Suppose I had two identical cubes, Cube A and Cube B, of volume V moving toward each other with velocity v, say on the x axis, as viewed by a third party observer, C on the x axis as well.

I know that transverse length contraction does not occur as shown in a proof by contradiction given in Morin Ch.11 Problem 1. However, take B's frame; B sees A contracted along the direction of motion (longitudinal contraction). B is somehow able to measure the volume of A.

My question is: what would this volume V' be? Would it still be V? If so, this must mean that with one of the directions which has contracted will cause a transverse direction to expand? Or is V' less than V, and only contraction occurs?

Thank you!

  • $\begingroup$ Are the objects cube when at rest or are they cubes when moving with $v$ as observed by C ? $\endgroup$
    – M111
    Commented Jan 11, 2018 at 20:16
  • $\begingroup$ Objects are cubic when at rest. $\endgroup$
    – Ac711
    Commented Jan 11, 2018 at 22:39

1 Answer 1


The volume observed by B would be less than the volume observed by C. This is precisely because the length along the direction of motion is contracted but in transverse directions it is not.

Assuming in the rest frame A & B are cubes with side length $a$ and volume $V_0$. Then from C's frame (in which they are moving at velocity $\nu$ towards each other) their lengths along the direction of motion would be contracted to $\frac{a}{\gamma}$ where $\gamma = \frac{1}{\sqrt{1 - (\nu /c)^2}}$. The volumes of each object in this frame is $$V = \frac{a^3}{\gamma}$$

In the frame of B, A seems to approach it with velocity $\mu$. Where the addition of velocity formulae gives us a value of $\mu$ as $$\mu = \frac{2\nu}{1 + (\nu /c)^2}$$ With $\mu > \nu$ (classically $\mu = 2 \nu$)

Correspondingly $$\gamma ' = \frac{1}{\sqrt{1 - (\mu /c)^2}} > \gamma$$ In this frame the dimensions of B are $\frac{a}{\gamma'} \times a \times a$ So the volume from A frams is $$V' = \frac{a^3}{\gamma'} < V$$

To get a better feel for of this,

For two cubes of unit volume in their rest frames, moving towards each other with $\nu = 0.6c$ and $\gamma = 1.25$. the volume becomes $V = 0.8$

Velocity of A observed by B is $\mu \approx 0.882c $ The corresponding $\gamma' = 2.12 $ and the volume as observed by A is $V' = 0.47$


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