How Does Observer Motion Affect Calculated Work in Relativity?

[ravi#]

I have seen this calculation on web
Let consider that old man displaced the cart from pole A to Pole B on platform. Observers are on platform S & in train S', moving with velocity –V then (let, X-axis is parallel to train direction)
1) When AB (displacement) parallel to the direction of train velocity.
Then, for observer on platform:-
so, Fy = F z = 0 , dx =d(AB)
& Work done W = Fx . dx ----------(1)
for observer on train :-
F’x = Fx – ( v/c² . Fy. Uy)/(1-V .Ux/c²)
F'x= Fx as Fy =0 by force transformation equation.
Here, as one meter in X- direction in S-frame is equal to 1/ γ meter in S’- frame in X-direction
& dx' =d(AB)' = dx/ γ where γ = (1-V²/C²) ^–0.5
So, W' = F'x . dx' = Fx . dx/ γ = W/ γ
So, W' = W/ γ
Case 2 :- When AB perpendicular to velocity of train
for observer on platform :-
Fx = Fz = 0 dy = d(AB) & dx =0
Work done W = Fy. dy
for observer in train :-
F’y = (Fy/ γ) / (1-V .Ux/c²) = Fy/ γ as Ux=0
& dy'=dy as it is perpendicular to V & dx' =dx/ γ = 0
Work done W' = F'y. dy' = (Fy/ γ) . dy = W/ γ
W' = W/ γ
Case 3:-Consider that old man pull the cart on platform from pole A to pole B which is not perpendicular to train velocity in straight line AB.
Fx, Fy, dx, dy are forces & displacement on the platform in X & Y direction then
For observer on platform:-
Work done W = Fx.dx + Fy dy
For observer in train :-
F’x = Fx – ( v/c² . Fy. Uy)/(1-V .Ux/c²)
F’y = (Fy/ γ) / (1-V .Ux/c²) ---------from transformation equation.
W’ = F’x.dx’ + F’y dy’
W’ = {Fx – ( v/c² . Fy. Uy)/(1-V .Ux/c²) } .dx’ + {(Fy/ γ) / (1-V .Ux/c²) } . dy’
Here, as one meter in X- direction in S-frame is equal to 1/ γ meter in S’- frame in X-direction
So, dx’ = dx/ γ & dy’=dy &
If m =(1-V .Ux/c²) then
W’ = (1/[m. γ]) .{Fx .dx-(Fx. V/c² . Ux. dx) - (Fy. v/c² . Uy. dx) + Fy.dy}
W’=(1/[m. γ]) .{Fx .dx(1-V .Ux/c²) + Fy . (dy - v/c² . Uy. dx) }
W’=(1/[m. γ]) .{Fx .dx(1-V .Ux/c²) + Fy . dt (Uy - v/c² . Uy. Ux) }
W’=(1/[m. γ]) .{Fx .dx.(1-V .Ux/c²) + Fy . dt .Uy. (1-V .Ux/c²) }
W’=(m/[m. γ]) .{Fx .dx+ Fy . dt .Uy}
W’=(m/[m.r]) .{Fx .dx+ Fy . dy}
W’=1/ γ.{Fx .dx+ Fy . dy} = 1/ γ . W
Or W’= W/ γ
This clear shows that in all cases W' = W/ γ
So, you call generally that in all cases W' = W/y
Means, energy consumed in doing work decreases by increasing frame velocity

I have check the mathematics. It is not wrong but final result is against S.R. as energy consumed decreases as frame velocity increases.
What is wrong in above calculation?

 

[A.T.]

Means, energy consumed in doing work decreases by increasing frame velocity

Work is frame dependent even in classical mechanics.

 

[ravi#]

Yes, I know but work done or energy consumed in doing work should increase as frame velocity increases as E=y.Eo.
but in above calculation it decreases.

 

[Ibix]

Your use of ##dx=\gamma dx'## implies that the cart is simultaneously at A and B in the primed frame. This is not consistent with your problem statement (and not plausible). You need to use the full Lorentz transform. I get that, in general, ##W=\gamma (W_0-F_xvT)##. If I understand your concern correctly, I think that answers it.

If you wish to post more maths, please use LaTeX. It is very difficult to read maths posted in plain text.

Edit: Apologies, forgot to define ##T=dx/u_x=dy/u_y##.

 

[ravi#]

Your transformation equation will not work here as old man applying force & consuming energy will remain on platform & only observer are changing
I will give example.
For Observer on platform :-let, displacement is in X-direction only.
force applied by old man Fx = 1 N
displacement dx = 1 m & time require dt = 1 second
then work done dW= 1N x 1 m = 1 joule
Means, old man will consumed 1 joule of energy for this work done for observer on platform & he will look tired.
For train rider :- As per your transformation equation
W=y.(Wo - Fx.V.T)
if train velocity V = 1 m/s
Work done W = y. (1-1 x 1 x 1) =0
mean, according to you, for observer in train, energy consumed by old man on platform is zero & he will look fresh.
This is only true when old man will be in train & applying force on platform then only you can add train velocity in this calculation.

 

[Ibix]

You are ignoring the force the man is exerting on the platform. In other words, your problem description is incomplete, so it is hardly surprising that you get incomplete answers.

 

[Ibix]

I suspect my last isn't quite correct. Mechanics is a pain and I need pen, paper, and more time than I've got right now to be certain I'm visualising this correctly.

In the mean time, it may be instructive for you to consider this: your train perspective is analogous to someone on the ground watching someone running on a treadmill. Clearly they make no progress (so F.dx=0) but they are clearly burning energy to do so.

 

[jartsa]

In train frame:

Work done by old man is force times the distance between the poles, where distance length-contracts, so work transforms as 1/gamma.

Edit:
The energy in the old man available for pushing carts must transform as 1/gamma, to avoid a situation where the old man starts to feel super-energetic, which would break the principle of relativity.

Edit2:
Another way to calculate, in train frame:

Power that goes into the man from the platform: speed of the platform * force

Power that comes out of the man and goes into the cart: speed of the man * force

Power produced by the man: difference of power in and power out, which is proportional to the speed difference of the man and the platform, and the speed difference transforms as 1/gamma², or something like that.

Energy produced by the man: power * time, where power transforms as I guessed above, and time transforms as gamma.
This is more or less irrelevant:

In train frame and assuming a slow speed of cart:

Work done by platform is power times time, where time dilates, and power is force times speed of the platform, so work transforms as speed*gamma.

 

[Ibix]

@jartsa - remember that the poles are moving in the train frame. The separation between the poles is ##\gamma dx##, but they do some or all of the traveling in anything except the platform frame.

@ravi# - I think I'm correct in #6. In the case where the train is traveling at the same speed as the cart the force exerted on the cart does no work. But the man is also exerting a force on the floor, and the soles of his feet are moving in this frame. This force does work in this case. The apparent problem you pointed out is due to neglecting this force.

 

[jartsa]

This would be much more intuitive if a cart was pushed from the rear of an airplane to the front of the plane by an air hostess, while the plane travels 10 km ahead.

Total work done to the cart: force * (10 km + plane length)
Work done to the cart by the plane: force * 10 km
Work done to the cart by the air hostess: force * plane length

Right?

At higher flying speed the work done by the plane is larger, because the distance traveled by the plane during the pushing is longer. And the work done by the stewardess is smaller because the plane is shorter.

Right?

 

[Ibix]

I think you are mixing frames. Replace the stewardess with a light clock, and try to analyse the light clock from the premise that the forward going light pulse travels a shorter distance in the frame where the plane is moving than the one where it's stationary.

I agree that there is a powerful intuition that you ought to be able to treat the situation the way you want to. If I stand on the platform and watch someone bouncing a ball on the floor of a train carriage, I see it bouncing vertically whether the train is moving or not. But this is my brain cheating me - it's using the train frame to describe things in the train and the platform frame to describe things outside it. It won't do for formal analysis, and that goes double when you go beyond Newton.

The formal analysis is that the force the stewardess applies to the trolley does more work the faster the plane travels becausetthe distance it moves through increases due to the motion of the plane. At the same time, the work done by the force between the stewardess' shoes and the floor increases for the same reason. So the stewardess has to work no harder on the moving plane.

I haven't done that analysis in full relativistic terms. There may be factors of gamma or similar in the work done by the stewardess - but in that case they must also be present in the transformed assessment of her available energy. You are welcome to try the analysis...

 

[A.T.]

Total work done to the cart: force * (10 km + plane length)
Work done to the cart by the plane: force * 10 km
Work done to the cart by the air hostess: force * plane length

Right?

Not right.

If we assume the cart has no rolling friction, the plane does no work on the cart.

If you want to include rolling friction, then you need to name that force differently than the push of the stewardess, and its also in the opposite direction (negative work).

 

[jartsa]

I think you are mixing frames. Replace the stewardess with a light clock, and try to analyse the light clock from the premise that the forward going light pulse travels a shorter distance in the frame where the plane is moving than the one where it's stationary.

I agree that there is a powerful intuition that you ought to be able to treat the situation the way you want to. If I stand on the platform and watch someone bouncing a ball on the floor of a train carriage, I see it bouncing vertically whether the train is moving or not. But this is my brain cheating me - it's using the train frame to describe things in the train and the platform frame to describe things outside it. It won't do for formal analysis, and that goes double when you go beyond Newton.

The formal analysis is that the force the stewardess applies to the trolley does more work the faster the plane travels becausetthe distance it moves through increases due to the motion of the plane. At the same time, the work done by the force between the stewardess' shoes and the floor increases for the same reason. So the stewardess has to work no harder on the moving plane.

I haven't done that analysis in full relativistic terms. There may be factors of gamma or similar in the work done by the stewardess - but in that case they must also be present in the transformed assessment of her available energy. You are welcome to try the analysis...

I can say for sure I did not make that kind of error.

But let's ask what is, in ground frame, the work done by an airplane passenger that crawls a distance d from his seat to the aisle, using proper force F.

Using a force transformation formula and E=F*d we get:
E = F / gamma * d

 

[ravi#]

According to me Work done in any frame = force in that frame x (total displacement - displacement when force is not applied)
& displacement when force is not applied is inertial displacements of the frames in above event that should be avoided from calculation.
So, resultant displacement will be distance between A & B poles of the platform in between which old man displaced the cart on platform.
So, according to me, calculation given in post 1 is not wrong.
For train rider length of platform & all displacement on it in X-direction will get contracted.

 

[A.T.]

According to me Work done in any frame = force in that frame x (total displacement - displacement when force is not applied)

Why not just force x displacement ?

 

[Dale]

According to me Work done in any frame = force in that frame x (total displacement - displacement when force is not applied)

Work is just force times displacement. The minus sign is not necessary, nor is it even well defined.

Are you familiar with four-vectors?

 

[ravi#]

Displacement is the vector & vector can be addition & subtraction of two vectors.
I fully agree with A.T.
For train rider:-When displacement is parallel to X-axis. Observer see that old man displaced the cart from pole A to pole B on platform & he see that d(AB) get contracted but force remain same (as per transformation equation)
work done by old man = F'x . d(AB)'= Fx. dx/y = W/y
Mean, calculation given in post 1 is not wrong.
We can not add frame displacement in calculation. (If it is done old man can become superman if frame velocity is very high)

 

[Dale]

We can not add frame displacement in calculation. (If it is done old man can become superman if frame velocity is very high)

Not only can you include the frame displacement, you must. If it is not done then energy is not conserved. This is far worse than the old man outputting a lot of energy.

 

[ravi#]

Then work done by old man on the platform observered by observer in train will be totally different & depend on velocity of train
for example
Case 1:-For Observer on platform :-let, displacement is in X-direction only.
force applied by old man Fx = 1 N
displacement dx = 1 m & time require dt = 1 second
then work done dW= 1N x 1 m = 1 joule
Means, old man will consumed 1 joule of energy for this work done for observer on platform & he will look tired.
Case 2:-For train rider :- Work done by old man on platform will be
As per your transformation equation W=y.(Wo - Fx.V.T)
if train velocity V = 1 m/s Work done W = y. (1-1 x 1 x 1) =0 joule
old man will not be tired for observer in train.
Case 3:- For train rider :- As per your transformation equation W=y.(Wo - Fx.V.T)
if train velocity V = -10 m/s Work done W = y. (1+1 x 10 x 1) =11 y joule
old man will be much tire for observer in train.
Now, this work done is used to glow light on platform then for observer on train
Case 1:-for rest train observer, light luminous is deem as work done is 1 joule.
Case 2:-for train rider with velocity V= 1m/s, there is no light as work done 0 joule.
Case 3:-for train rider with velocity V= -10m/s, light luminous is more as work done is 11 joule.

This is WRONG. By changing frame velocity or observer velocity, we can not consumed or produce more energy.

 

[A.T.]

Case 2:-For train rider :- Work done by old man on platform will be
As per your transformation equation W=y.(Wo - Fx.V.T)
if train velocity V = 1 m/s Work done W = y. (1-1 x 1 x 1) =0 joule
old man will not be tired for observer in train.

You forgot the work the man does on the ground, which is moving in the train-frame.

 

[Ibix]

You forgot the work the man does on the ground, which is moving in the train-frame.

...as I pointed out in #6.

 

[Dale]

@ravi# I agree with both A.T. and Ibix who have identified where the error is. One thing that I would like to add is that this is not a problem limited to relativity. You can make the same mistake and resolve it the same way using Newtonian physics.

 

[ravi#]

This is not the problem because man can not displaced the cart on friction less surface in any case. In all cases man apply the force on the ground ultimately through him to displace the cart. So, in the calculation of work done, force applied by old man on cart or force transmitted by him to ground both are same.
(This is similar to force on one end of string get transmitted to other end)
So, force is not the problem, problem is displacement.
Can we add frame displacement in work done calculation of old man? I think not & calculation given in post 1 is true.
(because work is happen on platform & just observed by different observers. So, this frame velocity have not any contribution for work done on platform. If it is added we get completely wrong results)
Work done by old man for train rider W' = F'x . d(AB)' = Fx . d(AB)/y

So calculation in post 1 is true but ...that will create further problem. I want to discuss that problems after this issue of displacement is settle down.

 

[ravi#]

Let consider that old man displaced the cart from pole A to Pole B on platform. Observers are on platform S & in train S', moving with velocity –V then
(let, X-axis is parallel to train direction)
Case 1 :- When AB perpendicular to velocity of train
for observer on platform :-
Fx = Fz = 0, dy = d(AB), & dx =0
Work done W = Fy. dy
for observer in train :-
F’y = (Fy/ γ) / (1-V .Ux/c2) = Fy/ γ as Ux=0 ( as per transformation of force)
& dy'=dy as it is perpendicular to V & dx' =dx/ γ = 0
Work done W' = F'y. dy' = (Fy/ γ) . dy = W/ γ
W' = W/ γ
This shows that in that case, even displacement is perpendicular to frame velocity. Work done or energy consumed decreases as frame velocity increases

 

[Ibix]

Here's an idea: listen to the advice you keep getting here from A.T., DaleSpam and me.

Transform the displacements properly. Take into account both the force between the man and the cart and the man and the platform. See what you get.

@jartsa - I see what you were getting at now. Apologies. However, I think your argument only works for the case where the force is parallel to the relative motion of the frames.

 

[Ibix]

Apologies for the tone of the above - it's grumpier than it needs to be. However, the fact remains that you are ignoring the advice of three people wearing mentor and science advisor badges. Certainly we make mistakes (at least I do) but I would suggest that three of us telling you the same thing is a strong hint that we're on to something.

You seem to be relying on an incorrect intuition that one can simply subtract out the frame motion. You can get away with that in Newtonian mechanics (although I won't guarantee it always works) because of the simple functional form of the Galilean transforms, but not in relativity, not in general. You need to work this out formally - then we can discuss the implications.

 

[Dale]

This is not the problem because man can not displaced the cart on friction less surface in any case. In all cases man apply the force on the ground ultimately through him to displace the cart.

That is not correct. You can certainly have the cart on a frictionless surface and the man right next to it on a surface with friction or pushing against a wall or something. The problem is not friction.

The problem (which you seem reluctant to acknowledge) is that the system you are describing is not isolated. The man+cart system interacts with the ground. In any frame where the ground is moving the interaction with the ground does work and exchanges energy. Failure to account for that work and energy leads to the problem you are facing.

So, in the calculation of work done, force applied by old man on cart or force transmitted by him to ground both are same.

Assuming that the man does not accelerate, yes. However, that is not the crucial thing. The crucial thing is that the net force of the man+cart system on the ground is non-zero. This means that the man+cart system interacts with the ground and work and energy can be transferred to the ground.

 

[ravi#]

I want to ask three questions. In each cases in above event
When force & displacement AB is perpendicular to frame velocity
1) Who is applying the force?
Answer:- definitely old man applying the force.
for observer on platform :- force = Fy
for observer in train :- force = Fy/γ (detail calculation is given)
2) What is forced displacement?
Answer :-For both observer displacement = dy
3) What is work done ?
Answer:-
for observer on platform :- Work done W = Fy .dy
for observer in train :- Work done W'= Fy/γ . dy = W/y

When force & displacement AB is parallel to frame velocity
1) Who is applying the force?
Answer:- definitely old man applying the force.
for observer on platform :- force = Fx
for observer in train :- force = Fx
2) What is forced displacement?
Answer :-
for observer on platform :- displacement = dx
for observer in train :- displacement = dx/γ
3) What is work done ?
Answer:-
for observer on platform :- Work done W = Fx .dx
for observer in train :- Work done W'= Fx . dx/γ = W/yThis all calculation proves that work done decreases when frame velocity increases. Mean's energy consumed by old man decreases when frame velocity increases.
This is against SR.
Consumed energy should increase with more frame velocity (not decrease).

 

[jartsa]

Consumed energy should increase with more frame velocity (not decrease).

What is the problem if consumed energy decreases? Relativistic people become overweight because working is too easy for them? Well yes that is a problem, which can be solved by noting that also contained energy decreases with more frame velocity.

But you say consumed energy should increase. Why?

 

[jartsa]

Let's say we have a particle gun that pushes a particle for time one second with force one Newton, and gives the particle speed 0.99 c. Gamma is seven at speed 0.99 c.

Then the particle enters slowly into second identical particle gun, which is moving almost at the same speed as the particle. The second gun does the same thing as the first one, just taking 14 seconds instead of one, because of time dilation. (That 14 seconds is an approximation, I can't calculate the exact number )

Ok so I try to calculate:

c = 1

First gun's effect on particle:
momentum: 1 s * 1 N =1 Ns
energy: momentum*speed=1Ns*0.99=0.99Ns

Second gun's effect on particle:
momentum: 14 s * 1 N = 14 Ns
energy: momentum*speed=14 Ns * about 1 = about 14Ns

Second gun's kinetic energy:
change of kinetic energy: change of momentum*speed=14 Ns * 0.99 = 13.86 Ns

You see, almost all energy that the second gun gave to the particle came from the kinetic energy of the gun. The gun used only 0.14 Ns non-kinetic energy to push the particle, about 1/7 of what the first gun used.

 

[Ibix]

I want to ask three questions. In each cases in above event
When force & displacement AB is perpendicular to frame velocity
1) Who is applying the force?
Answer:- definitely old man applying the force.

Which force? There are two on your man - the reaction from the platform and the reaction from the cart. You don't include both in your analysis so your calculations are not correct in general.

for observer on platform :- force = Fy
for observer in train :- force = Fy/γ (detail calculation is given)
2) What is forced displacement?
Answer :-For both observer displacement = dy

This is not correct for the train frame because you are neglecting the motion of the poles in this frame.

3) What is work done ?
Answer:-
for observer on platform :- Work done W = Fy .dy
for observer in train :- Work done W'= Fy/γ . dy = W/y

This is not correct in the train frame because you are not including both the forces acting on the man, and you are neglecting the motion of the poles. Do you know how to use the Lorentz transforms to do this correctly?

When force & displacement AB is parallel to frame velocity
1) Who is applying the force?
Answer:- definitely old man applying the force.
for observer on platform :- force = Fx
for observer in train :- force = Fx
2) What is forced displacement?
Answer :-
for observer on platform :- displacement = dx
for observer in train :- displacement = dx/γ
3) What is work done ?
Answer:-
for observer on platform :- Work done W = Fx .dx
for observer in train :- Work done W'= Fx . dx/γ = W/y

This analysis is not correct, for the same reasons as your first. In this case I believe that your errors cancel out, but that's really just luck.

This all calculation proves that work done decreases when frame velocity increases. Mean's energy consumed by old man decreases when frame velocity increases.
This is against SR.
Consumed energy should increase with more frame velocity (not decrease).

The statement you underlined and keep repeating does not appear to be generally true. Can you provide a reference for it?

 

[Dale]

This issue is not a relativistic issue so I will just do the Newtonian calculations. The relativistic ones are more complicated, but the same approach works

When force & displacement AB is parallel to frame velocity
1) Who is applying the force?
Answer:- definitely old man applying the force.
for observer on platform :- force = Fx
for observer in train :- force = Fx

Which force? Assuming that there is no friction between the ground and the cart then there are four forces of interest, the force of the man on the ground, the force of the ground on the man, the force of the man on the cart, and the force of the cart on the man.

Let's focus on the two forces exerted by the man, since the others are the equal and opposite reaction forces. Let's call ##F_g## the force that the man exerts on the ground and ##F_c## the force that the man exerts on the car.

2) What is forced displacement?
Answer :-
for observer on platform :- displacement = dx
for observer in train :- displacement = dx/γ

In the ground frame, for ##F_g## the displacement is 0 and for ##F_c## the displacement is ##\Delta x##

In the train frame, for ##F_g## the displacement is ##v \Delta t## where ##v## is the speed of the train, and for ##F_c## the displacement is ##v \Delta t + \Delta x##.

3) What is work done ?
Answer:-
for observer on platform :- Work done W = Fx .dx
for observer in train :- Work done W'= Fx . dx/γ = W/y

In the ground frame the work done by ##F_g## is ##F_g \cdot 0##, and the work done by ##F_c## is ##F_c \cdot \Delta x##.

In the train frame the work done by ##F_g## is ##F_g \cdot (v \Delta t)##, and the work done by ##F_c## is ##F_c \cdot (v \Delta t + \Delta x)##.

If the center of mass of the man does not accelerate, then ##F_g = -F_c## so that the net work done by the man is ## F_c \cdot \Delta x + F_g \cdot 0 = F_c \cdot \Delta x## in the ground frame and ##F_c \cdot (v \Delta t + \Delta x) + F_g \cdot (v \Delta t) = F_c \cdot (v \Delta t + \Delta x) - F_c \cdot (v \Delta t) = F_c \cdot \Delta x ## in the train frame. The second term in this is the reason that you cannot ignore the interaction with the ground.

 

[jartsa]

This is almost the same post as post #10, just the one error corrected, and the last sentence added.

Let's say a kid in a spacecraft throws a ball towards the front of the craft. During the motion of the kid's hand the spacecraft travels 1 light year. And the kid's hand travels 1 light year + 0.001 m. It's a very Lorentz contracted hand, you see.

Total work done to the ball: F * (1 light year + 0.001 m)
Energy lost by the spacecraft because of the throwing: F * (1 light year)
Energy lost by the kid because of the throwing must be: F * 0.001 m (very small)

I'm quite sure this is what ravi# means.

 

[ravi#]

Honorable Dale, Fg is the force applied by old man on the ground by foot. Remember that he is not standing at one place. He is pushing the cart & applying force Fg on ground & move on platform ground. So, Work done by Fg = Fg. dx & dx is not zero but this Fg effect get nullify. I explain it below.
You are true there are three forces. force is applied by old man on cart is Fx, force applied by old man on ground = -Fx & force applied by ground on old man = Fx
Resultant force = Fx-Fx+Fx =Fx=force applied by old man on cart.
When you apply force on ground, ground also apply the same force in reverse direction. So, ultimate force is null. So, remaining force is only force applied by old man on cart.
So, in this work done calculation force applied by old man on cart is only consider & reaction by ground is ignore as generally done in physics.
Ask one question, Who is losing energy?
Answer:- In this total event only old man is losing energy. Means' he is only doing work.

 

[A.T.]

When you apply force on ground, ground also apply the same force in reverse direction.

The same is true for the force on the cart.

So, ultimate force is null.

What is "ultimate force"? If you mean "net force" then you have some basic misconceptions about Newtonian mechanics:
https://www.lhup.edu/~dsimanek/physics/horsecart.htm