I Derivation of E=mc2 from Four-Vector Definitions in Special Relativity

[Ziang]

In SR, fundamental concepts velocity, acceleration, force, momentum were defined as four-vectors.
https://en.wikipedia.org/wiki/Four-vector
Does someone show me a derivation of the equation E = mc² from those definition?

 

[Ibix]

In your link, see the last equation in the "four momentum" section. Consider the special case where velocity is zero and hence the three momentum $\mathbf p=0$.

 

[Ziang]

I mean how the equations E = γm_oc² and E² = c²pp + (m_oc²)² come form four vector definitions of fundamental concepts in SR?

 

[SiennaTheGr8]

If you take the four-momentum as a given, then you already have the second of your equations, because mass in special relativity is defined as the magnitude of the four-momentum:

$(mc^2)^2 = E^2 - (pc)^2$.

As for the relation $E = \gamma mc^2$, consider that the four-momentum is just the four-velocity scaled by mass:

$(E/c , \mathbf{p}) = \mathbf{P} = m \mathbf{V} = m (\gamma c, \gamma \mathbf{v} )$,

so $E = \gamma mc^2$ and $\mathbf{p} = \gamma m \mathbf{v}$.

 

[Ibix]

What didn't you understand about the derivation given in the wiki article? Apart from "the four position" which you shouldn't really regard as a four vector (the four velocity is fine) it seemed ok to me.

 

[Ziang]

If you take the four-momentum as a given, then you already have the second of your equations, because mass in special relativity is defined as the magnitude of the four-momentum:

$(mc^2)^2 = E^2 - (pc)^2$.

So the energy-momentum relation above is not a consequence but it is by definition.

 

[SiennaTheGr8]

So the energy-momentum relation above is not a consequence but it is by definition.

That depends on what your starting point is!

You asked how to derive those equations from the four-vectors given on the Wikipedia article you linked to. That's the question I answered.

If instead you had asked how those equations were originally derived in the early 20th century, the answer would be different.

 

[Ziang]

I would like to make my question clearer.
In Newtonian mechanics, they derived directly the kinetic energy from 3D-vector definitions of velocity v = ds/dt, acceleration a = dv/dt, F = ma, p = mv, W = F*x.

Now in SR, the fundamental concepts were defined as four-vectors.
Velocity U = dX/dτ
Acceleration A = dU/dτ
P = m_oU
F =dP/dτ
W = F*X
and γ = dt/dτ

My question is can we derive the mass-energy relation directly from 6 definitions above only?

 

[Ibix]

Yes. It's the modulus of P.

 

[Ziang]

Please show me.
Thanks,

 

[Ibix]

Please show me.

It's the last equation in the section on the four momentum in the wiki article you linked!

 

[Ibix]

A bit more explicitly, the modulus-squared of a four vector is its time like component squared minus the sum of its spatial components squared. This is invariant. Thus it must be equal to $m_0c^2$ from the definition of U in the rest frame.

 

[SiennaTheGr8]

I would like to make my question clearer.
In Newtonian mechanics, they derived directly the kinetic energy from 3D-vector definitions of velocity v = ds/dt, acceleration a = dv/dt, F = ma, p = mv, W = F*x.

Now in SR, the fundamental concepts were defined as four-vectors.
Velocity U = dX/dτ
Acceleration A = dU/dτ
P = m_oU
F =dP/dτ
W = F*X
and γ = dt/dτ

My question is can we derive the mass-energy relation directly from 6 definitions above only?

Are you familiar with the Minkowski dot product? If you have the four-vectors $\mathbf{Q} = (Q_t , \mathbf{q})$ and $\mathbf{W} = (W_t , \mathbf{w})$, then:

$\mathbf{Q} \cdot \mathbf{W} = Q_t W_t - (\mathbf{q} \cdot \mathbf{w})$.

Additionally, the relationship between the magnitude of a four-vector and its components is given by the Minkowski dot product of the four-vector with itself. For example:

$\mathbf{Q} \cdot \mathbf{Q} = Q_t^2 - (\mathbf{q} \cdot \mathbf{q}) = Q^2$

(where $Q$ is the magnitude of $\mathbf{Q}$).

Apply all that to the four-momentum $\mathbf{P} = (E, \mathbf{p}c)$, whose magnitude is $P = mc^2$:

$\mathbf{P} \cdot \mathbf{P} = E^2 - (\mathbf{p}c \cdot \mathbf{p}c) = P^2 = (mc^2)^2$.

 

[Ziang]

In the link, they write
P = (m_oγc, m_oγu)
then they write E = γm_oc² with no an explanation ?
I know it is Einstein's famous equation. But how the equation E = γm_oc² come from the 6 definitions that I mentioned.

 

[exmarine]

Quote the OP.

I am not as qualified to answer this as some others on this forum, but perhaps I can help, as I have wondered and studied the same question. E=mc^2 originally (I think) came from a second little (3 page) paper Einstein also wrote in 1905. I’ll try to attach the translation I have to this message – I assume that is ok to do. In this paper, he refers back to Section 8 of his big SR paper earlier in 1905 where he derived an expression for the transformation of energy of a “system of plane waves of light”. He then shows that the release of energy in the form of radiation from a body decreases it mass “exactly like the kinetic energy of the electron” (in Section 10 of his big SR paper). But you need to read and study his exact words rather than my paraphrase, etc.

Regards.

PS. I assume you have access to his big SR paper? I'll attach it too just in case.

 

[SiennaTheGr8]

In the link, they write
P = (m_oγc, m_oγu)
then they write E = γm_oc² with no an explanation ?
I know it is Einstein's famous equation. But how the equation E = γm_oc² come from the 6 definitions that I mentioned.

Perhaps this will satisfy you. (This is basically what Einstein did, but he wasn't using the four-vector formalism.)

Start with the time component of the four-momentum: $\gamma m c^2$.

Now do a binomial expansion of it (remember that $\gamma = (1 - \beta^2)^{-1/2}$, where $\beta = v / c$):

$mc^2 \left(1-\beta^2\right)^{-1/2} = mc^2 + mc^2 \dfrac{1}{2} \, \beta^2 + mc^2 \dfrac {3}{8} \, \beta^4 + mc^2 \dfrac{5}{16} \, \beta^6 + mc^2 \dfrac{35}{128} \, \beta^8 + \dots$

That second term is $.5 mc^2 \beta^2 = .5 mv^2$, which is just the Newtonian kinetic energy. All the speed-dependent terms to the right must be higher-order contributions to the kinetic energy (not easily measured at non-relativistic speeds), and the first term $mc^2$ must be an energy contribution that doesn't depend on speed at all (it's the invariant energy $E_0$). The invariant energy term and the kinetic energy terms must sum up to the total energy, which we label $E$.

So $E = \gamma mc^2$.

 

[Ibix]

Just to add to what @SiennaTheGr8 said, yes, Einstein is essentially asserting that $\gamma m_0c^2$ is energy on the basis that it's similar to the Newtonian expression for energy. You really need to do an experiment to show that it is actually relevant to the real world. Bertozzi did this and filmed it in a very 1950s style - it's easily findable on YouTube.

 

[Ziang]

Thank you all,
The binomial expression worked well. But it is possible for the exist of other expressions that have the term 0.5mv².
For example, the following expression has that term:
KE = 0.5mv²(β⁰ + β¹ + β² + β³ + ... )

 

[PeterDonis]

it is possible for the exist of other expressions that have the term 0.5mv2.

What does this have to do with the topic of this thread?

 

[Ibix]

Thank you all,
The binomial expression worked well. But it is possible for the exist of other expressions that have the term 0.5mv².
For example, the following expression has that term:
KE = 0.5mv²(β⁰ + β¹ + β² + β³ + ... )

You can make up infinitely many expressions that look like Newtonian kinetic energy to leading order. But they don't follow from Einstein's postulates or any equivalent system, and they won't in general match experiment even approximately (edit: outside the Newtonian regime, anyway). It would be basically numerology (algebrology?) rather than scientific reasoning.

 

[Ziang]

What does this have to do with the topic of this thread?

Sienna say Einstein's expression has the term 0.5mv². So my question is why it is the right one because other expressions have that term too.
I figured out the other one from four-vector definitions

 

[SiennaTheGr8]

Thank you all,
The binomial expression worked well. But it is possible for the exist of other expressions that have the term 0.5mv².
For example, the following expression has that term:
KE = 0.5mv²(β⁰ + β¹ + β² + β³ + ... )

Are you asking if the appearance of $.5mv^2$ in the binomial expansion of $\gamma mc^2$ might just be a coincidence, so that the other terms don't really have anything to do with "energy"?

The thing to keep in mind is that energy is mainly a useful quantity insofar as it's conserved. That $.5 mv^2$ appeared in the binomial expansion is extremely suggestive. But ultimately the "justification" for calling $\gamma mc^2$ "total energy" is experimental: is it a conserved quantity? Particle accelerators prove every day that it is.

 

[PeterDonis]

Sienna say Einstein's expression has the term 0.5mv2. So my question is why it is the right one because other expressions have that term too.

Because the 0.5mv^2 term is not the only one you have to get right. You have to get all of the terms right. Isn't that obvious?

 

[Ibix]

Sienna say Einstein's expression has the term 0.5mv². So my question is why it is the right one because other expressions have that term too.
I figured out the other one from four-vector definitions

That's not the point. The point is that the quantity $\gamma m_0c^2$ emerges from the maths, so is automatically interesting. And in the case where v<<c it looks similar to what Newton called energy. So it's at least plausible that it's a more broadly applicable expression that Newton merely approximated. Then you go to experiment to see of it predicts the energy/velocity relation correctly.

 

[Ziang]

I figured out the other one from four-vector definitions

Let me show you my work. Let me use I as a symbol of integrate
KE = I(FdX) = mI(AdX) = mI(UdU) = 0.5mU²
Replace U = γv,
I get KE = 0.5mv²(β⁰ + β¹ + β² + β³ + ... )

 

[SiennaTheGr8]

Let me show you my work. Let me use I as a symbol of integrate
KE = I(FdX) = mI(AdX) = mI(UdU) = 0.5mU²
Replace U = γv,
I get KE = 0.5mv2(β⁰ + β¹ + β² + β³ + ... )

First, are your X, A, and U supposed to be three-vectors or four-vectors?

Second, the relation $\mathbf{f} = m\mathbf{a}$ (three-vectors) does not hold in special relativity.

 

[Ziang]

First, are your X, A, and U supposed to be three-vectors or four-vectors?

Second, the relation $\mathbf{f} = m\mathbf{a}$ (three-vectors) does not hold in special relativity.

They are four-vectors F, A, U, X that I found on wiki.
https://en.wikipedia.org/wiki/Four-force

 

[SiennaTheGr8]

It looks like you're assuming that familiar equations from Newtonian mechanics still work in special relativity. Whether you're talking about three-vectors or four-vectors, that's a bad assumption.

Now, it turns out that the work–energy principle does still hold in special relativity:

$d E_k = \mathbf{f} \cdot d \mathbf{r}$

(where $\mathbf{f}$ is the force three-vector, and $d \mathbf{r}$ is the infinitesimal three-displacement).

However, there's no four-vector equivalent to this (that I'm aware of).

Also, the four-velocity $\mathbf{V}$ is not equivalent to $\gamma \mathbf{v}$. Where did you get that from? The quantity $\gamma \mathbf{v}$ is the celerity (or proper velocity, though I very much dislike that term).

 

[Ibix]

Also note that if the approach can be made to work it will yield the work done by the force, i.e. the kinetic energy $(\gamma-1)mc^2$, which does not include the rest mass energy.

 

[Ziang]

It looks like you're assuming that familiar equations from Newtonian mechanics still work in special relativity. Whether you're talking about three-vectors or four-vectors, that's a bad assumption.

Now, it turns out that the work–energy principle does still hold in special relativity:

$d E_k = \mathbf{f} \cdot d \mathbf{r}$

(where $\mathbf{f}$ is the force three-vector, and $d \mathbf{r}$ is the infinitesimal three-displacement).

However, there's no four-vector equivalent to this (that I'm aware of).

Also, the four-velocity $\mathbf{V}$ is not equivalent to $\gamma \mathbf{v}$. Where did you get that from? The quantity $\gamma \mathbf{v}$ is the celerity (or proper velocity, though I very much dislike that term).

1. In my work, I assume dE = F dX,
where F and X are four vectors. If my assumption is wrong, please show me the definition of work in four-vectors.

2. About U --> γv, I do the same way as they transfer P to p :)

 

[SiennaTheGr8]

There is no four-vector "version" of work.

Well, I suppose you could take the Minkowski product of the four-force and the four-displacement and think of it as the spacetime analogue of "work," but for the case of constant mass it will always equal zero (unless I've done something wrong):

Four-force (for constant mass): $\mathbf{F} = (\gamma \mathbf{f} \cdot \mathbf{v}/c , \gamma \mathbf{f})$

Infinitesimal four-displacement: $d\mathbf{R} = (c \, dt, d \mathbf{r})$

$\mathbf{F} \cdot d\mathbf{R} = (\gamma \mathbf{f} \cdot \mathbf{v}/c)(c \, dt) - \gamma \mathbf{f} \cdot d \mathbf{r} = \gamma \mathbf{f} \cdot d \mathbf{r} - \gamma \mathbf{f} \cdot d \mathbf{r} = 0$.

Not a very interesting quantity.

If there's a spacetime "analogue" of energy $E$, it would have to be the invariant energy $E_0$ (aka mass). Consider that the three-momentum (using $c=1$) is $\mathbf{p} = E \mathbf{v}$, and that the four-momentum is $\mathbf{P} = E_0 \mathbf{V}$.

If you're interested in learning more about relativistic mechanics, I suggest consulting a textbook!

 

[Ibix]

There is no four-vector "version" of work.

Well, I suppose you could take the Minkowski product of the four-force and the four-displacement and think of it as the spacetime analogue of "work," but for the case of constant mass it will always equal zero (unless I've done something wrong):

Four-force (for constant mass): $\mathbf{F} = (\gamma \mathbf{f} \cdot \mathbf{v}/c , \gamma \mathbf{f})$

Infinitesimal four-displacement: $d\mathbf{R} = (c \, dt, d \mathbf{r})$

$\mathbf{F} \cdot d\mathbf{R} = (\gamma \mathbf{f} \cdot \mathbf{v}/c)(c \, dt) - \gamma \mathbf{f} \cdot d \mathbf{r} = \gamma \mathbf{f} \cdot d \mathbf{r} - \gamma \mathbf{f} \cdot d \mathbf{r} = 0$.

Not a very interesting quantity.

But it does include a direct statement that the change in the zeroth component of the four momentum is the change in energy. Which is relevant here.

 

[Ziang]

If you're interested in learning more about relativistic mechanics, I suggest consulting a textbook!

In textbooks, they derive the mass-energy relation by figuring out the work
dE = fdx
where f = dp/dt and p = γmv.
My question is that if the force f > 0 "violate" the SR range, because SR deal with inertial moving objects only?

 

[SiennaTheGr8]

SR does not only deal with inertial motion. It can handle acceleration just fine.

 

[Ibix]

Force is just much harder to use outside Newtonian physics. For example, the 3-force and the 3-acceleration aren't necessarily parallel. Generally, people seem to use a Lagrangian or Hamiltonian approach instead.

 

[Ziang]

SR does not only deal with inertial motion. It can handle acceleration just fine.

So is it ok to say that according to SR, the potential energy of an object m at a height h is PE = mgh, because g is as same as acceleration?

 

[SiennaTheGr8]

So is it ok to say that according to SR, the potential energy of an object m at a height h is PE = mgh, because g is as same as acceleration?

SR is fundamentally inconsistent with Newtonian gravity, so in general it's best to neglect gravity altogether in SR (though there may be circumstances in which you can overlook the inconsistency and get some "good enough" results that aren't quite right; I'm not sure whether your question falls into this category).

To deal with gravity in a way that's consistent with SR requires GR.

 

[Ibix]

So is it ok to say that according to SR, the potential energy of an object m at a height h is PE = mgh, because g is as same as acceleration?

No. As Sienna says, you can't handle gravity in SR, you need GR.

Gravity is usually used as a straightforward example of a force in Newtonian physics, but it is extremely complicated in relativity. You can handle electromagnetism within the framework of SR if you want an example of a force.

 

[Ziang]

SR is fundamentally inconsistent with Newtonian gravity, so in general it's best to neglect gravity altogether in SR (though there may be circumstances in which you can overlook the inconsistency and get some "good enough" results that aren't quite right; I'm not sure whether your question falls into this category).

Yes, I would like to see how SR potential energy look like in simple cases like constant gravitational acceleration. "Good enough" is good enough. :)

 

[pervect]

Yes, I would like to see how SR potential energy look like in simple cases like constant gravitational acceleration. "Good enough" is good enough. :)

I think you'll need tensor methods usually associated with GR to work this out. I'll take a stab at how it goes. Unfortunately I can't gurantee I won't make errors.

If we take "constant gravitational acceleration" to be the Rindler metric, then there is at least a conserved notion of total energy. There are several variants of the RIndler metric one might use, I would use

*CORRECTION*

$$-\left(1+gz\right)^2 \left(\frac{dt}{d\tau}\right)^2 + \left(\frac{dx}{d\tau}\right)^2 + \left(\frac{dy}{d\tau}\right)^2+\left(\frac{dz}{d\tau}\right)^2$$

This replaces the Minkoskii metric for an inertial observer

$$-dt^2 + dx^2 + dy^2 + dz^2$$

I've used geometric units for both the Rindler and Minkowskii metrics, which is how I avoid having the constant c (the speed of light) appear in the metric tensor.

If we let $t=x^0, x=x^1, y=x^2, z=x^3$ we have the energy momentum 4-vector $E^i$ and it's co-vector $E_i$. They are related by the usual method of raising and lowering indices with the metric tensor, i.e. $E^i = g^{ij} E_j$ and $E_i = g_{ij} E^j$.

The metric tensor, and the associated need for vectors and coverctors, make the formulation a bit more complicated if one isn't familiar with tensor methods already.

The only component of the metric tensor that's neither zero nor unity is $g_{00}$ which is a negative number with the chosen sign convention, $g_{00} = -(1+gz)^2$. The corresponding inverse metric has $g^{00} = -1/(1+gz)^2$.

We have a static space-time, so we can use the same techniques to define various conserved quantities, of which there is the conserved energy $E_0$, the covector component of the energy-momentum 4-vector, and two momenta, $E_1$ and $E_2$. $E_3$ won't be conserved. By "conserved" I mean that $E_0, E_1, E_2$ are all constant for an object with no forces acting on it, i.e. an object following a space-time geodesic.

There may be some sign issues remaining here, but regardless of the best choice of sign, the above are constants of motion.

We can then find that $E^0 = m \frac{dt}{d\tau} = g^{00} E_0$ so we wind up with $E^0 = -\frac{E_0}{(1+gz)^2}$ with $E_0$ being some constant of motion that we call "energy". I'm not aware of any scheme to break up total energy into "kinetic energy" and "potential energy" in this formulation, but the expression for total energy is of some interest.

The other formula we mainly need is that $E^i E_i = -m^2$, the tensor version of $E^2 = (pc)^2 + m^2$. Using this relationship, we can find, for instance the value of $dz/d\tau$ given the conserved constants of motion of the object representing "energy" $E_0$, "x-momentum" $E_1$ and "y-momentum" $E_2$. "z-momentum", isn't conserved, but we can calculate $\frac{dz}{d\tau}$ knowing the other conserved quantities $E_0, E_1, E_2$, the invariant mass $E^i E_i$, and the z-coordinate of the object much as we might calculate the velocity of a falling object knowing it's energy, x & y momentum, and it's current height in the Newtonian case.

 

[PeterDonis]

We have a static space-time, so we can use the same techniques to define various conserved quantities

[Note: corrections made corresponding to those made by @pervect in his post.]

I think it's worth discussing a bit the details of how this is done. I'll run through the details with the "energy" conserved quantity, since that's the one of interest here.

The "energy" conserved quantity exists because we have a timelike Killing vector field $K^\nu$, which in the coordinate chart you've chosen is simply the vector $(1, 0, 0, 0)$. The corresponding conserved quantity $E$ (no index because it's a Lorentz scalar), by Noether's Theorem, is then $E = - g_{\mu \nu} P^{\mu} K^{\nu}$, where $P^{\mu}$ is the 4-momentum of an object which is undergoing geodesic motion, and the minus sign is because of the $- + + +$ metric signature convention you're using. Writing this out, we get

$$
E = \left( 1 + g z \right)^2 P^0 = \left( 1 + g z \right)^2 m \frac{dt}{d\tau}
$$

We can invert this to get

$$
P^0 = \frac{E}{\left( 1 + g z \right)^2}
$$

Lowering the index then gives $P_0 = - E$. However, I think it's still worth keeping these two quantities conceptually distinct--one is the 0 component of the 4-momentum, which will change if you change coordinate charts, while the other is a conserved Lorentz scalar that remains the same in any chart. For example, it's a good exercise to re-do the computations I did above in the Minkowski chart; you will find that $P^0$ and $P_0$ look very different in terms of $E$ along the same geodesic.

I'm not aware of any scheme to break up total energy into "kinetic energy" and "potential energy" in this formulation

The way to do it is to pick a "zero point" for potential energy. For this case, an obvious such point is $z = 0$--which is actually a hyperplane. The reason this is an obvious choice is that, throughout this hyperplane, the metric is actually Minkowski, which is a nice property for the "zero point" of potential energy to have. Then the conserved energy $E$ of an object on a geodesic that is momentarily at rest at some $z$ in these coordinates will be $m \left( 1 + g z \right)^2$, and since the object is momentarily at rest, all of this energy is potential energy; you can then trace the object's motion along the geodesic and see how the potential energy is converted to kinetic energy as it falls. (Note that this computation also makes it clear that in this formulation, the rest energy $m$ is part of the potential energy.)

 

[pervect]

I think it's worth discussing a bit the details of how this is done. I'll run through the details with the "energy" conserved quantity, since that's the one of interest here.

The "energy" conserved quantity exists because we have a timelike Killing vector field $K^\nu$, which in the coordinate chart you've chosen is simply the vector $(1, 0, 0, 0)$. The corresponding conserved quantity $E$ (no index because it's a Lorentz scalar), by Noether's Theorem, is then $E = - g_{\mu \nu} P^{\mu} K^{\nu}$, where $P^{\mu}$ is the 4-momentum of an object which is undergoing geodesic motion, and the minus sign is because of the $- + + +$ metric signature convention you're using. Writing this out, we get

$$
E = \left( 1 + g z \right) P^0 = \left( 1 + g z \right) m \frac{dt}{d\tau}
$$

We can invert this to get

$$
P^0 = \frac{E}{1 + g z}
$$

Lowering the index then gives $P_0 = - E$. However, I think it's still worth keeping these two quantities conceptually distinct--one is the 0 component of the 4-momentum, which will change if you change coordinate charts, while the other is a conserved Lorentz scalar that remains the same in any chart. For example, it's a good exercise to re-do the computations I did above in the Minkowski chart; you will find that $P^0$ and $P_0$ look very different in terms of $E$ along the same geodesic.
The way to do it is to pick a "zero point" for potential energy. For this case, an obvious such point is $z = 0$--which is actually a hyperplane. The reason this is an obvious choice is that, throughout this hyperplane, the metric is actually Minkowski, which is a nice property for the "zero point" of potential energy to have. Then the conserved energy $E$ of an object on a geodesic that is momentarily at rest at some $z$ in these coordinates will be $m \left( 1 + g z \right)$, and since the object is momentarily at rest, all of this energy is potential energy; you can then trace the object's motion along the geodesic and see how the potential energy is converted to kinetic energy as it falls. (Note that this computation also makes it clear that in this formulation, the rest energy $m$ is part of the potential energy.)

Note that I made an unfortunate typo for the RIndler metric in my original post, which I've since corrected.

 

[Ziang]

Thank you all,
I know GR is the best to deal with gravity. But Sienna said SR can deal fine with acceleration. So let us assume constant g is as same as constant acceleration, to figure out what is the potential energy of an object m at a height h, using SR, please.

 

[SiennaTheGr8]

Thank you all,
I know GR is the best to deal with gravity. But Sienna said SR can deal fine with acceleration. So let us assume constant g is as same as constant acceleration, to figure out what is the potential energy of an object m at a height h, using SR, please.

It sounds like you want an answer that reconciles SR with Newton's theory of gravitation, but no such thing exists.

You can just use the $mgh$ calculation from Newtonian physics -- as far as I know, SR doesn't "modify" that at all. But what if the object is traveling so fast relative to the Earth that SR is needed to describe the relative motion? does that make the Newtonian calculation less accurate?

I'm going to guess that the answer is yes, and that you then need GR to get a good handle on the physics. But I'm out of my depth here. (And it's probably worth mentioning that the very concept of energy conservation is complicated in GR, so for all I know the question might be ill-posed).

 

[PeterDonis]

Note that I made an unfortunate typo for the RIndler metric in my original post, which I've since corrected.

Ah, yes, good catch. I'll update my post accordingly.

 

[Herman Trivilino]

But Sienna said SR can deal fine with acceleration. So let us assume constant g is as same as constant acceleration, to figure out what is the potential energy of an object m at a height h, using SR, please.

Think about what you're asking here. The premise is that SR can be used to analyze and understand acceleration. You cannot conclude that it's therefore possible to use SR to analyze and understand everything that involves an acceleration.

In other words, the quantity $mgh$ contains the factor $g$ which is the free fall acceleration. SR involves acceleration. It doesn't follow that $mgh$ can be understood using SR.

In fact, calling $mgh$ the potential energy is an approximation, valid only in the Newtonian realm, and even then valid only when the gravitational field is constant.

 

[Ziang]

You can just use the $mgh$ calculation from Newtonian physics -- as far as I know, SR doesn't "modify" that at all. But what if the object is traveling so fast relative to the Earth that SR is needed to describe the relative motion? does that make the Newtonian calculation less accurate?

The potential energy may not be mgh.
Let us consider the simplest case only. The object m is dropped freely from a height h.
We start with
dE = fdx (as they do in the derivation of mass-energy relation in textbooks)
but for now, x = h, and f = mγ³a = mγ³g
So PE = mgI(γ³dh)
This PE is larger than mgh.

 

[SiennaTheGr8]

The $E$ in your equation $dE = \mathbf{f} \cdot d \mathbf{x}$ is NOT potential energy. It's kinetic energy: https://en.wikipedia.org/wiki/Work_(physics)#Work–energy_principle

 

[Ziang]

The $E$ in your equation $dE = \mathbf{f} \cdot d \mathbf{x}$ is NOT potential energy. It's kinetic energy: https://en.wikipedia.org/wiki/Work_(physics)#Work–energy_principle

After the object m dropped, it will move down with the constant acceleration a during the trip. It is as same as it is pulled by a force f = mγ³a.

 

[PeterDonis]

It is as same as it is pulled by a force f = mγ3a.

No, it isn't, because the frame you are working in is not an inertial frame. In an inertial frame there is no "g"; in order to "simulate" gravity in SR, you have to use a non-inertial frame. An earlier post by @pervect, with some additional details added in a post by me, give the basics of how the simplest version of such a frame--Rindler coordinates--would work, including how potential energy is defined. I suggest reading those posts carefully.