# I Work done transformation

1. Jan 7, 2016

### 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?

2. Jan 7, 2016

### A.T.

Work is frame dependent even in classical mechanics.

3. Jan 8, 2016

### 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.

4. Jan 9, 2016

### 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 $T=dx/u_x=dy/u_y$.

Last edited: Jan 9, 2016
5. Jan 11, 2016

### 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.

6. Jan 11, 2016

### 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.

7. Jan 11, 2016

### 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.

8. Jan 11, 2016

### 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.

Last edited: Jan 11, 2016
9. Jan 11, 2016

### 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 travelling in anything except the platform frame.

@ravi# - I think I'm correct in #6. In the case where the train is travelling 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.

10. Jan 11, 2016

### 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?

Last edited: Jan 11, 2016
11. Jan 12, 2016

### 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...

12. Jan 12, 2016

### A.T.

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).

13. Jan 12, 2016

### jartsa

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

14. Jan 15, 2016

### 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.

15. Jan 15, 2016

### A.T.

Why not just force x displacement ?

16. Jan 15, 2016

### Staff: Mentor

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

Are you familiar with four-vectors?

17. Jan 16, 2016

### 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)

18. Jan 16, 2016

### Staff: Mentor

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.

Last edited: Jan 16, 2016
19. Jan 18, 2016

### 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.

20. Jan 18, 2016

### A.T.

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

21. Jan 18, 2016

### Ibix

...as I pointed out in #6.

22. Jan 18, 2016

### Staff: Mentor

@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.

23. Jan 20, 2016

### 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.

24. Jan 20, 2016

### 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

25. Jan 20, 2016

### 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.