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123

5:25 AM

Hello World....

123

6:25 AM

Hi @JohnRennie Sir

John Rennie

@123 hi :-)

123

@JohnRennie Pls give me useful ideas about energy as physical quantity in term of mathematics and importance in different application.

This question very important for me. Pls help..............

John Rennie

@123 that doesn't have an answer, because the term energy is used to mean so many different things that it doesn't have a single meaning.

123

6:41 AM

@JohnRennie Sir, I need detailed answer one by one for every single application. Pls help me out if you have time.

I don't need single answer. I need different answer for different application also in term of mathematics. in term of newtonian energy and energy used in lagrangian.

John Rennie

Mathematically the energy is associated with the symmetry of the system.

123

Symmetry in time or position. As in Noether's theorem?

John Rennie

From the Lagrangian we can calculate a quantity called the action:

Action (physics)

In physics, action is an attribute of the dynamics of a physical system from which the equations of motion of the system can be derived through the principle of stationary action. Action is a mathematical functional which takes the trajectory, also called path or history, of the system as its argument and has a real number as its result. Generally, the action takes different values for different paths. Action has dimensions of energy⋅time or momentum⋅length, and its SI unit is joule-second. Action is only of interest when the total energy of the system is conserved. == Introduction == Hamilton...

Noether's theorem tells us that if the action has a symmetry called time shift symmetry then there will be an associated conserved quantity, and that conserved quantity is the energy.

123

I have read the history of action quantity and in lagrangian. Also i am studied lagrangian mechanics still don't understand by heart fully yet.

John Rennie

So that is the fundamental meaning of energy.

123

6:49 AM

@JohnRennie Hm.... Good. Means rotational symmetry e.g. in case of uniform gravitational field?

Sarabsri mt

Do you need examples where energy can be used. Like in Newtonian terms , energy of a particle , thermodynamic system , ideal gases , in that way . Also , formula is force * displacement. So like work done by a body.

123

@Sarabsrimt Yes pls share importance of energy idea in different application

which can not be determined by newton's laws

John Rennie

@123 no, the symmetry related to conservation of energy is time translation symmetry:

Time translation symmetry

Time translation symmetry or temporal translation symmetry (TTS) is a mathematical transformation in physics that moves the times of events through a common interval. Time translation symmetry is the hypothesis that the laws of physics are unchanged, (i.e. invariant) under such a transformation. Time translation symmetry is a rigorous way to formulate the idea that the laws of physics are the same throughout history. Time translation symmetry is closely connected, via the Noether theorem, to conservation of energy. In mathematics, the set of all time translations on a given system form a Lie group...

Sarabsri mt

@123 I have studied mostly application which derived from Newton’s laws. But maybe quantum mechanics ideas you can use.

123

@JohnRennie Aahh.. Does it means if we transform time from one reference frame to another it is same?

Physics Meta

8:49 AM

0

Q: A downvoted question vs. a very well-received duplicate version of the same question!

Some months ago I asked a question entitled The “Satellite Paradox”: Twin paradox in orbiting satellites, and a duplicate version entitled Special Relativity in Free Fall has recently appeared. I really cannot understand why my question is downvoted despite a clear and detailed explanation, where...

discussion

Mohammad Javanshiry

9:10 AM

@JohnRennie Hello John. Do you think there is a clear resolution to my question?

John Rennie

@MohammadJavanshiry hi :-)

It's because the equation we all know for time dilation $t' = t/\gamma$ only works when both observers are in inertial frames.

And your orbiting observers are not in inertial frames.

Calculating the time dilation between the two orbiting observers would be very hard as you'd need to work in the non-inertial rest frame of one of the observers and that's hard. Obviously it can be done, even if only by a numerical calculation, but I don't know how to do it.

@MohammadJavanshiry do you know what the Rindler metric is?

Mohammad Javanshiry

But I think the observers themselves are inertial, and instead, the frame attached to each observer is not inertial. If the satellites reach each other, I think it is only SR that rules time dilation. I think the Rindler metric explains the uniform acceleration in GR, does not it?

Sorry, I am slightly slow in typing.

John Rennie

Yes, the Rindler metric describes the spacetime geometry for an accelerating observer. The key feature of the Rindler metric is that relative to the Rindler observer time runs faster above them and slower below them.

So we get not just time dilation but also time acceleration i.e. time running faster.

So suppose you had a clock orbiting a Rindler observer in a circle (at non-relativistic speed), then the clock would run faster when it was above them and slower below them.

Overall the changes would balance out and the clock would run at the same rate as the observer's clock i.e. around one complete orbit there would be no net time dilation.

Something like this happens for the two orbiting observers in your example.

i.e. the relative time dilation between them varies around the orbit but overall averages out to zero.

Mohammad Javanshiry

9:25 AM

Yes, but how do these time dilation and acceleration compensate for each other so that after the satellites are united there is no net time dilation using GR demonstrations? It seems that the precise calculations are hard. Do you have any idea about the calculations?

Indeed, I think that GR-based calculations become obscure for these practical experiments.

John Rennie

We can easily show the elapsed time between the observers meeting is the same for both observer. The elapsed time for an observer is just the length of the observer's world line calculated using the metric. And in this case that's the Schwarzschild metric.

This is a straightforward calculation and indeed I've done it in answers on this site. Give me a moment and I'll find an example ...

Mohammad Javanshiry

Yes, thank you.

John Rennie

@MohammadJavanshiry Here:

53

A: Is gravitational time dilation different from other forms of time dilation?

No, gravitational time dilation is no different to other forms of time dilation. They all stem from the invariance of the line element. If we choose some coordinates, $x^i$, then the line element is given by: $$ ds^2 = g_{ab}dx^adx^b \tag{1} $$ where the matrix $g_{ab}$ is called the metric te...

The lengths of the world lines of the two orbiting observers between their meetings is the same for both - this should be obvious from the symmetry. So both observers have to measure the same elapsed time between their meetings.

But what this doesn't calculate is how the time dilation varies between meetings. I've never done this calculation but my guess is that around one orbit the time dilation changes during the orbit but the changes cancel to give zero time dilation around a complete orbit.

Mohammad Javanshiry

Let me read it precisely. I will send you a message. I have to go now. Thanks.

Physics Meta

-1

Q: Why aren't homework problems accepted?

What are we afraid of? People might actually learn how to do an actual calculation? It seems like people would rather talk in generalities than do calculations. Anyway, calculations are fun. I don't understand the no homework rule. Unless someone has done no work on the problem.

discussion homework-and-exercises

Slereah

10:28 AM

"You've earned the "Taxonomist" badge (Create a tag used by 50 questions) for the "closed-timelike-curve" tag."

I am proud but also mildly embarrassed, because I misspelt it when I first typed it in

I meant to write closed-timelike-curves

ery cerious matherfunker

1:25 PM

hey I just dropped out of high school

where do I start physics lol

Charlie

high school

schn

6:07 PM

Maybe obvious, but when expressing the refractive index as a complex number, is this kind of the same thing as expressing the wave number as a complex number? In Griffith’s Introduction to Electrodynamics, only the wave number is introduced as a complex number, where the imaginary part $\kappa$ supposedly would not equal the imaginary part of the complex refractive index, which is also sometimes expressed as $\kappa$.

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