# Why is there a need for the concept of energy if we have the concept of momentum?

We have two concepts that are energy and momentum. To me, momentum is more fundamental than energy and I think that momentum was the thing which we wanted to discover as energy.

Now momentum can describe several things that that energy does and if it is not able to describe it then it can be somehow extended to describe that thing.

For example, momentum can not describe the potential energy let's say due to the gravitational field of Earth on an object but it can be easily twisted to be able to describe it.

Momentum can describe quantum stuff as well. Also, we know that momentum is conserved just as energy.

In short, I want to know the physical difference b/w momentum and energy.

• Can you provide a bit of your background? Are you studying physics for the first time? If so, at the high school level or university level? Oct 11, 2020 at 16:17
• @DavidWhite Yes, I am a new Physics learner at the school level or It can be called as university level for other countries as the syllabus in my country covers most of physics at the school level and I love logic and according to me everything in physics should be explained on basis of logic. Oct 11, 2020 at 16:26
• Logic can only be applied after the core "rules" are defined. This is even true for mathematics. At some point, a few definitions and/or facts must be taken on faith, and derivations, conclusions, etc., are based on those core facts. Oct 11, 2020 at 16:29
• @user921307 "most of physics". Really? You'd be surprised how much physics you'd still be able to learn and discover, even after 20 years at university. Oct 12, 2020 at 1:31
• It may be good to point out that this was core of Vis viva (proposed by Leibniz) and momentum (proposed by Newton) controversy. Oct 12, 2020 at 16:17

Here is a thought experiment that should convince you that it's necessary to consider energy and not just momentum. Suppose you stand on a train track in between two identical trains traveling at the same speed. One of them approaches you from your left, and the other from your right. The trains are moving in such a way that they will both collide with you at the same instant.

Since the trains have the same mass and speed but are traveling in opposite directions, they carry zero total momentum. When they hit you, they will not tend to accelerate you to the left or to the right — you will stay right in place. In spite of their lack of net momentum, the trains do carry kinetic energy. When they collide with you, they will transfer some of this kinetic energy to your body. Anyone watching will clearly see the effect of this energy, although you yourself will probably not be able to observe it.

• Subtly passive-aggressive. I love it. Oct 11, 2020 at 22:29
• Good example. Also implied by this experiment is the fact that momentum is a directional vector and can essentially cancel out to zero - when the trains collide, they will stop. Energy, on the other hand, is a scalar quantity - there's no way to have negative kinetic energy that will somehow cancel out. One train doesn't simply absorb the other's KE like it does momentum, the KE has to be dissipated in some other form. Oct 12, 2020 at 13:24
• Why does this answer remind me of XKCD 123?.. Oct 12, 2020 at 21:27
• Cute, but does this hold up when you consider that a human body is not just one part, but consists of multiple small parts which do not experience a simultaneous zero total momentum impact? Oct 14, 2020 at 17:14
• The impact you are talking about can be described with help of momentum. Just take mod of momentum and you will get the same feeling as energy. Hence why I didn't chose your answer as satisfactory. This I clarified because I thought some people would think that since your answer is aggressive hence why I would not have chosen but this is not the cause. Dec 29, 2021 at 21:49

Already some good answers here, but also let's add the following important idea which no one has mentioned yet.

Suppose two particles are in a collision. The masses are $$m_1$$, $$m_2$$, the initial velocities are $${\bf u}_1$$ and $${\bf u}_2$$, the final velocities are $${\bf v}_1$$ and $${\bf v}_2$$. Then conservation of momentum tells us $$m_1 {\bf u}_1 + m_2 {\bf u}_2 = m_1 {\bf v}_1 + m_2 {\bf v}_2.$$ That is a useful and important result, but it does not completely tell us what will happen. If the masses and the initial velocities are known, for example, then there would be infinitely many different combinations of $${\bf v}_1$$ and $${\bf v}_2$$ which could satisfy this equation.

Now let's bring in conservation of energy, assuming no energy is converted into other forms such as heat. Then we have $$\frac{1}{2}m_1 u^2_1 + \frac{1}{2}m_2 u^2_2 = \frac{1}{2}m_1 v^2_1 + \frac{1}{2} m_2 v^2_2.$$ Now we have some new information which was not included in the momentum equation. In fact, in a one dimensional case these two equations are sufficient to pin down the final velocities completely, and in the three-dimensional case almost completely (up to rotations in the CM frame; see below). This shows that energy and momentum are furnishing different insights, both of which help to understand what is going on. Neither can replace the other.

There are plenty of other things one might also say. The most important is the connection between energy and time on the one hand, and between momentum and position on the other, but other answers have already mentioned that. It may also interest you to know that the two most important equations in quantum theory are a relationship between energy and development in time (Schrodinger's equation) and a relationship between momentum and position (the position, momentum commutator).

Further info

The general two-body collision can be analysed in the CM frame (variously called centre of mass frame; centre of momentum frame; zero momentum frame). This is the frame where the total momentum (both before and after the collision) is zero. The conservation laws fix the sizes but not the directions of the final velocities in this frame, except to say that the directions are opposite to one another.

• Nitpick: even if you specify the plane that two final $\mathbf{v}$ vectors lie in, you only have three equations and four unknowns (two components each). So it's not quite true that the only freedom left in the solution is a rotation about the total momentum direction; you generally have to specify something about the final velocities as well to get a complete solution. (In undergrad collision problems, this is often the angle of one of the $\mathbf{v}$ vectors.) Oct 13, 2020 at 15:46
• @MichaelSeifert Yes, thanks for that; I was a bit hasty. I have corrected my answer. Oct 13, 2020 at 17:50

To me, momentum is more fundamental than energy ... In short, I want to know the physical difference b/w momentum and energy.

In modern physics the most fundamental facts have to do with symmetries. From Noether’s theorem any differentiable symmetry in the laws of physics corresponds to a conserved quantity.

Momentum is the conserved quantity associated with spatial translation symmetry. In other words, the laws of physics are the same here and there, therefore there is a corresponding conserved quantity which we call momentum.

Energy is the conserved quantity associated with time translation symmetry. In other words, the laws of physics are the same yesterday and today, therefore there is a corresponding conserved quantity which we call energy.

In modern physics neither time nor space is prioritized over the other, but together they are unified in a single overall framework called spacetime. In that framework energy and momentum are the timelike and spacelike parts of a single overall conserved quantity called the four-momentum which is a four-dimensional vector $$(E/c,p_x,p_y,p_z)$$. It is incorrect to assert the preeminence of either over the other. They are non-redundant conserved quantities.

In short, I want to know the physical difference b/w momentum and energy.

Momentum is used when you have temporal (time related) information regarding the system. Whereas energy is used in the absence of the temporal information related to the system.

When you'd learn Noether's Theorem you'll find out that conservation of momentum emerges from space related symmetries whereas that of energy emerges from time related ones.

In short these are quite different quantities and one cannot be replaced by the other. Neither can you claim fundamentalness of one over the other.

Your impression that "momentum can describe several things that that energy does and if it is not able to describe it then it can be somehow extended to describe that thing" is a reflection of something real and deep. When you move to the framework of general relativity, in which space and time are combined to form the 4-dimensional manifold "spacetime", then momentum and energy also combine into a 4-dimensional vector, given the name four-momentum:

$$p^\mu=(E,p_x,p_y,p_z)$$

So, in a sense, energy "really is" momentum in the time dimension.

However, working in this framework is a waste of effort in any context where Newtonian physics are "good enough"—you just wind up doing extra math to get the same result. And one of the really useful ways Newtonian mechanics is simpler than relativistic mechanics is that you can treat energy and momentum as independent quantities. Don't get out the tensor calculus if you don't need it.

To me, momentum is more fundamental than energy and I think that momentum was the thing which we wanted to discover as energy.

Just like your other question, you still have your concept of definitions backwards. It's not like someone said "let's come up a definition of momentum" but accidentally came across energy instead. These ideas didn't exist until trying to describe important aspects of physical phenomena.

It was observed that the quantity $$m\mathbf v$$ was useful in certain situations, so it was given the name "momentum". It was observed that the quantity $$\frac12mv^2$$ was useful in other situations, so it was given the name "kinetic energy". Etc.

You seem to think the quantity came after the definition, but it's the other way around.

• I think you are not able to understand my question. I am not saying that definitions comes first. I am saying that is energy useful to define or in other words can we use just the momentum to describe all things that we describe with the help of energy. Just like we don't need to define change in force per unit change in force because we can describe it in other ways. See the answer I marked as best answer to understand my question. Oct 11, 2020 at 20:59
• And regarding the previous question, I was questioning the usefulness of force times displacement over force times time. And since energy is takes as heart of physics hence I asked that shouldn't work be force times time because I see its clear significance. Oct 11, 2020 at 21:02

We never “discovered” energy or momentum. They are inventions: merely mathematical constructs that help us understand the natural laws.

Our current understanding of the stage where physical laws play out tells us that we in fact live on a 4-dimensional manifold, for which a chart could be given by the coordinate system $$\xi = (t,x,y,z)$$. The first coordinate describes time, the other three are space coordinates.

Our intuition tells us that space is homogeneous and isotropic (see Cosmological principle), meaning that physical laws should be same everywhere. This is a $$\textit{symmetry}$$: physical laws should remain invariant no matter which part of the universe we live in. A consequence of this symmetry is conservation of momentum. Similarly, another intuition is that the physical laws should be invariant no matter when we decide to test it. This time symmetry leads to the conservation of energy. Both of these result as a consequence of Noether’s theorem, as other answers have mentioned.

In the language of Special Relativity, one is allowed to define therefore a canonically conjugate 4-vector called the 4-momentum : $$p^\mu = (E,p_x,p_y,p_z)$$. A direct consequence of the fact that both time symmetry and space symmetry holds in our local (both in the sense of space and time) universe, is that the length of this vector is equal to $$m^2$$: where $$m$$ is the rest mass of the particle dictated by such physical laws.

You might find it curious that for massless objects (like photon), energy and momentum are the same thing. For massive objects (like electron), energy and momentum are inherently different quantities.

Another important difference between energy and momentum is from mathematics. Energy is a scalar quantity: it remains invariant under coordinate transformations, and is essentially just a real number. But momentum is a vector quantity: its dimensions depend on the dimension of the spacetime under consideration, and are covariant under coordinate transformations. Logistically, it is simpler to deal with invariants (scalars) rather than covariants/contravariants (vectors). Which is why in topics such as statistical mechanics that deal with large number of particles, we prefer the formalism of density of states rather than keeping track of a 6-D phase space for each particle, represented by $$(x,y,z,p_x,p_y,p_z)$$.

Conservation of energy can be applied in situations involving potential energy, such as a ball rolling down a slope, or a frictionless pendulum. In these situations, it is difficult to apply conservation of momentum because you would have to include the entire earth in your momentum calculation.

Conservation of momentum can be applied in situations which involve dissipative forces, such as non-elastic collisions, or situations in which chemical energy is released, such as a rocket. In these situations, it is difficult to apply conservation of energy because it is difficult to determine the energy "lost" due to the dissipative forces or "gained" from chemical reactions.

So momentum and energy are both useful concepts in Newtonian mechanics. You cannot replace one with the other.

I think energy quantifies how much work can be done, or has been done. If you lift a block, then you have done some amount of work, and one would say you expended that much energy, but there's no question of momentum. So, perhaps energy helps in quantifying stuff that has happened or can happen, while momentum is about things in the present? (Also, energy is a scalar, so that helps with algebra).

• If the block was lifted, then it can very well be described based on the change in momentum in some other object which lifted it. Oct 11, 2020 at 15:05
• Ok, how about this, let the block fall from zero velocity, what is its velocity going to be when it reaches the ground? Won't this necessitate the use of kinetic and potential energies? My point being that the change in momentum is related to the force and how long (in time and distance) it acts etc. Oct 11, 2020 at 15:15

Now momentum can describe several things that that energy does and if it is not able to describe it then it can be somehow extended to describe that thing.

For example, momentum can not describe the potential energy let's say due to the gravitational field of Earth on an object but it can be easily twisted to be able to describe it.

Have you actually tried to do this? I would think it would not be possible to extend momentum in this way, not without reforming fundamental assumptions about mechanics. I will come back to this.

To answer your question, the formal answer is that conservation of energy and momentum are consequences of distinct symmetries of the interaction. We can stick to classical, Newtonian mechanics to see hints of this. From Newton's second law,

$$F = m \frac{\textrm{d}v}{\textrm{d}t} \implies \int F\,\textrm{d}t = \int m \,\textrm{d}v = \Delta p,$$

and with a simple use of the chain rule,

$$F = m \frac{\textrm{d}v}{\textrm{d}t} = m \frac{\textrm{d}v}{\textrm{d}x}\frac{\textrm{d}x}{\textrm{d}t} = mv \frac{\textrm{d}v}{\textrm{d}x} \implies \int F \,\textrm{d}x = \int mv \,\textrm{d}v = \Delta T.$$

(the quantity $$\int F \,\textrm{d}t$$ is called the impulse of the force, and $$\int F \,\textrm{d}x$$ is called the work done by the force.)

Even here you can see that the conservation of kinetic energy $$T$$ and momentum $$p$$ depend on different characteristics of the force $$F$$. Let's take this further.

Given a force $$F(x,t)$$ that has explicit time dependence, if we can find a sort of generating function of the force $$F$$, called the potential $$U(x,t)$$, such that $$F = - \partial{U}/\partial{x}$$, then

$$\int F \,\textrm{d}x = -\int \frac{\partial U}{\partial x}\textrm{d}x = - \int \left( \textrm{d}U - \frac{\partial U}{\partial t}\textrm{d}t \right) = - \Delta U + \int \frac{\partial U}{\partial t}\textrm{d}t.$$

So,

$$\Delta E := \Delta (T + U) = \int \frac{\partial U}{\partial t}\textrm{d}t,$$

and the total energy $$E$$ will be conserved for all times if $$\partial{U}/\partial{t} = 0$$; or, in other words, if $$U$$ has no explicit time dependence, and hence if $$F = - \partial{U}/\partial{x}$$ has no explicit time dependence. Thus, we see that the conservation of energy is a statement about the dependence of the interaction (force) on time alone. (Note here the primacy that $$U$$ plays and not $$F$$ directly: $$F$$ is merely a "consequence" of $$U$$.) Most fundamental forces in nature do indeed have no explicit time-dependence, and thus conserve total energy.

We can, theoretically, follow the same logic with conservation of momentum. If we can find another generating function $$G(x,t)$$ such that $$F = -\partial{G}/\partial{t}$$, then, following the same steps as above, we will find that:

$$\Delta (p + G) = \int \frac{\partial G}{\partial x}\textrm{d}x,$$

and that the quantity $$(p+G)$$ will be conserved everywhere if $$\partial{G}/\partial{x} = 0$$. Or, in other words, if $$G$$ has no explicit $$x$$-dependence, and hence $$F$$ has no explicit $$x$$-dependence. However, no fundamental forces that are purely time-dependent are known. No forces, for example, behave like $$F = \sin{t}$$ as opposed to, say, $$F = - k x$$. Thus, the quantity $$G$$ plays no significant role in physics. Indeed, the only force that will conserve both energy $$E$$ and the quantity $$(p+G)$$ must have no explicit time or space dependence, and thus must be $$F = 0$$, in which case $$U = 0 = G$$ anyway. So, we talk of conservation of momentum only, and this will occur whenever $$\int F \,\textrm{d}t = 0$$. The stronger condition that momentum is conserved at all times requires that $$F(x) = 0$$, and if we understand $$F(x)$$ to be a condition upon space, then the conservation of momentum is a statement about the spatial dependence of $$F$$. For closed systems, Newton's third law, which says that two interacting bodies experience "equal and opposite" forces, ensures that the total impulse $$\int F\,\textrm{d}t$$ will always cancel since such interactions always occur in cancelling pairs of internal forces. In the language of symmetries, this would be the argument that Newton's third law guarantees that forces have no explicit absolute $$x$$-dependence since the only spatial quantity that can be inherently common to two distinct bodies can only be some function of their mutual separation. Thus, for such closed systems, the total momentum will always be conserved since the system as a whole has no absolute $$x$$-dependence.

Thus, to conclude, it cannot be possible for conservation of energy and conservation of momentum to be dependent since they are statements about the characteristics of the interaction with respect to distinct mechanical parameters (i.e., space and time). To be able to describe one in terms of the other would be tantamount to saying something like that space and time are mechanically, fundamentally connected, which would be a strange mechanical world. Even Quantum-mechanically, the details may be different, but the broad relationship between energy and momentum, and time and space are consonant.

A more rigorous treatment of the above ideas will involve the Lagrangian and Hamiltonian formalisms, which you may want to look in to for more details. These will naturally also lead on to quantum mechanics and Noether's theorem.

Conservation of energy implies conservation of momentum and conservation of mass in the context of classical mechanics. Suppose $$N$$ particles interact with each other elastically so that total kinetic energy is conserved:

$$\sum_{j=1}^{N} m_i \vec{v}_i^2 = \sum_{j=1}^{N} m_i \vec{u}_i^2 \tag{1}$$

where the $$\vec{v}_i$$ are the initial velocities, the $$\vec{u}_i$$ are the final velocities and the $$m_i$$ are the masses of the particles. We demand that (1) is valid in all inertial frames. In another frame that moves w.r.t. the original frame with velocity $$\vec{w}$$ the initial and final velocities of the particles are $$\vec{v}_i-\vec{w}$$ and $$\vec{u}_i-\vec{w}$$, respectively. We thus have:

$$\sum_{j=1}^{N} m_i \left(\vec{v}_i-\vec{w}\right)^2 = \sum_{j=1}^{N} m_i \left(\vec{u}_i-\vec{w}\right)^2$$

Expanding the squares and using (1) yields:

$$\left(\sum_{j=1}^{N} m_i \vec{v}_i - \sum_{j=1}^{N} m_i \vec{u}_i\right)\cdot\vec{w} = 0$$

Since $$\vec{w}$$ is arbitrary, it follows that momentum is conserved.

• $\vec{v}\cdot\vec{w}=0 \nRightarrow \vec{v}=0$ for arbitrary $\vec{w}$. It's not true that conservation of momentum follows from conservation of energy since they depend on different symmetry conditions, even classically. Oct 14, 2020 at 15:07

Let me try to answer this with the help of a simple example and just the equations of motion.

In the diagram above we have got two objects with mass and velocity annotated as shown, the direction of motion of these objects is from left to right and we can figure out that momentum $$\left( p \right)$$ of both these objects are the same, now to the right of the diagram from the line drawn there is a force field that exerts $$1$$ Newton of force at each point from right to left. Now with the information we have, let's try to find out the distance they travel in the force field till they come to stop as the force acting is opposite to the direction of the objects motion.

For the top object, the negative acceleration $$a = \frac{1 N}{1 kg} = 1 \frac{m}{s^{2}}$$

With the help of equation $$v^{2} = u^{2} + 2 a s$$

$$0 = \left( 100 \right)^{2} + 2 \cdot \left( -1 \right) \cdot s \implies s = 5000 m$$

So the distance to stop for the top object in the force field is $$5000$$ metres.

Now for the bottom object, the negative acceleration $$a = \frac{1 N}{10 kg} = 0.1 \frac{m}{s^{2}} \implies 0 = \left( 10 \right)^{2} + 2 \cdot \left( -0.1 \right) \cdot s \implies s = 500 m$$

The distance to stop for the bottom object in the force field is $$500$$ metres.

Now we clearly see that even though both objects had the same momentum, it took different distances for the force field to stop them, this quantity of how much something is required to bring them to stop is Kinetic Energy or just energy, which momentum alone cannot describe.