momentum is also conserved, so maybe we could equally well have defined that as energy, since it is also conserved with time. I think this is where the bit about 'energy corresponds to an invariance of the system with respect to time translation' is important. And to show if something fits this definition, we would need to be more rigorous than simply showing it is conserved with time.
     
  9. Nov 17, 2011 #8
    BruceW:
    Thanks for your suggestions and comments. I write little because my English is very bad.
    In the previous example, the horizontal momentum ([itex]mv_x[/itex] and [itex]mv_z[/itex]) is conserved since [itex]a_x = 0[/itex] and [itex]a_z = 0[/itex]. However, in other cases, [itex]\vec{a}[/itex] can be constant, but [itex]a_x[/itex], [itex]a_y[/itex] and [itex]a_z[/itex] can be different to zero.
    Consequently, I will incorporate your suggestions in my next post.
     
  10. Nov 17, 2011 #9
    In classical mechanics, if we consider the motion of a particle of mass [itex]m[/itex], then

    [tex]m=constant[/tex][tex]\vec{v}=d\vec{r}/dt[/tex][tex]\vec{a}=d\vec{v}/dt[/tex][tex]\vec{j}=d\vec{a}/dt[/tex][tex]\ldots[/tex]
    Definition of Momentum [itex](\vec{M})[/itex]

    [tex]\vec{M} \; = \int_a^b m\;\vec{a}\;dt \; = \int_a^b m\;d\vec{v} \; = \Delta \; m\;\vec{v}[/tex][tex]If \quad \vec{a} = 0 \: \; \rightarrow \; \: \int_a^b m\;\vec{a}\;dt = 0 \: \; \rightarrow \; \: m\;\vec{v} = constant \: \; \rightarrow \; \: \vec{P} = constant[/tex]
    Definition of Momentum 2 [itex](\vec{M}_2)[/itex]

    [tex]\vec{M}_2 \; = \int_a^b m\;\vec{j}\;dt \; = \int_a^b m\;d\vec{a} \; = \Delta \; m\;\vec{a}[/tex][tex]If \quad \vec{j} = 0 \: \; \rightarrow \; \: \int_a^b m\;\vec{j}\;dt = 0 \: \; \rightarrow \; \: m\;\vec{a} = constant \: \; \rightarrow \; \: \vec{P}_2 = constant[/tex]
    Definition of Work [itex](W)[/itex]

    [tex]W \; = \int_a^b m\;\vec{a} \cdot d\vec{r} \; = \int_a^b m\;\frac{d\vec{v}}{dt} \cdot \vec{v}\;dt \; = \Delta \; {\textstyle \frac{1}{2}}\;m\;\vec{v}^2[/tex][tex]If \quad \vec{a} = constant \: \; \rightarrow \; \: \int_a^b m\;\vec{a} \cdot d\vec{r} = \Delta \; m\;\vec{a} \cdot \vec{r} \: \; \rightarrow \; \: {\textstyle \frac{1}{2}}\;m\;\vec{v}^2 + \left(- \; m\;\vec{a} \cdot \vec{r}\right) = constant \: \; \rightarrow \; \: T + V = constant[/tex]
    Definition of Work 2 [itex](W_2)[/itex]

    [tex]W_2 \; = \int_a^b m\;\vec{j} \cdot d\vec{v} \; = \int_a^b m\;\frac{d\vec{a}}{dt} \cdot \vec{a}\;dt \; = \Delta \; {\textstyle \frac{1}{2}}\;m\;\vec{a}^2[/tex][tex]If \quad \vec{j} = constant \: \; \rightarrow \; \: \int_a^b m\;\vec{j} \cdot d\vec{v} = \Delta \; m\;\vec{j} \cdot \vec{v} \: \; \rightarrow \; \: {\textstyle \frac{1}{2}}\;m\;\vec{a}^2 + \left(- \; m\;\vec{j} \cdot \vec{v}\right) = constant \: \; \rightarrow \; \: T_2 + V_2 = constant[/tex]

    If [itex]\vec{a}[/itex], [itex]\vec{j}[/itex], [itex]\ldots[/itex] are not constant but [itex]\vec{a}[/itex], [itex]\vec{j}[/itex], [itex]\ldots[/itex] are functions of [itex]\vec{r}[/itex], [itex]\vec{v}[/itex], [itex]\ldots[/itex] respectively, then the same final result is obtained; even if Newton's second law were not valid (even if [itex]\sum \vec{F} \ne m\;\vec{a}[/itex])
     
  11. Nov 17, 2011 #10

    BruceW

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    Hmm, whenever you write \vec{a} or anything else instead of a, it does a weird box thing instead of doing a vector line above the letter like it should. It makes all the writing difficult to read. Is this some setting on my computer that is wrong, or I dunno?
     
  12. Nov 17, 2011 #11

    BruceW

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    I need to sort out my settings, or this is just going to annoy me in other threads. It must be some settings my browser has..
     
  13. Nov 17, 2011 #12
    BruceW: I believe that the equations are better shown in the new post,
    Physics Forums > Physics > Classical Physics > Definitions of Momentum and Work
     
  14. Nov 17, 2011 #13

    dextercioby

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    Do you mind explaining this part ?
     
  15. Nov 18, 2011 #14
    ok, dextercioby:

    In classical mechanics, if we consider the motion of a particle of mass [itex]m[/itex] and with acceleration [itex]\vec{a}[/itex], then
    [tex]{\vphantom{\int_a^b}}\vec{a} = \vec{a}[/tex][tex]\int_a^b \vec{a} \cdot d\vec{r} = \int_a^b \vec{a} \cdot d\vec{r}[/tex]
    [tex]\int_a^b \vec{a} \cdot d\vec{r} = \Delta \; {\textstyle \frac{1}{2}}\vec{v}^{\, 2}[/tex]
    Now, if [itex]\vec{a} \ne constant[/itex] but [itex]\vec{a}[/itex] is function of [itex]\vec{r}[/itex]; for example, if [itex]\vec{a} = k \; \vec{r}[/itex], where [itex]{k} = constant[/itex], then

    [tex]\int_a^b \vec{a} \cdot d\vec{r} = \int_a^b k \; \vec{r} \cdot d\vec{r}[/tex]
    [tex]\int_a^b \vec{a} \cdot d\vec{r} = \Delta \; {\textstyle \frac{1}{2}} k \; \vec{r}^{\, 2}[/tex]
    Therefore,

    [tex] \Delta \; {\textstyle \frac{1}{2}}\vec{v}^{\, 2} = \Delta \; {\textstyle \frac{1}{2}} k \; \vec{r}^{\, 2}[/tex]
    [tex] \Delta \; {\textstyle \frac{1}{2}}\vec{v}^{\, 2} - \Delta \; {\textstyle \frac{1}{2}} k \; \vec{r}^{\, 2} = 0[/tex]
    [tex] m \; \left( \Delta \; {\textstyle \frac{1}{2}}\vec{v}^{\, 2} - \Delta \; {\textstyle \frac{1}{2}} k \; \vec{r}^{\, 2} \right) = 0[/tex]
    [tex] \Delta \; {\textstyle \frac{1}{2}} \, m \, \vec{v}^{\, 2} - \Delta \; {\textstyle \frac{1}{2}} \, m \, k \; \vec{r}^{\, 2} = 0[/tex]
    [tex] \Delta \; T + \Delta \; V = 0[/tex]
    [tex] T + V = constant[/tex]
    where:

    [tex]T = {\textstyle \frac{1}{2}} \, m \, \vec{v}^{\, 2}[/tex][tex]V = - \; {\textstyle \frac{1}{2}} \, m \, k \; \vec{r}^{\, 2}[/tex]
    Consequently, the same (final) result is obtained, that is:

    [itex] T + V = constant[/itex]

    Even if Newton's second law were not valid, that is:

    If [itex] \Sigma \; \vec{F} = m \, \vec{a}[/itex] then [itex] T + V = constant[/itex]; but if [itex] \Sigma \; \vec{F} \ne m \, \vec{a}[/itex] then also [itex] T + V = constant[/itex]
    [itex]{\vphantom{k}}[/itex]
     
  16. Nov 18, 2011 #15

    dextercioby

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    I still don't see the proof of <assume ΣF⃗ ≠ma⃗ , then also T+V=constant>
     
  17. Nov 18, 2011 #16
    In the previous example, I do not assume that: [itex]\Sigma \; \vec{F} \ne m \, \vec{a}[/itex], neither that: [itex]\Sigma \; \vec{F} = m \, \vec{a}[/itex]

    If we consider a particle with acceleration [itex]\vec{a} = 0 = constant[/itex] and whit velocity [itex]\vec{v} = 0 = constant[/itex] relative to a inertal frame S, then

    [tex]\Sigma \; \vec{F} = m \, \vec{a}[/tex]
    [tex]{\textstyle \frac{1}{2}} \, m \, \vec{v}^{\, 2} - \; m \, \vec{a} \cdot \vec{r} = constant[/tex]
    [tex] T + V = constant[/tex]

    But, from a non-inertial frame S´with acceleration [itex]\vec{a}_{s´} = constant[/itex] relative to the inertial frame S,

    [tex]\Sigma \; \vec{F} \ne m \, \vec{a}[/tex]
    [tex]{\textstyle \frac{1}{2}} \, m \, \vec{v}^{\, 2} - \; m \, \vec{a} \cdot \vec{r} = constant[/tex]
    [tex] T + V = constant[/tex]
    There is no (real) force acting on the particle, then [itex]\Sigma \; \vec{F} = 0[/itex]
    [itex]{\vphantom{k}}[/itex]
     
  18. Nov 18, 2011 #17

    BruceW

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    In Lagrangian mechanics (which is a particular form of classical mechanics), Energy is conserved as long as the potential doesn't depend explicitly on time. And momentum is conserved as long as the potential doesn't depend on the absolute position of particles.

    So it is possible to have energy conserved, even though momentum is not conserved. For example, we can say for a single particle, the potential is [itex]V=-mgz[/itex] and its KE is [itex]\frac{1}{2}mv^2[/itex] In this case, energy is conserved, but vertical momentum is not conserved. Of course, this is only a model, and we could instead make a model including the earth, in which case momentum would be conserved. But you can see the point that we can conceive of systems where energy is conserved but momentum is not.

    So I agree that we don't need Newton's second law to have a system where energy is conserved. But this doesn't mean energy is conserved in all conceivable systems (unless we define energy to be a quantity that is conserved in all conceivable systems).
     
  19. Nov 19, 2011 #18
    Reformulation:

    In classical mechanics, if we consider the motion of a particle of mass [itex]m[/itex], then

    Definition of Impulse [itex]\vec{I}[/itex]

    [tex]\vec{I} \; = \int_a^b m \, \vec{a} \; dt \; = \Delta \; m \, \vec{v}[/tex]
    [tex]\vec{I} \; = \Delta \; \vec{P}[/tex]
    where:

    [tex]\Delta \; \vec{P} \; = \Delta \; m \, \vec{v}[/tex]

    If [itex]\; \vec{a} = 0[/itex]

    [tex]\rightarrow \; \; \Delta \; \vec{P} \; = 0[/tex]
    [tex]\rightarrow \; \; \vec{P} \; = constant[/tex]

    Definition of Work [itex]W[/itex]

    [tex]W \; = \int_a^b m \, \vec{a} \cdot d\vec{r} \; = \Delta \; {\textstyle \frac{1}{2}} \, m \, \vec{v}{\: ^2}[/tex]
    [tex]W \; = \Delta \; T + \Delta \; V \; = \int_a^b m \, \vec{a}_{n\vec{r}} \cdot d\vec{r}[/tex]
    where:

    [tex]\Delta \; T = \Delta \; {\textstyle \frac{1}{2}} \, m \, \vec{v}{\: ^2}[/tex]
    [tex]\Delta \; V = - \int_a^b m \, \vec{a}_{\vec{r}} \cdot d\vec{r}[/tex]
    [tex]\vec{a} = \vec{a}_{\vec{r}} + \vec{a}_{n\vec{r}}[/tex]
    [tex]\vec{a}_{\vec{r}} \; \; \; is \; \; function \; \; of \; \; \; \vec{r}[/tex]
    [tex]\vec{a}_{n\vec{r}} \; \; \; is \; \; not \; \; function \; \; of \; \; \; \vec{r}[/tex]

    If [itex]\; \vec{a}_{n\vec{r}} = 0[/itex]

    [tex]\rightarrow \; \; \Delta \; T + \Delta \; V \; = 0[/tex]
    [tex]\rightarrow \; \; T + V \; = constant[/tex]

    If [itex]\; \vec{a} = 0[/itex]

    [tex]\rightarrow \; \; \Delta \; T \; = 0[/tex]
    [tex]\rightarrow \; \; T \; = constant[/tex]
     
    Last edited: Nov 19, 2011
  20. Nov 21, 2011 #19

    BruceW

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    So you're saying that if the acceleration is a function of position, then the energy of the system is constant? I don't think this is enough.

    In Lagrangian mechanics, if we have:
    [tex]\frac{\partial V}{\partial t} = 0 [/tex]
    Then the energy is conserved (since we are assuming the kinetic energy is 1/2 mv^2). Here's a rough way to show it, starting with the Euler-Lagrange equation (for a particle in 1D):

    [tex] L = \frac{1}{2}m \dot{x}^2 - V [/tex]
    (Where L is the Lagrangian of the system and the dot on the x means the derivative of x with respect to time). And the Euler-Lagrange equation for this system is:
    [tex]\frac{\partial L}{\partial x} = \frac{d}{dt} \frac{\partial L}{\partial \dot{x}}[/tex]
    I don't think there is a way to derive the Euler-Lagrange equation from anything more fundamental, so I'm just going to have to assume it is true here. Now, I'll also assume that the potential depends only on time and position, (not on speed), and do the calculation for the Euler-Lagrange equation:
    [tex]- \frac{\partial V}{\partial x} = m \ddot{x} \ \text{ (equation 1)} [/tex]
    (Where I also assumed that the mass did not depend on time). We can use this equation later. For now, the energy is:
    [tex]E = T + V = \frac{1}{2}m \dot{x}^2 + V [/tex]
    And so differentiating the energy with respect to time will tell us how the energy of the system depends on the position and time coordinates of the particle:
    [tex]\frac{dE}{dt} = \frac{\partial E}{\partial \dot{x}} \frac{d \dot{x}}{dt} + \frac{\partial E}{\partial x} \frac{dx}{dt} + \frac{\partial E}{\partial t} [/tex]
    (Which is the full differentiation using partial derivatives). Now, we can use our definition of E, to get:
    [tex]\frac{dE}{dt} = m \dot{x} \ddot{x} + \frac{\partial V}{\partial x} \dot{x} + \frac{\partial V}{\partial t}[/tex]
    And we can use the useful equation from before (equation 1), to put in something more useful for the middle term:
    [tex]\frac{dE}{dt} = m \dot{x} \ddot{x} - m \ddot{x} \dot{x} + \frac{\partial V}{\partial t} [/tex]
    So two of the terms disappear, and we end up with:
    [tex]\frac{dE}{dt} = \frac{\partial V}{\partial t}[/tex]
    Which means that energy is conserved as long as the potential is not explicitly time dependent. (And I also assumed that the mass was constant and that the potential did not depend on the speed of the particle).

    This result can be extended to 3 dimensions, and for multiple particles. But the Euler-Lagrange equation for several particles in 3D is more complicated, so I don't have the energy to write that all out!
     
    Last edited: Nov 21, 2011
  21. Nov 22, 2011 #20

    This equation is the definition of potential energy [itex]V[/itex], where [itex]\vec{a}_{\vec{r}}[/itex] is a function of [itex]\vec{r}[/itex].


    If the acceleration [itex]\vec{a}_A[/itex] of a particle A is a function of their position [itex]\vec{r}_A[/itex], then the energy [itex]( \, T_A + V_A \, )[/itex] is conserved.



    Example A (for a particle in 1D)


    If [itex]a_x = k \; x^0[/itex], where [itex]k = constant = 0[/itex], then

    [tex]\Delta \; V = - \int_a^b m \, k \; x^0 \; dx \; = - \Delta \; m \, k \; x \; \; \; \rightarrow \; \; \; V = - \; m \, k \; x[/tex]
    Since [itex]a_x[/itex] is a function of [itex]x[/itex], then

    [tex]T + V = constant[/tex][tex]{\textstyle \frac{1}{2}} \; m \; v_x^2 - \; m \, k \; x = constant[/tex]
    Therefore, in this example A, the energy [itex]( \, T + V \; \, )[/itex] is conserved, and [itex]V \; [/itex] is constant (since [itex]k = constant = 0[/itex])



    Example B (for a particle in 1D)


    If [itex]a_x = k \; x^0[/itex], where [itex]k = constant \ne 0[/itex], then

    [tex]\Delta \; V = - \int_a^b m \, k \; x^0 \; dx \; = - \Delta \; m \, k \; x \; \; \; \rightarrow \; \; \; V = - \; m \, k \; x[/tex]
    Since [itex]a_x[/itex] is a function of [itex]x[/itex], then

    [tex]T + V = constant[/tex][tex]{\textstyle \frac{1}{2}} \; m \; v_x^2 - \; m \, k \; x = constant[/tex]
    Therefore, in this example B, the energy [itex]( \, T + V \; \, )[/itex] is conserved, but [itex]V \; [/itex] is not constant (since [itex]k = constant \ne 0[/itex], then [itex]a_x \ne 0[/itex], therefore [itex]x[/itex] is not constant)
     
    Last edited: Nov 23, 2011
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