# Contributions to total energy in different areas of Physics

• B
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Classically, the total energy of a particle/system ##E = T+V##. This is the usage seen in the Schrödinger equation, since E is an eigenvalue of ##\hat{H}## and ##\hat{H} = \hat{T} + \hat{V}##.

Then in special relativity, the total energy ##E## of a particle/system becomes the rest energy plus the kinetic energy, i.e. ##E = \sqrt{(mc^{2})^{2} + (pc)^{2}} = \gamma mc^{2} = (mc^{2}) + (\frac{1}{2}mv^{2} + \mathcal{O}(v^{4}))##. I understand that this is valid for free particles (i.e. if the potential is constant, set it arbitrarily to zero for the particle). I would have thought then that the full relation would be ##E = \sqrt{(mc^{2})^{2} + (pc)^{2}} + V##, but apparently this is incorrect because it is not a covariant equation?

And secondly, there are some formulae like the Planck-Einstein ##E=hf## relation (which, for example's sake might apply to an electron), where I can't decide whether ##E=T+V## or ##E=E_{0} + T## applies.

I wondered whether someone could help to clarify this? Thank you.

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PeroK
Homework Helper
Gold Member
Classically, the total energy of a particle/system ##E = T+V##. This is the usage seen in the Schrödinger equation, since E is an eigenvalue of ##\hat{H}## and ##\hat{H} = \hat{T} + \hat{V}##.

Then in special relativity, the total energy ##E## of a particle/system becomes the rest energy plus the kinetic energy, i.e. ##E = \sqrt{(mc^{2})^{2} + (pc)^{2}} = \gamma mc^{2} = (mc^{2}) + (\frac{1}{2}mv^{2} + \mathcal{O}(v^{4}))##. I understand that this is valid for free particles (i.e. if the potential is constant, set it arbitrarily to zero for the particle). I would have thought then that the full relation would be ##E = \sqrt{(mc^{2})^{2} + (pc)^{2}} + V##, but apparently this is incorrect because it is not a covariant equation?

And secondly, there are some formulae like the Planck-Einstein ##E=hf## relation (which, for example's sake might apply to an electron), where I can't decide whether ##E=T+V## or ##E=E_{0} + T## applies.

I wondered whether someone could help to clarify this? Thank you.
You have to distinguish between relativistic and non-relativistic theories. Classical Mechanics and Quantum Mechanics are non-relativistic.

You can look up the relativistic Lagrangian for a free particle and the Klein-Gordon equation if you want to see a basic attempt to combine QM with SR.

The full story is, of course, QFT.

• etotheipi
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You have to distinguish between relativistic and non-relativistic theories. Classical Mechanics and Quantum Mechanics are non-relativistic.
So in the case of a classical photon it might then seem correct to say ##E = T + V = hf##, though ##V## is often zero.

But I've come across questions about e.g, the change in wavelength of photons as they move through gravitational fields, so we might say that if it "gains 100J of GPE" (if we give it an effective mass) then ##hf## reduces by 100J. But this seems different to the ##E = T+V## relation just above, since now we're just taking ##hf+V## to be constant!

PeroK
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Gold Member
So in the case of a classical photon it might then seem correct to say ##E = T + V = hf##, though ##V## is often zero.

But I've come across questions about e.g, the change in wavelength of photons as they move through gravitational fields, so we might say that if it "gains 100J of GPE" (if we give it an effective mass) then ##hf## reduces by 100J. But this seems different to the ##E = T+V## relation just above, since now we're just taking ##hf+V## to be constant!
This post has a significant confusion of ideas and concepts.

I'm not sure there is a such a thing as a classical photon. Classically, light is described as an EM wave and there is no concept of a photon as a particle subject to EM potential.

Relativistically, changes in wavelength of a photon do not reflect any inherent change in the photon itself. Instead, redshift and blueshift (whether "velocity-based" or "curved-spacetime-based") are a function of the relationship between the source and the receiver.

• etotheipi
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2019 Award
This post has a significant confusion of ideas and concepts.

I'm not sure there is a such a thing as a classical photon. Classically, light is described as an EM wave and there is no concept of a photon as a particle subject to EM potential.

Relativistically, changes in wavelength of a photon do not reflect any inherent change in the photon itself. Instead, redshift and blueshift (whether "velocity-based" or "curved-spacetime-based") are a function of the relationship between the source and the receiver.
I think this was in a BPhO question a few years back where you had to calculate something to do with how the wavelength changed as a photon escaped a planet, and the way you were supposed to do it was equate ##hf = mc^{2}## to define an effective mass, and then let ##hf + mgh = \text{constant}##. Seems quite sketchy, though.

It might be the case that the effect is real (Pound-Rebka?) but the Physics is far too complicated to put on a paper like that. Because I agree with what you say, and photons having GPE seems wrong.

PeroK
Homework Helper
Gold Member
I think this was in a BPhO question a few years back where you had to calculate something to do with how the wavelength changed as a photon escaped a planet, and the way you were supposed to do it was equate ##hf = mc^{2}## to define an effective mass, and then let ##hf + mgh = \text{constant}##. Seems quite sketchy, though.

It might be the case that the effect is real (Pound-Rebka?) but the Physics is far too complicated to put on a paper like that. Because I agree with what you say, and photons having GPE seems wrong.
I suppose it depends how you want to learn physics. I'd be slightly concerned that you're looking at many areas at once, but not necessarily digesting them. You need to avoid dilettanteism!

• etotheipi
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2019 Award
I suppose it depends how you want to learn physics. I'd be slightly concerned that you're looking at many areas at once, but not necessarily digesting them. You need to avoid dilettanteism!
Didn't know that was a word! I think part of the problem is that examiners tend to squeeze bits of quantum and relativity onto the spec but then don't go into nearly enough depth for anything to be coherent, and this just means that misconceptions are much more likely to come up, this thread being an example of one such misconception...!

I'd be much happier if high school physics was more focused around solely Classical Physics so that they could go a lot further with mechanics/optics/E&M/astro etc. and leave quantum/relativity for university, where they could be taught properly from the ground up. But hey ho!

• weirdoguy and PeroK