# B Does the mass of an object increase in a gravitational field?

Gold Member
From a frame of reference outside of a gravitational field, does the mass of an object near a gravitational field increase?

Related Special and General Relativity News on Phys.org

#### timmdeeg

Gold Member
No it doesn't. Mass is invariant which means it is not coordinate dependent.

Gold Member
No it doesn't. Mass is invariant which means it is not coordinate dependent.
Just to be clear, I'm not referring to rest mass. Since time dilates and mass increases for an object that has a velocity relative to an observer, it lead me to wonder if mass also increases (since time dilates) for an object in a gravitational field relative to an observer outside that field.

#### Ibix

mass increases for an object that has a velocity relative to an observer,
Please don't say that. "Mass" means invariant mass. Say "relativistic mass" if you must use the term; better to call it total energy (divided by $c^2$ if you like).

How are you planning to measure the mass of this thing from far away?

Gold Member
Please don't say that. "Mass" means invariant mass. Say "relativistic mass" if you must use the term; better to call it total energy (divided by $c^2$ if you like).

How are you planning to measure the mass of this thing from far away?
Sorry, I will refer to it as relativistic mass then.

To measure the relativistic mass, I would measure the time of oscillation of a clock spring for example relative to my watch. Since the watch fulcrum would be moving more slowly, it would indicate it's mass would have to have increased so as not to violate m=f/a since the force available in the spring itself would be constant.

Gold Member

#### Ibix

So you are going to combine a time measurement in one place with a measurement of the spring constant (which includes a dimension of time) in another. That won't produce anything meaningful.

Gold Member
So you are going to combine a time measurement in one place with a measurement of the spring constant (which includes a dimension of time) in another. That won't produce anything meaningful.
It seems meaningful in this regard: From the point of view of the observer in a strong gravitational field (near a black hole for example) according to her watch the Earth is rotating too fast. So fast in fact according to the force law it should be shedding it's oceans, unless it's relativistic mass had decreased proportionally to the amount of time dilation.

edit: Actually I meant it would have to increase its relativistic mass to hold the oceans in at such a high rotational velocity.
edit again: forget that last edit, it seems the mass must decrease. My point is, something has to give to prevent stresses from changing, either the mass or the force.

Last edited:

#### Ibix

In this planet experiment you are comparing a gravitational acceleration where G is defined using time local to the planet to a centripetal acceleration caclulated using your watch. You know that your watch and the clocks on the planet (used to define G) run at different rates. But you are ignoring this and attributing the resultant mismatch in measurements to a change in mass. It's no more a change in mass than it would be if I calculated the centripetal acceleration in metres per minute squared, the gravitational acceleration in metres per second squared, and blamed the resulting factor of 3600 on a mass change.

So neither the mass nor the force has changed. You're just measuring the force wrongly because you are not accounting for the way your instruments work.

Gold Member
In this planet experiment you are comparing a gravitational acceleration where G is defined using time local to the planet to a centripetal acceleration caclulated using your watch. You know that your watch and the clocks on the planet (used to define G) run at different rates. But you are ignoring this and attributing the resultant mismatch in measurements to a change in mass.
Thank you for this clarification. Doesn't this indicate that if I correct this error and use my time to determine the value of G, then it would be different than the (planets) local value of G in which case it is F that changes along with t in the force equation when all measurements are performed using only my time and that it is this then that prevents the planet from ripping apart?

#### Ibix

Again, you are just using funny time units. They're consistent this time, so things will work out. The numbers are different, but they'd be different if you switched to using minutes instead of seconds.

It's not recommended to do this in relativity because the ratio of my clock rate to yours isn't generally well defined in curved spacetime. You get away with it here because it's possible to write the Schwarzschild metric in a time-independent way. In general, though, you need to make all measurements (time and otherwise) locally or derive the local measurements from your remote observations.

Gold Member
Again, you are just using funny time units. They're consistent this time, so things will work out. The numbers are different, but they'd be different if you switched to using minutes instead of seconds.

It's not recommended to do this in relativity because the ratio of my clock rate to yours isn't generally well defined in curved spacetime. You get away with it here because it's possible to write the Schwarzschild metric in a time-independent way. In general, though, you need to make all measurements (time and otherwise) locally or derive the local measurements from your remote observations.
I'm grateful for your help in getting this sorted out for me. The reason I needed to get this clear is because I'm giving a short philosophical talk on "The Flow of Time" and am trying to predict questions I may get. Central to my talk will be that time as we know it is the relationship between things that change and according to Leibniz (paraphrasing) "Time can only be defined as a sequence of events" and further argued "There is no such thing as an autonomous time" (Source: Carlo Rovelli in "The Order of Time").

Since time is only an order of events without an autonomous time it could be argued the the rate of change of the universe as a whole might be undefinable. The question I predict will come up might be along the lines of "If the universe as a whole changed too quickly, wouldn't things just fly apart?" I suspected that something in Newtons force law must change in order for things to remain stable hence my question about whether it might be the mass that changes. I was planning on using the example of an observer near a black whole as an example.

#### Klystron

Gold Member
If you have time before your lecture for research, Lee Smolin and James Gleick have written popular books on time that you could skim at the library. Please consider following the mentors' suggestions: mass particularly change in mass may not be the best concept to help you explain "Flow of Time".

Look again at post #4 from Ibix. Big hint: Energy may be a useful concept to help understand time, rather than mass. Thanks.

Gold Member
If you have time before your lecture for research, Lee Smolin and James Gleick have written popular books on time that you could skim at the library.
Thank you for suggesting that. I read "Time Reborn" about a year ago, and again about 2 months ago and probably need to read it a third time. One thing I was disappointed in was how he only fleetingly mentioned relativity of simultaneity in relation to his new theory. It seems they are at odds with each other and I was looking for a good resolution to that in the book. I admire Smolin very much. I think he is a maverick thinker. I also enjoyed his book "The Trouble With Physics" which I also read twice. With regard to Gleick, are you referring to his book "Time Travel"?

Please consider following the mentors' suggestions: mass particularly change in mass may not be the best concept to help you explain "Flow of Time".

Look again at post #4 from Ibix. Big hint: Energy may be a useful concept to help understand time, rather than mass. Thanks.
Absolutely!! Believe me, I always take seriously the mentor's suggestions. Sometimes I feel like I'm talking to Einstein himself when I ask questions in this forum.

I'm hoping I will not have to dive that deep in my talk, but with Ibex's post #9 as a guide ("In this planet experiment you are comparing a gravitational acceleration where G is defined using time local to the planet...") I can see that G, being defined in terms of time changes using the observer's (near the black hole) time. And this resolves the problem of why the planets don't disintegrate if they are spinning faster.

With regard to your suggestion: "Energy may be a useful concept to help understand time, rather than mass", could you elaborate a little? How can calling mass energy via E=mc^2 help with explaining the flow of time?

I'm excited about my talk but still have a lot of research to do.

#### PeterDonis

Mentor
To measure the relativistic mass
The term "relativistic mass" is not correct for the (hypothetical, it doesn't actually happen--see below) effect you are describing a measurement of. "Relativistic mass" refers specifically to an effect of relative motion. The mass inside the gravitational field in your scenario is at rest relative to the observer well outside the field. So any effect on mass due to a gravitational field, such as you are asking about (and there actually isn't any--see below) would not be correctly described as "relativistic mass".

Since the watch fulcrum would be moving more slowly, it would indicate it's mass would have to have increased so as not to violate m=f/a since the force available in the spring itself would be constant.
This is not correct. Force "redshifts" the same way energy does, i.e., by the time dilation factor, so the slower motion is due to the reduced force, not to any change in mass.

From the point of view of the observer in a strong gravitational field (near a black hole for example) according to her watch the Earth is rotating too fast.
You are making the common mistake of throwing around different scenarios without first getting a proper understanding of the original one. Throwing in the black hole plus the rotating Earth is too much. You need to get a handle on the simplest case--a non-rotating source of gravity and a test mass in its gravitational field--first.

#### Tomas Vencl

From a frame of reference outside of a gravitational field, does the mass of an object near a gravitational field increase?
Maybe there is such an higher order effect :
https://en.wikipedia.org/wiki/Frame-dragging
"Static mass increase is a third effect noted by Einstein in the same paper.[6] The effect is an increase in inertia of a body when other masses are placed nearby. While not strictly a frame dragging effect (the term frame dragging is not used by Einstein), it is demonstrated by Einstein that it derives from the same equation of general relativity. It is also a tiny effect that is difficult to confirm experimentally."

Gold Member
The term "relativistic mass" is not correct for the (hypothetical, it doesn't actually happen--see below) effect you are describing a measurement of. "Relativistic mass" refers specifically to an effect of relative motion. The mass inside the gravitational field in your scenario is at rest relative to the observer well outside the field. So any effect on mass due to a gravitational field, such as you are asking about (and there actually isn't any--see below) would not be correctly described as "relativistic mass".
OK, great. Thank you for clarifying that.

This is not correct. Force "redshifts" the same way energy does, i.e., by the time dilation factor, so the slower motion is due to the reduced force, not to any change in mass.
Interesting. This would correlate with what Idex said in post #9 about G changing due to time dilation. So it does seem things "pony up" to keep Newtonian laws working in the way I had hoped they would.

You are making the common mistake of throwing around different scenarios without first getting a proper understanding of the original one. Throwing in the black hole plus the rotating Earth is too much. You need to get a handle on the simplest case--a non-rotating source of gravity and a test mass in its gravitational field--first.
I'm sorry, but I don't understand this. Without the black hole plus the rotating Earth I have no question. I have to use the two together as that is my question. Just to be clear, my core question now (not my original question) is this: An observer in a gravitational field notes that the Earth (outside the strong gravitational field of the observer) is spinning too fast (or a watch is ticking too fast). This observation is a real event. i.e. Time is actually moving faster and is a physical change that can be verified upon meeting up and comparing watches. That being the case, I find Newtonian laws would be violated unless something else also changed. If this were not the case, the Earth would fly apart from spinning too fast. I originally thought it might be Mass that changed, but this error has been corrected. It seems instead that it is the force that changes which includes the force in a watch spring and the force due to G and in any other force. The so called "redshift" in the force. Is this correct?

#### PeterDonis

Mentor
This would correlate with what Idex said in post #9 about G changing due to time dilation.
G doesn't change due to time dilation. I had missed that post of @Ibix, I'll need to take a look at it and respond separately.

it does seem things "pony up" to keep Newtonian laws working in the way I had hoped they would.
Locally, yes.

An observer in a gravitational field notes that the Earth (outside the strong gravitational field of the observer) is spinning too fast (or a watch is ticking too fast).
What do you mean, "too fast"? How does the observer in the gravitational field know this?

Time is actually moving faster and is a physical change that can be verified upon meeting up and comparing watches.
The difference in elapsed time on watches of people who followed different paths through spacetime does not mean "time is actually moving faster". It just means their paths through spacetime had different lengths. It's no different from two people taking two routes from New York to Los Angeles and finding different elapsed distances on their odometers when they meet again: that doesn't mean "distance is actually moving faster" for one of them.

That being the case, I find Newtonian laws would be violated unless something else also changed.
Nope. Locally the laws are the same everywhere. But you are trying to apply local laws to situations that aren't local. That doesn't work.

I originally thought it might be Mass that changed, but this error has been corrected. It seems instead that it is the force that changes which includes the force in a watch spring and the force due to G and in any other force.
Neither the mass nor the force changes locally. But you are trying to describe things that are happening not locally, but distant from you. And you can't expect such a description to follow all of the same rules as a local description.

#### PeterDonis

Mentor
G is defined using time local to the planet
G isn't defined using anyone's local notion of time. It's a universal constant. So the explanation you are offering, that the distant observer's observations should be attributed to a change in G, does not work.

#### PeterDonis

Mentor
It seems instead that it is the force that changes which includes the force in a watch spring and the force due to G and in any other force. The so called "redshift" in the force. Is this correct?
Not for the question you are asking now. The answer to the question you are asking now is that the spin rate of the Earth that matters for determining the Earth's structure and whether it will hold together is the spin rate local to the Earth, not the apparent spin rate seen by a distant observer.

You appear to have a misconception about the laws of physics and the principle of relativity. The principle of relativity, in the context of GR, is that the laws of physics locally are the same. You are leaving out the "locally" and trying to plug numbers based on the observations of a distant observer into laws that apply locally. That doesn't work. (@Ibix made this same point to you back in post #7. You need to think about it carefully.)

#### Nugatory

Mentor
An observer in a gravitational field notes that the Earth (outside the strong gravitational field of the observer) is spinning too fast (or a watch is ticking too fast). This observation is a real event. i.e. Time is actually moving faster and is a physical change that can be verified upon meeting up and comparing watches.
You have to be careful here - there are some some dubious assumptions hidden in the innocent-sounding words "real observation" and "too fast".

First, we need to be clear about exactly what we are observing: Suppose we position a strobe light on the surface of the earth, calibrated to flash once every revolution (according to an observer at rest relative to the center of the earth and close enough to the earth to also be considered "outside the string gravitational field". Now we, deep inside the gravitational field, will observe the time at which these flashes arrive. That is, our "real observation" is of the proper time along our worldline between the receipt of successive flashes, not the proper time the earth takes to make one rotation.

We conclude from this observation (it helps if the transmitted light signals include the time on earth at which they are sent) that the proper time the earth takes to make one rotation is greater than the proper time between our receipt of successive flashes. But how does that have to mean that the earth is spinning "too fast", or equivalently that too little time is passing between revolutions? We have two time intervals. One is longer than the other, but we can interpret this as the longer one being too long, the shorter one being too short, or (best) both of them being just right for what they are.

Note that if we're going to bring the clocks back together again the whole thing is just a variant of the twin paradox, with the same resolution: the earth isn't spinning too fast, it's spinning at the same rate for a longer time.

#### PeterDonis

Mentor
The so called "redshift" in the force.
I should clarify what this means. Suppose you are sitting far away from a massive body like the Earth, and you want to measure the mass of an object deep inside the gravitational field of that body. You propose to measure it this way: assemble a mechanical linkage that will let you push on the mass from a distance by transmitting the force you exert to the mass. Then you measure the acceleration of the object by observing its motion and timing it with your own clock, and plug that and the force into $m = F / a$ to get the mass.

Since the object is time dilated with respect to you, the object's motion will appear slow, so $a$ will be smaller than what an observer next to the mass would measure. But the force $F$ that you have to apply will also be smaller by the same redshift factor, so you will still get the same ratio $F / a$ for the mass. The force will be smaller because, as it gets transmitted down the mechanical linkage, the linkage itself, which, heuristically, "wants" to fall in the gravitational field, will increase the force that finally gets applied locally to the mass. Or, to put it another way, an observer sitting next to the mass, at the bottom of the linkage, will measure a force that is larger, by the redshift factor, than the force you apply at the top of the linkage.

(Note that all of this is still just for this particular case; it is not a general argument that everything a distant observer observes gets adjusted by the same factor, nor is it a general argument that you can plug distant observations into the same laws that apply locally. You can't.)

#### Ibix

G isn't defined using anyone's local notion of time. It's a universal constant. So the explanation you are offering, that the distant observer's observations should be attributed to a change in G, does not work.
I'm attributing it to a change of units in G, I think. If I am a distant observer using time on my wristwatch to analyse something happening at some more-or-less constant altitude far below me in a Schwarzschild metric, that's the same as an observer local to the experiment using a watch that ticks Schwarzschild coordinate time at that altitude, surely? It's just funny time units.

Edit: and you can only get away with it in stationary metrics, and as long as the experiment isn't changing r coordinate. And many other caveats, I'm sure.

#### PeterDonis

Mentor
I'm attributing it to a change of units in G, I think
No, it isn't.

If I am a distant observer using time on my wristwatch to analyse something happening at some more-or-less constant altitude far below me in a Schwarzschild metric, that's the same as locally using a watch that ticks Schwarzschild coordinate time, surely?
No.

It's just funny time units.
No, it's different curves through spacetime having different lengths. The units measuring the length along each curve are the same.

Consider the analogy I made in response to @Buckethead in post #18: two people travel from New York to Los Angeles along different routes. Does the fact that their odometers register different elapsed distances at the end mean they were using different units to measure distance?

#### Ibix

No, it's different curves through spacetime having different lengths. The units measuring the length along each curve are the same.
I think that's a matter of interpretation of the original question. My reading of #5 is basically that Buckethead is looking down at a mass on a spring and using his own watch to time its oscillations. He isn't correcting for redshift, which is how he can say the spring is moving slowly.

I completely agree that you can (should!) use the same units to measure the interval between ticks of Buckethead's watch and cycles of the spring. But he isn't doing so - if he were doing then he'd be using the spring's proper time and would have no grounds to complain that the spring was oscillating at any unusual rate. You seemed to me to be making this point yourself in #18 when you asked What do you mean, "too fast"?
Consider the analogy I made in response to @Buckethead in post #18: two people travel from New York to Los Angeles along different routes.
But I don't think that's a precise analogy to Buckethead's experiment. Buckethead's worldline and that of the spring never meet. It's more like Buckethead walking a line of constant latitude near the pole and wondering how he's staying ahead of a sprinter on the equator. What he really should be doing is using metre rules on the equator to measure the distance covered by the sprinter. But what he's doing is more like projecting the ends of his own metre rules along lines of constant longitude down to the equator. Which, in this highly symmetric case, would be a perfectly valid procedure if only he didn't call the projected length of his ruler a metre.

"Does the mass of an object increase in a gravitational field?"

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving