Weight of a Moving Body

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What is the weight of a moving Body?

Consider a particle, of proper mass m₀, moving in a uniform gravitational field that is aligned parallel to the z-direction. Restrict the motion to the z = 0 plane. Define the weight, W, of the body as the magnitude of the force required to constrain the body in the z = 0 plane. Determine the weight of the moving body.

Heuristic approach

Consider a box containing a gas of N particles that interact only with the walls of the box. We assume the mass of the walls is negligible compared to the mass of the “box + gas” system. For simplicity, let the gas be 2-dimensional by restricting motion of the particles to sliding friction free on the bottom of the box. Let the box be a rest on a weight scale and let the total momentum of all the particles as well as the box zero. Let the proper mass of the kth particle be m_0k. The rest energy of the box, E_box, is the sum of the energies of the particles in the box, i.e.

The weight of the box equals the passive gravitational mass, M_g, of the box times g, the local gravitational acceleration due to gravity at z = 0. According to the equivalence principle, the passive gravitation mass of an object particle equals it’s inertial mass, i.e. M_g = M_i . Also, according to the equivalence of inertial mass and energy content we then we have E_box = M_i c². Therefore E_box= M_pc². Therefore

This reduces to

The mass of the box is therefore the sum of the relativistic masses, m_k º g_km_0k, of the particles. The weight of the box must be the sum of the weights of the individual particles, I.e.

Since this relation must hold in all cases of non-interacting particles then it follows that it must hold in the special case where all particles have the same proper mass and are moving at the same speed. The total weight will then be the weight of the box. This must also equal the sum of the weights of all the particles inside the box. We therefore have

It therefore follows that

We conclude that the passive gravitational mass, m_p, of an object is defined as m_p º w_p/g. We therefore find that

However this is just the relativistic mass m of the particle. So we find the expected relation

Eq.  (8) says gravitational mass = inertial mass = relativistic mass.

Straightforward Calculation

Now solve the same problem using the methods of general relativity.  Define the following terms

The components of 1-form corresponding to the 4-force on a particle can be expressed as

Eq. 9 can be written as

This relation implies the following definition of the gravitational force, G, as

Also now the following definitions

where f_total is he total force on the particle and F_ex is he external force required to support the particle, constraining it to move in the z = 0 plane.

Eq. (10) then becomes

For the body to be at rest in a gravitational field (i.e. remaining at rest with respect to the surface of the Earth) then the total force must vanish. For a time orthogonal space-time (i.e. g_0i = 0 ) the space-time interval  ds² = c²dt² may be written as

where ds² = g_jkdx^j dx^k . The gravitational potential is defined as F(r) º (g₀₀ - 1)c²/2. . Then

We have set m_p º gm₀ = m  and v² º (ds/cdt)². If the metric has constant spatial components (i.e. g_ij,k = 0) then we can place Eq.

This will be true for an accelerating frame of reference and according to the equivalence principle the gravitational field is always locally equivalent to an accelerant frame of reference. The metric for a uniformly accelerating frame of referenced is in this form

F has the value 

Evaluated at z = 0 gives

Thus it is seen that if work is done on a particle the kinetic energy will increase. This physically manifests itself in an increased speed. The increased speed causes an increased amount of force to support the particles in the box and therefore there is an increased passive gravitational mass of the box. That’s why E = mc²!

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