Conservation of Energy in different inertial frames

  • Thread starter bartieshaw
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today my dynamics lecturer illustrated a situation to us where energy did not appear to be conserved.

The situation involves observing the motion of a mass sliding down a slope, height 'h' above the ground from rest

initially in the rest frame, we see;

initially,
PE = mgh
KE = 0

as the mass starts from rest (we are assuming the force required to start the motion of the mass is insignificant)

and finally (at the bottom of the slope),
PE = 0
KE = 1/2 mv^2

NB. i cannot work out how to make a diagram and post it, but imagine the mass is sliding down the slope to the right.


using this we can claculate the simple equation to determine the final velocity of the mass (to the right)

v = (2gh)^(1/2)

This was simple to comprehend, but now our lecturer asked us to imagine observing the motion from another inertial frame moving to the right (relative to the rest frame) at velocity, v = (2gh)^(1/2)

when drawing this, we see the mass now appears to initially be be moving to the left with velovity, v = (2gh)^(1/2), and when the mass reaches its maximum velocity (v = (2gh)^(1/2) to the right) at the bottom, the mass appears to be at rest.

thus when observing from this frame we see the following occuring

initially
PE = mgh
KE = 1/2 mv^2 = mgh

hence, initial energy, E=2mgh

and finally (at the bottom of the slope)
PE = 0
KE = 0

this obviously seems to disagree with the conservation of energy and im sure there is something wrong with what i have described, but i simply do not know

can anyone explain what is really going on here...?
 

mezarashi

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Actually this is not an issue of the conservation of energy. This is an issue known as invariance. Energy is always conserved, but it is not invariant meaning that depending on your frame of reference you will measure it differently. Any object travelling at the same velocity as you will appear to have no kinetic energy right? But you will find that whatever reference frame you use, the conservation laws will hold. When you step into the realm of relativity you will also find other quantities:

mass: not conserved but invariant
charge: conserved and invariant
velocity: neither conserved nor invariant
 

Doc Al

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Your instructor is just pulling your leg.
bartieshaw said:
thus when observing from this frame we see the following occuring

initially
PE = mgh
KE = 1/2 mv^2 = mgh

hence, initial energy, E=2mgh
You must include the KE energy of the Earth, since it's moving in this frame.

and finally (at the bottom of the slope)
PE = 0
KE = 0
Figure out the work done on the Earth during the ball's motion. You'll find that the work done on the Earth will exactly equal the missing energy. Energy is conserved!

(Hint: How far does the Earth move during the time that the ball falls? Realize that the average horizontal speed of the ball is only v/2, so that the Earth moves twice as far as the ball.)
 
i dont understand....
bartieshaw...
do u understand?
 
Yeah, i do now, thank you for answering
 
Still unsure

I sort of understand what you are saying, but can't seem to be prove it...???
 
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So which is right? Is the statement that energy is invariant correct, or is it right to say energy is conserved even during changing frames?

Is energy invariant or not?

Also, why is it called invariant? Doesn't that mean in does not vary??
 

Doc Al

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leright said:
So which is right? Is the statement that energy is invariant correct, or is it right to say energy is conserved even during changing frames?
Huh? Who said energy was invariant?

Yes, energy is conserved in either frame.
 
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Doc Al said:
Huh? Who said energy was invariant?

Yes, energy is conserved in either frame.
n/m. nobody said it was invariant. :p I was lacking sleep when I wrote that.
 
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bartieshaw said:
when drawing this, we see the mass now appears to initially be be moving to the left with velovity, v = (2gh)^(1/2), and when the mass reaches its maximum velocity (v = (2gh)^(1/2) to the right) at the bottom, the mass appears to be at rest.

thus when observing from this frame we see the following occuring

initially
PE = mgh
KE = 1/2 mv^2 = mgh

hence, initial energy, E=2mgh

and finally (at the bottom of the slope)
PE = 0
KE = 0
There's your mistake. Should be [tex]KE = -1/2 mv^2[/tex]. KE has a vector, and at the start of the experiment, it is opposite to your "moving" reference frame.
 

Gokul43201

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WhyIsItSo said:
There's your mistake. Should be [tex]KE = -1/2 mv^2[/tex]. KE has a vector, and at the start of the experiment, it is opposite to your "moving" reference frame.
KE is a scalar, not a vector!
 
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How so, given the velocity term?


PS: (im not arguing with your statement, i think i know why, i just cant be sure :P)
 
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Because the velocity term appears in the expression for KE within a scalar product:

[tex]T = \frac{m|\vec{v}|^2}{2} = \frac{m\vec{v}\cdot\vec{v}}{2}[/tex]

i.e. it's the length squared. It's the only way to turn a pair of vectors into a scalar - by the scalar/inner (sometimes known as dot) product. And by definition, the inner product of two vectors is equal to or greater than 0 (the equality holding for the zero vector only).
 
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So confused???

:confused: Can someone please clarify this? What is actually happening? Thanks
 
Look at it this way, change the math the way I suggested and the mystery disappears.

Kinetic Energy is the result of motion which has a vector. You cannot ignore the direction the force is applied in and expect the OP situation to make any mathematical sense.
 

Gokul43201

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WhyIsItSo said:
Look at it this way, change the math the way I suggested and the mystery disappears.
It only appears to. The math as suggested by you is wrong...and it wouldn't work for any other velocity of the reference frame (w.r. to the earth frame).


Kinetic Energy is the result of motion which has a vector. You cannot ignore the direction the force is applied in and expect the OP situation to make any mathematical sense.
The KE itself is a scalar and the situation makes perfect mathematical sense if you do the math right.

Pete, see post #3.
 
Work Done on the Earth?

Ok I understand the concept now. So with the addition of the work done by the earth energy is conserved. But how do I find the work done on the Earth so that I can prove that energy has been conserved?...
 

Doc Al

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To properly account for the KE, you must consider all the forces acting. In particular, the sliding mass and slope exert forces on each other.

In the usual frame of reference, where the slope is stationary, this contact force--the normal force--does no work.

But viewed from a moving frame, the slope moves and thus "work" is being done by those forces on the slope and on the mass. Since they move at different speeds, the work done on each is not the same. But the net change in KE of all objects (earth and sliding mass) will exactly balance the change in gravitational PE.
 

Gokul43201

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Proof that changing the refererence inertial frame doesn't change the truth of energy conservation:

First, we have the "rest frame", the frame where we accept the truths of momentum and energy conservation (for the problem in the OP, this is the earth frame). In this frame,

[tex]P_0 = \sum m_iv_i = const [/tex]

[tex]E_0 = (1/2)\sum m_iv_i^2 + \sum U_i = const [/tex]

Now consider a frame moving a speed v (v<<c), relative to this frame. In the new frame,

[tex]E_v = (1/2)\sum m_i(v_i-v)^2 + \sum U_i = E_0 - (1/2)\sum m_iv_i^2 + (1/2)\sum m_i(v_i-v)^2 = E_0 - 2vP + (1/2)\sum m_iv^2 [/tex]

All the terms in the RHS are constants, and hence, the total energy is conserved in the new frame as well.

Note: All squares and products of velocities (or velocities with momenta) are to be understood as scalar (dot or inner) products.
 
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Doc Al

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Excellent, Gokul. That's the easiest way to convince oneself that energy is conserved in all frames, since it avoids having to consider the details of the forces involved.
 

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