7
$\begingroup$

How does curved space explain why a denser object of the same shape and volume feels heavier?

$\endgroup$
4
  • $\begingroup$ are you taking a scenario in which you are holding two objects in which one of them is denser ? $\endgroup$
    – Ankit
    Sep 29 '20 at 3:39
  • 1
    $\begingroup$ Just to warn you, some of the answers here are fine but many of them show that whoever wrote them knows very little about this area of physics and they are getting it all in a muddle. I didn't have the heart to go through putting downvotes since I have also provided an answer (which agrees with the correct ones, of course, but is shorter and I hope clearer). $\endgroup$ Oct 12 '20 at 22:54
  • $\begingroup$ @AndrewSteane I find your comment very mean. In your answer you jump, at the very crucial point, from general relativity to Newtonian mechanics. Well thank you very much, I don't see how such a move is less muddled than others. $\endgroup$
    – pglpm
    Oct 13 '20 at 6:08
  • $\begingroup$ @pglpm That 'jump' is a standard procedure in GR. You can call it a use of a local inertial frame, or an appeal to the equivalence principle, or simply how observations are calculated. $\endgroup$ Oct 13 '20 at 9:14

11 Answers 11

4
$\begingroup$

The weight of an object is the amount of force that has to be provided to prevent the object from moving towards some other nearby object such as planet Earth.

If you have two objects of different mass, both near planet Earth, then in the absence of some force to stop them doing it, both will follow the same trajectory if they start out from the same place with the same speed. In other words, they have the same acceleration. This is because they both follow the straightest possible line through spacetime, and there is just one such line (for given starting conditions) so both will follow it.

The line they follow approaches Earth more and more rapidly.

Now if you want to prevent either object from following that line, then you are going to have to provide a force, such as the force from your hand, which is ultimately an electromagnetic force (with some quantum mechanics involved too). You will have to provide more force on the more massive object, because you are trying to remove the same amount of acceleration for both objects, and you can use Newton's second law, $f = m a$. Yes it really is as simple as that. In General Relativity, laws such as this one apply to motions involving low relative velocity and small regions of spacetime. So since you have to provide more force to steer the more massive object away from its freefall line, it will have more weight.

(Added note for readers wishing to know how Newton's law popped in here. To calculate the force observed by a given observer, one adopts a local inertial frame which is momentarily at rest relative to that observer. In this frame the laws of physics are just the same as in all other local inertial frames, and are not affected by gravitation. In particular you have ${\bf f} = d {\bf p}/dt$ and ${\bf p} = \gamma m {\bf v}$ where $\gamma=1$ for an object at rest in the frame.)

$\endgroup$
2
  • 1
    $\begingroup$ Nice answer. i read it and again ang again. It made me realize that my answer was not completely correct and needed clarification. You made me realize that gravity, spacetime curvature just accelerates both objects (approximately at the same rate), but the heavier object (more stress energy) is the one that has more inertia. That is the ultimate answer to the question I believe, inertia. When we hold both objects in our hands, we just feel one of them heavier because that one has more inertia (because it is more massive/more stress-energy), gravity just accelerates them. Thank you so much! $\endgroup$ Oct 13 '20 at 5:21
  • $\begingroup$ I'd like to add a precisation about local inertial frames & laws of physics. Tthe laws of physics can still be different from Newtonian mechanics in a local inertial frame. For example, if a body is emitting or absorbing heat, its 4-momentum is not collinear with its 4-velocity. This means that the force won't generally be collinear with $\mathrm{d}\pmb{v}/\mathrm{d}t$, the acceleration of the body. So even in a local inertial frame we can observe phenomena at variance with Newtonian (thermo)mechanics, even if such effecst may be negligible in most cases. $\endgroup$
    – pglpm
    Oct 22 '20 at 8:36
1
$\begingroup$

GR says that gravity is a fictitious force, which exists only because you chose a noninertial frame of reference. GR defines a noninertial frame as one that isn't free-falling.

So the fact that gravitational forces are proportional to mass is explained simply because noninertial forces, in the Newtonian approximation, are always proportional to mass. For example, centrifugal and Coriolis forces are proportional to mass, as is the fictitious force you feel in an accelerating elevator.

The reason that fictitious forces are always proportional to mass in the Newtonian approximation is that they produce the same acceleration on all test particles, and in the Newtonian approximation we have $F=ma$.

It is not true in GR, outside the Newtonian approximation, that gravitational forces are always proportional to mass. For example, the radiation reaction force on one of the stars in a binary star system is proportional to the square of the star's mass. This is the same as in the case of electromagnetic radiation, where the radiation reaction force is proportional to the square of the charge.

$\endgroup$
1
  • $\begingroup$ Re, "The reason that [gravitational mass is proportional to inertial mass] in the Newtonian approximation is..." No reason at all. Newton observed that relationship. Nothing in Newton's theory explains it. But, the explanation is trivial in General Relativity: The force that Newton called "gravity" is just a manifestation of inertia (see some of the other answers here for more details on that.) $\endgroup$ Oct 13 '20 at 12:42
1
$\begingroup$

This will be a very loose, hand wavy explanation.

Massive objects curve spacetime. Objects freely falling in curved spacetime follow a "straight" trajectory, much like they do in flat spacetime. But "straight" doesn't mean what you might think.

A common analogy is driving on the curved surface of the Earth. If you curve left or right, you drive in a circle. But if you go "straight" you also drive in a circle all the way around the Earth. Because Earth is curved, so is a "straight" path.

A "straight" path on Earth is the least curved of all possible paths on Earth. The curvature of the path is determined by the curvature of Earth. All "straight" paths have the same curvature.

This "straight" path that follows the curved Earth is different than a truly straight path. The curvature of Earth is so small that you usually don't notice the difference. But if you could go truly straight, you would notice after a few miles.

General Relativity is something like this. Freely falling objects in curved spacetime follow the straightest possible curved path. The curvature near Earth is strong enough that you notice the difference between the "straight" path in curved spacetime and the straight path in flat spacetime right away.

Near Earth, the Earth determines how curved spacetime is. It isn't obvious, but all objects follow a path of the same curvature. You can see a hint of this if you throw a big and small rock at the same speed and direction. They follow the same trajectory at the same speed.


It isn't obvious what a curved or "straight" path in spacetime means. One way to think of it is that we travel in the 3 space dimensions like usual, and the time dimension toward the future. The time direction is something like perpendicular to all the space dimensions. At ordinary speeds, the future-ward speed of everything is about 1 second per second.

If you see something moving fast, it doesn't go as far into the future in one of your seconds. It's clock runs slower than yours. Unless you are measuring time very accurately, this isn't noticeable unless the object is traveling near the speed of light. But it is a real, tiny, effect even at ordinary speeds.

People talk about spacetime because space and time are more closely related than you might expect. In spacetime, 186,000 miles of distance is just a far as 1 second of time. So in a sense, we are traveling into the future at about 186,000 miles/second.


Let's consider a couple examples of paths in spacetime near Earth.

Toss a small rock. It lands a few feet away in distance and 1 second away in time. This means the highest point of the trajectory was 16 feet. The space part of the trajectory is sharply curved. But keep in mind that the rock landed the equivalent of 186,000 miles away in time. The curvature of a trajectory 16 feet high and 186,000 miles long is extremely close to flat. You can calculate the radius of a circle like that is about 1 light year.

This means the gravity from Earth is extremely weak. When astronomers talk about strong gravity, they means something like a black hole, where an object might be traveling at relativistic speeds 1 second after being dropped. Never the less, it is plenty strong enough for us.


Another point is that if you toss a big and small rock together, they follow the same trajectory in spacetime. They rise to the same height and land the same distance and time away.

Getting back to classical physics, this means they have the spatial trajectory and take the same amount of time to follow it. They follow a parabola with the same velocity and acceleration. The acceleration of gravity is the same for big and small rocks.


Consider a second example. You point a rifle just slightly upward from horizontal and fire a bullet. We do this in a very large vacuum chamber so air doesn't slow the bullet. We chose the angle so the bullet rises to a maximum height of 16 feet. It follows a very flat curve and lands 1 second later about a mile away.

Once again it isn't obvious, but this trajectory has the same curvature as the rock. It is clearly almost the same. The other end is a mile away in distance and 186,000 miles away in time. You might think that it is slightly longer than the rock's trajectory because a mile is longer than a few feet.

But it isn't for two reasons. First, time isn't really a space dimension perpendicular to all the others. The rule to calculate the total length of a path in spacetime is slightly different from Pythagoras.

Second, the bullet moved faster than the rock. So it didn't travel quite as far into the future as the rock. That is to say the bullet is slightly younger that another bullet that didn't get fired. This too changes the length of the path in spacetime, and changes the curvature of the trajectory. It works out that both have the same curvature.


Switching topics somewhat, there are different notions of what you might mean when you say a denser object of the same size is heavier. All really mean more massive.

Inertial mass: A more massive object is harder to push around, harder to accelerate. Like a big truck needs a bigger engine to get up to speed than a motorcycle.

Active gravitational mass: A more massive object curves space-time more than a less massive object. It deflects objects farther from the trajectory they would have in flat spacetime. This means that Earth generates stronger gravitational forces than the Moon.

Passive gravitational mass: A more massive object is attracted to Earth with a larger force than a less massive object.

These three notions of mass are conceptually different. You can measure the mass of an object by accelerating it, letting it attract a mass, or watching as a mass attracts it. Physicists have found no deep reasons why the three answers must come out the same. But even extremely sensitive experiments always show they do. This is one of the deep mysteries of physics. We can't show they are the same. We have to assume it. This assumption is used in the foundation of Newton's laws and to derive General Relativity.

It is also part of the explanation of why a more massive object is heavier.


Suppose you are holding large and small rocks. You could drop them and let them follow their "straight" trajectories toward Earth. If you do, they accelerate downward at the same rate, $g$.

But you don't. You exert upward forces on them, forcing them away from this trajectory toward ones that remain a constant distance from Earth. To do this, the upward force must be the same as the force of gravity. The gravitational force is proportional to the mass.

$$F_{gravity,1} \propto m_1 g = F_{inertial,1}$$

$$F_{gravity,2} \propto m_2 g = F_{inertial,2}$$

So a denser rock is heavier than a less dense rock.


Perhaps the biggest takeaway form this is that even though GR is the true and correct answer, this is a lot easier.

$$F = \frac{GmM_{Earth}}{r_{Earth}^2}$$

$\endgroup$
2
  • $\begingroup$ If an object moves through space time do we perceive its momentum (which is of course proportional to mass) as weight? $\endgroup$
    – releseabe
    Sep 29 '20 at 2:04
  • $\begingroup$ No. Weight is a force. $\endgroup$
    – mmesser314
    Sep 29 '20 at 3:11
1
$\begingroup$

How does curved space explain why a denser object of the same shape and volume feels heavier?

The space we see around us is flat within our measurement accuracies. When going to energies of special relativity space-time four vectors are involved, but still flatness is withing the Lorenz transformation descriptions.

What tells us that space is curved? The answer is :General relativity, at large masses dominates and introduces space time curvature.

What is general relativity?:

is the geometric theory of gravitation published by Albert Einstein in 1915 and is the current description of gravitation in modern physics. General relativity generalizes special relativity and refines Newton's law of universal gravitation, providing a unified description of gravity as a geometric property of space and time or four-dimensional spacetime. In particular, the curvature of spacetime is directly related to the energy and momentum of whatever matter and radiation are present.

Einstein's equation relates the curvature of space to the energy momentum tensor $T_{μν}$.

eineq

The result is : if there is no matter to supply energy and momentum so that a tensor can be defined , there is no space curvature, it is flat.

So the first level answer to your question is: the existence of mass/energy creates the curvature. The larger the masses the larger the curvature.

So a denser object will give a stronger energy-momentum tensor and greater curvature according to the mathematics of general relativity.

$\endgroup$
6
  • $\begingroup$ Anna, shouldn't the sentence "What tells us that space is curved?" be more precisely about space-time? $\endgroup$
    – Alchimista
    Sep 29 '20 at 12:09
  • $\begingroup$ @alchimista yes, thank you $\endgroup$
    – anna v
    Sep 29 '20 at 13:26
  • $\begingroup$ It was a genuine question. Otherwise I would just suggested an edit. The fact that you didn't edit let me confused. I mean to whose acquainted it is clear that the space under discussion is a 4 dimensional one, but that it is not true to myself that I do my best to grasp some basic content. Again say, while discussing the Eddington et al. observation during the famous eclipse, what the Sun was curving? Space or spacetime? It is an example in which one can see even the xyz space as curved, but I otherwise tend to see the classical space flat even near the Sun edges. Thx for the next answ. $\endgroup$
    – Alchimista
    Sep 29 '20 at 15:55
  • $\begingroup$ @Alchimista As far as know ,when masses are large enough to give measurable effect both space and time are affected/curved i.e. not euclidian. The curving of light behind large stars for example is detected in space.en.wikipedia.org/wiki/Gravitational_lens $\endgroup$
    – anna v
    Sep 29 '20 at 16:48
  • $\begingroup$ Yes thanks that is right what I was asking for $\endgroup$
    – Alchimista
    Sep 30 '20 at 12:05
1
$\begingroup$

I have always found the bowling ball in a trampoline model of gravity to be a poor analogy. Instead, think of gravity more like a low pressure system in the weather, with the pressure a measure of spacetime. The slower is time, the lower the pressure. Nothing in this analogy disagrees with relativity, it is simply a better way to visualize the issue.

An interesting thing about pressure is that it allows for all the behaviors of gravity. For example, first imagine if you pulled three pieces of cork under water, fighting against the pressure of the water: 1kg, 10kg and 100kg. It would obviously be much more difficult to pull the 10 and 100 kg pieces of cork under than the 1 kg piece. Just like attempting to lift the 10 and 100 kg pieces is more difficult than the 1 kg piece. Second, if you imagine that the larger pieces are shaped to have the same water resistance as the smaller piece (ie. they are bullet shaped) and if you pulled all three down to 100 metres and released, they would all rise to the top with the same acceleration, exactly as if you had taken them 100 metres in the air and dropped them down.

So think of why one object is heavier as because you are fighting against greater pressure in lifting it.

$\endgroup$
1
$\begingroup$

Basically Andrew Steane's answer is correct for cases where the objects are relatively small (little stress-energy relative to Earth), I would just like to give clarification about another effect, in the case of objects that have stress-energies that are comparable to that of Earth.

You are specifically asking about why a more massive object feels heavier.

It is very important to understand that it is stress-energy, not mass that creates the effects of gravity (including spacetime curvature).

Now your question is really, why does an object, with more stress-energy, feel heavier, let's say, here on Earth?

For the sake of argument, let's suppose you are holding in one hand a feather, and in the other hand a mini-black hole (disregard other effects).

Why do I feel that the feather is light, while the mini-black hole is extremely heavy?

You are saying that gravity is just spacetime curvature, and both the feather and the mini-black hole are trying to follow geodesics, that is, they both try to move along a path towards the center of Earth. You are saying that that path is determined by the static gravitational field of the Earth.

If they just move along a geodesic path (determined by Earth), why does one feel heavier?

The answer is, on the one hand, inertia (because one object is more massive, it has more inertia) and on the other hand, that both objects have their own gravitational effects, which is determined by their own stress-energy.

Now the former effect (inertia) is the main one here, when we deal with relatively small objects, that have relatively little stress-energy compared to Earth. The latter effect (own gravitational field) only becomes detectable when the objects' stress-energy is of comparable scale to Earth's. That is why I am choosing a mini-black hole for the example, so we can see both effects are at play.

  1. Inertia

This is the main effect that we experience when we talk about relatively small objects, that have relatively little stress-energy (compared to Earth). In this case we can disregard the latter effect (the small objects' own static gravitational field).

  1. Objects' own static gravitational fields

This effect is only detectable at the level of objects comparable the stress-energy of the Earth itself.

The mini-black hole has much more stress-energy then the feather, thus it creates a static gravitational field around itself that is way stronger then the feather's.

This static gravitational field of the mini-black hole effects the Earth just the same way as the Earth's effects the mini-black hole. This mini-black hole might seem small, but it is mighty.

Since the mini-black hole has much more stress-energy, then the feather, it is making the Earth accelerate towards it just the same way the Earth is making both objects accelerate towards Earth. In the case of the feather this effect is so tiny, that it is not detectable at all.

So when you are trying to hold the mini-black hole and the feather at the same distance from Earth, you are trying to work against the static gravitational field of all these objects.

So the inaccuracy introduced by ignoring the motion of the Earth is roughly one part in a trillion trillion, far beyond the sensitivity of any measuring device that exists (or can even be imagined) today.

Don't heavier objects actually fall faster because they exert their own gravity?

And here is a fact, yes, contrary to popular belief, the mini-black hole will accelerate towards the Earth faster then the feather and it will reach the surface sooner then the feather (if you let it go, disregard atmosphere). This is caused by the fact that the mini-black hole has more stress-energy and creates a stronger static gravitational field around itself that effects the Earth itself.

Please note:

  1. gravity is spacetime curvature, not just spatial curvature

  2. the effects of gravity in your case are created by a mutual relationship between the feather and the Earth or the mini-black hole and the Earth, both objects act on each other

So next time you hold a feather and a mini-black hole (please don't), please understand that one of them feels so much heavier, because the effects are due to a phenomenon that is created by a mutual interaction between the feather and the Earth, or the mini-black hole and the Earth, and the mini-black hole has so much more stress-energy. It is just that usually in the case of relatively small (little stress energy) objects, we can disregard the contribution of the small objects' own static gravitational fields.

So the answer to your question is mainly two effects:

  1. inertia, this is the ultimate answer to your question. At the level of relatively small (little stress energy) objects, this is what is causing one of them to feel heavier. If one object is more massive (has more stress-energy), we will proportionately feel it heavier. Gravity in this case is simply spacetime curvature, that accelerates both objects, approximately at the same rate. The more massive object feels heavier because it has more inertia, and you are trying to stop it from moving along the geodesic.

  2. small objects own static gravitational field, that will cause them to accelerate at different rates towards Earth. this effect becomes detectable only at the level of objects that have comparable level of stress-energy relative to Earths'.

$\endgroup$
0
$\begingroup$

I can't think of a neat, intuitive way, how it does explain that. One problem is that "heavy" is a non-relativistic notion, so you have to translate the equations of motion of general relativity back to Newton's forces in an absolute frame of reference.

One way to get to "heavy" could be:
The Einstein-equation relates space-time-curvature to the energy-momentum-tensor, which includes the energy-(i.e. mass-)density. If you solve the Einstein-equations and find the space-time-metric you can compute the motion of an object in that space-time (the motion is then completely determined by the space-time-metric), see e.g. https://physics.stackexchange.com/a/135236/275840
And the motion of that object can, in some frame of reference (for instance, you standing on earths surface), be broken down to an acceleration in that frame, which in turn would act as a force on something in its way trying to stop it (e.g. you, hence you feel it being "heavy").

$\endgroup$
0
$\begingroup$

Gravity effects all objects the same regardless of mass. An object that is travelling on its natural path through space and time is in free fall. Free falling objects all travel along the same path regardless of their mass, as was demonstrated by Galileo (unless they are so massive that we need to take their own gravity into account). The reason some objects feel heavier than others is because it takes more energy to deflect them from their natural path.

For example, an object resting on the surface of the earth is not travelling along its natural path (it's not in free fall). The earth has to exert a force to prevent the object from falling through. For more massive the objects, the surface of the earth has to exert a greater force.

$\endgroup$
0
$\begingroup$

If we are only dealing with weights in a localized place at the surface of the earth, it is possible to use the principle of equivalence: the physical properties are the same as being in a spaceship in the outer space with an acceleration $g$.

Heavier objects here on earth would be equally heavier there. It is easy to see that the acceleration is the same for all objects, because the ship is really accelerated. Of course, as $F = ma$, more mass means more weight.

The metric at the surface of earth is such that the covariant acceleration is $g$ for a body at rest, as explained here.

$\endgroup$
0
$\begingroup$

For a classical answer, can’t really go past Wheelers summary of GR:

Matter (energy density) tells space-time how to curve

Curved space-time tells matter how to move.

On Earth, if the objects have much less energy density than the earth (basically every case in daily life) then local curvature (acceleration $g$) is essentially the same, but you have to do more work (Greater force x same distance) to lift the more massive (=more inertia) object = feels heavier.

Theoretically, though, if an object on earth had similar energy density to the earth (i.e. small black hole) then it will actually noticeably curve local space-time. Then the earth will also accelerate toward it, so you have to do a lot++ more work (against the objects own curvature) to lift it compared to any other daily object.

$\endgroup$
2
  • $\begingroup$ I like the spirit of your answer. But in this case isn't the curvature mainly due to the Earth, rather than the small object under consideration? The latter can be considered as a test mass that doesn't modify the background curvature. So the curvature is the same for the test object with higher density and the one with lower density – which is what perplexes the OP. $\endgroup$
    – pglpm
    Oct 13 '20 at 6:17
  • $\begingroup$ Well, I was trying to condense what ultimately took Andrew Steane and Arpad Szendrei together a lot of text to correctly outline. But was a bit slapdash, so…edited $\endgroup$ Oct 13 '20 at 11:12
0
$\begingroup$

In general relativity, gravitational force doesn't exist. So we cannot think of "heaviness" or "weight" as "the force exerted from Earth to the body" you're considering. So a preliminary question is: how do we define "heaviness" here?

In your case we have a background metric and curvature coming from the energy-momentum-stress distribution of the earth, and negligibly affected by the body you consider. Therefore, if the body were in free fall it would have the same 4D worldline, independently of its density, shape, volume (as long as these are within such limits as to allow us to consider the body as a test mass).

In this case we can define "heaviness" or "weight" as the 4-force necessary to keep the test body at rest in a frame in which the Earth is (on average) at rest. If the body is at rest in such a frame, then its worldline is not a geodesic. This means that the body has a 4-acceleration, that is, its 4-momentum changes (in terms of covariant derivative) along its worldline – it doesn't stay "parallel to itself" along the worldline. (Remember that 4-momentum is perceived – it can be decomposed – as ordinary momentum and mass-energy in a specific reference frame, but such perception and decomposition depends on the frame; similary, 4-force can be decomposed as ordinary force and work+heating.)

According to Einstein's equations, such 4-acceleration is only possible if a 4-force – what we're defining as "heaviness" – acts on the body, and it turns out that such 4-force does depend on the rest-mass of the body. For this reason if the object has greater mass it will "feel heavier". A similar reasoning could also been made for the "feeling of inertia".

In general, the amount of 4-force $\pmb{f}$ necessary to deflect a test body from its geodesic will depend on the background curvature (the body feels heavier on a more massive planet, which gives rise to greater curvature), expressed in the covariant derivative $\nabla$; and on the rest-mass $m$ of the body (more massive bodies feel heavier), which is equal to the "4-length" of the 4-momentum $\pmb{p}$ of the body. If the body is not emitting or absorbing heat, we have $\pmb{p}=m\pmb{u}$, where $\pmb{u}$ is the 4-velocity of the body, having unit "4-length". In this case the force satisfies $\pmb{f} = \nabla_{\pmb{u}}(m\pmb{u})$, which can be derived from the Einstein equations (this formula is related to the "$F=ma$" of Newtonian mechanics, which appears in other answers). In your example the "$\nabla_{\pmb{u}}\pmb{u}$ part" is the same for the denser and less dense objects (same non-geodesic worldline, same background curvature), but the "$m$ part" is larger for the denser object.

If the body is emitting or absorbing heat, its 4-momentum and its 4-velocity are actually non-collinear; this is usually a negligible effect in Newtonian contexts.

References

Four-momentum, 4-acceleration, 4-force can be found in different places in

  • Misner, Thorne, Wheeler: Gravitation (Freeman 1973),

for example chaps 6 and 13.

The derivation of $\pmb{f} = \nabla_{\pmb{u}}(m\pmb{u})$ for test bodies from Einstein's equations isn't trivial. See for example:

For the non-collinearity of 4-velocity and 4-momentum see for example

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.