I have read this problem / exercise (I don't remember the exact numbers, so I put my own) in William Burke's "Applied Differential Geometry" and it has always puzzled me regarding the relation between inertial and gravitational mass:

A car with a rest mass of 1000kg is speeding at 0.9c over a bridge, which is able to bear loads of up to 2000kg. Is the bridge going to collapse?

Proposed solution:

According to special relativity, the inertial mass of the car with respect to the rest frame of the bridge will be 2294kg. If, according to the equivalence principle, inertial mass equals gravitational mass, shouldn't the bridge collapse?

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    $\begingroup$ Everything within roughly a mile of the park is leveled, and a firestorm engulfs the surrounding city. So the bridge would certainly collapse. $\endgroup$
    – rob
    Aug 31, 2018 at 8:43
  • $\begingroup$ @rob: What if there were no air resistance? $\endgroup$ Aug 31, 2018 at 8:46
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    $\begingroup$ @user7777777 The linked analysis (firestorm, death, etc.) also applies if the car touches the bridge. If the car doesn't touch the bridge, it's hard to see why it would collapse. ... If the question really is about whether the inertial mass $m$ or the total mass-energy $\gamma m$ determines the "weight" to be supported, you might imagine instead parking a 1000 kg flywheel on the bridge and then spinning it up so that its rim is moving near $c$. $\endgroup$
    – rob
    Aug 31, 2018 at 9:02
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    $\begingroup$ @rob: "Gedanken"-experiments in relativity are idealized and are meant to probe the principles of the theory. Almost all of them, including Einstein's, would collapse, if we were to assume them taking please in a realistic experimental framework. However, I assume that you understand the point that Burke was trying to drive home. $\endgroup$ Aug 31, 2018 at 9:11
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    $\begingroup$ This question really boils down to two things : (1) are gravitational and rest mass the same and (2) is relativistic mass "real". I'd direct you to this question which itself links to a few others. $\endgroup$ Aug 31, 2018 at 12:48

1 Answer 1


Let's take the question seriously, even if as it is phrased it is totally devoid of physical meaning. We can use the Principle of Equivalence as follows. Put yourself in a reference frame in free vertical fall on Earth. Furthermore, assume there is no bridge, and that the car driver relies on its speed to jump over the river.

The question now becomes: how much will the car fall during the jump? In the free-falling frame, the car will not fall at all: it will move horizontally, as in this frame gravity is absent. So, reverting to an earth-bound frame, we can safely assert that the car will have a vertical acceleration equal to $g$ (independent of its mass and speed). Let $s$ be the bridge span: then the car will fall $h=(g s^2)/(2 v^2)$.

Just for fun, let's put some numbers. Speed $v$ is given; as to $s$, the longest suspension bridge on Earth does not reach 2000 m, and the longest one under construction barely exceeds this length. So we may take $s=2000\rm m$. We find $h=2.7\times10^{-10}\rm m$: the size of an atom. The driver would hardly care of such a drop.

Anyhow, to answer the question: if you want to cancel the drop, you will have to apply a vertical force causing an acceleration opposite to gravity. The SR formula relating force to transverse acceleration is $F=m\gamma a$, and the force the bridge must apply to the car to sustain it is $m\gamma g$, greater than the maximum load the bridge can bear.

Would it mean that the bridge would collapse? Not at all! The maximum load is a static one, whereas the stress due to the car would last $s/v=0.74\mu\rm s$. Surely the bridge would not even notice the passage of the car.

Someone could doubt about the above reasoning, arguing that we should get in the full GR machinery. This I can't do here, but the result would be the same.


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