I How Does Observer Motion Affect Calculated Work in Relativity?

[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

 

[Dale]

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

This is a slightly different scenario from what I had in mind, but your analysis is not correct.

The displacement at the contact patch for $F_g$ is always 0 (in the ground frame) under the usual assumptions that the ground is rigid and his feet are not slipping. So in the ground frame the work done by $F_g$ is $F_g\cdot 0$. It is not the displacement of the center of mass which is important, it is the displacement of the contact patch. That is 0 for $F_g$

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.

This is a very common error for beginning physics students. The force applied by the man on the ground and the force applied by the ground on the man act on two different objects. You never add them. They cannot be added to get a resultant force because they don't act on the same object. This is a common misunderstanding of Newton's 3 rd law, but all of the subsequent reasoning is wrong.

Ask one question, Who is losing energy?
Answer:- In this total event only old man is losing energy. Means' he is only doing work.

That is true in the ground frame. In other frames the ground could be losing energy, or even the cart could lose energy.

 

[ravi#]

Honorable Dale if you are right then
In frame of train observer, person in resting room on platform, slipping back on his seat is also doing work because his body is applying force Fx on seat in back ward direction & there is displacement V.dt.
Can you say that train rider observered that this slipping man in rest room is loosing his energy & look tired because he is doing work Fx. (V.dt) with relative to it.
Honorable Dale, there are infinite number of balance forces present on platform frame. They are not doing any work in any inertial frame & loose energy in any frame. Other wise whole system dies down automatically.

Only one unbalance force is present on platform , that is given below
Just consider only cart, Force applied on it in horizontal direction is force applied by old man & that unbalance force create acceleration, displacement on platform.
This force is only responsible for change in state of motion of body(cart) on platform.
So, I think in this complete event the only force applied by old man on cart has to be consider . That force is only doing work & making old man tired.

 

[A.T.]

there are infinite number of balance forces present on platform frame. They are not doing any work in any inertial frame

Balanced forces can do work.

 

[ravi#]

Can you say that
Slipping man in rest room is loosing his energy with relative to train rider & look tired because he is doing work Fx. (V.dt) with relative to it. where Fx is force applied by him on the back of chair.

 

[ravi#]

This can be done in following way also
Honorable A.T. you are true, cart apply the same force on old man but in this event there are two actors
One is old man:- on this old man, two forces are acting in horizontal direction -Fx is force by cart & Fx is the force by ground.
so, resultant force in X-direction = Fx-Fx = 0
On Cart :- There is only one force is acting i.e. Force applied by old man
so, resultant force in X-direction = Fx
As this force is unbalance, this do work on cart & displace it in the direction of force.
One thing must remember that old man is doing work. No one other is doing work on old man.
(It is not necessary that old man move on platform, he can pull the cart by rope also.)

 

[A.T.]

Slipping man in rest room is loosing his energy with relative to train rider

Is that rest room on the train?

where Fx is force applied by him on the back of chair.

Why is there a horizontal force? Is the train accelerating?

 

[Dale]

Honorable Dale if you are right then
In frame of train observer, person in resting room on platform, slipping back on his seat is also doing work because his body is applying force Fx on seat in back ward direction & there is displacement V.dt.

Yes. That is how work is defined.

Can you say that train rider observered that this slipping man in rest room is loosing his energy & look tired because he is doing work Fx. (V.dt) with relative to it.

Draw a free body diagram of the man and identify all of the forces acting. For each force calculate the displacement. Then determine the NET work.

Honorable Dale, there are infinite number of balance forces present on platform frame. They are not doing any work in any inertial frame & loose energy in any frame. Other wise whole system dies down automatically.

You should actually calculate it sometime and see. Your intuition is wrong here. This is why it is important to do homework problems.

Just consider only cart, Force applied on it in horizontal direction is force applied by old man & that unbalance force create acceleration, displacement on platform.
This force is only responsible for change in state of motion of body(cart) on platform.

Yes. For the cart the force from the man is the only force responsible for the change in the state of motion. You can see that by drawing the free body diagram.

So what is the change in the KE of the cart? In the ground frame the initial KE is $0.5 m 0^2 = 0$, and the final KE is $0.5 m (\Delta v)^2$ so the change in KE is $0.5 m (\Delta v)^2 = F \Delta x$

In the train frame the initial KE of the cart is $0.5 m v^2$ and the final KE is $0.5 m (v+\Delta v)^2$ so the change in KE is $0.5 m (v+\Delta v)^2-0.5 m v^2 = 0.5 m (\Delta v)^2 + m v \Delta v = F \Delta x + F v \Delta t = F (\Delta x + v \Delta t)$

As you said, the force of the man on the cart is the only force responsible for changing that state of motion. Therefore the force of the man is the one that did work $F(\Delta x + v\Delta t)$, which is the total displacement in the train frame as I described in my previous post.

Your suggested method for calculating the "work" violates the conservation of energy because the cart gains more KE than the "work" done on it.

That force is only doing work & making old man tired.

If you want to determine the "tiredness" of the man then you obviously need to consider the forces acting on the man rather than the ones acting on the cart. It is obviously insufficient to consider only the forces acting on the cart when making conclusions about the man.

 

[Dale]

Since the OP is clearly not interested in learning, and since this thread has become repetitive, and since this thread is about a mistake in Newtonian physics rather than relativity, it is now closed.