Sol65

Solution
Week 65 (12/8/03)
Relativistic cart
Ground frame (your frame): In your frame, the force at your feet is responsible
for changing the momentum of the cart-plus-sand-inside system. Let’s label this
system as “C”.
To find the dp/dt of C, let’s determine how fast the mass of C increases. We
claim that the rate of mass increase is γσ. This can be seen as follows. Assume
that C has a mass M at a given time. A mass σ dt falls into the cart during a time
dt. The energy of the resulting C is γM + σ dt (we’ll drop the c’s here), while the
momentum is still γMv. Using E2 = p2 + m2, we see that the resulting mass equals
M = (γM + σ dt)2 − (γMv)2 ≈ M2 + 2γMσ dt, (1)
where we have dropped the second-order dt2 terms. Using the Taylor series
√
1 + ≈
1 + /2, we may approximate M as
M ≈ M 1 +
2γσ dt
M
≈ M 1 +
γσ dt
M
= M + γσ dt. (2)
Therefore, C’s mass increases at a rate γσ.1 Intuitively, this rate of increase must
certainly be greater than the nonrelativistic result of “σ”, because heat is generated
during the collision, and this heat shows up as mass in the final object.
Having found the rate at which the mass increases, we see that the rate at which
the momentum increases is (using the fact that v is constant)
dp
dt
= γ
dM
dt
v = γ(γσ)v = γ2
σv. (3)
Since F = dp/dt, this is the force that you exert on the cart. Therefore, it is also
the force that the ground exerts on your feet (because the net force on you is zero).
Cart frame: The sand-entering-cart events happen at the same location in the
ground frame, so time dilation says that the sand enters the cart at a slower rate
in the cart frame; that is, at a rate σ/γ. The sand flies in at speed v, and then
eventually comes at rest on the cart, so its momentum decreases at a rate γ(σ/γ)v =
σv. This is the force that your hand applies to the cart.
If this were the only change in momentum in the problem, then we would have
a problem, because the force on your feet would be σv in the cart frame, whereas
we found above that it is γ2σv in the ground frame. This would contradict the fact
that longitudinal forces are the same in different frames. What is the resolution of
this apparent paradox?
1
This result is easier to see if we work in the frame where C is at rest. In this frame, a mass σ dt
comes flying in with energy γσ dt, and essentially all of this energy shows up as mass (heat) in the
final object. That is, essentially none of it shows up as overall kinetic energy of the object, which
is a general result for when a small object hits a stationary large object.
1

The resolution is that while you are pushing on the cart, your mass is decreasing.
You are moving with speed v in the cart frame, and mass is continually being
transferred from you (who are moving) to the cart (which is at rest). This is the
missing change in momentum we need. Let’s be quantitative about this.
Go back to the ground frame for a moment. We found above that the mass of
C increases at rate γσ in the ground frame. Therefore, the energy of C increases at
a rate γ(γσ) in the ground frame. The sand provides σ of this energy, so you must
provide the remaining (γ2 − 1)σ part. Therefore, since you are losing energy at this
rate, you must also be losing mass at this rate in the ground frame (because you are
at rest there).
Now go back to the cart frame. Due to time dilation, you lose mass at a rate
of only (γ2 − 1)σ/γ. This mass goes from moving at speed v (that is, along with
you), to speed zero (that is, at rest on the cart). Therefore, the rate of decrease in
momentum of this mass is γ((γ2 − 1)σ/γ)v = (γ2 − 1)σv.
Adding this result to the σv result due to the sand, we see that the total rate of
decrease in momentum is γ2σv. This is therefore the force that the ground applies
to your feet, in agreement with the calculation in the ground frame.
2

Sol65

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