Energy and reference frames

  • #71
Dale said:
In this case 180 is the energy of the flashlight in the primed frame and 156 is the momentum of the flashlight in the primed frame after emitting the light.

the total energy of the light is not the density of the energy flow. Energy flow describes how much energy is moving from one location to another. This is the actual amount of energy in the fields. The relationship between the energy in the field, the energy flow, and the work done on matter is given by Poynting's theorem. I like this textbook: https://web.mit.edu/6.013_book/www/chapter11/11.2.html

In particular, see chapter 11 equation 10 for the energy in the fields. They use ##W## for the energy density, so if you integrate ##W## over all space then you get the total energy with $$W=\frac{\epsilon_0}{2}\vec E \cdot \vec E + \frac{\mu_0}{2} \vec H \cdot \vec H$$in free space.
Thank you for the detailed explanation. To summarize, do I understand the process correctly now? For the reference frame of a photocell, a moving flashlight has more energy than at rest. The increase in the observed energy is related to the movement of the flashlight. This leads to the fact that for the reference frame of the photocell, the light of the flashlight has a higher total energy, that is, a greater amount of electric and magnetic energy. When the light of the flashlight hits the photocell (passes into another frame of reference), the total energy of the light decreases for the frame of reference of the photocell. Thus, the illumination of the photocell for a period of 2 times longer is due to the energy associated with the movement of the flashlight.
 
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  • #72
Ivan Nikiforov said:
So, let the primed frame be the rest frame of the photovoltaic cells and let the unprimed frame be the rest frame of the flashlight. Let the velocity ##v## be in the ##x'## direction and the light be in the ##-y## direction. Analyze it as a pulse of energy ##\Delta E## in the unprimed frame emitted every ##\Delta t##. Use units where ##c=1##. Let the photovoltaics have an initial mass ##M## and let the flashlight have an initial mass ##m##. Let the distance between the flashlight and the photovoltaics be negligible, and set the origin ##t=t'=0## when the flashlight passes the leading edge of the photovoltaics. Use units where ##c=1##. Let ##m## and ##M## be large enough that the acceleration due to light pressure is negligible.

At a time ##t=n \ \Delta t## a total of ##n## light pulses have been emitted for a total light four-momentum ##P=(E,p_x,p_y,p_z)## of $$P_{light}=\left(n\Delta E,0,-n\Delta E,0\right)$$ So the flashlight in time ##n\Delta t## has gone from $$P_{flashlight,initial}=(m,0,0,0)$$ to $$P_{flashlight,final}=(m-n\Delta E,0,n \Delta E,0)$$ and the photovoltaics have gone from $$P_{photovoltaics,initial}=\left(\frac{M}{\sqrt{1-v^2}},-\frac{M}{\sqrt{1-v^2}},0,0 \right)$$ to $$P_{photovoltaics,initial}=\left(\frac{M}{\sqrt{1-v^2}}+ n \Delta E,-\frac{M}{\sqrt{1-v^2}},-n \Delta E,0 \right)$$

Now, the ##n##-th light pulse in the primed frame occurs at ##t'=n \Delta t/\sqrt{1-v^2}## (and also at position ##t' v##). In the primed frame the light has a total four momentum $$P'_{light}=\left(\frac{n\Delta E}{\sqrt{1-v^2}},\frac{vn\Delta E}{\sqrt{1-v^2}},-n\Delta E,0\right)$$ So the flashlight in time ##n \Delta t/\sqrt{1-v^2}## has gone from $$P'_{flashlight,initial}=\left( \frac{m}{\sqrt{1-v^2}},\frac{mv}{\sqrt{1-v^2}},0,0 \right)$$ to $$P'_{flashlight,final}=\left( \frac{m-n\Delta E}{\sqrt{1-v^2}},\frac{vm-vn\Delta E}{\sqrt{1-v^2}}, n\Delta E,0 \right)$$ and the photovoltaics have gone from $$P'_{photovoltaics,initial}=(M,0,0,0)$$ to $$P'_{photovoltaics,final}=\left(M + \frac{n \Delta E}{\sqrt{1-v^2}},\frac{vn \Delta E}{\sqrt{1-v^2}},-n \Delta E,0 \right)$$

So, as you can see, energy is conserved in both frames, but the amount of energy transferred is different as well as the amount of time over which the transfer happens.
 
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  • #73
Ivan Nikiforov said:
Thanks for the comment. It will take me quite a lot of time to comprehend the information and materials on the links that you provided. I will definitely do it. But for now, I can say the following. It seems to me that in the model I specified, there is no need to move the load. The load is attached to long rails and is a single unit. Electric current in rails travels at the speed of light in metal. I also don't quite understand the presence of wave processes and related impedances. It is assumed that a direct current flows in the electrical circuit, and it begins to flow even before the capsule starts moving.

If your generator is stationary, I don't understand how you think you can avoid a moving load. I'm going to continue to talk about the case with a stationary generator, and a load moving along a pair of wires, which are arranged to form a transmission line so we can (attempt to) analyze it.

You may have to dig around to get a good textbook or article on transmission lines. Online, seems to be reasonable at first glance, I don't have a really solid reference unfortunately. It may be a digression, but hopefully an interesting one.

You can get the same results via circuit theory and it's actually a common introduction to the topic of transmission lines. The fields ultimately turn out to be what's physically important, but one can also think of it in circuit terms. In these terms, the wires have a distributed inductance (per unit length) and capacitance (per unit length). If one takes this distributed model, and replaces it with a "lumped" model with a ladder network of series inductances and parallel capacitances, one can solve the resulting circuit and find that the solution consists of a superposition of waves, one in each direction (say, left and right). There is also an important quantity of the line, it's characteristic impedance, which relates the current in the wave to the voltage in the wave.
 
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  • #74
Dale said:
So, as you can see, energy is conserved in both frames, but the amount of energy transferred is different as well as the amount of time over which the transfer happens.
Thanks! I am very grateful to you for taking the time to perform the calculations and explain everything in detail. I learned a lot of new things. I am sure that the proof you have received will be useful for other forum participants as well.
 
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  • #75
pervect said:
If your generator is stationary, I don't understand how you think you can avoid a moving load. I'm going to continue to talk about the case with a stationary generator, and a load moving along a pair of wires, which are arranged to form a transmission line so we can (attempt to) analyze it.

You may have to dig around to get a good textbook or article on transmission lines. Online, seems to be reasonable at first glance, I don't have a really solid reference unfortunately. It may be a digression, but hopefully an interesting one.

You can get the same results via circuit theory and it's actually a common introduction to the topic of transmission lines. The fields ultimately turn out to be what's physically important, but one can also think of it in circuit terms. In these terms, the wires have a distributed inductance (per unit length) and capacitance (per unit length). If one takes this distributed model, and replaces it with a "lumped" model with a ladder network of series inductances and parallel capacitances, one can solve the resulting circuit and find that the solution consists of a superposition of waves, one in each direction (say, left and right). There is also an important quantity of the line, it's characteristic impedance, which relates the current in the wave to the voltage in the wave.

Thank you for your comment! To be honest, I did not even imagine that wave processes take place in this system. Thanks to you, I have significantly expanded my knowledge. Thanks a lot again!
 
  • #76
Ivan Nikiforov said:
Please excuse me, I do not fully understand the mechanism of the process related to the frames. I would appreciate it if you could provide a specific conclusion - is the original statement true or not? What will be the ratio for energy?

I don't know if there was any mechanism. We just used one frame, and then we used another frame.


When we launch a sack full of photon-gas using a catapult, then the mechanism that increases the energy of the photon gas is the catapult mechanism. (Energy comes from a spring)

And now the same launch using another frame:

When we slow down a moving sack full of photon-gas using a moving catapult, then then the mechanism that decreases the energy of the photon gas is the catapult mechanism. (energy goes to increasing the kinetic energy of the catapult)
 
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  • #77
Ivan Nikiforov said:
View attachment 356754 As I understand it, there is no difference at which point of the photocell the flashlight shines during its movement, since the process in the photocell takes place at the speed of light, the maximum possible and constant for any reference frame.
The velocity-arrow at the flashlight in your drawing suggests, that the angle between the light-direction and the velocity-direction is 90° in the restframe of the photocell. This angle is frame-dependent (aberration, see i.e. light clock).

Then the light-energy in the restframe of the photocell is smaller by the factor ##1/\gamma## than in the restframe of the flashlight (=redshift).

However, if the angle between the light-direction and the velocity-direction would be 90° in the restframe of the flashlight, then the light-energy in the restframe of the photocell would be bigger by the factor ##\gamma## than in the restframe of the flashlight (=blueshift).
 
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  • #78
Sagittarius A-Star said:
if the angle between the light-direction and the velocity-direction would be 90° in the restframe of the flashlight, then the light-energy in the restframe of the photocell would be bigger by the factor γ than in the restframe of the flashlight (blueshift).
@Ivan Nikiforov note that this is what I assumed and this was indeed the result that I found.
 
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  • #79
Sagittarius A-Star said:
The velocity-arrow at the flashlight in your drawing suggests, that the angle between the light-direction and the velocity-direction is 90° in the restframe of the photocell. This angle is frame-dependent (aberration, see i.e. light clock).

Then the light-energy in the restframe of the photocell is smaller by the factor ##1/\gamma## than in the restframe of the flashlight (=redshift).

However, if the angle between the light-direction and the velocity-direction would be 90° in the restframe of the flashlight, then the light-energy in the restframe of the photocell would be bigger by the factor ##\gamma## than in the restframe of the flashlight (=blueshift).
Thank you for your comment. The article on the link and the animation are very informative.
 
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  • #80
Dale said:
@Ivan Nikiforov note that this is what I assumed and this was indeed the result that I found.
You have an excellent command of the mathematical apparatus and are very well versed in this scientific field. It is a great honor for me to communicate with you and learn from you. Thank you.
 
  • #81
Ivan Nikiforov said:
Thank you for the detailed explanation. To summarize, do I understand the process correctly now? For the reference frame of a photocell, a moving flashlight has more energy than at rest. The increase in the observed energy is related to the movement of the flashlight. This leads to the fact that for the reference frame of the photocell, the light of the flashlight has a higher total energy, that is, a greater amount of electric and magnetic energy. When the light of the flashlight hits the photocell (passes into another frame of reference), the total energy of the light decreases for the frame of reference of the photocell. Thus, the illumination of the photocell for a period of 2 times longer is due to the energy associated with the movement of the flashlight.

Let me replace photocell by a photon-gas bag, into which the light emitted by the flashlight gets trapped . (it's simpler that way)


Using bag frame:

High energy object (light) collides with a parked bag. Bag does nor budge, because of its high mass (a simplification). Collision is inelastic, because the bag and the light stick together. Collision energy = light energy = increase of energy of bag.


Using flashlight frame:

"Normal" energy object (light) collides with a moving bag. Bag does nor change its velocity, because of its high mass (a simplification). Collision is inelastic, because the bag and the light stick together. Collision energy = light energy + (change of momentum of bag * velocity of bag) = increase of energy of bag.


So bag-frame observer and flashlight-frame observer both say that energy of the light inside the bag increased quite a lot, and they agree about the amount. (Bag was originally empty)
 
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  • #82
jartsa said:
Let me replace photocell by a photon-gas bag, into which the light emitted by the flashlight gets trapped . (it's simpler that way)


Using bag frame:

High energy object (light) collides with a parked bag. Bag does nor budge, because of its high mass (a simplification). Collision is inelastic, because the bag and the light stick together. Collision energy = light energy = increase of energy of bag.


Using flashlight frame:

"Normal" energy object (light) collides with a moving bag. Bag does nor change its velocity, because of its high mass (a simplification). Collision is inelastic, because the bag and the light stick together. Collision energy = light energy + (change of momentum of bag * velocity of bag) = increase of energy of bag.


So bag-frame observer and flashlight-frame observer both say that energy of the light inside the bag increased quite a lot, and they agree about the amount. (Bag was originally empty)
Thanks for the comment. Now I understand well that for the reference frame A and for the reference frame B, there is an increase in the energy of light on the photocell. Will there be a difference in the duration of the illumination of the photocell (the flashlight in frame B works for 1 hour, but the illumination of the photocell in frame A takes place for 2 hours)? I'm sorry if I didn't understand something. I just came to the conclusion that the difference in lighting duration is directly related to the difference in light energy, which in turn is a consequence of the movement of the flashlight.
 
  • #83
jartsa said:
So bag-frame observer and flashlight-frame observer both say that energy of the light inside the bag increased quite a lot, and they agree about the amount
They do not generally agree about the amount of energy increase. I believe that you are thinking about the mass. They do agree about the increase in the invariant mass.
 
  • #84
Dale said:
They do not generally agree about the amount of energy increase. I believe that you are thinking about the mass. They do agree about the increase in the invariant mass.

Oh. But maybe in the special case of a flashlight shining to the direction perpendicular to its motion they agree?

In a photon-gas bag frame the energy from the flashlight is increased increased by gamma(v).
In a flashlight frame when the light emitted by the flashlight is stored into a bag and the bag is then accelerated to speed v, the energy of the light is increased by gamma(v).

Oh yes, a terminology question: When I say "electromagnetic energy of a moving photon cloud", should I then include the kinetic energy of the cloud? And when I say "energy collected by a moving photocell", should I then include the kinetic energy of the collected energy?

[mentors’ note: edited to fix broken bbcode]
 
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  • #85
jartsa said:
QUOTE=["Dale, post: 7240762, member: 43978"]
The opening square bracket goes to the left of the "QUOTE=" rather than to the right.

i.e. [QUOTE="Dale, post: 7240762, member: 43978"] which renders as:
Dale said:
They do not [...]
 
  • #86
jbriggs444 said:
The opening square bracket goes to the left of the "QUOTE=" rather than to the right.

i.e. [QUOTE="Dale, post: 7240762, member: 43978"] which renders as:
Corrected by application of mentorly superpowers
 
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  • #87
jartsa said:
When I say "electromagnetic energy of a moving photon cloud", should I then include the kinetic energy of the cloud? And when I say "energy collected by a moving photocell", should I then include the kinetic energy of the collected energy?
Yes. If you are talking about the energy collected according to a particular frame then that includes the kinetic energy collected according to that frame. You could talk about something like the mass of fuel generated or consumed by a fuel cell if you wanted something frame independent.
 
  • #88
An additional aspect: The time-dilation and the kinetic energy of a moving object depend generally on the Einstein-synchronization convention of the inertial reference-frame.
 
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