Force at relativistic speed

  • #26
DrGreg
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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.)
A few comments on this
  1. The constancy of the proper acceleration was hardly used. My proof still applies if you delete the first two equations for ##x## and ##t##, replace ##\alpha \tau## by ##\varphi## (rapidity) throughout, and observe that ##d\varphi/d\tau## is the proper acceleration ##\alpha##, an invariant. What this shows is that for a particle of constant mass accelerating in a straight line, all inertial frames agree on the magnitude of the 3-force acting on the particle (even if the force varies over time).
  2. It seems to be "just coincidence" that this is so; there doesn't seem to be any obvious reason for it, it's just something that drops out of the maths.
  3. The result may well be false if the mass isn't constant or if the motion isn't along a straight line.
  4. I can't claim credit for this result; I've seen it before (but it was so long ago that I can't remember if my method of proof was used).
 
  • #27
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A few comments on this
  1. The constancy of the proper acceleration was hardly used. My proof still applies if you delete the first two equations for ##x## and ##t##, replace ##\alpha \tau## by ##\varphi## (rapidity) throughout, and observe that ##d\varphi/d\tau## is the proper acceleration ##\alpha##, an invariant. What this shows is that for a particle of constant mass accelerating in a straight line, all inertial frames agree on the magnitude of the 3-force acting on the particle (even if the force varies over time).
  2. It seems to be "just coincidence" that this is so; there doesn't seem to be any obvious reason for it, it's just something that drops out of the maths.
  3. The result may well be false if the mass isn't constant or if the motion isn't along a straight line.
  4. I can't claim credit for this result; I've seen it before (but it was so long ago that I can't remember if my method of proof was used).
Oh good. From the book "Basic Relativity" by Richard A Mould (Chapter 8 section 8.2)
".....at every instant of time the force acting on the particle in the S' system will be f' = f......."
I have been strugglling for a long time trying to find out how the author got that.
Thanks.
 
  • #28
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In the spirit of the first post, let's say a fully tanked train starts accelerating from zero speed, train's proper acceleration is a, and force on the rails is F in the rail frame.

Later the train is coasting at constant speed 0.87 c, and somehow the train has been tanked full again. Now the train again starts accelerating with proper acceleration a, and we ask what is the force on the rails in rail frame.

The right answer is F/gamma, that's my opinion, I don't know if I could somehow calculate that answer or not.

A passenger feels the same force pushing her at both speeds. Let us transform that force to rail frame: It's just the same force in the rail frame. So the whole train is pushed by force F at both speeds, in both frames. Correction: almost the whole train is pushed by force F, the fuel that is burned is a somewhat problematic thing in this regard.
 
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  • #29
DrGreg
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In the spirit of the first post, let's say a fully tanked train starts accelerating from zero speed, train's proper acceleration is a, and force on the rails is F in the rail frame.

Later the train is coasting at constant speed 0.87 c, and somehow the train has been tanked full again. Now the train again starts accelerating with proper acceleration a, and we ask what is the force on the rails in rail frame.

The right answer is F/gamma, that's my opinion, I don't know if I could somehow calculate that answer or not.

A passenger feels the same force pushing her at both speeds. Let us transform that force to rail frame: It's just the same force in the rail frame. So the whole train is pushed by force F at both speeds, in both frames. Correction: almost the whole train is pushed by force F, the fuel that is burned is a somewhat problematic thing in this regard.
If the fuel consumption is negligibly small, my analysis still applies and the answer is ##F##.

To include the fuel consumption, a slight modification to my method gives an answer $$
f = m \alpha + \frac{dm}{d\tau} \tanh \alpha \tau = m \alpha + v \frac{dm}{d\tau} = m \alpha + \gamma v \frac{dm}{dt}
$$
Initially, when ##v=0##, ##f=m\alpha=F##.
 
  • #30
PeroK
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Oh good. From the book "Basic Relativity" by Richard A Mould (Chapter 8 section 8.2)
".....at every instant of time the force acting on the particle in the S' system will be f' = f......."
I have been strugglling for a long time trying to find out how the author got that.
Thanks.
You can also crank this out from the velocity and acceleration transformations for motion in 1D:

##v = \frac{v'+ V}{1 + Vv'/c^2}##

##a = \frac{a'}{\gamma^3 (1 + Vv'/c^2)^3}##

Where the unprimed frame is the ground frame and the primed frame is a frame moving to the right at velocity ##V##. And if we let the primed frame be the frame where the train is instantaneously at rest (##V = v, \ v' = 0##):

##a = \frac{a'}{\gamma^3}##

The force in the ground frame is:

##F = \frac{dp}{dt} = \gamma^3 ma = ma' = F'##

That's for any force ##F## or ##F'## and, in particular, where there is constant proper acceleration.
 
  • #31
PeroK
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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?
In general, for two dimensional motion, the forces on a particle transform according to:

##F'_x = \frac{F_x - V/c^2(\vec{v}.\vec{F})}{1 - v_xV/c^2}, \ F'_y = \frac{F_y}{\gamma(1 - v_xV/c^2)}##

And, if we take the primed frame to be the frame in which the particle is instantaneously at rest, the velocity to be ##(v, 0, 0)## and the force to be ##F = (F_x, F_y, 0)##, then:

##F'_x = \frac{F_x(1 - v^2/c^2)}{1 - v^2/c^2} = F_x, \ F'_y = \frac{F_y}{\gamma(1 - v^2/c^2)} = \gamma F_y##

The particle measures a larger force in the y-direction in its frame than is measured in the ground frame.
 
  • #32
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Let's study what effect a short acceleration of train has on the momentum of rails which have rest mass m:

Train frame : in time t, rails gain momentum: gamma m delta v

Rail frame: in time gamma t, rails gain momentum: m delta v

Force on rails in the rail frame is quarter of the force on rails in the train frame, when gamma is 2.

Edit: delta v refers to velocity change of rails. Oh but delta v is not the same in both frames ... well that brings the "quarter" back to my previous guess "half" , I hope. :smile:

Oh and "train frame" means the frame where the train was at rest when the acceleration started. And "rail frame" means the frame where the rails were at rest when the acceleration started.
 
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  • #33
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A more rigorous version of the previous post, this may be as rigorous as I can get: :wink:

Let's calculate momentum change of rails that have rest mass ##m## and undergo a small velocity change ##\Delta v##, in rail frame, when a train accelerates for a short time ##t##, in train frame.

I will use ##k = \frac{1}{\gamma}##

Frame where train was at rest originally, or "train frame":
because ##\Delta v## is small, we can say that in this frame rails change velocity by ## k \Delta v##
$$ \Delta p = \gamma m * k \Delta v $$


Frame where rails are at rest originally, or "rail frame":
$$ \Delta p' = m \Delta v $$


Now let us calculate forces:

Frame where train was at rest originally, or "train frame":
$$ F=\frac{ \Delta p}{t} = \frac{\gamma m * k \Delta v}{t} = \frac{m \Delta v}{t}$$

Frame where rails were at rest originally, or "rail frame":
$$ F'=\frac{ \Delta p'}{\gamma t} = \frac{ m \Delta v}{\gamma t} $$

Comparing the forces in the two frames:
$$F/F' = \gamma $$
 
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  • #34
PeterDonis
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Let's study what effect a short acceleration of train has on the momentum of rails
This seems to me to be a different version of the scenario than everyone else is analyzing. DrGreg's analysis implicitly assumes that the rails are at rest in a fixed inertial frame. That implies either that the rails have infinite rest mass (or at least a rest mass so much larger than the train's that the total impulse added by the train's acceleration does not produce any appreciable change in velocity--this would be the case if the rails were attached to a planet the size of Earth, for example), or that some other force is being exerted on the rails that cancels the reaction force of the train on the rails (for example, the rails could be attached to a rocket whose engine is providing the force).
 
  • #35
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This seems to me to be a different version of the scenario than everyone else is analyzing. DrGreg's analysis implicitly assumes that the rails are at rest in a fixed inertial frame. That implies either that the rails have infinite rest mass (or at least a rest mass so much larger than the train's that the total impulse added by the train's acceleration does not produce any appreciable change in velocity--this would be the case if the rails were attached to a planet the size of Earth, for example), or that some other force is being exerted on the rails that cancels the reaction force of the train on the rails (for example, the rails could be attached to a rocket whose engine is providing the force).

I managed to avoid any calculation of the momentum of the train. Maybe that was the key to not making the same error as everybody else. :smile:

We want to know the force on the rails, so we calculate the momentum change of the rails, which rails are really massive in my scenario.
 
  • #36
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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?
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.
 
  • #37
PeterDonis
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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.
 
  • #38
PeterDonis
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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##.
 
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  • #39
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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.)

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.
 
  • #40
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Am I correct in thinking that y'all are assuming a flat Earth?
 
  • #41
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Am I correct in thinking that y'all are assuming a flat Earth?
Well I assume a flat Earth.
 
  • #42
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A hot wheel rolls on a cool surface. The wheel loses heat and mass to the surface. How is momentum conserved in this case?
 
  • #43
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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"
 

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