# Why "time part" represents energy in Four-momentum?

I was going through Spacetime Physics by Taylor and Wheeler and came to a point where they said, and I quote,

In what follows we find that momenergy is indeed a four-dimensional arrow in spacetime, the momenergy 4-vector (Box 7-1). Its three "space parts" represent the momentum of the object in the three chosen space directions. Its "time part" represents energy. The unity of momentum and energy springs from the unity of space and time.

This part feels too abrupt for me and I am looking for a more elaborated explanation.

Here is a link to that chapter.

• $p^0=\gamma m c=\frac{E}{c}$ Commented Mar 9 at 13:08
• Presumably, you seek an answer using the storyline that you are following, as opposed to another viewpoint that develops relativity in a [possibly completely-] different way. What did you think about the preceding subsection 7.1, particularly the Question and Answer? Commented Mar 14 at 0:38

By Noether's theorem every continuous symmetry has a corresponding conserved quantity.

Energy is the conserved quantity corresponding to translation symmetry in the time direction of space-time. That's why energy is found in the "time part".

Likewise, the momentum in the $$x$$ direction is the conserved quantity corresponding to translation symmetry in the $$x$$ direction of space-time. That's why $$p_x$$ is found in the $$x$$ part.

In quantum physics, we have the formulas $$p_x = -i\hbar \frac{\partial}{\partial x}$$ and $$E = i\hbar \frac{\partial}{\partial t},$$ which also are related to this.

• An explicit reference to Noether's theorem might be helpful here, to explain what you mean by "related to". Commented Mar 9 at 23:59

$$\vec{P}=(E,p_x,p_y,p_z)$$ of an object are said to be the components of a 4-vector $$\vec{P}$$ because their values transform like a 4-vector when the object is boosted or rotated.

For example, suppose the initial object with energy and momentum $$\vec{P}$$ is boosted by $$\lambda=\tanh^{-1}\left(\frac{v}{c}\right)$$ in the $$x$$-direction. You will measure the new energy and momentum $$\vec{P}\,'$$. $$\begin{bmatrix} E' \cr p_x' \cr p_y' \cr p_z' \end{bmatrix} =\begin{bmatrix} \cosh\lambda & \sinh\lambda & 0 & 0 \cr \sinh\lambda & \cosh\lambda & 0 & 0 \cr 0 & 0 & 1 & 0 \cr 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} E \cr p_x \cr p_y \cr p_z \end{bmatrix}$$ Energy and momentum are said to be a 4-vector because their measured values transform like this with a 4$$\times$$4 matrix. Time and space displacements $$\vec{X}=(t,x,y,z)$$ are also said to be a 4-vector because their values also transform like this with a 4$$\times$$4 matrix.

Nothing to add to md2perpe's very good answer about time-translation, but you can also see that the zeroth component of the four-momentum is, to lowest order, (classical) kinetic energy plus rest energy. The four-velocity is defined as: $$u^\mu = \frac{d x^\mu}{d\tau} = \left( \begin{array}{c} \frac{c}{\sqrt{1-\beta^2}}\\ \frac{v_x}{\sqrt{1-\beta^2}}\\ \frac{v_y}{\sqrt{1-\beta^2}}\\ \frac{v_z}{\sqrt{1-\beta^2}}\\ \end{array} \right)$$ where $$\beta^2 = v^2/c^2 = \frac{1}{c^2} \left( v_x^2 + v_y^2 + v_z^2\right)$$. Defining the four-momentum as $$p^\mu = m u^\mu$$, we can then Taylor expand its first component about $$\beta = 0$$, to find: \begin{align} p^0 & = \frac{mc}{\sqrt{1-\beta^2}} \\ & = mc + \left[ - \frac{1}{2} \frac{m c \left(-2 \beta \right)}{\left( 1 - \beta^2\right)^{3/2}} \right]_{\beta = 0} \left(\beta - 0 \right) \\ & \quad \quad + \frac{1}{2} \left[ \left(\cdots\right)(\beta)^2 + \frac{mc}{\left(1 - \beta^2\right)^{3/2}}\right]_{\beta=0} \left(\beta - 0\right)^2 + O\left(\beta \right)^3 \\ & = \frac{1}{c}\left[ mc^2 + \frac{1}{2} m v^2 \right] + O\left(\beta\right)^3 \end{align} So defining $$E = c p^0$$, we have: $$E = m c^2 + \frac{1}{2} m v^2 + \cdots$$

In classical mechanics, E is canonically conjugate to the time and p is canonically conjugate to the position. Consider a 4-vector made of some canonical variables, ($$u_1$$, $$u_2$$, $$u_3$$, $$u_4$$). It can evolve according to the Hamiltonian $$H(u_i)$$ within its own frame.

Now view it in a frame that is travelling at velocity $$\vec{v}$$ with respect to the frame in which the canonical variables $$u_i$$ are measured. The Lorentz transform transforms the canonical variables to $$\vec{u^\prime}$$. In this reference frame, the Hamiltonian is $$H(\vec{u^\prime})$$ but because of the law that physical laws remain the same within each reference frame, the Hamiltonian takes the same form with respect to $$\vec{u}$$ and $$\vec{u^\prime}$$ for each $$i$$ in $$u_i$$

Since Hamiltons can be expressed in terms of canonically conjugate variables, as an alternative, within any reference frame, that means that the 4-vector ($$w_1$$, $$w_2$$, $$w_3$$, $$w_4$$) is another possible four vector that obeys the Lorentz symmetry, such that physical laws remain the same in transformed reference frames, if $$\vec{w}$$ is canonically conjugate to $$\vec{u}$$ for each $$i$$ in $$u_i$$.

E is canonically conjugate to t and p is canonically conjugate to x so (E, $$\vec{p}$$) is a four vector since (t, $$\vec{x}$$) is a four vector.