# What is Three Force?

• I
• Kashmir
In Newtonian physics mass cannot be created or destroyed, so in a closed system ##m## cannot change. In a system that can lose mass, there is no general conservation of momentum and ##\vec F = \frac{d\vec p}{dt}## does not in general... hold.

#### Kashmir

We define four force as one which satisfies : ##\mathbf{f}=m \mathbf{a}## where ##\mathbf{a}=\frac{d \mathbf{u}}{d \tau}##.
The quantities in bold face are four vectors.

Hartle pg 88 defines a three force as :
##\frac{d \vec{p}}{d t} \equiv \vec{F}##
Where ##\vec{p}=\frac{m \vec{V}}{\sqrt{1-\vec{V}^2}}## is the three momentum.

Is this three force what we measure? I.e The force that we used in Newtonian mechanics?

Kashmir said:
Is this three force what we measure? I.e The force that we used in Newtonian mechanics?
Essentially, yes.

• Kashmir and topsquark
PeroK said:
Essentially, yes.
Not sure I agree.

The spatial components of the four force are ##d\vec p/d\tau##, and at low speed where Newtonian physics is valid ##\tau\approx t##. So you could argue for the three force or the four force being the generalisation of Newtonian force. A force-measuring apparatus will always be at rest with respect to the object to which the force is applied, so will always measure in its (instantaneous) rest frame which will again give you either the three- or four-force (at least its spatial components) in that frame because they are the same in that frame.

But four force and four acceleration are parallel, which three-force and three-acceleration are not. In that sense, four force is a better generalisation (and four vectors are always more useful in relativity).

• topsquark and vanhees71
It's better to write equations of motion in terms of covariant quantities, i.e.,
$$m \mathrm{d}_{\tau}^2 x^{\mu}=K^{\mu}.$$
The Minkowski force, ##K^{\mu}##, then is a four-vector, ##m## is the scalar (invariant mass), and ##\tau## the proper time of the particle. The Minkowski force is necessarily subject to the constraint
$$K^{\mu} \mathrm{d}_{\tau} x^{\mu}=0.$$
The most simple example is the motion of a particle in an external electromagnetic field (neglecting the notorious radiation reaction, which is a very delicate subject)
$$K^{\mu}=\frac{q}{c} F^{\mu \nu} \mathrm{d}_{\tau} x^{\nu}.$$

• dextercioby and topsquark
vanhees71 said:
It's better to write equations of motion in terms of covariant quantities, i.e.,
$$m \mathrm{d}_{\tau}^2 x^{\mu}=K^{\mu}.$$
The Minkowski force, ##K^{\mu}##, then is a four-vector, ##m## is the scalar (invariant mass), and ##\tau## the proper time of the particle. The Minkowski force is necessarily subject to the constraint
$$K^{\mu} \mathrm{d}_{\tau} x^{\mu}=0.$$
The most simple example is the motion of a particle in an external electromagnetic field (neglecting the notorious radiation reaction, which is a very delicate subject)
$$K^{\mu}=\frac{q}{c} F^{\mu \nu} \mathrm{d}_{\tau} x^{\nu}.$$
Okay, but if you look at where the OP has reached with the Hartle book, this is beyond the level of mathematics presented at this stage.

• Kashmir, topsquark and vanhees71
Sorry, I don't know the book by Hartle. It seems to overcomplicate things though.

vanhees71 said:
Sorry, I don't know the book by Hartle. It seems to overcomplicate things though.
It's actually an introduction to gravity at undergraduate level. Quite basic.

• PeroK
But then, why is he introducing the old-fashioned way of point-particle mechanics in this non-covariant form? It's very confusing.

vanhees71 said:
But then, why is he introducing the old-fashioned way of point-particle mechanics in this non-covariant form? It's very confusing.
I don't know.
I think its safe to think of 3 Force being same as the Force used in Newtonian Mechanics.

Kashmir said:
I don't know.
I think its safe to think of 3 Force being same as the Force used in Newtonian Mechanics.
A lot of care is needed. Rigid bodies are a common simplification in Newtonian physics but they can't work in relativity, which makes forces a much less useful quantity for anything except point particles.

• vanhees71, dextercioby and Kashmir
Whether "three-force" is the Newtonian force is a little subtle for another reason: usually (in my experience), Newtonian force is defined as ##\vec F = m \vec a## as opposed to ##\vec F = \dot{ \vec p }##, the difference being important (in Newtonian physics) only in a variable-mass scenario.

A benefit of choosing ##m \vec a## is that ##\vec F## remains a Galilean invariant even when ##\dot{m} \neq 0##. A downside (arguably) is that it de-emphasizes the primacy of momentum. Another downside is that it makes the transition to SR slightly bumpier, since in SR it really only makes sense to use ##\vec F = \dot{\vec p}##, even when ##m## doesn't change.

SiennaTheGr8 said:
Whether "three-force" is the Newtonian force is a little subtle for another reason: usually (in my experience), Newtonian force is defined as ##\vec F = m \vec a## as opposed to ##\vec F = \dot{ \vec p }##, the difference being important (in Newtonian physics) only in a variable-mass scenario.

A benefit of choosing ##m \vec a## is that ##\vec F## remains a Galilean invariant even when ##\dot{m} \neq 0##. A downside (arguably) is that it de-emphasizes the primacy of momentum. Another downside is that it makes the transition to SR slightly bumpier, since in SR it really only makes sense to use ##\vec F = \dot{\vec p}##, even when ##m## doesn't change.
In Newtonian physics mass cannot be created or destroyed, so in a closed system ##m## cannot change. In a system that can lose mass, there is no general conservation of momentum and ##\vec F = \frac{d\vec p}{dt}## does not in general hold.

PeroK said:
In Newtonian physics mass cannot be created or destroyed, so in a closed system ##m## cannot change.
A closed system's mass can't change in SR, either (though maybe you're talking about the sum of the masses of a system's constituents?).

PeroK said:
In a system that can lose mass, there is no general conservation of momentum and ##\vec F = \frac{d\vec p}{dt}## does not in general hold.
Yes, an open system's momentum can change, but whether ##\vec F = \dot{\vec p}## holds in general is only a matter of definition, isn't it?

SiennaTheGr8 said:
Yes, an open system's momentum can change, but whether ##\vec F = \dot{\vec p}## holds in general is only a matter of definition, isn't it?
It's fairly clear that if sand is leaking out of a truck, the mass (hence momentum) of the "truck" is decreasing in the absence of any external force.

• SiennaTheGr8
PeroK said:
It's fairly clear that if sand is leaking out of a truck, the mass (hence momentum) of the "truck" is decreasing in the absence of any external force.
If you've defined force as ##\dot{\vec p}##, then ##\vec F \neq \vec 0## for the truck here. Whether that ##\vec F## constitutes an "external force" is semantics—not an unimportant consideration, but not a physical one.

I think Appendix C in the following document is relevant: https://scholar.harvard.edu/files/david-morin/files/cmappendices.pdf

Anyway, perhaps we should stop before we veer too off-topic?

SiennaTheGr8 said:
If you've defined force as ##\dot{\vec p}##, then ##\vec F \neq \vec 0## for the truck here.
Well, that's nonsense. There is no force. A force is physical.
SiennaTheGr8 said:
Whether that ##\vec F## constitutes an "external force" is semantics—not an unimportant consideration, but not a physical one.
Forces are physical.
SiennaTheGr8 said:
I think Appendix C in the following document is relevant: https://scholar.harvard.edu/files/david-morin/files/cmappendices.pdf
It seems clear to me that ##F = \frac{dp}{dt}## only works with "variable" mass is when you keep track of all the mass. Morin couldn't be clearer on this point:

Remarks: It is still true that F = dp/dt in this second example, provided that you let F be total
force, and let p be the total momentum. In this example, F is the only force. However, the total
momentum consists of both the sand in the cart and the sand that has leaked out and is falling
through the air.4 A common mistake is to use F = dp/dt, with p being only the cart’s momentum.
The leaked sand still has momentum.

There is an simple example that demonstrates why F = dp/dt doesn’t work when p refers only to the
cart. Choose F = 0, so that the cart moves with constant speed v. Cut the cart in half, and label
the back part as the “leaked sand” and the front part as the “cart.” If you want the cart’s p to have
dp/dt = F = 0, then the cart’s speed must double if its mass gets cut in half. But this is nonsense.
Both halves simply continue to move at the same rate.

• Lnewqban
Sorry, didn't give it a thorough read before posting (at work; remembered reading it a few years ago so quickly dug it up to share).

I'll take another look when I have time. I certainly could be barking up the wrong tree!

Ibix said:
A lot of care is needed. Rigid bodies are a common simplification in Newtonian physics but they can't work in relativity, which makes forces a much less useful quantity for anything except point particles.
I'll keep that in mind that there is more to it than my simplistic thinking. However at my level I guess this level of sophistication suffices. When I'll move on to the next level from Hartle I'm sure I'll ask and discuss it again with better understanding.

Thank you :)

PeroK said:
In Newtonian physics mass cannot be created or destroyed, so in a closed system ##m## cannot change. In a system that can lose mass, there is no general conservation of momentum and ##\vec F = \frac{d\vec p}{dt}## does not in general hold.
To be picky, I searched and found the (Google©) meaning of "closed system" to be: "A closed system is a natural physical system that does not allow transfer of matter in or out of the system", but it's obvious that if, for example, you don't consider the mass that leaves "the system", then you're not very "conservative" to begin with. That's why normally the Reynolds Transport Theorem is used.

I don't know, but if you consider a control volume, then in Newtonian, S-R or G-R is there an scenario in which momentum isn't conserved?

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• dextercioby
Is Morin not putting the cart before the horse here (sorry for the pun)?

PeroK said:
There is an simple example that demonstrates why F = dp/dt doesn’t work when p refers only to the
cart. Choose F = 0, so that the cart moves with constant speed v. Cut the cart in half, and label
the back part as the “leaked sand” and the front part as the “cart.” If you want the cart’s p to have
dp/dt = F = 0, then the cart’s speed must double if its mass gets cut in half. But this is nonsense.
Both halves simply continue to move at the same rate.

What I mean is: if you define ##\vec F = \dot{m} \vec v + m \dot{\vec v}##, then ##\vec F = \vec 0## gives ##\dot{m} \vec v = - m \dot{\vec v}## (and not necessarily ##\dot{\vec v} = \vec 0##); likewise, ##\dot{\vec v} = \vec 0## gives ##\vec F = \dot{m} \vec v## (and not necessarily ## \vec F = \vec 0 ##).

So with that definition, you can't "choose" ## \vec F = \vec 0 ## and ## \dot{ \vec v } = \vec 0 ## for a (moving) object whose mass changes. The sum ## \dot{m} \vec v + m \dot{\vec v} ## is simply what it is.

I still don't see why it would be impossible to define Newtonian force as ## \dot{\vec p} ##. If one did, it would of course be useful to also have a word just for the ##m \vec a## part. Nothing would change about how one goes about solving problems. You'd just have some different terminology and symbols.

Anyway, I'm really not trying to advocate for anything. My main point is just that the "discrepancy" between the Newtonian ##\vec F = m \vec a## and the relativistic ## \vec F = \dot{ \vec p } ## adds some subtlety to the OP's question, which was: is relativistic three-force just the Newtonian force?

I suppose one could define the relativistic three-force as ##\vec F = m \dot{ \vec w } ##, with ## \vec w = \gamma \vec v ##. But in SR, an additional argument in favor of ## \vec F = \dot{ \vec p } ## is that the four-force ## \frac{d \vec P}{d \tau} = \frac{d m}{d \tau} \vec V + \frac{d \vec V}{d \tau} m ## requires the varying-mass term in order for its magnitude to be Lorentz-invariant (i.e., ## \frac{d \vec V}{d \tau} m ## by itself is not generally a four-vector). And the three-vector ## \gamma \dot{ \vec p } ## is the spatial component of that four-vector.

You can define force however you like but not if you want to retain Newton's laws of motion.

As I said, all this confusion is only, because for some reason relativistic point-particle mechanics is often treated even in modern textbooks in a way as if we hadn't understood it much better since Einstein's seminal paper of 1905.

Simply work with covariant (invariant) quantities, i.e., four-vectors and -tensors. Start with the definition of four-velocity (rather than the very complicated non-covariant quantity ##\mathrm{d} \vec{x}/\mathrm{d} t##):
$$u^{\mu}=\mathrm{d}_{\tau} x^{\mu},$$
where
$$\mathrm{d} \tau^2=\frac{1}{c^2} \eta_{\mu \nu} \mathrm{d} x^{\mu} \mathrm{d} x^{\nu}.$$
This implies
$$u_{\mu} u^{\mu}=c^2.$$
Then it's clear that momentum should simply be the spatial part of a four-vector. So it's suggestive to use
$$p^{\mu}=m u^{\mu}=m \frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau}.$$
Here ##m## is the usual mass, occurring also in Newtonian mechanics, and it's a Lorentz scalar (very much simpler than any confusing idea of speed-dependent relativistic masses!). That this is the same quantity as in Newtonian mechanics is clear, because Newtonian mechanics becomes valid for ##|\vec{v}|\ll c##, and ##\vec{p}## in this limit becomes indeed ##\vec{p}=m \mathrm{d} \vec{x}/\mathrm{d} t##.

The definition of the four-momentum then implies
$$p_{\mu} p^{\mu}=m^2 c^2=\text{const}.$$
First of all this implies that
$$p^0=\frac{E}{c} = \sqrt{m^2 c^2 + \vec{p}^2},$$
which implies that ##E## is the energy of the particle, including the "rest energy" ##E_0=E|_{\vec{p}=0}=m c^2##, and this is a convenient "energy zero" convention, because then ##p^0=E/c## is the time-like component of the four-momentum vector.

Second, this "onshell condition" implies that
$$p_{\mu} \mathrm{d}_{\tau} p^{\mu}=0,$$
and then the natural formulation of an equation of motion for a relativistic particle is
$$\mathrm{d}_{\tau} p^{\mu}=m \mathrm{d}_{\tau}^2 x^{\mu}=K^{\mu},$$
where ##K^{\mu}## must fulfill the constraint ##p_{\mu} K^{\mu}=0##. This implies that for a given force ##K^{\mu}(x,u)## this is an equation of motion for three indpendent degrees of freedom, i.e., as in Newtonian mechanics.

SiennaTheGr8 said:
What I mean is: if you define ##\vec F = \dot{m} \vec v + m \dot{\vec v}##, then ...
... force is not even frame independent. It's zero in the rest frame of an unaccelerating object and non-zero in a frame where the object is moving and decomposing into separate parts.

Why not just admit that Morin is correct instead of trying to redefine physics to cover up a mistake that Morin himself describes as "common"?

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• vanhees71
SiennaTheGr8 said:
What I mean is: if you define ##\vec F = \dot{m} \vec v + m \dot{\vec v}##, then ...
PeroK said:
... force is not even frame dependent.

To clarify (so that I don't misunderstand you): did you mean "not even frame independent?"

SiennaTheGr8 said:
To clarify (so that I don't misunderstand you): did you mean "not even frame independent?"
Yes. That was a typo, which I'm happy to admit was a mistake on my part!

• vanhees71
Right. So, I acknowledged that in my very first post in this thread:

SiennaTheGr8 said:
A benefit of choosing ##m \vec a## is that ##\vec F## remains a Galilean invariant even when ##\dot{m} \neq 0##.

PeroK said:
Why not just admit that Morin is correct instead of trying to redefine physics to cover up a mistake that Morin himself describes as "common"?

I'm not trying to redefine anything, and I don't think I've made the mistake you're attributing to me. I'm saying that if you defined ##\vec F = \dot{\vec p}## instead of ## \vec F = m \vec a ## (which is clearly a question of semantics, not physics), then you'd have to stick to it. You couldn't also let ##\vec F = m \vec a## sneak in.

If I'm reading Morin correctly (which I might not be), then I think that he makes that error in his "Remarks" section: on the one hand, he's supposing that we're using ##\vec F = \dot{\vec p}## for the cart alone; on the other hand, he declares upfront that ## \vec F = \vec 0 ##, but then describes a scenario in which ## \vec a = \vec 0 ## and ## \dot{m} \neq 0 ## for a moving object. Can't have it both ways, and the reason I think he's letting ##\vec F = m \vec a## sneak in is this sentence: "Choose ##F = 0##, so that the cart moves with constant speed ##v##."

This is all tangential to my point that OP's question about the relationship between relativistic three-force and the Newtonian force is a little nuanced. I'd be happy to drop the matter or move it to another thread.

• PeroK
It would help very much, if you could precisely describe, which situation with your cart you are discussing and whether it's about a Newtonian or a relativistic treatment of this example!

SiennaTheGr8 said:
I suppose one could define the relativistic three-force as ##\vec F = m \dot{ \vec w } ##, with ## \vec w = \gamma \vec v ##. But in SR, an additional argument in favor of ## \vec F = \dot{ \vec p } ## is that the four-force ## \frac{d \vec P}{d \tau} = \frac{d m}{d \tau} \vec V + \frac{d \vec V}{d \tau} m ## requires the varying-mass term in order for its magnitude to be Lorentz-invariant (i.e., ## \frac{d \vec V}{d \tau} m ## by itself is not generally a four-vector). And the three-vector ## \gamma \dot{ \vec p } ## is the spatial component of that four-vector.

Oh, the bolded part here was completely wrong and not what I was thinking of before I wrote it.

What I had in mind was: the ##\frac{d m}{d \tau} \vec V## term is a four-vector too (not that ##\frac{d \vec V}{d \tau} m## isn't a four-vector). So while the Newtonian variable-mass term ##\frac{d m}{dt} \vec v## is frame-dependent, the ##\frac{d m}{d \tau} \vec V## term has a Lorentz-invariant magnitude and therefore is frame-independent.

(I don't seem to be able to edit that post. I'd like to cross it out and correct there also.)