# Thread: The Energy - Momentum Equation vs the Energy - Mass Equation

1. First, introduce the energy  momentum equation E² = p²c² + (m0c²)².
Next, just think it in natural way.

1. If the energy  momentum equation reflects the stationary situation, then, momentum p naturally equals to zero. Then, we got E² = 0 + (m0c²)², namely: E = m0c². It can be denoted exactly as E0 = m0c². This is the energy - mass equation in stationary situation;

2. If the energy  momentum equation reflects the dynamic situation, then, momentum p ≠ 0.
Transform the energy  momentum equation E² = p²c² + (m0c²)² into p²  E² / c² = - m0²c²,
- m0²c² = m0²v² / (1  v² / c²)  m0²c² / (1  v² / c²),
Because m² = m0² / (1  v² / c²), then, - m0²c² = m²v²  m²c² = p²  E² / c²,
Because m²v² = p², then,  m²c² =  E² / c²,
Then E² = m²c ^4, namely: E = mc². This is the energy  mass equation in dynamic situation.

Since the energy  momentum equation E² = p²c² + (m0c²)² is generally applicable (to any particle), the stationary situation E0 = m0c² as well as the dynamic situation E = mc² is generally applicable (to any particle) too.

Below is the transformation in counter way:
E = mc² is Dynamic mass  energy relationship in nature. Its a basic natural property. (Dynamic means the particle is moving and hints it has momentum);
Next square both sides: E² = m²c ^4, namely: m²c² = E² / c², then, add a negative mark on both side:  m²c² =  E² / c²;
Add two redundant items m²v² on both sides: m²v²  m²c² = m²v²  E² / c²;
Because m²v² = p², then, m²v²  m²c² = p²  E² / c²;
Because m = γm0, (pay attention here, it just means that you can consider the magnitude of the moving mass m is γm0, but not means the particle be rest), square both sides: m² = m0² / (1  v² / c²);
Then, m0²v² / (1  v² / c²)  m0²c² / (1  v² / c²) = p²  E² / c²;
A mathematical calculation: m0²v² / (1  v² / c²)  m0²c² / (1  v² / c²) = - m0²c²;
Then, - m0²c² = p²  E² / c²;
Transform it in math, then, E² = p²c² + (m0c²)².
Its the so called energy  momentum equation.

If its the stationary situation, v = 0, so, no math game can be played. Then, its just an energy - mass equation in stationary situation: E0 = m0c².

Liqiang Chen
Oct 6 ,2020  2.

3. Non – relativistic mass problem:

If the non – relativistic mass is employed, the so called energy – momentum equation is:
E² = p²c² + (mc²)²

Then:
1. If the energy – momentum equation reflects the stationary situation, then, momentum p naturally equals to zero. Then, we got E² = 0 + (mc²)², namely: E² = m²c^4.
2. If the energy – momentum equation reflects the dynamic situation, then, p = mv, so:
E² = p²c² + (mc²)² = m²v²c² + (mc²)² = m² (v² + c²) c².
That means the energy – mass relationship is not the same in different inertial frames.
If the relativistic mass is employed, no matter according to the analysis in the OP of this thread or the original idea of Einstein:
1. If it is the stationary situation, E0 = m0c².
2. If it is the dynamic situation, E = mc².
That means the energy – mass relationship is the same (namely c²) in different inertial frames.

(Why the employment of dynamic / relativistic mass is reasonable? Details see the thread “The mass – space equation” )  4. There shouldn’t be two kinds of energy – mass relationship in nature:
1. Einstein’s energy – mass equation;
2. Such representation of “for a massless particle, its energy E = pc”

The energy – momentum equation is just a tricky math game as illustrated above. So, such representation of “for a massless particle, its energy E = pc” is also a tricky math game. The energy of a particle should be calculated according to the energy – mass equation in physics. If you consider a particle is “massless”, then, its energy will be zero. Such “massless - energyless” particle is theoretically meaningless in physics. It does not exist in nature.  5. Originally Posted by htam9876 There shouldnt be two kinds of energy  mass relationship in nature:
1. Einsteins energy  mass equation;
2. Such representation of for a massless particle, its energy E = pc
The universe doesn't care what you think should be.

If you consider a particle is massless, then, its energy will be zero.
Clearly incorrect since gamma rays are energetic and massless.  6. @ Bufofrog:
How do you measure the mass of gamma rays directly? Weigh it?   7. Originally Posted by htam9876 @ Bufofrog:
How do you measure the mass of gamma rays directly? Weigh it? Look at this article. Specifically Look at experimental checks on photons mass.  8. I don't think anyone understand what is a photon... Light trap phenomenon:
Tiny molecules can be pulled to the center of laser light beam. Why? If released photon has mass, we can explain it. The (mass) density of released photons at the center is larger than that on the side, then, the gravity generated by released photon at the center is larger than that on the side.  9. I show you how to weigh a photon: employ the Earth.
We throw out a small stone. It moves in curve because it has energy and mass. So, if light can travel in curve nearby the Earth, photon has both energy and mass.  10. Originally Posted by htam9876 I show you how to weigh a photon: employ the Earth. We throw out a small stone. It moves in curve because it has energy and mass.
It will only move in a curve in a gravitational field, otherwise it would travel in a straight line. The curve that the object follows is a result of the initial velocity not the mass of the object. Neglecting friction a 1 kg mass and a 100 kg mass shot from level ground at 100 m/s at a 45 degree angle will both fly the same distance.
So, if light can travel in curve nearby the Earth, photon has both energy and mass.
General Relativity tells us that a massless particle will follow a curved geodesic in a gravitational field. Therefore your experiment will not work, it will not tell you anything about the mass.  11. @ Bufofrog:
GR is just “ a method of description” for gravity, but not equals to gravity.   12. Originally Posted by htam9876 @ Bufofrog:
GR is just “ a method of description” for gravity, but not equals to gravity. That seems pretty meaningless to me.  13. Originally Posted by Bufofrog  Originally Posted by htam9876 @ Bufofrog:
GR is just “ a method of description” for gravity, but not equals to gravity. That seems pretty meaningless to me.
Maybe htam is trying to say that GR is a model for what we observe and not the thing itself?

Like Markus often says ;Do not confuse the map for the terrain.....  14. Originally Posted by geordief  Originally Posted by Bufofrog  Originally Posted by htam9876 @ Bufofrog:
GR is just  a method of description for gravity, but not equals to gravity. That seems pretty meaningless to me.
Maybe htam is trying to say that GR is a model for what we observe and not the thing itself?
Which would be irrelevant, since Bufofrog didn't argue that GR was "equals (sic) to gravity." He in fact expicitly uses GR as a model for what we observe, which is fundamentally what science is about. It is more in the domain of philosophy to argue about what thus-and-so "really is."  15. @geordief:
Interesting comment.  16. Originally Posted by htam9876 @ Bufofrog:
GR is just “ a method of description” for gravity, but not equals to gravity. Regardless, what I stated shows that your idea (if light can travel in curve nearby the Earth, photon has both energy and mass) for detecting mass won't work.

You have made statements that I have disagreed with, why have you not responded?  17. Maybe htam is trying to say that GR is a model for what we observe and not the thing itself?

I have 2 responses to that.

1. Why should we have to guess what he is trying to say? None of us can read minds.

2. The post (whatever he is trying to say) is just to change the subject about the mass of a photon so that he doesn't have admit he doesn't know what he is talking about.

The second point is a rather classic troll tactic, we will have to see a few more posts to see what htam's purpose here is.  18. Originally Posted by Bufofrog Maybe htam is trying to say that GR is a model for what we observe and not the thing itself?

I have 2 responses to that.

1. Why should we have to guess what he is trying to say? None of us can read minds.

2. The post (whatever he is trying to say) is just to change the subject about the mass of a photon so that he doesn't have admit he doesn't know what he is talking about.

The second point is a rather classic troll tactic, we will have to see a few more posts to see what htam's purpose here is.
That sounds reasonable.

Foreign language bar.

So ,the back of my mind is wondering whether there is sometimes a language problem.

Honesty of purpose is a pre requisite for posting on forums and this ,as you say should become clear.  19. "So ,the back of my mind is wondering whether there is sometimes a language problem."
@geordief:
NO bad comment. Other problems: I nearly can't access any link especially Wiki, Youtude, etc. I am in another world. As I said in another thread, I fail attaching files...  20. “General Relativity tells us that a massless particle will follow a curved geodesic in a gravitational field.”
The energy – mass equation (SR) tells us if a particle is “massless”, then, its energy will be zero. But sounds that the author of GR and SR is the same person.
In another site, an American attached an article in my thread and told me that photon’s mass is impossible to be measured. But it’s the affair of that site. So, I wouldn’t move it here. I encountered all kinds of opinions.
As it shows, this is an “alternative idea” column. This kind of column is good for the development of science. The spirit of science should be exploration never stop its feet.
………………………………………
Okay, as the title shows this thread’s theme is SR rather than GR. I’ll start another thread to discuss GR.
Thank you, guys.  21. First, introduce the energy – momentum equation E² = p²c² + (m0c²)².

This equation tells us the energy equivalency of a mass and it's associated kinetic energy. What you have shown is that if there is no mass then there will be no energy from that 0 amount of mass. That is obviously true. This equation does not say that a massless particle has zero energy.
The mass-energy equivalence equation can be used to see how much energy a photon must have to produce a mass. A photon with an energy level of 1.022 Mev can produce an electron and a positron with no KE. Higher energy photons can produce an electron and positron with KE. The photon can undergo pair production only with a sufficient energy level.
You're conjecture that a massless particle has no energy is clearly incorrect.  22. The energy – momentum equation is just a tricky math game as illustrated above.

No, it really isn't! It is a ground breaking equation that clearly, concisely and accurately shows the relationship between mass and energy.  23. Below is some opinions I gathered in another site. I move it here for reference and show what other physicists care and think.

Abstract from PAllen’s post:
“You are doing simple, well understood manipulations showing equivalence of different relativistic formulas. One thing you introduce is m versus m0, which is generally considered an obsolete notion.”

Abstract from Dale's post:
“Relativistic mass is a concept that has been abandoned by the scientific community now for several decades. I wouldn’t recommend using it.”

Abstract from Vanhees71's post:
“Einstein abandoned quite quickly the idea of relativistic masses. If you want to introduce this very confusing idea that the mass depends on the velocity of a particle, then you must introduce not only one such quantity but at least two, i.e., "transverse and longitudinal mass". This is utmost confusing and thus not done anymore by physicists using the special or general theory of relativity in contemporary research. The reason is that relativistic physics is most simply expressed in terms of tensor (and in high-energy physics also spinor) notation, where you only work with invariant quantities, i.e., tensors and spinors (or their covariant components). Einstein's "energy-mass equation" in modern form reads… For massless "particles" (there are no such things in nature though) m=0, and that's it. Here, m is the invariant mass and a scalar, and that's the only notion of mass which makes sense in a manifestly covariant formalism, and that's how it's done in modern formulations of relativistic theories.”
…………………………………
All these three physicists actually care the problem of “Relativistic mass”.
But mass is a most basic physical quantity such as space and time. Why space and time can be relativistic while mass not? Such most basic physical quantity as space, time and mass all be relativistic is reasonable.  Bookmarks
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