Force due to acceleration and time flowing differently

[HansH]

True, the he feels normal force mg upwards.

The concept "feeling" a force is very fuzzy.

ok so what is it you wanted to reach with your question? my knowledge of basic physics? that is pre university level with a mark 9 for physics and 2 years of physics during the study of 4 years of electical engineering. Dutch HTS which could be compared to batchelor I believe.

 

[malawi_glenn]

ok so what is it you wanted to reach with your question? my knowledge of basic physics? that is pre university level with a mark 9 for physics and 2 years of physics during the study of 4 years of electical engineering. Dutch HTS which could be compared to batchelor I believe.

Rather what the concept of "feeling a force" means. It is fuzzy.

 

[vanhees71]

Fair point. So what would you suggest as the origin of force as a concept? It remains the rate of change of four momentum, and fouromentum is a normalised tangent to the worldline, but why is that normalisation of interest?

Indeed, "a force", seems to be best defined by the equation of motion of a particle in an external field,
$$\mathrm{d}_{\tau} p^{\mu}=F^{\mu}(x,p), \quad \mathrm{d}_{\tau} x^{\mu}=\frac{1}{m} p^{\mu},$$
where $m$ is the invariant mass. The "Minkowski force" $F$ must obey the contraint
$$p_{\mu} F^{\mu}=0,$$
such that the equation of motion is consistent with
$$p_{\mu} p^{\mu}=m^2 c^2 =\text{const}.$$
Of course, all this even holds within GR. Then you only have to use the covariant derivative wrt. $\tau$,
$$\mathrm{D}_{\tau} p^{\mu} = \mathrm{d}_{\tau} p^{\mu} + {\Gamma^{\mu}}_{\rho \sigma} p^{\rho} \mathrm{d}_{\tau} x^{\sigma}.$$
Note that then on the right-hand side you only have the forces but not gravity, which is contained on the left-hand side of the equation. In GR gravity is not a force to begin with of course. Perhaps it's anyway best to first concentrate on SRT only, excluding the discussion of gravity at this stage...

 

[HansH]

Rather what the concept of "feeling a force" means. It is fuzzy.

not sure how that relates to the answer the topic question. What actually happens is not fuzzy but well defined.

 

[malawi_glenn]

not sure how that relates to the answer the topic question

Topic question used the phrasing "feeling a force".

 

[vanhees71]

Feelings are anyway not subject of the natural sciences (except in brain research, maybe ;-)).

 

[PeterDonis]

gravity gives a force downwards

Not in relativity, no. In relativity gravity is not a force. It's just spacetime geometry. In relativity the Earth pushes up on you, causing you to have nonzero proper acceleration. That's why you feel a force (an upward force) when you are at rest on the Earth's surface.

 

[PeroK]

not sure how that relates to the answer the topic question. What actually happens is not fuzzy but well defined.

One of the things the principle of relativity implies is that an acceleration phase has no material or physical effect on a particle. This is equivalent there being "no such thing as a state of absolute rest".

It's clear that in your knowledge of physics the concept of absolute rest and absolute velocity is deeply embedded. And, everything you try to ask or understand about relativity is distorted by this misconception.

 

[HansH]

Topic question used the phrasing "feeling a force".

Ok I see. I did not want to talk about my feelings and especially not when it makes the topic diverge instead of converge as it now seems to do. So shall I change it to 'experiencing or what is the best word'? so let's discuss the essential parts.

 

[PeterDonis]

he feels a force F=m*a pointing to the left pushing him in his chair to the back

No. He feels a force F = ma to the right because the car is accelerating to the right.

or the person this feels like a force backwards as he resists to the acceleration

No, it feels like a force in the direction of the proper acceleration. The Earth pushes up on you giving you an upward proper acceleration, so you feel an upward force. The car pushes you to the right giving you a rightward proper acceleration, so you feel a rightward force.

 

[PeterDonis]

what would you suggest as the origin of force as a concept? It remains the rate of change of four momentum with respect to proper time

With the bolded addition above, this answers the question.

fouromentum is a normalised tangent to the worldline

No, 4-velocity is the normalized tangent vector. 4-momentum is not normalized; its norm is the invariant mass, not a constant number.

 

[vanhees71]

<Pedantic mode on> The Lorentzian fundamental form doesn't induce a metric, because it's not positive definite. For a classical particle, $p_{\mu} p^{\mu}=m^2 c^2=\text{const}$ with $m$ the (invariant) mass of the particle and $c$ the speed of light.

 

[Dale]

So shall I change it to 'experiencing or what is the best word'? so let's discuss the essential parts.

Perhaps talk about measurements with an accelerometer. That is why I brought in the second of the two suggested references, the one which determines the metric in a radar-coordinates non-inertial frame using an accelerometer as the input.

 

[HansH]

I don't know, what you mean by "one does not cause the other".

That came from Ibix in #14 I assume indicating the possible relation being the main subject.

 

[malawi_glenn]

Not in relativity, no.

Which is why I wrote "forget about relativity for a while" or something along those lines

experiencing

Being exerted on.

 

[PeterDonis]

Which is why I wrote "forget about relativity for a while" or something along those lines

I'm not sure that's a good idea. The OP is asking about relativity, and the treatment of gravity in Newtonian mechanics vs. relativity is very different, so I'm not sure trying to start with a Newtonian treatment will be helpful.

If the scenario were modified to eliminate gravity (for example, taking place out in deep space far away from all gravitating bodies), then starting with a Newtonian treatment might work better.

 

[Dale]

The Lorentzian fundamental form doesn't induce a metric, because it's not positive definite

Physicists nevertheless call it a metric, even knowing that technically it is not

 

[malawi_glenn]

Physicists nevertheless call it a metric, even knowing that technically it is not

Not all of them ;)

 

[vanhees71]

Given the confusion of the OP, I think we should be a bit more pedantic. Our sloppy physicists' slang is not helpful!

 

[PeterDonis]

Given the confusion of the OP, I think we should be a bit more pedantic. Our sloppy physicists' slang is not helpful!

I'm not sure harping on the pedantic difference between a metric and a pseudometric is going to be the silver bullet that resolves the OP's confusion. IMO it's more likely to increase it.

 

[Dale]

Given the confusion of the OP, I think we should be a bit more pedantic. Our sloppy physicists' slang is not helpful!

Understood. That is where we differ. Given the confusion of the OP, I think the OP should study the basics some more and then come back to this question. I don't think being pedantic will resolve the confusion, but I do think studying the basics will.

 

[vanhees71]

That's of course true ;-). I still think that sloppy slang adds to the confusion. When I started to study relativity in high school this particular example of calling an indefinite quadratic form "a metric" took me quite some time to sort out again.

 

[HansH]

You don't accelerate frames. You accelerate objects. You can have a frame that is not an inertial frame, but talking about accelerating a frame doesn't make sense; accelerating a frame would mean the frame changes with time, but a frame already has to include time, so talking about it changing with time doesn't make sense.

Also, you can't .....

I see that I have to be much more accurate in my descriptions to prevent that the discussion drives away from the subject. as I understand An inertial frame of reference is one in which the motion of a particle not subject to forces is in a straight line at constant speed. so what I mean is then probably 2 rockets in space with an observer in each rocket going with relative speed =0 close to each other. Then they both synchronze their clocks and then one rocket accelerates for a certain amount of time with say 1g. so this gives a force of 1 Newton per kg mass at that mass during this amount of time. then after some time the same rocket accelerates for (I assume) a same amount of time with -1g so that after that there is no speed difference between the 2 rockets but only a certain distance in between. now they come together again by slow acelleration en de-aceleration and after that compare their watches. This means there is now a time difference between their watches related to the amount of time and acceleration that one of them acellerated. if I would use the double accellaration or the double time while acelerating finally the difference between their clocks would also be different. so therefore I suppose there could be an uderlying reason why both are related. (at least that both are related is clear and could be calculated)

 

[Vanadium 50]

Perhaps you should get a bit more comfortable with the basics of SR before moving on to accelerations?
Just a friendly tip.

yes of course, but I cannot prevent that the question pops up in my head at the wrong moment.

No, but you can't expect to start in the middle rather than the beginning and understand. That's why we call it the "beginning".

 

[HansH]

Understood. That is where we differ. Given the confusion of the OP, I think the OP should study the basics some more and then come back to this question. I don't think being pedantic will resolve the confusion, but I do think studying the basics will.

That sounds as a good compromise. For sure we differ as I am an engineer interested in the physics from pure interest relarted to topics I never studied so far (ok you have to make choices in your life) and most of you are professional physics guys. Only point for me is that I have a question about a possible relation that I see but no idea if or what underlying theory supports that, so what basics should I study then? looks like a chicken and egg. so there I need your advice. What I can do is study the links Dale gave and see if that makes sense, as I think that was a good advice.

 

[PeterDonis]

as I understand An inertial frame of reference is one in which the motion of a particle not subject to forces is in a straight line at constant speed

This definition is correct, but note that its actual meaning is different in relativity than in Newtonian physics. In Newtonian physics gravity is a force, but in relativity it is not. So in Newtonian physics a frame in which you, standing on the surface of the Earth, are at rest can be treated as an inertial frame, since the net force on you is zero (downward force of gravity exactly balances upward force of the Earth's surface pushing on you). But in relativity this is not the case; this frame is not inertial, because gravity is not a force so the only force on you is the Earth's surface pushing upward. In relativity, an inertial frame in your vicinity would be one in which a freely falling object, such as a rock dropping in vacuum, would be at rest.

one rocket accelerates for a certain amount of time with say 1g. so this gives a force of 1 Newton per kg mass at that mass during this amount of time. then after some time the same rocket accelerates for (I assume) a same amount of time with -1g

Time defined by what? By the rocket's clock? That's the simplest answer, but note that it is not the only possible answer, and other answers result in a different scenario being specified. (That would not be the case in Newtonian physics, which has absolute time, but there is no absolute time in relativity.)

after that there is no speed difference between the 2 rockets but only a certain distance in between. now they come together again by slow acelleration en de-aceleration and after that compare their watches. This means there is now a time difference between their watches related to the amount of time and acceleration that one of them acellerated

This is true, but it does not mean that acceleration "causes a difference in the rate of time flow". The fact that in this particular scenario, it happens to be possible to use the acceleration to compute the difference between the watches does not mean it is always possible to do that.

The only general rule that works is the geometric one: the difference between the watches is due to the difference in arc length along the two worldlines between the two meeting events. The particular scenario you describe just happens to be set up so that the difference in arc lengths, and hence the difference between the watches, has a definite relationship to the acceleration of one of the objects (and the time by that object's clock that the acceleration lasts). But that won't always be true.

 

[HansH]

No, but you can't expect to start in the middle rather than the beginning and understand. That's why we call it the "beginning".

The point is: if you are i a forest where you do not know the route, how do you then know where the beginning is? that only know the one that knows the route.

 

[PeroK]

The point is: if you are i a forest where you do not know the route, how do you then know where the beginning is? that only know the one that knows the route.

Most textbooks on SR start at the edge of the forest, not in the middle.

 

[Dale]

so what basics should I study then?

First, you should get comfortable with non-inertial frames in classical physics, and more importantly be comfortable using the Lagrangian for classical physics. Then you should study special relativity until you are comfortable with special relativity in inertial frames, and especially special relativity using the geometrical concept of four-vectors. It would also be helpful to have a passing familiarity with tensors.

With that background you will be prepared for understanding the nuances of the answers to this question.

if I would use the double accellaration or the double time while acelerating finally the difference between their clocks would also be different

I can produce scenarios where this is not true. You can have double the acceleration with the same difference between clocks.

 

[HansH]

The particular scenario you describe just happens to be set up so that the difference in arc lengths, and hence the difference between the watches, has a definite relationship to the acceleration of one of the objects (and the time by that object's clock that the acceleration lasts). But that won't always be true.

but then you agree that there is a relation. but as I understood so far we cannot explain that. probably the link of Dale can but I first need to check that (64 messages checking takes also time)

 

[PeterDonis]

but then you agree that there is a relation

There is a relation in this particular scenario. But that doesn't mean there is always such a relation in all scenarios. The question is whether you are just trying to understand this particular scenario, with the knowledge that what you learn about this particular scenario will not generalize to many other scenarios, or whether you want to learn the general rules that can be used in all scenarios.

 

[PeterDonis]

as I understood so far we cannot explain that

Depends on what you mean by "explain". Is there an "explanation" that will satisfy your current intuitions at your current level of knowledge? Quite possibly not. But that means you need to retrain your intuitions and increase your level of knowledge. It doesn't mean there is no explanation at all.

 

[HansH]

With that background you will be prepared for understanding the nuances of the answers to this question.

But as most people in this discussion have done these steps, I assume they should be able to draw a conclusion and translate that to a level that can be understood without first studying all the details. But I now understand that that is not possible or the answer is already given probably but too detailed for me?

 

[Dale]

I assume they should be able to draw a conclusion and translate that to a level that can be understood without first studying all the details

Your assumption is incorrect in this case, as evidenced by the disagreement on un-nuanced responses.

I know you want the shortcut, but I don't think there is one here. If I understand your question correctly then the second paper I provided answers it in what I think is the most basic way possible.

 

[PeterDonis]

translate that to a level that can be understood without first studying all the details

I gave you such a translation in post #61:

the difference between the watches is due to the difference in arc length along the two worldlines between the two meeting events.

"Difference in arc length" should be a geometrical concept that you can grasp.