Force at relativistic speed

  • #1
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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?
 

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  • #2
PeterDonis
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The quick answer is 500N, i.e., the force felt by the train (which by assumption is constant, since its proper acceleration is constant) divided by the gamma factor corresponding to the train's velocity relative to the rails (which is 2 for a speed of 0.87 c).

The easiest way I know of to understand why this is the answer is to observe that the force felt by the train is equal to the rate of change of the train's momentum in its momentarily comoving inertial frame, i.e., with respect to the train's proper time. In the rest frame of the rails, the train's proper time runs slow by the gamma factor; so a change in the train's momentum that takes 1 unit of the train's proper time will take gamma units of the rails' time. That is, the rate of change with respect to the rails will be the rate of change with respect to the train, divided by gamma.
 
  • #3
HallsofIvy
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But the force the train is exerting on the rail is perpendicular to the direction of motion. I don't see why gamma should come into this.
 
  • #4
PeterDonis
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the force the train is exerting on the rail is perpendicular to the direction of motion
The train's weight is, but that is not how I was interpreting the OP's question. I was interpreting the OP's question as asking about the force that is accelerating the train, which must be in the horizontal direction, i.e., parallel to the rails, since the train is accelerating horizontally.

Of course that raises the obvious question: how, if at all, does the train's weight change as it accelerates? This has been discussed on previous PF threads, but it's been a while; IIRC the answer we arrived at was that, somewhat counterintuitively, the train's weight increases by the factor gamma. I'll see if I can find any of those previous threads.
 
  • #5
Mister T
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But the force the train is exerting on the rail is perpendicular to the direction of motion.
No, it isn't. There's a component of that contact force parallel to the motion. That's the one being discussed because that's the one responsible for the train's acceleration.
 
  • #6
pervect
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I think the problem is pretty straightforward using the concept of proper velocity. My impression from previous discussions is that Jartsa doesn't use this concept. But I'll go through an analysis anyway as I'm getting different answers than some of the other posters to this thread.

We solve the problem by considering the conservation of momentum.

To solve the problem, we introduce the time coordinate t for time in the track frame, and the position coordinate x for position in the track frame. We introduce x mainly to define the proper velocity ##u##, which is the rate of change of proper distance (x) with respect to proper time ##\tau##, ##u = dx/d\tau##. u is one component of the 4-velocity - but for this post I will only talk about the proper velocity and not the 4-velocity. We'll also introduce the non-proper velocity v = dx/dt, though we won't use it very much.

The proper acceleration ##\alpha## is ##\alpha = du/d\tau## by the definition of proper acceleration.

We can write the momentum of the train in the track frame as ##p = m u = \gamma m v##, ##\gamma = 1/\sqrt{1-(v/c)^2}##. The 3-force on the train is dp/dt. We know that in the track frame, the 3-force on the track is equal and opposite to the 3-force on the train, due to the conservation of momentum. Any momentum gained by the train must come from the track, this exchange of momentum is what is meant by the concept of force.

Thus we can write the 3-force f on the track as:
$$
f = \frac{d \,(m u)}{dt} = m \frac{du}{dt} = m \frac{du}{d\tau} \frac{d\tau}{dt} = \frac{m \alpha}{\gamma}
$$
where ##\alpha## is the proper acceleration as we noted previously.

If one wants to find ##u(\tau)## or ##v(t)## for constant proper acceleration, the solution is well known and I won't go through it here unless someone has a question. The well-known solution (see for instance http://math.ucr.edu/home/baez/physics/Relativity/SR/Rocket/rocket.html)is [Broken] that ##u = \sinh \alpha \tau##, which implies ##\gamma(\tau) = \cosh \alpha \tau##.

Note that the sci.physics.faq webpage just gives the solution, it doesn't derive it. My approach to deriving the solution would be based on using 4-velocitie (as in MTW's textbook). I've found through experience that people don't seem to be willing to learn about 4-velocities in order to answer their questions, so I won't bother to go through such a soultion unless someone is actually interested in it.
 
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  • #7
Drakkith
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But the force the train is exerting on the rail is perpendicular to the direction of motion. I don't see why gamma should come into this.
Is it? I thought the force mentioned was the force accelerating the train, which is parallel to the direction of motion, right?
 
  • #8
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Thank you guys. So force felt by rails decreases as gamma increases. In the rail frame that decreasing force is the force that is accelerating the train (or maybe not, as we will find out later).

I guess we are all assuming the composition of the train stays constant. No change of amount of fuel occurs. One way to achieve that is a device that shoots small fuel pellets at the train at speed close to the train's speed. Overhead electric wires I don't consider good for this purpose, it's a bit like aiming a laser beam at the train.

So at very high speed it's the pellet shooter that is doing almost all the work when a train is accelerated.

The pellet shooter also exerts a large reaction force on the ground it's standing on.

I would guess the recoil force is 500N when the speed of the train is 0.87 c.

As rails are fixed on the ground and pellet shooter is fixed on the ground, ground feels a 1000N force when the speed of the train is zero and also when the speed of the train is 0.87 c. So every second the ground gains the same amount of momentum, and the train gains the same amount of opposite momentum.
 
  • #9
PeterDonis
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I guess we are all assuming the composition of the train stays constant.
Yes, that was assumed. One way to realize this would be to put an electrified third rail between the two rails the train runs on, and transfer power to the train that way.

Overhead electric wires I don't consider good for this purpose, it's a bit like aiming a laser beam at the train.
I'm not sure I understand what you mean by this. Overhead wires might have trouble making good contact, but a third rail, as I described above, should work fine.

I would guess the recoil force is 500N when the speed of the train is 0.87 c.
500N is the force felt by the rail because of the train pushing on it in order to accelerate. By Newton's Third Law, 500N is also the force the rail exerts back on the train, evaluated in the rail frame.
 
  • #10
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I'm not sure I understand what you mean by this. Overhead wires might have trouble making good contact, but a third rail, as I described above, should work fine.
I suspect that the energy might push the train. Basically a power plant that delivers energy to a train feels a force away from the train, and the train feels a force away from the power plant, assuming straight power lines. Even if you don't buy this, just to be really sure, let's deliver the energy as pellets. :smile: And let's say pellets are fully converted to energy in the train, so that I can say that it's just pure energy that is shot at the train by a pellet shooter.

500N is the force felt by the rail because of the train pushing on it in order to accelerate. By Newton's Third Law, 500N is also the force the rail exerts back on the train, evaluated in the rail frame.
Ah but it doesn't work like that at all. :smile: At very high speeds trains accelerate by accelerating some energy in the backwards direction. In the train frame what the energy of the fuel does is to increase the kinetic energy of the ground, and the force felt on the train is the result of accelerating said energy from train speed to ground speed, like a rocket.

Seriously. I have thought about it, and something like that seems to be happening.

Increasing the kinetic energy of the ground involves pushing the ground, but the pushing consists mostly of pushing the kinetic energy, ground will not feel any of that part of the pushing.

1 MW photon rocket at speed 0.99 c, in ground frame: Pushes itself with a small force, pushes the ground that the photons hit with a very small force.

1 MW train at speed 0.99 c, in rail frame: Pushes itself with small force, pushes the rails with a very small force.
 
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  • #11
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Note that the sci.physics.faq webpage just gives the solution, it doesn't derive it. My approach to deriving the solution would be based on using 4-velocitie (as in MTW's textbook). I've found through experience that people don't seem to be willing to learn about 4-velocities in order to answer their questions, so I won't bother to go through such a soultion unless someone is actually interested in it.
The simplest self-contained derivation of the constant acceleration equations I have seen can be found here. If you know a shorter way, I for one would be interested to see it.
 
  • #12
DrGreg
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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.)

The proper acceleration [itex]\alpha[/itex] is [itex]\alpha = du/d\tau[/itex] by the definition of proper acceleration.
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.)
 
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  • #13
DrGreg
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The simplest self-contained derivation of the constant acceleration equations I have seen can be found here. If you know a shorter way, I for one would be interested to see it.
Long ago, I once gave a derivation in this thread.

(There's a typo in equation 5: delete the symbol ##v_0##.)
 
  • #14
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Long ago, I once gave a derivation in this thread.

(There's a typo in equation 5: delete the symbol ##v_0##.)
Thanks, still trying to work my way through it . . .

I can't help feeling (units) that the ##dx## term equation 5 should also be divided by c rather than ##c^2##.

Is there another typo re. eqns. 7 & 8 (I suspect you must mean substituting 5 & 7 into 4 at the start of the second post (rather than 2 & 3) but I don't see the pattern of what you are doing here yet).
 
  • #15
PeterDonis
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I reckon that (surprisingly) the answer is 1000 N
Hm, you're right. :wideeyed:

Just to be sure, I'll first write down what the proper acceleration 4-vector looks like. It has components:

$$
\frac{d\gamma}{d\tau} = \alpha \sinh \alpha \tau = \alpha \gamma v
$$
$$
\frac{d\gamma v}{d\tau} = \alpha \cosh \alpha \tau = \alpha \gamma
$$

The magnitude of this vector is clearly ##\alpha##, so it is indeed the proper acceleration. And multiplying it by ##m## gives the 4-force, whose components are ##dE / d\tau## and ##dp / d\tau##. Then dividing by ##\cosh \alpha \tau = \gamma = dt / d\tau## gives ##dE / dt## and ##dp / dt##, the latter of which is the 3-force in the rail frame, as derived by DrGreg.

So where did my reasoning in post #2 go wrong? Well, I said:

the force felt by the train is equal to the rate of change of the train's momentum in its momentarily comoving inertial frame, i.e., with respect to the train's proper time
I then adjusted the "rate of change" part by ##\gamma## to account for the change from the train's proper time to the time in the rail frame. But I forgot to take into account that the train's momentum is also frame-dependent! In the MCIF, the train's momentum after a small impulse has been applied is smaller by a factor of ##\gamma## than it is in the rail frame. So a given impulse applied over a time that is shorter by ##\gamma## gives a result that is smaller by ##\gamma##; the two factors of ##\gamma## cancel, leaving the force the same.

So what does change between the train frame and the rail frame? Obviously what changes is ##dE / dt##, i.e., the rate of change of the train's energy. In the train's MCIF, the 4-force is purely spatial; it doesn't change the train's energy at all (only its momentum). But in the rail frame, as DrGreg pointed out, the 4-force has a temporal component:

$$
\frac{dE}{dt} = m \alpha v
$$

This is a manifestation of what used to be thought of as "relativistic mass"--the force applied to the train, as seen in the rail frame, goes more and more into increasing the train's mass (a better word might be "inertia" here, since it's really "resistance to increase in velocity" that is going up) instead of its velocity, as its velocity gets closer to the speed of light. Another way of seeing this is to calculate the train's 3-acceleration in the rail frame:

$$
\frac{dv}{dt} = \frac{dv / d\tau}{dt / d\tau} = \frac{\alpha / \cosh^2 \alpha \tau}{\cosh \alpha \tau} = \frac{\alpha}{\gamma^3}
$$

This obviously goes to zero as ##v## approaches ##1## (and ##\gamma## increases without bound).

Sorry to beat this to death, but I wanted to make up for having given the wrong answer earlier. :oops:
 
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  • #16
PeterDonis
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I suspect that the energy might push the train.
Of course it does; applying a force to the train takes energy. We have been assuming that the energy is stored in the rails, or something attached to the rails, so that it is at rest in the rail frame. It then gets converted, as seen in the rail frame, to the energy of the train, at the rate ##dE / dt## which I gave in my previous post. Note that ##dE / dt = f v## in the train frame, where ##f## is the 3-force ##dp / dt##; this is exactly what we would expect for the power required from whatever energy source is pushing the train.
 
  • #17
DrGreg
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Thanks, still trying to work my way through it . . .

I can't help feeling (units) that the ##dx## term equation 5 should also be divided by c rather than ##c^2##.
Oops, yes you are right. It's ##(v_0/c)## that should be deleted!

Is there another typo re. eqns. 7 & 8 (I suspect you must mean substituting 5 & 7 into 4 at the start of the second post (rather than 2 & 3) but I don't see the pattern of what you are doing here yet).
Actually, you substitute (2) and (3) into (5) and (6). (4) isn't used at all at this step.

P.S. Just noticed I previously spotted most of these typos in post #28 of that same thread!)
 
  • #18
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Oops, yes you are right. It's ##(v_0/c)## that should be deleted!

Actually, you substitute (2) and (3) into (5) and (6). (4) isn't used at all at this step.

P.S. Just noticed I previously spotted most of these typos in post #28 of that same thread!)
Cheers, I should be OK with the rest now, might go and read that whole thread too ;)
 
  • #19
pervect
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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.)
I can't argue with your derivation, 1000 newtons it is.
 
  • #20
A.T.
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I can't argue with your derivation, 1000 newtons it is.
Aside from the derivations, isn't it there a simple argument for this: The forward force, that a strain gauge would measure at the axle, and the proper acceleration an accelerometer on the train would measure, are both frame invariant and proportional to each other.
 
  • #21
PeterDonis
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The forward force, that a strain gauge would measure at the axle
Is the strain gauge at rest relative to the axle, or relative to the rails?

If it's the former, yes, the force and proper acceleration are obviously proportional (and the constant of proportionality is the rest mass ##m## of the train). But the force that's being asked about is the latter--what a strain gauge at rest relative to the rails would measure. That is a different invariant from the proper acceleration/strain gauge at rest relative to the axle invariant, because the two strain gauges are in different states of motion. DrGreg's derivation shows why these two different invariants have the same numerical value.
 
  • #22
A.T.
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Is the strain gauge at rest relative to the axle, or relative to the rails?
Isn't the force on the track equal but opposite to the force on the train, per Newton's 3rd?
 
  • #23
PeterDonis
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Isn't the force on the track equal but opposite to the force on the train, per Newton's 3rd?
In a given inertial frame, yes. But that in itself doesn't mean the magnitude of the 3-force in one frame has to be the same as the magnitude of the 3-force in a different frame. DrGreg's derivation shows that, in this case, it is.
 
  • #24
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In a given inertial frame, yes. But that in itself doesn't mean the magnitude of the 3-force in one frame has to be the same as the magnitude of the 3-force in a different frame.
If two objects exert equal but opposite forces on each other, don't the instantaneous inertial rest frames of the objects have to agree on the magnitude of those forces? If not, how would you for example determine which of the two frames measures the higher magnitudes, given that all inertial frames are equivalent?
 
  • #25
PeterDonis
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how would you for example determine which of the two frames measures higher magnitudes, given that all inertial frames are equivalent?
The 4-force points in a particular spacelike direction; that direction picks out a preferred frame in which the 4-force is purely spatial (no time component). In DrGreg's derivation, that frame is the MCIF of the train.

"All inertial frames are equivalent" requires that the laws of physics be the same when expressed in all inertial frames. The law in question, Newton's third law, says that the magnitudes of the forces--train on rail and rail on train--have to be equal in the frame in which the law is expressed. That law, in itself, does not require that the magnitudes in one frame be the same as the magnitudes in another; the magnitudes could change when we transformed from one frame to another, as long as they both changed by the same amount, so they were still equal in the new frame.

To infer, just from "all inertial frames are equivalent", that the magnitudes of the forces must be the same in all frames, we would have to show that there is nothing that breaks the symmetry between all frames, in the particular scenario being analyzed. But that is not the case here: as above, the 4-force has to point in some particular spacelike direction, and that breaks the symmetry.

I think the reason that this breaking of the symmetry does not make the magnitudes of the forces in different frames unequal, in this particular case, is that the preferred frame--the train's MCIF--is not fixed; it changes as the train accelerates. So the spacetime direction of the 4-force changes.
 
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