# 4-momentum in GR

In recently closed thread titled Conservation of energy in GR, 4-velocity of popped up baseball along geodestic ##(u^0(r),u^1(r),0,0)## where ##x^1=r,x^2=\theta,x^3=\phi##, is derived.

In SR, 4-momentum of ball is ##m(u^0,u^1,0,0)## in contravariant component and ##m(u_0,u_1,0,0)## in covariant component. The two component may differ only in sigature.

In GR, the two componets differ in value also because ##g_{ik}\neq\eta_{ik}## anymore.

My questions are

1. In GR how can we derive energy of the ball using the two components ?

2. For energy-momentum, not 4-vector but two rank tensor ##T^{ik}## seems familiar in GR. Shall I move from 4-vector to 2 rank tensor in GR ? The two ways are compatible in SR/GR ?

## Answers and Replies

Orodruin
Staff Emeritus
Homework Helper
Gold Member
2021 Award
1. In GR how can we derive energy of the ball using the two components ?
Defined how and according to whom? There are many things you might mean by "energy of the ball".

2. For energy-momentum, not 4-vector but two rank tensor TikTikT^{ik} seems familiar in GR. Shall I move from 4-vector to 2 rank tensor in GR ? The two ways are compatible in SR/GR ?
The use of 4-momentum and the energy-momentum tensor is the same in GR as it is in SR. A point mass has a 4-vector describing its 4-momentum and a continuum has a rank 2 tensor describing its energy, momentum, and internal stresses.

PeterDonis
Mentor
how can we derive energy of the ball using the two components ?

Your question here has the same problem that got a previous thread of yours closed: you need to be more specific about what definition of "energy" you are talking about.

Thanks. As a beginner, I am not aware of varieties of energy in GR. I should appreciate it if you show some references I can learn about it.

PeterDonis
Mentor
As a beginner, I am not aware of varieties of energy in GR. I should appreciate it if you show some references I can learn about it.

All of them have already come up in this and other threads. A quick list off the top of my head:

(1) The energy of an object as measured by an observer at rest in a local inertial frame. This works (within the confines of the LIF) like energy in SR.

(2) The invariant mass of an object. Note that this also works in SR (and in GR as well, it's just the norm of the 4-momentum).

(3) The energy at infinity of an object moving geodesically in a stationary spacetime.

(4) The stress-energy tensor (or particular contractions of it with chosen vectors, or integral quantities derived from it).

The problem is that none of these appear to match what you are asking about in the OP. So we need to know what you are asking about in the OP: what do you mean by "energy of the ball using the two components"? Note that this doesn't even make sense, as you state it, in SR, let alone GR. You also ask about the stress-energy tensor, but you're not clear about why you would want to "move" to it, or what "moving" to it even means.

sweet springs
DrGreg
Gold Member
The 4-momentum vector is conserved in GR in the sense that its time-derivative is zero, for an inertial particle, and is also locally conserved in particle collisions. "Time-derivative" here means the appropriate coordinate-independent derivative ##\mathrm{D}\mathbf{P} / \mathrm{d}\tau##, the absolute or intrinsic derivative with respect to proper time, which in coordinate form is$$\frac{\mathrm{D}P^\alpha}{\mathrm{d}\tau} = \frac{\mathrm{d}P^\alpha}{\mathrm{d}\tau} + \Gamma^\alpha{}_{\beta\gamma} \, P^\beta \frac{\mathrm{d}x^\gamma}{\mathrm{d}\tau}$$or, for the covector,
$$\frac{\mathrm{D}P_\beta}{\mathrm{d}\tau} = \frac{\mathrm{d}P_\beta}{\mathrm{d}\tau} - \Gamma^\alpha{}_{\beta\gamma} \, P_\alpha \frac{\mathrm{d}x^\gamma}{\mathrm{d}\tau}$$In general an individual component ## P^\alpha ## is not conserved unless ## \Gamma^\alpha{}_{\beta\gamma} \, P^\beta \, \mathrm{d}x^\gamma / \mathrm{d}\tau = 0 ##, and, similarly, component ##P_\beta## is not conserved unless ## \Gamma^\alpha{}_{\beta\gamma} \, P_\alpha \, \mathrm{d}x^\gamma / \mathrm{d}\tau = 0 ##.

In a stationary spacetime, it's possible to choose a coordinate system (aligned to the timelike Killing vector field) in which all of the components of the metric are independent of the time coordinate ##t## and it can be proved in this case that ## \Gamma^\alpha{}_{t\gamma} \, P_\alpha \, \mathrm{d}x^\gamma / \mathrm{d}\tau = 0 ##, so that ##P_t## is conserved. Note that it is the covariant component ##P_t## that is conserved, not the contravariant component ## P^t ##. It therefore makes sense to describe ##P_t## as "energy". If ##t## is appropriately scaled, this is the "energy at infinity" referred to earlier in this thread. It includes both kinetic energy and gravitational potential energy (as well as rest mass energy).

The same priniciple applies to spacetimes with a spacelike Killing field, which can give rise to conserved linear momentum ## P_z ## or angular momentum ## P_\phi ##.

PAllen and sweet springs
Thanks PeterDonis
(2) The invariant mass of an object. Note that this also works in SR (and in GR as well, it's just the norm of the 4-momentum).
I got it. stevendaryl suggested it in the previous thread.
(3) The energy at infinity of an object moving geodesically in a stationary spacetime.
PeterDonis by Killing vector and stevendaryl by geodesic equation taught me in the previous thread.
(1) The energy of an object as measured by an observer at rest in a local inertial frame. This works (within the confines of the LIF) like energy in SR.
Question 1 in my OP is about it. Is this energy given by contravariant component and its pair covariant in IFR or covariant component and its contravariant pair in IFR? They give different results. Or such covariant contravariant pair is a wrong idea ?
(4) The stress-energy tensor (or particular contractions of it with chosen vectors, or integral quantities derived from it).
This is concerning Question 2. Is energy momentum tensor the easiest and the most convenient way to describe energy-momentum in GR simply by writing down product of 4-velocity components ,like ##\epsilon u^i u^j ## or so, needless to regard various energy cases like (1)-(3) ?

Last edited:
PAllen
"1) The energy of an object as measured by an observer at rest in a local inertial frame. This works (within the confines of the LIF) like energy in SR."
Specifically to relate this to @sweet springs follow up question, the contraction of a particle's 4-momentum with an instrument's 4-velocity gives the total energy of the particle as measured by that instrument. Equivalently, this is the energy in an LIF in which the instrument is at rest.

sweet springs
PAllen
"1) The energy of an object as measured by an observer at rest in a local inertial frame. This works (within the confines of the LIF) like energy in SR."
Specifically to relate this to @sweet springs follow up question, the contraction of a particle's 4-momentum with an instrument's 4-velocity gives the total energy of the particle as measured by that instrument. Equivalently, this is the energy in an LIF in which the instrument is at rest.
Let me add that the value of the time component of 4 momentum in general coordinates (even with restriction of 1 timelike and 3 spacelike) has no particular meaning in GR. If you want the energy as measured by an instrument 'at rest' in such general coordinates, you have to compute a unit vector in the timelike basis direction and contract with the 4-momentum.

Last edited:
sweet springs
Thank you PAllen.

Actually in the case of baseball popup, 4-velocity along geodesic are:

$$u^0=\frac{\sqrt{1-\frac{2M}{r_{max}}}}{1-\frac{2M}{r}}$$

$$u_0=\sqrt{1-\frac{2M}{r_{max}}}$$.

$$u^1=\sqrt{\frac{2M}{r}-\frac{2M}{r_{max}}}$$

$$u_1=\frac{\sqrt{\frac{2M}{r}-\frac{2M}{r_{max}}}}{1-\frac{2M}{r}}$$

Say ##U^i## as 4-vector of instrument in LIF of SR,

Taking an inner product of contravariant 4-velocity of instruments in LIF, say

$$U^i=(1,0,0,0)$$

$$u_iU^i=u_0=\sqrt{1-\frac{2M}{r}}$$

Taking an inner product with covariant 4-velocity of instruments in LIF, say

$$U_i=(1,0,0,0)$$

$$u^iU_i=u^0=\frac{\sqrt{1-\frac{2M}{r_{max}}}}{1-\frac{2M}{r}}$$

Or say ##U^i## as 4-vector of insryment in GR,

$$U^i=(\frac{1}{\sqrt{g_{00}}},0,0,0)$$

$$U_i=(\sqrt{g_{00}},0,0,0)$$

$$u^iU_i=u_iU^i=\sqrt{\frac{1-\frac{2M}{r_{max}}}{1-\frac{2M}{r}}}$$

It seems that the last line is energy of case (1) PeterDonis listed. Is 'unit timelike basis vector' you mentioned is above
$$U^i=(\frac{1}{\sqrt{g_{00}}},0,0,0)$$ or do I have to do calculation in another way?

PeterDonis
Mentor
taking an inner product of contravariant 4-velocity of instruments in LIF

The vectors and components you are using are not in the LIF; they are in global Schwarzschild coordinates. In an LIF, the 4-velocity of a free-falling object at rest in the LIF (which is the usual case for this kind of problem) is ##(1, 0, 0, 0)##. And the timelike Killing vector field no longer has components ##(1, 0, 0, 0)## in an LIF (it can't, because the energy at infinity still has to be a constant frame-independent value).

Thanks PeterDonis. Using terms in post#10, you say u is in Schwartzushild coordinate and U is in LIF of SR. In the last calculation I take U in Schwartzshild coordinate also. Are such a treatment and thus derived result of energy all right ?

PeterDonis
Mentor
Using terms in post#10, you say u is in Schwartzushild coordinate and U is in LIF of SR.

The way you have written ##u##, it is in Schwarzschild coordinates, yes. I'm not sure what you mean by ##U##. If you mean it to be a vector in the LIF, then your formulas are incorrect, because you can't mix components in different coordinate charts. In Schwarzschild coordinates, the 4-velocity of a measuring device momentarily at rest at radius ##r## is ##\left( 1 / \sqrt{1 - 2M / r}, 0, 0, 0 \right)##. That is what you should plug into the formula ##u_i U^i## in order to obtain an inner product, using the ##u_i## components you are using.

Thanks to your teachings, now I can tell what I thoght in the previous closed thread as follows

Making use of the case of ball popup for presentation of my thought,
energy type (1),
$$\sqrt{\frac{1-\frac{2M}{r_{max}}}{1-\frac{2M}{r}}}$$,
has properties that
-the ball holds it. and
-Along trajectory it changes so does not conserve.

Energy type (3),
$$\sqrt{1-\frac{2M}{r_{max}}}$$
has properties that
-It is enegy of the system as a whole. No holders in the system are identified, and
-It is constant. Along geodesic it conserves.

Hoping that
-holders of any energy should be identified, and
-total energy conserbvation law stands,

I would combine these two energies in the way that

$$energy\ type(1) \ +\ \phi \ =\ energy\ type(3)$$

where ##\phi## is a entity intoroduced and called potential energy.
Actually
$$\phi\ =:\ \sqrt{1-\frac{2M}{r_{max}}}\ (\ 1-\frac{1}{\sqrt{1-\frac{2M}{r}}}\ )\ \ <\ 0$$

In SR, such ##\phi## exists for examples as elastic energy of spring attached to the ball, interaction energy of charged ball and electro magnetic field. They have holders(spring, field, etc. ) and distribution of potential energy is identifiable. Local energy transfer flow between
##\phi## and the ball are clearly discribed.

In analogy from this SR case, this ##\phi## in GR could be also called as gravitational potential energy. However its holder or spacial distribution with invariance under coordinate transformation can never be identified.
Rogorousy ##\phi## is a fake product. It is a subtract of two different type enegies so it is very artificial invention in analogy with potential energy of electromagnetism.

In approximation of 2m<<r
$$Energy\ type\ (1)\ \approx 1-\frac{M}{r_{max}}+\frac{M}{r}$$
$$Energy\ type\ (3)\ \approx 1-\frac{M}{r_{max}}$$
$$\phi\approx -\frac{M}{r}$$

$$Energy\ type\ (1)\ \approx 1-\frac{M}{r_{max}}+\frac{M}{r} = 1 + \frac{v^2}{2}$$ i.e. $$\frac{v^2}{2}={-\frac{M}{r_{max}}+\frac{M}{r}}$$
Thanks.

Last edited:
I would go farther though with less conviction.
In the ball popup case, the ball energy of type(1) decreases and potentian energy ##\phi## increases as the ball goes up. May I say without referring ##\phi## that even the ball energy of type(1) decreases, "energy" of the ball, not conserved energy of type(3) which is constant and whose owner cannot be identified, is conserved ? Here "energy" of the ball I am thinking comes from the integral of energy-momentum tensor in region of the ball. Covariant derivative form of energy-momentum conservation law
$$T^{ik}_{\ :k}=0$$ mention that no flow in or out of the region the ball occupying happens during its movement in covariant derivative divergence sense. So its volume integral, I mentioned it as "energy" of the ball above, is conserved in this sense. Then we do not need artificial ##\phi# to explain decrease of energy of type(1) as ball goes up. Its decrease is conservation. Decrease of energy of type(1) as the ball goes up is compatible with conservation of "energy" of the ball in GR. It stands as the law of inertial motion or conservation of energy of the ball in Newtonian mechanics and SR stand.

Last edited:
stevendaryl
Staff Emeritus
I would go farther though with less conviction.
In the ball popup case, the ball energy of type(1) decreases and potentian energy ##\phi## increases as the ball goes up. May I say without referring ##\phi## that even the ball energy of type(3) decreases, "energy" of the ball, not conserved energy of type(3) which is constant and whose owner cannot be identified, is conserved ? Here "energy" of the ball I am thinking comes from the integral of energy-momentum tensor in region of the ball. Covariant derivative form of energy-momentum conservation law
$$T^{ik}_{\ :k}=0$$ mention that no flow in or out of the region the ball occupying happens during its movement in covariant derivative divergence sense. So its volume integral, I mentioned it as "energy" of the ball above, is conserved in this sense. Then we do not need artificial ##\phi# to explain decrease of energy of type(1) as ball goes up. Its decrease is conservation. Decrease of energy of type(1) as the ball goes up is compatible with conservation of "energy" of the ball in GR. It stands as the law of inertial motion or conservation of energy of the ball in Newtonian mechanics and SR stand.

I'm not sure that this distinction helps, but there is a distinction between local conservation laws and global constants of the motion. The stress-energy tensor ##T_{ik}## is locally conserved in any spacetime. However, that doesn't lead to a "constant of the motion" except in special cases.

In a spacetime where the metric is independent of the time coordinate, the corresponding t-component of momentum, ##p^t \equiv \frac{dt}{d\tau}## is a constant of the motion. Locally, all 4 components of the 4-momentum are conserved, but the other three don't correspond to constants of the motion.

sweet springs
PeterDonis
Mentor
energy type (1)
has properties that
-the ball holds it

No, that's not correct, because this energy is the energy of the ball as measured by a static observer at radius ##r##, so it's not just a property of the ball alone; it's a property of the ball + the static observer. It's not "conserved" because at each different radius ##r## there is a different static observer, so this energy at two different values of ##r## refers to two different things.

Energy type (3)
It is enegy of the system as a whole. No holders in the system are identified

This is not correct either. The ball is a "holder" of this energy. It is true that this energy is a conserved quantity because of a property of the spacetime as a whole--that it has a timelike Killing vector field--but it is the 4-momentum of the ball that determines its value.

I would combine these two energies

This has no physical meaning and I don't know why you are doing it. You were already shown in the previous thread how the GR analogue of Newtonian gravitational potential energy is defined, and this isn't it.

PeterDonis
Mentor
I would go farther though with less conviction.

It's good that you have less conviction, because you are simply going off into personal speculation here. Please review the forum rules.

Covariant derivative form of energy-momentum conservation law

Nothing that you have talked about has anything to do with this. You are talking about a ball in geodesic motion in a vacuum; there is no stress-energy anywhere. Note that the stress-energy tensor was energy type (4) in my post. It has nothing to do with energy types (1) or (2) or (3).

Nothing that you have talked about has anything to do with this. You are talking about a ball in geodesic motion in a vacuum; there is no stress-energy anywhere. Note that the stress-energy tensor was energy type (4) in my post. It has nothing to do with energy types (1) or (2) or (3).

Does the ball material density ##\epsilon##(+ pressure? and with defining region where the ball material distribute?) multiplied by
$$u^0u^0,u^0u^1,u^1u^1$$
,value of which are given from post #10, give us contravariant component stress-energy tensor description of the ball in motion ?

PeterDonis
Mentor
Does the ball material density ##\epsilon## (+ pressure? and with defining region where the ball material distribute?) multiplied by
$$u^0u^0,u^0u^1,u^1u^1$$
,value of which are given from post #10, give us contravariant component stress-energy tensor description of the ball in motion ?

No. First, treating the ball as a continuous extended object means throwing away all of the formulas we've used so far; all of those formulas assume that the ball is a point-like test object, with negligible effect on the spacetime geometry. Giving the ball a stress-energy tensor means it now has a non-negligible effect on the spacetime geometry.

Second, treating the ball as a continuous extended object means it no longer has a single 4-velocity ##u##. The 4-velocity can be different at different points within the ball. Also, in addition to just energy density and pressure, the stress-energy includes momentum and shear stress. So this kind of model is much more complicated (which is why physicists don't use it in cases where it can be avoided, like cases where the ball can be treated as a test object with no effect on the spacetime geometry).

You were already shown in the previous thread how the GR analogue of Newtonian gravitational potential energy is defined, and this isn't it.
In your post#30 in the previous thread "Conservation of energy in GR", you show the potential energy mgH for the ball is at the top h=H and still. How do you get potential energy with the ball in motion at h<H ? In other words I would like to know how you estimate kinetic energy of the ball in GR to subtract it from mgH in order to get potential energy for the ball in motion.

PeterDonis
Mentor
In your post#30 in the previous thread "Conservation of energy in GR", you show the potential energy mgH for the ball is at the top h=H and still. How do you get potential energy with the ball in motion at h<H ?

I told you in the previous thread: solve the geodesic equation. Or, if you accept that the Newtonian approximation is valid for this case, just use the Newtonian solution, which tells you ##v## as a function of ##h < H## and easily confirms that ##m v^2 / 2 + m g h## is constant.

Note that, even if you use the full GR geodesic equation, solving it is easier than it might seem, since we are talking about purely radial motion. The key is to properly understand what quantity in the math corresponds to the Newtonian ##v##, which in turn requires you to properly understand what ##v## represents, physically.

f you accept that the Newtonian approximation is valid for this case, just use the Newtonian solution, which tells you vv as a function of h<Hh < H and easily confirms that mv2/2+mghm v^2 / 2 + m g h is constant.
Again from your post#30 in the previous thread
$$\sqrt{1-\frac{2M}{R+H}}=potential\ energy\ +kinetic\ energy\ + 1(mass)$$
LHS you give as a constant of motion I divided to three parts in RHS that meets Newton Mechanics with mass in SR. Newtonian approximation way, potential energy =gh, kinetic energy=1/2 v^2 is a case satisfying the above relation.
Does rigorous GR has no definite way to devide energy to P.E. and K.E. (and even mass)? If so, what kind of policy is used to derive Newtonian way of division ? I show my idea in post#14.

Last edited:
all of those formulas assume that the ball is a point-like test object, with negligible effect on the spacetime geometry.
Point-like test object is simple and so a good one to ingestigate. Let us take an electron for example.
$$T^{ik}=m\delta^{(4)}(x-x_e)u^iu^k$$ where ##x_e## in ##\delta## function is 4-coordinate of the electron, m is mass of electron.
May I say like that?
Giving the ball a stress-energy tensor means it now has a non-negligible effect on the spacetime geometry.
Very high speed electron(s) become source of gravity.

PAllen
Very high speed electron(s) become source of gravity.
In isolation, no they do not. Any electron is ultrareletivistic in velocity in some frame, but curvature is frame and coordinate independent. An electron interacting with a body relative to which it has high speed will “produce gravity”. More accurately, the stress energy of the system will correspond via the EFE to a metric showing curvature in this region.

In isolation, no they do not. Any electron is ultrareletivistic in velocity in some frame, but curvature is frame and coordinate independent
Thank PAllen for correction.
So I would say as for contribution of an electron to geometry, even m is so small, it has Schwartzushild metric around itself which has tiny difference from Minkowsky's. Tiny but not zero stress energy tensor of electron shall be nececessary as RHS of Einstein equation.

PAllen
Thank PAllen for correction.
So I would say as for contribution of an electron to geometry, even m is so small, it has Schwartzushild metric around itself which has tiny difference from Minkowsky's. Tiny but not zero stress energy tensor of electron shall be nececessary as RHS of Einstein equation.
Not quite. In a pure classical treatment, an electron would be represented as a Kerr-Newman BH because it has both spin and charge.

sweet springs
PeterDonis
Mentor
May I say like that?

You're still missing the point. If the ball is a test object, it has no stress-energy tensor; it's stress-energy is zero. Anything with nonzero stress-energy changes the geometry of spacetime. That means the metric is no longer the Schwarzschild metric, so everything we have done up to now is no longer valid.

Not quite. In a pure classical treatment, an electron would be represented as a Kerr-Newman BH because it has both spin and charge.
Awesome. Thanks.

PeterDonis
Mentor
from your post#30 in the previous thread

√1−2MR+H=potential energy +kinetic energy +1(mass)​

No, that's not from post #30 in my previous thread. That's something you made up yourself.

What I said in post #30 in the previous thread is that ##m \sqrt{1 - 2M / (R + H)}## is the energy at infinity of the ball. Then I used the Newtonian approximation to show where ##m g H## comes from in that approximation. Nowhere did I write down a formula like the one you wrote above.

Does rigorous GR has no definite way to devide energy to P.E. and K.E. (and even mass)?

I've already told you how to do this. Several times. Are you not reading my posts?

I show my idea in post#14.

Your post #14 mixes together different definitions of energy and hence can't be correct as it stands. If you want to derive the GR equivalent of the Newtonian formula ##m v^2 / 2 + m g h##, I've already told you how to do this. Several times. The way I've told you is pretty straightforward if you make use of the fact that the energy at infinity (your energy type 3 in post #14) is a constant of the motion; you don't even have to solve the full geodesic equation. (Some textbooks call this method, using constants of the motion to simplify the solution, the "effective potential" method.)

PeterDonis
Mentor
I would say as for contribution of an electron to geometry, even m is so small, it has Schwartzushild metric around itself which has tiny difference from Minkowsky's. Tiny but not zero stress energy tensor of electron shall be nececessary as RHS of Einstein equation.

Not quite. In a pure classical treatment, an electron would be represented as a Kerr-Newman BH because it has both spin and charge.

If you want to analyze an electron as a source of gravity, yes, this is what you would do. But again, as I said in post #28, if you do this, everything we have discussed up to now about the geodesic motion of an object in the Schwarzschild geometry is invalid. We would have to start from scratch and analyze things using the spacetime geometry including the effects of the electron (or the ball, or whatever object you want to use). Which is way, way beyond the scope of a "B" level thread or even an "I" level thread.

PAllen
You're still missing the point. If the ball is a test object, it has no stress-energy tensor; it's stress-energy is zero. Anything with nonzero stress-energy changes the geometry of spacetime. That means the metric is no longer the Schwarzschild metric, so everything we have done up to now is no longer valid.

I do not catch your comment as for to my post#24 on energy-momentum tensor of a particle.
A particle cannot form energy-stress or energy-momentum ,I think they refer the same, tensor ? In
https://www.reed.edu/physics/courses/Physics411/html/page2/files/Lecture.19.pdf
I find the formula (19.5).

PeterDonis
Mentor
I do not catch your comment as for to my post#24 on energy-momentum tensor of a particle.

The issue is not how you would define the stress-energy tensor of a point particle. The issue is that if you want to talk about "energy" in terms of the stress-energy of a point particle, or a baseball, or anything else, that is a different topic from the topic we have been discussing, which is the energy at infinity of a test object moving geodesically in Schwarzschild spacetime. As I've said before, you need to decide what specific topic you want to discuss; trying to mix them together is just increasing your confusion.

What I said in post #30 in the previous thread is that m√1−2M/(R+H)m1−2M/(R+H)m \sqrt{1 - 2M / (R + H)} is the energy at infinity of the ball. Then I used the Newtonian approximation to show where mgHmgHm g H comes from in that approximation.
So my question in other words is how Newtonian approximation, e.g. K.E=1/2 m v^2 v for what ? still to what ? observer should be staying still at the same radius or free-falling? coordinate time or proer time?, is introduced and authorized in GR.
$$K.E.=\frac{1}{2}mv^2$$ in Newtonian Mechanics
$$E=m\frac{1}{\sqrt{1-v^2}}$$ in SR, But we are in GR. How do you allow or not allow them to use formula in GR as approximation suggesting such and such observation conditions.
If we have GR formula of mass+kinetic energy, Newton and SR formula will be given from it by approximation. Do we have such a formula ? If we do not have it, how we can admit Newton and SR formula as they are?

Last edited:
PeterDonis
Mentor
If we have GR formula of mass+kinetic energy, Newton and SR formula will be given from it by approximation. Do we have such a formula ?

The simple "B" level answer? Yes, there is such a formula. (I thought that was already clear from my previous answers, but I guess not.) I've already told you, a number of times now, that you find out what it is by solving the geodesic equation. You can use the fact that the energy at infinity is a constant of the motion to simplify the calculation.

Now go do that, and start a new thread (probably at the "I" level at least; solving the geodesic equation is probably beyond "B" level) if you have questions about it. This thread is closed.

Sorcerer