How Does Relativistic Speed Affect the Force Exerted by a Train on Rails?

[Wee-Lamm]

I came up with this problem, which is non-trivial at least for me:
Train accelerates with constant proper acceleration. When the speed of the train relative to rails is zero, rails feel the train exerting 1000N force on the rails.
What force do rails feel the train exerting on the rails when the speed of the train is 0.87 c, relative to the rails?

Perhaps you asked the question in the wrong order?

I read that you are saying the train weighs 1000N while at rest, and want to know how much it will weigh once it reaches a speed of 0.87c.

If I did translate your question correctly, I suggest the rails would still measure a downward pressure from the train at 1000N.

 

[PeterDonis]

We want to know the force on the rails, so we calculate the momentum change of the rails

I don't think you've done that correctly (I'll respond to that in a separate post), but in any case that wasn't my point. My point was that you are using different assumptions from everyone else. So you can't compare your answer to everyone else's answer; you're comparing apples and oranges.

 

[PeterDonis]

because $\Delta v$ is small, we can say that in this frame rails change velocity by kΔv

No, you can't. The constraint is not that the ratio of velocities is $\gamma$. The constraint is that momentum has to be conserved when calculated in any particular frame. So in the rail frame, the momentum change of the rails must be equal and opposite to the momentum change of the train.

So if the rails have rest mass $M$ and the train has rest mass $m$ (note that it's a good idea not to use the same symbol for both rest masses, as you did), and the rails change velocity by $\Delta V$ in the rail frame (the frame where the rails are at rest before the velocity change), and the train changes velocity by $\Delta v$, then momentum conservation gives $M \Delta V = \gamma' m \left( v + \Delta v \right) - \gamma m v$, where $\gamma$ is the gamma factor for velocity $v$ and $\gamma'$ is the gamma factor for velocity $v + \Delta v$, where $v$ is the train's initial velocity in this frame. So the rails will have final velocity $- \Delta V$ and the train will have final velocity $v + \Delta v$.

Expanding out the momentum change of the train and assuming $\Delta v << v$, we get

$$
m \left( v + \Delta v \right) \left[ 1 - \left( v + \Delta v \right)^2 \right]^{- \frac{1}{2}} - m v \left( 1 - v^2 \right)^{- \frac{1}{2}} = \frac{1}{2} m \left( 2 v^2 \Delta v + v \Delta v^2 + \Delta v + v^2 \Delta v + 2 v \Delta v ^2 + \Delta v^3 \right) \approx \frac{1}{2} m \Delta v \left( 1 + 3 v^2 \right)
$$

So we have $\Delta V = \frac{m}{M} \Delta v \frac{1}{2} \left( 1 + 3 v^2 \right)$.

If we then transform into the train frame (the frame in which the train is at rest before the velocity change), then each of these final velocities gets transformed using the velocity addition formula. This gives, for the train:

$$
\frac{\Delta v}{1 - v \Delta v}
$$

For the rails:

$$
\frac{v + \Delta V}{1 + v \Delta V} = \frac{v + \frac{m}{M} \Delta v \frac{1}{2} \left( 1 + 3 v^2 \right)}{1 + \frac{m}{M} v \Delta v \frac{1}{2} \left( 1 + 3 v^2 \right)}
$$

If we assume that $m << M$ and $\Delta v << 1$, then we are left with a simple change in speed of $\Delta v$ for the train and zero for the rails in both frames. But in any case, the relationship between the two velocity changes is certainly not a simple factor of $\gamma$.

 

[jartsa]

I disagree with the answer of 500 N. I reckon that (surprisingly) the answer is 1000 N -- the force in the rail frame remains constant if we assume the train's mass remains constant.

For a particle undergoing constant proper acceleration $\alpha$ in a straight line, we have, in inertial coordinates in units where $c = 1$,$$\begin{align*}
t &= \frac{1}{\alpha} \sinh \alpha \tau \\
x &= \frac{1}{\alpha} \cosh \alpha \tau \\
\gamma &= dt/d\tau = \cosh \alpha \tau \\
\text{celerity} &= dx/d\tau = \sinh \alpha \tau \\
v &= \frac{dx}{dt} = \frac{dx/d\tau}{dt/d\tau} = \tanh \alpha \tau \\
p &= \gamma m v = m \sinh \alpha \tau \\
f &= \frac{dp}{dt} = \frac{dp/d\tau}{dt/d\tau} = \frac{ m \alpha \cosh \alpha \tau}{ \cosh \alpha \tau} = m \alpha
\end{align*}$$assuming $m$ is constant. (That assumption would be wrong if the train is burning fuel carried by the train.)

That's one of the reasons why I don't like calling $dx/d\tau$ "proper velocity" and would rather use the alternative name "celerity". It's not true that $(d/d\tau) (\text{celerity})$ is proper acceleration, as the above calculation shows. (What is true is that proper acceleration is the magnitude of the 4-force, but for objects not at rest in the coordinate system, the 4-force has a temporal component as well as spatial components.)

That's a force on the train in the rail frame, not a force on the rails in the rail frame.

What is the force on the rails in the rail frame when the train picks up its fuel from the ground? (fuel energy=mc²)

It's 1000N. 500 N is needed for accelerating the train, and 500 N is needed for accelerating the fuel from 0 to 0.87 c. If for a while the train stops picking up fuel, the force on the rails decreases to 500 N in rail frame.

How do I know the force is 1000 N in that case where I claim it's 1000 N?

Well that situation is equivalent to a situation where superman stands on the rails and pushes the train, because in both scenarios train's rest mass does not change and train's kinetic energy changes the same way. And it was calculated by DrGreg that superman exerts 1000 N force on the train and 1000 N force on the rails.

 

[Hornbein]

Am I correct in thinking that y'all are assuming a flat Earth?

 

[jartsa]

Am I correct in thinking that y'all are assuming a flat Earth?

Well I assume a flat Earth.

 

[jartsa]

A hot wheel rolls on a cool surface. The wheel loses heat and mass to the surface. How is momentum conserved in this case?

 

[jartsa]

A hot wheel rolls on a cool surface. The wheel loses heat and mass to the surface. How is momentum conserved in this case?

The less massive wheel rolls faster, that's how momentum is conserved. But the surface feels no force.

So if we reheat the wheel every now and then, without changing its speed, the wheel keeps accelerating while the force on the surface is always zero.

This is maybe somehow related to my original question which concerned force felt by a rail when a train is accelerating. Post #1:

Train accelerates with constant proper acceleration. When the speed of the train relative to rails is zero, rails feel the train exerting 1000N force on the rails.

What force do rails feel the train exerting on the rails when the speed of the train is 0.87 c, relative to the rails?

That question is missing the important information about the energy source of the train. So we can answer: "the force on the rails is zero, with suitable design, ha ha"