# Pushing with a lorentz contracting stick

If I use a stick to push and accelerate an object, my hand pushes one end of the stick distance $x$, while the other end of the stick pushes the object distance $y$.

Distance $y$ is smaller than distance $x$, because of Lorentz contraction of the stick.

My hand does work $Fx$.

Work $Fy$ is done on the object.

Energy $F \cdot(\text{Lorentz contraction of the stick})$ seems to disappear.

So I'm asking, what happens to the "missing" energy?

EDIT: In this thought experiment pushing causes the object and the stick to accelerate, which causes the stick to Lorentz-contract. In extreme case the length of the stick becomes zero, which means my hand moved a distance of the stick's length kind of unnecessarily. Shorter stick saves energy.

EDIT2: I noticed that "lost" energy approaches zero, when force approaches zero. This suggests the energy loss is linked to deformation of the stick.

EDIT3: This very simple problem may be very difficult to understand, so I ask this way: A good push rod is rigid. Relativity says rigid push rods don't exist. So what kind of energy goes into a push rod, that is as rigid as relativity allows, when we use the push rod, using moderate force, and the speed that the push rod is accelerated to, is relativistic?

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Why would y be less than x? Lorentz contraction just makes the whole stick shorter. Perhaps I'm not understanding the setup properly? – Nathaniel Mar 5 '12 at 18:34
See edit ....... – kartsa Mar 6 '12 at 2:02
Edit2: thats cos $F=d(\gamma mv)/dt$. Recognize the $\gamma$? Same one in length contraction. – Manishearth♦ Mar 6 '12 at 7:59
I see. Interesting question. My intuition says it's $\text{work} = \text{force} \times \text{distance}$ that breaks down at relativistic speeds, but I'll have to think about it. – Nathaniel Mar 9 '12 at 15:08

## 2 Answers

$y=x$

For a constant pushing velocity, lorentz contraction is constant. It's just a smaller, rigid rod, solve classically.

V2:

The missing energy went into accelerating the stick, of course. I'm not sure if you even are allowed to use an accelerating situation in SR.

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See the edit .. – kartsa Mar 6 '12 at 2:02
@kartsa You forgot about energy lost in acceleration. See above. – Manishearth♦ Mar 6 '12 at 3:01
If the stick has mass, then there happens a force decrease in the stick. But I'm interested about the distance part of energy = force * distance – kartsa Mar 6 '12 at 8:04
I'm saying that aside from the difference in forces, you're increasing kinetic energy of the rod. That accounts for a loss of work done. You can't separately conserve types of energy. – Manishearth♦ Mar 6 '12 at 8:07
Is kinetic energy of a rod that was used to push an object to the speed 0.99 c, by using a force of 10 Newtons, larger than kinetic energy of a rod that was used to push an object to the speed 0.99 c by using a force of 5 Newtons?  If not, then the "lost" energy did not turn into kinetic energy of the rod. We agreed that there is a lost energy that is proportional to force, didn't we? – kartsa Mar 6 '12 at 11:09
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As velocity increases, the flow of energy through the stick becomes time dilated, which means there is an increasing amount of energy in the stick.

The type of energy is pressure energy.

When hand stops pushing, the object is still pushed by the other end of the stick.

If the stick is massless, and the pushing force decreases slowly, all energy from the stick goes into the object.

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